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Progress in Mathematics142

J. Donald Monk

Cardinal Invariants on Boolean AlgebrasSecond Revised Edition

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Progress in MathematicsVolume 142

For further volumes:http://www.springer.com/series/4848

Series EditorsHyman Bass, University of Michigan, Ann Arbor, USA Jiang-Hua Lu, The University of Hong Kong, Hong Kong SAR, China

Yuri Tschinkel, Courant Institute of Mathematical Sciences, New York, USAJoseph Oesterlé, Université Pierre et Marie Curie, Paris, France

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http://www.springer.com/series/4848

J. Donald Monk

Cardinal Invariantson Boolean Algebras

Second Revised Edition

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DOI 10.1007/978-3- - ISBN 978-3-0348- - ISBN 978-3-0348- - (eBook)

Springer Basel Heidelberg New York Dordrecht London

0729ISSN 0743-1643 ISSN 2296-505X (electronic)

0730 260348-0730 2

Library of Congress Control Number: 2014932538

Basel 20141st edition: © Birkhäuser Verlag 19962nd revised edition: © Springer

Mathematics Subject Classification (2010): 03E05, 03E10, 03E17, 03G05, 06-02, 06E05, 06E10, 54A25, 54D30,54G05, 54G12

partmentUniversit

e

Boulder, CO

DJ. Donald Monk

of Mathematicsy of Colorado

USA

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Contents

Foreword . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . vii

0 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1

1 Special Operations on Boolean Algebras . . . . . . . . . . . . . . . . 11

2 Special Classes of Boolean Algebras . . . . . . . . . . . . . . . . . . 35

3 Cellularity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 83

4 Depth . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 155

5 Topological Density . . . . . . . . . . . . . . . . . . . . . . . . . . . 219

6 π-Weight . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 237

7 Length . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 279

8 Irredundance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 291

9 Cardinality . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 311

10 Independence . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 335

11 π-Character . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 373

12 Tightness . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 387

13 Spread . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 405

14 Character . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 421

15 Hereditary Lindelof Degree . . . . . . . . . . . . . . . . . . . . . . . 437

16 Hereditary Density . . . . . . . . . . . . . . . . . . . . . . . . . . . 449

17 Incomparability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 461

18 Hereditary Cofinality . . . . . . . . . . . . . . . . . . . . . . . . . . 481

19 Number of Ultrafilters . . . . . . . . . . . . . . . . . . . . . . . . . 489

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vi Contents

20 Number of Automorphisms . . . . . . . . . . . . . . . . . . . . . . . 491

21 Number of Endomorphisms . . . . . . . . . . . . . . . . . . . . . . . 495

22 Number of Ideals . . . . . . . . . . . . . . . . . . . . . . . . . . . . 497

23 Number of Subalgebras . . . . . . . . . . . . . . . . . . . . . . . . . 499

24 Other Cardinal Functions . . . . . . . . . . . . . . . . . . . . . . . . 505

25 Diagrams . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 509

26 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 531

27 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 539

References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 551

Symbol Index . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 561

Subject Index . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 567

Name Index . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 571

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Foreword

This book is the successor of Cardinal functions on Boolean algebras (Birkhauser1990) and Cardinal invariants on Boolean algebras (Birkhauser 1996). It containsmost of the material of these books, and adds the following:

(1) Indication of the progress made on the open problems formulated in theearlier versions, with detailed solutions in many cases.

(2) Inclusion of some new cardinal functions, mainly those associated with con-tinuum cardinals.

The material on sheaves, Boolean products, and Boolean powers has been omitted,since these no longer play a role in our discussion of the cardinal invariants.

Although many problems in the earlier versions have been solved, many ofthem are still open. In this edition we repeat those unsolved problems, and addseveral new ones.

J. Donald Monk

Boulder, [email protected]

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0 Introduction

This book is concerned with the theory of certain natural functions k which assignto each infinite Boolean algebra A a cardinal number k(A) or a set k(A) of cardinalnumbers. The purpose of the book is to survey this area of the theory of BAs, givingproofs for a large number of results, some of which are new, mentioning most ofthe known results, and formulating open problems. Some of the open problems aresomewhat vague (“Characterize. . . ” or something like that), but frequently theseare even more important than the specific problems we state; so we have opted toenumerate problems of both sorts in order to focus attention on them.

The framework that we shall set forth and then follow in investigating car-dinal functions seems to us to be important for several reasons. First of all, thefunctions themselves seem intrinsically interesting. Many of the questions whichnaturally arise can be easily answered on the basis of our current knowledge ofthe structure of Boolean algebras, but some of these answers require rather deeparguments of set theory, algebra, or topology. This provides another interest intheir study: as a natural source of applications of set-theoretical, algebraic, ortopological methods. Some of the unresolved questions are rather obscure anduninteresting, but some of them have a general interest. Altogether, the study ofcardinal functions seems to bring a unity and depth to many isolated investigationsin the theory of BAs.

There are several surveys of cardinal functions on Boolean algebras, or, moregenerally, on topological spaces: See Arhangelskiı [78], Comfort [71], van Douwen[89], Hodel [84], Juhasz [75], Juhasz [80], Juhasz [84], Monk [84], Monk [90], andMonk [96]. We shall not assume any acquaintance with any of these. On the otherhand, we shall frequently refer to results proved in Part I of the Handbook ofBoolean Algebras, Koppelberg [89]. One additional bit of terminology: in a weakproduct

∏wi∈I Ai, we call an element a of type 1 iff {i ∈ I : ai �= 0}, the 1-support

of a, is finite, and of type 2 iff {i ∈ I : ai �= 1}, called the 2-support of a, is finite.

We have not attempted to give a complete history of the results mentionedin this book. The references can be consulted for a detailed background.

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© SpringerProgress in Mathematics , D

014

J.D. Monk Cardinal Invariants on Boolean Algebras: Second Revised Edition142 0348 0730 2

Basel 2

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2 Chapter 0. Introduction

Definition of the main cardinal functions considered

We defer until later the discussion of the existence of some of these functions; theydo not all exist for every BA.

Cellularity. A subset X of a BA A is called disjoint if its members are pairwisedisjoint. The cellularity of A, denoted by c(A), is

sup{|X | : X is a disjoint subset of A}.

Depth. Depth(A) is

sup{|X | : X is a subset of A well ordered by the Boolean ordering}.

Topological density. The density of a topological space X , denoted by d(X), isthe smallest cardinal κ such that X has a dense subspace of cardinality κ. Thetopological density of a BA A, also denoted by d(A), is the density of its Stonespace Ult(A).

π-weight. A subset X of a BA A is dense in A if for all a ∈ A+ there is an x ∈ X+

such that x ≤ a. The π-weight of a BA A, denoted by π(A), is

min{|X | : X is dense in A}.

This could also be called the algebraic density of A. (Recall that for any subset Xof a BA, X+ is the collection of nonzero elements of X .)

Length. Length(A) issup{|X | : X is a chain in A}.

Irredundance. A subset X of a BA A is irredundant if for all x ∈ X , x /∈ 〈X\{x}〉.(Recall that 〈Y 〉 is the subalgebra generated by Y .) The irredundance ofA, denotedby Irr(A), is

sup{|X | : X is an irredundant subset of A}.

Cardinality. This is just |A|. Sometimes we denote it by card(A).

Independence. A subset X of A is called independent if X is a set of free generatorsfor 〈X〉. Then the independence of A, denoted by Ind(A), is

sup{|X | : X is an independent subset of A}.

π-character. For any ultrafilter F on A, let πχ(F ) = min{|X | : X is dense in F}.Note here that it is not required that X ⊆ F . Then the π-character of A, denotedby πχ(A), is

sup{πχ(F ) : F is an ultrafilter of A}.

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Chapter 0. Introduction 3

Tightness. For any ultrafilter F on A, let t(F ) = min{κ : if Y is contained inUlt(A) and F is contained in

⋃Y , then there is a subset Z of Y of power at most

κ such that F is contained in⋃Z}. Then the tightness of A, denoted by t(A), is

sup{t(F ) : F is an ultrafilter on A}.

Spread. The spread of A, denoted by s(A), is

sup{|D| : D ⊆ Ult(A), and D is discrete in the relative topology}.

Character. The character of A, denoted by χ(A), is

min{κ : every ultrafilter on A can be generated by at most κ elements}.

Hereditary Lindelof degree. For any topological space X, the Lindelof degree ofX is the smallest cardinal L(X) such that every open cover of X has a subcoverwith at most L(X) elements. Then the hereditary Lindelof degree of A, denotedby hL(A), is

sup{L(X) : X is a subspace of Ult(A)}.

Hereditary density. The hereditary density of A, hd(A), is

sup{d(S) : S is a subspace of Ult(A)}.

Incomparability. A subset X of A is incomparable if for any two distinct elementsx, y ∈ X we have x �≤ y and y �≤ x. The incomparability of A, denoted by Inc(A), is

sup{|X | : X is an incomparable subset of A}.

Hereditary cofinality. This cardinal function, h-cof(A), is

min{κ : for all X ⊆ A there is a C ⊆ X with |C| ≤ κ and C cofinal in X}.

Number of ultrafilters. Of course, this is the same as the cardinality of the Stonespace of A, and is denoted by |Ult(A)|.

Number of automorphisms. We denote by Aut(A) the set of all automorphisms ofA. So this cardinal function is |Aut(A)|.

Number of endomorphisms. We denote by End(A) the set of all endomorphismsof A, and hence this cardinal function is |End(A)|.

Number of ideals of A. We denote by Id(A) the set of all ideals of A, so here wehave the cardinal function |Id(A)|.

Number of subalgebras of A. We denote by Sub(A) the set of all subalgebras ofA; |Sub(A)| is this cardinal function.

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Some classifications of cardinal functions

Some theorems which we shall present, especially some involving unions or ultra-products, are true for several of our functions, with essentially the same proof. Forthis reason we introduce some rather ad hoc classifications of the functions. Someof the statements below are proved later in the book.

A cardinal function k is an ordinary sup-function with respect to P if P isa function assigning to every infinite BA A a subset P (A) of P(A) so that thefollowing conditions hold for any infinite BA A:

(1) k(A) = sup{|X | : X ∈ P (A)};(2) If B is a subalgebra of A, then P (B) ⊆ P (A) and X ∩ B ∈ P (B) for any

X ∈ P (A).

(3) For each infinite cardinal κ there is a BA C of size κ such that there is anX ∈ P (C) with |X | = κ.

Table 0.1 lists some ordinary sup-functions.

Table 0.1: ordinary sup-functions

Function The subset P (A)

c(A) {X : X is disjoint}

Depth(A) {X : X is well ordered by the Boolean ordering of A}

Length(A) {X : X is linearly ordered by the Boolean ordering of A}

Irr(A) {X : X is irredundant}

Ind(A) {X : X is independent}

s(A) {X : X is ideal-independent}

Inc(A) {X : X is incomparable}

Given any ordinary sup-function k with respect to a function P and any infinitecardinal κ, we say that A satisfies the κ−k-chain condition provided that |X | < κfor all X ∈ P (A).

A cardinal function k is an ultra-sup function with respect to P if P is a func-tion assigning to each infinite BA a subset P (A) of P(A) such that the followingconditions hold:

(1) k(A) = sup{|X | : X ∈ P (A)}.(2) If 〈Ai : i ∈ I〉 is a sequence of BAs, F is an ultrafilter on I, and Xi ∈ P (Ai)

for all i ∈ I, then {f/F : f(i) ∈ Xi for all i ∈ I} ∈ P(∏

i∈I Ai/F).

All of the above ordinary sup-functions except Depth are also ultra-sup functions.

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For the next classification, extend the first-order language for BAs by addingtwo unary relation symbols F and P. Then we say that k is a sup-min function ifthere are sentences ϕ(F,P) and ψ(F) in this extended language such that:

(1) k(A) = sup{min{|P | : (A,F, P ) |= ϕ} : A is infinite and (A,F ) |= ψ}. Inparticular, for any BA A there exist F, P ⊆ A such that (A,F, P ) |= ϕ.

(2) ϕ has the form ∀x ∈ P(x �= 0 ∧ ϕ′(F, x)) ∧ ∀x ∈ F ∃y ∈ Pϕ′′(F, x, y).

(3) (A,F ) |= ψ(F)→ ∃x(x �= 0 ∧ ϕ′(F, x)).

Some sup-min functions are listed in Table 0.2, where μ(F) is the formulasaying that F is an ultrafilter.

Table 0.2: sup-min functions

Function ψ(F) ϕ(F,P)

π ∀xFx ∀x ∈ P(x �= 0) ∧ ∀x ∈ F ∃y ∈ P(x �= 0→ y ≤ x)

πχ μ(F) ∀x ∈ P(x �= 0) ∧ ∀x ∈ F ∃y ∈ P(y ≤ x)

χ μ(F) ∀x ∈ P(x �= 0 ∧ x ∈ F) ∧ ∀x ∈ F ∃y ∈ P(y ≤ x)

h-cof x = x ∀x ∈ P(x �= 0 ∧ x ∈ F) ∧ ∀x ∈ F ∃y ∈ P(y ≥ x)

A cardinal function k is an order-independence function if there exists a sentenceϕ in the language of (ω,<, ω, ω) such that the following two conditions hold:

(1) For any infinite BA A we have k(A) = sup{λ : there exists a sequence〈aα : α < λ〉 of elements of A such that for all finite G,H ⊆ λ such that(λ,<,G,H) |= ϕ we have

∏α∈G aα ·

∏α∈H −aα �= 0}.

(2) If λ is an infinite cardinal, (λ,<,G,H) |= ϕ, G′, H ′ ⊆ λ, and f is a one-to-onefunction from G ∪H onto G′ ∪H ′ such that for all α, β ∈ G ∪H , if α < βthen f(α) < f(β), then (λ,<,G′, H ′) |= ϕ.

Some order-independence functions are listed in Table 0.3.

Table 0.3: order-independence functions

Function ϕ

t ∀x ∈ G ∀y ∈ H (x < y)

hd ∃x ∈ G ∀y ∈ G (x = y) ∧ ∀x ∈ G ∀y ∈ H (x < y)

hL ∃x ∈ H ∀y ∈ H (x = y) ∧ ∀x ∈ G ∀y ∈ H (x < y)

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Algebraic properties of a single function

Now we go into more detail on the properties of a single function which we shallinvestigate. From the point of view of general algebra, the main questions are:what happens to the cardinal function k under the passage to subalgebras, ho-momorphic images, products, and free products? There are natural problems tooabout more special operations on algebras in general, or on Boolean algebras inparticular: what happens to k under weak products, amalgamated free products,unions of well-ordered chains of subalgebras, ultraproducts, dense subalgebras,subdirect products, moderate products, one-point gluing, Alexsandroff duplica-tion, and the exponential? The mentioned operations which are not discussed inthe Handbook will be explained in Chapter 1. There are also several special kindsof subalgebras where one can ask what happens to the functions when passingto such a special subalgebra. Many of these special subalgebras are discussed inHeindorf, Shapiro [94]. For ease of reference, we list here ones which we considerto be worthwhile to investigate in this context:

A ≤reg B: A is a regular subalgebra of B. (Handbook, page 21.)

A ≤rc B; A is relatively complete in B. (Handbook, page 123.)

A ≤π B: A is a dense subalgebra of B.

A ≤s B: B is a simple extension of A. (See Chapter 2.)

A ≤m B: B is a minimal extension of A. (See Chapter 2.)

A ≤mg B: B is minimally generated over A. (See Chapter 2.)

A ≤free B: B is a free extension of A. This means that B = A ⊕ F for some freeBA F .

A ≤σ B: A is σ-embedded in B. This means that A ≤ B, and for every b ∈ B, theideal {a ∈ A : a ≤ b} of A is countably generated.

A ≤proj B: A is projectively embedded in B. This means that there is a free BA Cand homomorphisms e : B → A⊕C and q : A⊕C → B such that q ◦e = IdBand e � A = q � A = IdA. See Koppelberg [89b], page 752. This is illustratedby the following diagram:

A ≤u B: every ultrafilter on A has at least two different extensions to ultrafilterson B.

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One may notice that several of the above functions, such as depth and spread,are defined as supremums of the cardinalities of sets satisfying some property P .So, a natural question is whether such sups are attained, that is, with depth as anexample, whether for every BA A there always is a subset X well ordered by theBoolean ordering, with |X | = Depth(A). Of course, this is only a question in caseDepth(A) is a limit cardinal. For such functions k defined by sups, we can define aclosely related function k′; k′(A) is the least cardinal such that there is no subsetof A with the property P . So k′(A) = (k(A))+ if k is attained, and k′(A) = k(A)otherwise.

Derived functions

From a given cardinal function one can define several others; part of our work isto see what these new cardinal functions look like; frequently it turns out thatthey coincide with others of our basic functions, but sometimes we arrive at a newfunction in this way:

kH+(A) = sup{k(B) : B is a homomorphic image of A}.kH−(A) = inf{k(B) : B is an infinite homomorphic image of A}.kS+(A) = sup{k(B) : B is a subalgebra of A}.kS−(A) = inf{k(B) : B is an infinite subalgebra of A}.kh+(A) = sup{k(Y ) : Y is a subspace of UltA}.kh−(A) = inf{k(Y ) : Y is an infinite subspace of UltA}.

dkS+(A) = sup{k(B) : B is a dense subalgebra of A}.dkS−(A) = inf{k(B) : B is a dense subalgebra of A}.

Note that kh+(A) and kh−(A) make sense only if k is a function which naturallyapplies to topological spaces in general as well as BAs. Any infinite Boolean spacehas a denumerable discrete subspace, and frequently kh− will take its value onsuch a subspace. Also note with respect to dkS+(A) and dkS−(A) that one couldconsider other kinds of subalgebras, as in the previous list of them.

Given a function defined in terms of ultrafilters, like character above, thereis usually an associated function l assigning a cardinal number to each ultrafilteron A. Then one can introduce two cardinal functions on A itself:

lsup(A) = sup{l(F ) : F is an ultrafilter on A}.linf(A) = inf{l(F ) : F is a non-principal ultrafilter on A}.

Another kind of derived function applies to cases where the function is definedas the sup of cardinalities of sets X with a property P , where P is such thatmaximal families with the property P exist (usually seen by Zorn’s lemma). Forsuch a function k, we define

kmm(A) = min{|X | : X is an infinite maximal family satisfying P};kspect(A) = {|X | : X is an infinite maximal family satisfying P}.

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We also consider the following two spectrum functions, which assign to each BAa set of cardinal numbers:

kHs(A) = {k(B) : B is an infinite homomorphic image of A}(the homomorphic spectrum of A)

kSs(A) = {k(B) : B is an infinite subalgebra of A}(the subalgebra spectrum of A)

It is also possible to define a caliber notion for many of our functions, in analogyto the well-known caliber notion for cellularity. Given a property P associatedwith a cardinal function, a BA A is said to have κ, λ,P-caliber if among any setof λ elements of A there are κ elements with property P. The property P is notnecessarily one used to define the function; thus for cellularity P is the finiteintersection property, while for independence it is, indeed, independence.

Comparing two functions

Given two cardinal functions k and l, one can try to determine whether k(A) ≤l(A) for every BA A or l(A) ≤ k(A) for every BA A. Given that one of these casesarises, it is natural to consider whether the difference can be arbitrarily large (aswith cellularity and spread, for example), or if it is subject to restrictions (aswith depth and length). If no general relationship is known, a counterexample isneeded, and again one can try to find a counterexample with an arbitrarily largedifference between the two functions. Of course, the known inequalities betweenour functions help in order to limit the number of cases that need to be consideredfor constructing such counterexamples; here the diagrams at the end of the bookare sometimes useful. For example, knowing that πχ can be greater than c, wealso know that χ can be greater than c.

Other considerations

In addition to the above systematic goals in discussing cardinal functions, thereare some more ideas which we shall not explore in such detail. One can compareseveral cardinal functions, instead of just two at a time. Several deep theoremsof this sort are known, and we shall mention a few of them. There is also alarge number of relationships between cardinal functions which involve cardinalarithmetic; for example, Length(A) ≤ 2Depth(A) for any BA A. We mention a fewof these as we go along.

One can compare two cardinal functions while considering algebraic oper-ations; for example, comparing functions k, l with respect to the formation of

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subalgebras. We shall investigate just two of the many possibilities here:

kSr(A) = {(κ, λ) : there is an infinite subalgebra B of A

such that |B| = λ and k(B) = κ};kHr(A) = {(κ, λ) : there is an infinite homomorphic image B of A

such that |B| = λ and k(B) = κ}.

These are called, respectively, the subalgebra k relation and the homomorphic krelation.

For each function k, it would be nice to be able to characterize the possiblerelations kSr and kHr in purely cardinal number terms.

Another general idea applies to several functions that are defined somehowin terms of finite sets; the idea is to take bounded versions of them. For example,independence has bounded versions: for any positive integer n, a subset X of a BAA is called n-independent if for every subset Y of X with at most n elements andevery ε ∈ Y 2 we have

∏y∈Y yεy �= 0. (Here x1 = x, x0 = −x for any x.) And then

we define Indn(A) = sup{|X | : X is n-independent}. It is interesting to investigatethis notion and its relationship to actual independence; and similar things can bedone for various other functions.

Special classes of Boolean algebras

We are interested in all of the above ideas not only for the class of all BAs, butalso for various important subclasses: complete BAs, interval algebras, tree alge-bras, and superatomic algebras, which are discussed in the Handbook. To a lesserextent we give facts about cardinal functions for other subclasses like all atomicBAs, atomless BAs, initial chain algebras, minimally generated algebras, pseudo-tree algebras, semigroup algebras, and tail algebras. In Chapter 2 we describesome properties of the special classes mentioned which are not discussed in theHandbook, partly to establish notation.

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1 Special Operations onBoolean Algebras

We give the basic definitions and facts about several operations on Boolean alge-bras which were not discussed in the Handbook.

We begin with some elementary but useful results concerning products.

Proposition 1.1. C is a homomorphic image of A×B iff C is isomorphic to A′×B′

for some homomorphic images A′ and B′ of A and B respectively.

Proof. We may assume that A and B are non-trivial.

⇐: obvious. ⇒: suppose that f is a homomorphism from A × B onto C.It suffices to show that C � f(1, 0) is a homomorphic image of A. Let I be amaximal ideal in A, and for any a ∈ A let g(a) = f(a, a/I) · f(1, 0). Clearly g is ahomomorphism from A into C � f(1, 0). To show that it is onto, let x ∈ C � f(1, 0).Say f(a, b) = x. Then

g(a) = f(a, a/I) · f(1, 0) = f(a, b) · f(1, 0) = x. �

Proposition 1.2. Let 〈Ai : i ∈ I〉 be a system of non-trivial BAs, with I infinite.Then C is a homomorphic image of

∏wi∈I Ai iff there is a system 〈Bi : i ∈ I〉 of

BAs such that ∀i ∈ I[Bi is a homomorphic image of Ai], and C ∼=∏w

i∈I Bi.

Proof. For brevity let D =∏w

i∈I Ai.

For ⇐, suppose that fi : Ai → Bi is a surjective homomorphism for eachi ∈ I. Define g : D →

∏wi∈I Bi by setting, for each a ∈ D, (g(a))i = fi(ai).

Clearly g is a homomorphism from D into∏w

i∈I Bi. To show that it is surjective,let b ∈

∏wi∈I Bi. For each i ∈ I choose ai ∈ Ai such that fi(ai) = bi, with ai = 0 if

bi = 0 and ai = 1 if bi = 1. Clearly a ∈ D and g(a) = b. Thus∏w

i∈I Bi∼= D/ker(g).

For ⇒, suppose that K is a proper ideal in D. For each i ∈ I let Li = {bi :b ∈ K}. Clearly Li is an ideal in Ai.

Case 1. There is an a ∈ K of type 2. Let F be the 2-support of a. Define f([b]K) =〈[bi]Li : i ∈ F 〉 for each b ∈ D. Clearly f is well defined, and is a homomorphismfrom D/K into

∏i∈F Ai/Li. It is one-one, for suppose that bi ∈ Li for all i ∈ F .

For each i ∈ F choose ci ∈ K such that bi = cii. Then b ≤ a+∑

i∈F ci ∈ K, and so

, ,OI 10.1007/978-3- - - _ ,


014


Basel 22

11

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12 Chapter 1. Special Operations on Boolean Algebras

b ∈ K. It maps onto∏

i∈F Ai/Li; for suppose that xi ∈ Ai for each i ∈ F . Extendto a function x ∈ D. Then f([x]K) = 〈[xi]Li : i ∈ F 〉, as desired. Let Bi = Ai/Li

for all i ∈ F , and let Bi be trivial for all i ∈ I\F .

Case 2. Every a ∈ K is of type 1. For each i ∈ I define χi ∈ D by setting, for eachj ∈ I,

χi(j) =

{1 if i = j,

0 otherwise.

Let J = {i ∈ I : χi /∈ K}. If J = ∅, thenK is a maximal ideal inD, and |D/K| = 2,giving the desired result, letting one Bi be a two element homomorphic image ofAi, the other Bis trivial.

Suppose that ∅ �= J and J is finite. Let M be the maximal ideal of D consist-ing of all elements of type 1. For each b ∈ D define f([b]K) = (〈[bi]Li : i ∈ J〉, [b]M ).Clearly f is a well-defined homomorphism from D/K into (

∏i∈J (Ai/Li)) × 2. f

is one-one: suppose that bi ∈ Li for all i ∈ J and b ∈ M . For each i ∈ J chooseci ∈ K such that bi = cii. Let F be the 1-support of b. Define d ∈ D by

d(i) =

⎧⎨⎩cii if i ∈ J ,

1 if i ∈ F\J ,0 otherwise.

Then d ∈ K and b ≤ d, so b ∈ K. Also, f maps onto (∏

i∈J Ai/Li) × 2. For,suppose that c ∈

∏i∈J Ai and ε ∈ 2. If ε = 1, extend c to b ∈ D by defining

b(i) = 1 for all i ∈ I\J . Clearly f([b]K) = (〈[ci]Li : i ∈ J〉, 1). If ε = 0, extend c tob ∈ D by defining b(i) = 0 for all i ∈ I\J . Clearly f([b]K) = (〈[ci]Li : i ∈ J〉, 0).Now we can let Bi = A/Li for all i ∈ J , Bi a two-element homomorphic image ofAi for some i ∈ I\J , and all other Bis trivial.

Suppose that J is infinite. For each b ∈ D define f([b]K) = 〈[bi]Li : i ∈ J〉.Clearly f is a well-defined homomorphism from D/K into

∏wi∈J Ai/Li. f is one-

one: suppose that bi ∈ Li for all i ∈ J . For each i ∈ J choose ci ∈ K such thatbi = cii. Since J is infinite, 1 /∈ Li for each i ∈ J , and bi ∈ Li for each i ∈ J , itfollows that b is of type 1. Let F be the 1-support of b. Define d ∈ D by

d(i) =

⎧⎨⎩cii if i ∈ J ,

1 if i ∈ F\J ,0 otherwise.

Then d ∈ K and b ≤ d, so b ∈ K. Also, f maps onto∏w

i∈J Ai/Li. For, supposethat c ∈

∏wi∈J Ai. If c is of type 2, extend c to b ∈ D by defining b(i) = 1 for all

i ∈ I\J . Clearly f([b]K) = 〈[ci]Li : i ∈ J〉. If c is of type 1, extend c to b ∈ D bydefining b(i) = 0 for all i ∈ I\J . Clearly f([b]K) = 〈[ci]Li : i ∈ J〉. Now we candefine Bi = Ai/Li for all i ∈ J , with the other Bis trivial. �

Concerning arbitrary products, we have the following simple result.

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Chapter 1. Special Operations on Boolean Algebras 13

Proposition 1.3. Let 〈Ak : k ∈ K〉 be a system of BAs, and 〈Ik : k ∈ K〉 a systemsuch that Ik is an ideal in Ak for all k ∈ K. Let J = {a ∈

∏k∈K Ak : ∀k ∈ K[ak ∈

Ik]}. Then J is an ideal in∏

k∈K Ak, and (∏

k∈K Ak)/J ∼=∏

k∈K(Ak/Ik).

Proof. Clearly J is an ideal in∏

k∈K Ak. Now for each a ∈∏

k∈K Ak and eachk ∈ K let (f(a))k = [ak]Ik . Clearly f is a homomorphism from

∏k∈K Ak onto∏

k∈K(Ak/Ik) with kernel J , so the result follows by the homomorphism theorem.�

However, the property of Proposition 1.2 does not extend to arbitrary products.In the following example we use the notation Finco(I) for the BA of finite andcofinite subsets of I.

Example 1.4. For each k ∈ ω let Ak = Finco(ω). Then ω2 is a subalgebra of∏k∈ω Ak. Let f be a homomorphism from ω2 onto P(ω)/fin, and extend f to a

homomorphism g from∏

k∈ω Ak into the completion of P(ω)/fin. Then there isan ideal J on

∏k∈ω Ak such that (

∏k∈ω Ak)/J is isomorphic to rng(g). Note that

rng(g) is atomless. But if 〈Ik : k ∈ ω〉 is any sequence of ideals on Finco(ω), then∏k∈ω(Ak/Ik) is atomic. Thus (

∏k∈ω Ak)/J is not isomorphic to

∏k∈ω(Ak/Ik).

Moderate products

This operation, due to Weese [80] and Gurevich [82] independently, is extensivelystudied in Heindorf [92], to whom the name is due. Suppose that 〈Ai : i ∈ I〉is a system of BAs; we assume that Ai is a field of subsets of some set Ji, andthat the Ji’s are pairwise disjoint. Furthermore, let B be an algebra of subsetsof I containing all of the finite subsets of I. For each b ∈ B let b =

⋃i∈b Ji. Set

K =⋃

i∈I Ji. For each b ∈ B, each finite F ⊂ I, and each a ∈∏

i∈F Ai, the set

b ∪⋃i∈F

ai

will be denoted by h(b, F, a). If F ∩ b = ∅ and 0 ⊂ ai ⊂ Ji for every i ∈ F , thenwe call (b, F, a) normal.

Proposition 1.5. Assume the above notation.

(i) For any b ∈ B, F ∈ [I]<ω, and a ∈∏

i∈F Ai we have h(b, F, a) = h(b′, F ′, a′),where b′ = b ∪ {i ∈ F : ai = Ji}, F ′ = {i ∈ F\b : ∅ ⊂ ai ⊂ Ji}, anda′ = a � F ′; moreover, (b′, F ′, a′) is normal.

(ii) If (b, F, a) is normal, then K\h(b, F, a) = h(I\(b∪F ), F, a′), where a′i = Ji\aifor all i ∈ F ; and (I\(b ∪ F ), F, a′) is normal.

(iii) h(b, F, a) ∩ h(b′, F ′, a′) = h(b′′, F ′′, a′′) for normal (b, F, a) and (b′, F ′, a′),where b′′ = b ∩ b′, F ′′ = (F ′ ∩ b) ∪ (F ∩ b′) ∪ {i ∈ F ∩ F ′ : ai ∩ a′i �= ∅}, and

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for any i ∈ F ′′,

a′′i =

⎧⎨⎩a′i if i ∈ F ′ ∩ b,

ai if i ∈ F ∩ b′,ai ∩ a′i if i ∈ F ∩ F ′ and ai ∩ a′i �= ∅;

and (b′′, F ′′, a′′) is normal.

(iv) If (b, F, a) and (b′, F ′, a′) are normal, then h(b, F, a) ⊆ h(b′, F ′, a′) iff b ⊆ b′,F ′ ∩ b = ∅, F ⊆ b′ ∪ F ′, and ∀i ∈ F ∩ F ′[ai ⊆ a′i].

(v) If (b, F, a) and (b′, F ′, a′) are normal and h(b, F, a) = h(b′, F ′, a′), then b = b′,F = F ′, and a = a′.

Proof. It is straightforward to check (i)–(iii). For (iv), suppose that (b, F, a) and(b′, F ′, a′) are normal.

First suppose that h(b, F, a) ⊆ h(b′, F ′, a′). Clearly b ⊆ b′ and F ′ ∩ b = ∅.Next, suppose that i ∈ F\b′. Then ai ⊆ h(b′, F ′, a′), so i ∈ F ′. Now suppose thati ∈ F ∩ F ′. Then ai ⊆ h(b, F, i) and i /∈ b′, so ai ⊆ a′i.

Second, suppose that b ⊆ b′, F ′∩b = ∅, F ⊆ b′∪F ′, and ∀i ∈ F ∩F ′[ai ⊆ a′i].Take any x ∈ h(b, F, a). If x ∈ b, then x ∈ b′ and hence x ∈ h(b′, F ′, a′). Supposethat i ∈ F and x ∈ ai. If i ∈ b′, then x ∈ h(b′, F ′, a′). Suppose that i /∈ b′.So i ∈ F\b′, and so i ∈ F ′. Since then i ∈ F ∩ F ′, we have ai ⊆ a′i, and sox ∈ h(b′, F ′, a′).

(v) follows from (iv). �

The BA of all sets h(b, F, a) is the moderate product of the Ai’s over B, and is

denoted by∏B

i∈I Ai.

Theorem 1.6. Suppose that 〈Ai : i ∈ I〉 is a system of BAs; each Ai is a field ofsubsets of some set Ji; the Ji’s are pairwise disjoint, and Finco(I) ≤ B ≤ P(I).

(i) For every i ∈ I we have Ji ∈∏B

i∈I Ai and Ai =(∏B

i∈I Ai

)� Ji.

(ii) B is isomorphic to a subalgebra of∏B

i∈I Ai; in fact, 〈b : b ∈ B〉 is an isomor-phic embedding.

(iii) If Finco(I) ≤ B ≤ C ≤ P(I), then∏B

i∈I Ai ≤∏C

i∈I Ai.

(iv)∏B

i∈I Ai can be embedded in∏

i∈I Ai.

(v)∏w

i∈I Ai∼=∏Finco(I)

i∈I Ai.

(vi) If I is finite, then B = P(I), and∏B

i∈I Ai∼=∏

i∈I Ai.

(vii) For each i ∈ I and each x ∈∏B

i∈I Ai let f(x) = x ∩ Ji. Then f is a homo-

morphism from∏B

i∈I Ai onto Ai.

(viii) Let g be the natural isomorphism of P(I) onto I2. Then

B∏i∈I

Ai∼=⟨

w∏i∈I

Ai ∪ g[B]

⟩∏

i∈I Ai

.

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(ix) If b ∈ [I]<ω, then∏

i∈b Ai∼=(∏B

i∈I Ai

)� b.

(x) Suppose that U ⊆∏B

i∈I Ai. Then U is an ultrafilter on∏B

i∈I Ai iff one of thefollowing holds:

(a) There exist an i ∈ I and an ultrafilter V on Ai such that U={h(b,F,a) :(b,F,a) is normal, and i ∈ b or (i /∈ b and i ∈ F and ai ∈ V )}.

(b) There is a nonprincipal ultrafilter W on B such that U = {h(b, F, a) :(b, F, a) is normal and b ∈ W}.

Proof. (i)–(iii) are clear by Proposition 1.5. For (iv), define (f(x))i = x ∩ Ji for

any x ∈∏B

i∈I Ai and each i ∈ I. We show that f is the desired embedding. For·, let normal (b, F, a), (b′, F ′, a′) be given, and let (b′′, F ′′, a′′) be as in Proposition1.5(iii). If i ∈ b′′, then

[h(b, F, a) ∩ h(b′, F ′, a′)] ∩ Ji = Ji = [h(b, F, a) ∩ Ji] ∩ [h(b′, F ′, a′) ∩ Ji].

If i ∈ F ′ ∩ b, then

[h(b, F, a) ∩ h(b′, F ′, a′)] ∩ Ji = Ji ∩ a′i = [h(b, F, a) ∩ Ji] ∩ [h(b′, F ′, a′) ∩ Ji].

If i ∈ F ∩ b′, then

[h(b, F, a) ∩ h(b′, F ′, a′)] ∩ Ji = ai ∩ Ji = [h(b, F, a) ∩ Ji] ∩ [h(b′, F ′, a′) ∩ Ji].

Finally, if i ∈ F ∩ F ′ and ai ∩ a′i �= ∅, then

[h(b, F, a) ∩ h(b′, F ′, a′)] ∩ Ji = ai ∩ a′i = [h(b, F, a) ∩ Ji] ∩ [h(b′, F ′, a′) ∩ Ji].

For −, recall Proposition 1.5(ii). If i ∈ b, then

[K\h(b, F, a)] ∩ Ji = ∅ = Ji\Ji = Ji\[h(b, F, a) ∩ Ji].

If i ∈ F , then

[K\h(b, F, a)] ∩ Ji = Ji\ai = Ji\[h(b, F, a) ∩ Ji].

Finally, if i ∈ I\(b ∪ F ), then

[K\h(b, F, a)] ∩ Ji = Ji = Ji\∅ = Ji\[h(b, F, a) ∩ Ji].

Clearly f is one-one; this finishes the proof of (iv).

In case B = Finco(I), this mapping is easily seen to be onto∏w

i∈I Ai, prov-ing (v).

(vi) clearly follows from (v).

For (vii), clearly f is a homomorphism from∏B

i∈I Ai into Ai. For each x ∈ Ai

we have x ∈∏B

i∈I Ai and f(x) = x. So f maps onto Ai.

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For (viii), we use the function f defined in the proof of (iv). Note that ifb ∈ B, then f(b) = g(b), and if F is a finite subset of I and a ∈

∏i∈I Ai, then

f(h(∅, F, a) = k, where

k(i) =

{ai if i ∈ F ,

∅ otherwise.

If (b, F, a) is normal, then h(b, F, a) = b∪h(∅, F, a). Hence {b : b ∈ B}∪{h(∅, F, a) :F ∈ [I]<ω, a ∈

∏i∈I Ai} generates

∏Bi∈I Ai. Clearly also the image of this set under

f generates the right side of the equation in (viii). Hence (viii) follows.

For (ix), define f(x) =⋃

i∈b xi for any x ∈∏

i∈b Ai; clearly this is the desiredisomorphism.

Finally, we consider (x). First suppose that U is an ultrafilter on∏B

i∈I Ai.

Case 1. There is an i ∈ I such that h({i}, ∅, ∅) ∈ U . Let V = {x ∈ A+i :

h(∅, {i}, {(i, x)}) ∈ U . Suppose that x ∈ V and x ⊆ y ⊆ Ji. Then

h(∅, {i}, {(i, x)}) ⊆ h(∅, {i}, {(i, y)}),

so h(∅, {i}, {(i, y)}) ∈ U . It follows that y ∈ V . Next suppose that x, y ∈ V . Nowh(∅, {i}, {(i, x)})∩ h(∅, {i}, {(i, y)}) = h(∅, {i}, {(i, x∩ y)}). Thus h(∅, {i}, {(i, x∩y)}) ∈ U , hence x ∩ y ∈ V . So V is a filter. Now let x ∈ Ai, and suppose thatx /∈ V . Thus K\h(∅, {i}, {(i, x)}) ∈ U . Now

K\h(∅, {i}, {(i, x)}) = h(I\{i}, {i}, {(i, Ji\x)}),

andh({i}, ∅, ∅) ∩ h(I\{i}, {i}, {(i, Ji\x)}) = h(∅, {i}, {(i, Ji\x)}),

so h(∅, {i}, {(i, Ji\x)}) ∈ U . It follows that Ji\x ∈ V . So V is an ultrafilter on Ai.

Now suppose that h(b, F, a) ∈ U with (b, F, a) normal. Suppose that i /∈ b.Now h(b, F, a) ∩ h({i}, ∅, ∅) ∈ U and hence this set is nonempty, so it follows thati ∈ F , and hence

h(b, F, a) ∩ h({i}, ∅, ∅) = h(∅, {i}, {(i, ai)});

hence ai ∈ V .

Conversely, suppose that (b, F, a) is normal. Suppose first that i ∈ b. Then wehave h({i}, ∅, ∅) ⊆ h(b, F, a), so h(b, F, a) ∈ U . Second, suppose that i /∈ b, i ∈ F ,and ai ∈ V . Then h(∅, {i}, {(i, ai)}) ∈ U . Then h(∅, {i}, {(i, ai)}) ⊆ h(b, F, a), soh(b, F, a) ∈ U .

Thus we have shown that Case 1 implies (a).

Case 2. There is no i ∈ I such that h({i}, ∅, ∅) ∈ U . Let W = {b ∈ B : h(b, ∅, ∅) ∈U}. Clearly W is a nonprincipal ultrafilter on B. Suppose that h(b, F, a) ∈ U with(b, F, a) normal. For each i ∈ F we have K\h({i}, ∅, ∅) = h(I\{i}, ∅, ∅) ∈ U , andhence

h(b, F, a) ∩⋂i∈F

h(I\{i}, ∅, ∅) = h(b, ∅, ∅)

is in U . So b ∈ W .

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Conversely, suppose that (b, F, a) is normal and b ∈ W . Since h(b, ∅, ∅) ⊆h(b, F, a), we have h(b, F, a) ∈ U .

This finishes the proof of ⇒.

For ⇐, first suppose that (a) holds; we want to show that U is an ultrafilter

on∏B

i∈I Ai. Suppose that h(b, F, a) ∈ U , h(b, F, a) ⊆ h(b′, F ′, a′), with (b, F, a)and (b′, F ′, a′) normal.

Case 1. i ∈ b. Then also i ∈ b′, so h(b′, F ′, a′) ∈ U .

Case 2. i /∈ b, i ∈ F , and ai ∈ V . If i ∈ b′, then h(b′, F ′, a′) ∈ U . Suppose thati /∈ b′. Then i ∈ F ′, so i ∈ F ∩ F ′ and hence ai ⊆ a′i. Hence a′i ∈ V and soh(b′, F ′, a′) ∈ U .

Next, suppose that h(b, F, a), h(b′, F ′, a′) ∈ U , with (b, F, a), (b′, F ′, a′) nor-mal. Then h(b, F, a) ∩ h(b′, F ′, a′) = h(b′′, F ′′, a′′) as in 1.5(iii). We consider somesubcases.

Subcase 2.1. i ∈ b ∩ b′. Then h(b′′, F ′′, a′′) ∈ U .

Subcase 2.2. i ∈ b\b′. Then i ∈ F ′ and a′i ∈ V . Now i ∈ F ′ ∩ b ⊆ F ′′ and a′′i = a′i,so h(b′′, F ′′, a′′) ∈ U .

Subcase 2.3. i ∈ b′\b. Similar to Subcase 2.2.

Subcase 2.4. i /∈ b ∪ b′. Then i ∈ F , ai ∈ V , i ∈ F ′, a′i ∈ V . So i ∈ F ∩ F ′ andai ∩ a′i ∈ V , hence ai ∩ a′i �= ∅. So h(b′′, F ′′, a′′) ∈ U .

Thus U is a filter. Clearly ∅ /∈ U . Finally, suppose that a normal (b, F, a) is given.Suppose that i /∈ b and (i /∈ F , or i ∈ F and ai /∈ V ). If i /∈ F , then i ∈ I\(b ∪ F )and hence K\h(b, F, a) ∈ U . Suppose that i ∈ F and ai /∈ V . Then Ji\ai ∈ V ,and hence K\h(b, F, a) = h(I\(b ∪ F ), F, a′) ∈ U , where a′j = Jj\aj for all j ∈ F .

Second, suppose that (b) holds. Clearly U is a proper filter. To show that it is anultrafilter, suppose that (b, F, a) is normal and h(b, F, a) /∈ U . Now K\h(b, F, a) =h(I\(b ∪ F ), F, a′) with a′i = Ji\ai for all i ∈ F . Since h(b, F, a) /∈ U , we haveb /∈ W , hence I\b ∈ W . Since W is nonprincipal, we have {i} /∈ W for all i ∈F , hence (I\{i}) ∈ W . Hence I\(b ∪ F ) = (I\b) ∩

⋂i∈F (I\{i}) ∈ W , and so

(K\h(b, F, a)) ∈ U . So U is an ultrafilter. �

Theorem 1.6(viii) suggests an alternative formulation of the notion of moderateproducts. Let I be an infinite set, 〈Ai : i ∈ I〉 a system of BAs, and Finco(I) ≤B ≤ P(I). Let g be the natural isomorphism from P(I) onto I2. Then we cantake the moderate product to be⟨

w∏i∈I

Ai ∪ g[B]

⟩∏

i∈I Ai

.

Note that here one does not need to assume that each Ai is an algebra of sets.

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Theorem 1.7. Assume the hypotheses of Theorem 1.6, and suppose that Ldef= {i ∈

I : |Ai| > 2} is infinite. Then∏B

i∈I Ai is not complete.

Proof. For each i ∈ L choose ai ∈ Ai such that 0 ⊂ ai ⊂ Ji. Suppose that∑i∈L ai exists in

∏Bi∈I Ai; say it is equal to h(b, F, c), where we may assume that

(b, F, c) is normal. Fix i ∈ L\F . Then ai ⊆ h(b, F, c) implies that i ∈ b. But thenh(b\{i}, F ′, c′) is still an upper bound, where F ′ = F ∪ {i} and c′ extends c withc′i = ai. Since h(b\{i}, F ′, c′) ⊂ h(b, F, c), this is a contradiction. �

It is clear that if each Ai is atomless, then so is∏B

i∈I Ai; similarly for each Ai

atomic. It is somewhat less trivial to check that the moderate product preservessuperatomicity:

Theorem 1.8. If each Ai is superatomic and also B is superatomic, then∏B

i∈I Ai

is superatomic.

Proof. For brevity write C =∏B

i∈I Ai. It suffices to show that if f is a homo-morphism from C onto a nontrivial BA D, then D has an atom. We consider twocases.

Case 1. f(Ji) �= 0 for some i ∈ I. Let f ′ = f � Ai. Clearly f ′ is a homomorphismfrom Ai onto D � f(Ji). Let f ′(ui) be an atom of D � f(Ji); this is possible sinceAi is superatomic. Clearly f ′(ui) is also an atom of D.

Case 2. f(Ji) = 0 for all i ∈ I. Note that h(b, F, a) = h(b, ∅, ∅)∪ h(∅, F, a) for anyb, F, a. Hence f(h(b, F, a)) = f(h(b, ∅, ∅)). For each b ∈ B, let k(b) = f(h(b, ∅, ∅).Clearly k is a homomorphism from B onto D, so D has an atom since B issuperatomic. �

An important use of moderate products in connection with homomorphisms hasbeen given by Koszmider [99]. We give details on his construction.

Proposition 1.9. If K is an ideal in∏B

i∈I Ai and i ∈ I, then {x ∩ Ji : x ∈ K} isan ideal in Ai.

Proof. Clearly {x ∩ Ji : x ∈ K} is closed under unions. Now suppose that y ∈ Ai

and y ≤ x ∩ Ji with x ∈ K. Clearly y ∈∏B

i∈I Ai and y = y ∩ Ji, so y is in the settoo. �

Note in fact that this ideal is merely K ∩ Ai.

Proposition 1.10. Suppose that C is an infinite homomorphic image of∏B

i∈I Ai,with I infinite. Then there is an infinite homomorphic image D of C which is alsoa homomorphic image of B or of some Ai.

Proof. Let K be the kernel of a homomorphism of∏B

i∈I Ai onto C. So∏B

i∈I Ai/Kis isomorphic to C and hence is infinite. For each i ∈ I let Li = K ∩ Ai. So byProposition 1.9, each Li is an ideal in Ai. We now consider several cases.

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Case 1. Ndef= {i ∈ I : Ji /∈ K} is finite, and ∀i ∈ I[Ai/Li is finite]. Let M =

〈K ∪ {Ji : i ∈ I, Ji /∈ K}〉id.(1)

∏Bi∈I Ai/M is infinite.

In fact, define f :∏B

i∈I Ai/K →∏B

i∈I Ai/M ×∏

i∈N (Ai/Li) by

f([a]K) = ([a]M , 〈[ai]Li : i ∈ N〉).First of all, f is well defined. For, if a ∈ K, then a ∈ M , and ai ∈ Li for eachi ∈ N . Moreover, f is an injection. For, if a ∈M and ai ∈ Li for each i ∈ N , thenthere is a b ∈ K such that a ⊆ b ∪

⋃i∈N Ji, so

a =

[a\⋃i∈N

Ji

]∪[a ∩

⋃i∈N

Ji

]

=

[a\⋃i∈N

Ji

]∪⋃i∈N

ai

⊆ b ∪⋃i∈N

ai

∈ K.

Thus, indeed, f is an injection. Hence (1) follows.

Now define g(b) = [h(b, ∅, ∅)]M for any b ∈ B. Then g is a homomorphism of

B into∏B

i∈I Ai/M , by Theorem 1.6(ii). For any element h(b, F, a) of∏B

i∈I Ai wehave [h(b, F, a)]M = [h(b, ∅, ∅)]M , since h(b, F, a)�h(b, ∅, ∅) =

⋃i∈F ai, and for all

i ∈ F , if i /∈ N then ai ⊆ Ji ∈ K, while if i ∈ N , then still ai ⊆ Ji ∈ M . Thus gmaps onto

∏Bi∈I Ai/M , and so

∏Bi∈I Ai/M is as desired.

Case 2. N is infinite, and ∀i ∈ I[Ai/Li is finite]. For each i ∈ N let Mi be amaximal ideal in Ai such that Li ⊆ Mi. Let P be the ideal generated by K ∪⋃

i∈N Mi. For each i ∈ N we have Ji /∈ P , and so∏B

i∈I Ai/P is infinite. Define

g(b) = [h(b, ∅, ∅)]P for all b ∈ B. Then g is a homomorphism of B into∏B

i∈I Ai/M ,by Theorem 1.6(ii). We claim that it is a surjection. Let h(b, F, a) be any element

of∏B

i∈I Ai, with (b, F, a) normal, and let c = b ∪ {i ∈ F : ai /∈Mi}. Then

h(b, F, a)�h(c, ∅, ∅) = (h(b, F, a)\h(c, ∅, ∅)) ∪ (h(c, ∅, ∅)\h(b, F, a))

=⋃i∈F

ai∈Mi

ai ∪⋃i∈F

ai /∈Mi

(Ji\ai).

Now if i ∈ F and ai ∈ Mi then ai ∈ P , while if i ∈ F and ai /∈ Mi, then(Ji\ai) ∈Mi ⊆ P . Hence g maps onto

∏Bi∈I Ai/P , as desired.

Case 3. There is an i0 ∈ I such that Ai0/Li0 is infinite. Let M be the ideal

generated by K ∪{⋃

i�=i0Ji

}, and define

g([h(b, F, a)]M ) = [h(b, F, a) ∩ Ji0 ]Li0.

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If h(b, F, a) ∈ M , then there is a c ∈ K such that h(b, F, a) ⊆ c ∪⋃

i�=i0Ji, hence

h(b, F, a) ∩ Ji0 ⊆ c, and so h(b, F, a) ∩ Ji0 ∈ (K ∩Ai0) = Li0 . So g is well defined.It clearly maps onto Ai0/Li0 , as desired. �

Proposition 1.11. Suppose that X is a subset of∏B

i∈I Ai with |X | = κ uncountableand regular, κ > |I|. Then there exist Y ∈ [X ]κ and a finite G ⊆ I such that 〈Y 〉is isomorphic to a subalgebra of (B � (I\G))×

∏i∈G Ai.

Proof. Let 〈Fξ : ξ < κ〉 ∈ κ([I]<ω), b ∈ κB and a with domain κ be such thataξ ∈

∏i∈Fξ

Ai for all ξ < κ, (bξ, Fξ, aξ) is normal, and

X = {h(bξ, Fξ, aξ) : ξ < κ}.

Choose Y ∈ [κ]κ and G ∈ [I]<ω such that Fξ = G for all ξ ∈ Y . Let

W =

{h(c,G, d) : c ∈ B, c ∩G = ∅, d ∈

∏i∈G

Ai

}.

This is a subalgebra of∏B

i∈I Ai by the above computation rules, and Y ⊆W . Nowdefine

f(h(c,G, d)) = (h(c, ∅, ∅), d)for each h(c,G, d) ∈ W . We claim that f is an isomorphism from W into (B �(I\G))×

∏i∈G Ai. In fact, f clearly preserves ·. For −,

f(−h(c,G, d)) = f(h(I\(c ∪G), G,−d)= (h(I\(c ∪G), ∅, ∅),−d)= (h(I\G, ∅, ∅) · h(I\c, ∅, ∅),−d)= (h(I\G, ∅, ∅) · −h(c, ∅, ∅),−d)= −f(h(c,G, d)). �

Let L be the set of all countable limit ordinals, and L2 the set of all countablelimits of elements of L. Let 〈cnα : α ∈ L2, n ∈ ω〉 be such that for all α ∈ L2,〈cnα : n ∈ ω〉 is strictly increasing, cofinal in α, with c0α = 0 and ci+1

α ∈ L for alli ∈ ω.

Let A and B be subalgebras of P(ω), with [ω]<ω ⊆ B. We construct 〈Cα :α ∈ L〉 by recursion. Each Cα will be a subalgebra of P(α). For each β ∈ L letfβ be a bijection from ω to [β, β + ω).

Define Cω = A. If Cα has been defined, with α ∈ L, let

Cα+ω = 〈Cα ∪ {fα[a] : a ∈ A}〉P(α+ω).

If β ∈ L2 and Cα has been defined for all α < β such that α ∈ L, let Cβ =∏Bi∈ω Dβ

i , with

Dβi = Cci+1

β� [ciβ , ci+1

β ).

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Here in general G � h = {g ·h : g ∈ G} for G a subalgebra of H and h ∈ H , withoutassuming that h ∈ G. Next, for each α ∈ L let Eα = Cα ∪ {a : ω1\a ∈ Cα}. ThusEα is a subalgebra of P(ω1). Finally, let

B∏c

A =

⟨⋃α∈L

Cα

⟩P(ω1)

.

Lemma 1.12. Assume the above notation. Then

(i) If α, β ∈ L and α < β, then Cα ⊆ Cβ.

(ii) If α, β ∈ L and α ≤ β, then α ∈ Cβ.

(iii) If α1, α2, β ∈ L and α1 < α2 ≤ β, then [α1, α2) ∈ Cβ.

(iv) If α, β ∈ L, α < β, define fβα(a) = a ∩ α for each a ∈ Cβ . Then fβα is ahomomorphism from Cβ onto Cα which is the identity on Cα.

(v) If α, β ∈ L and α < β, then Eα ⊆ Eβ, and there is a homomorphism fromEβ onto Eα which is the identity on Eα.

(vi)∏B

c A =⋃

α∈LEα.

(vii) For any α ∈ L, Cα+ω∼= Cα ×A.

(viii) L ⊆∏B

c A.

Proof. We prove (i)–(iv) simultaneously by induction on β. The case β = ω holdsvacuously. Now assume inductively that β > ω.

Case 1. β = γ + ω for some γ ∈ L. For (i), suppose that α ∈ L and α < β. ThenCα ⊆ Cγ by the inductive hypothesis, and Cγ ⊆ Cβ by construction.

For (ii), suppose that α ≤ β with α ∈ L. If α < β, then α ∈ Cα ⊆ Cβ by theinductive hypothesis. Obviously β ∈ Cβ .

For (iii), we consider two subcases.

Subcase 1.1. α2 < β. Then [α1, α2) ∈ Cγ by the inductive hypothesis, and Cγ ⊆ Cβ

by construction.

Subcase 1.2. α2 = β. Then [α1, γ) ∈ Cγ by the inductive hypothesis, and [γ, β) =fγ [ω] ∈ Cβ by construction. So [α1, β) = [α1, γ) ∪ [γ, β) ∈ Cβ .

For (iv), we have a ∩ α ∈ Cα for each a in the generating set of Cβ , and so(iv) holds.

Case 2. β ∈ L2. First we take (ii). Clearly β ∈ Cβ by construction. Now supposethat α < β. Choose i ∈ ω such that ciβ ≤ α < ci+1

β . Then by the inductive

hypothesis, α ∈ Cci+1β

, so [ciβ, α) ∈ Cci+1β

� [ciβ , ci+1β ) = Dβ

i . Clearly [0, ciβ) ∈ Cβ .

Hence α ∈ Cβ .

Now assume the hypotheses of (iii). By (ii), [α1, α2) = α2\α1 ∈ Cβ . Thus(iii) holds.

Now assume the hypotheses of (i), and suppose that a ∈ Cα. In particular,a ⊆ α. Choose i0 such that ci0β ≤ α < ci0+1

β . If i+1 ≤ i0, then a∩ ci+1β ∈ Cci+1

βby

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the inductive hypothesis on (iv), so a ∩ [ciβ , ci+1β ) ∈ Dβ

i . Also, [ci0β , α) ∈ C

ci0+1

βby

the inductive hypothesis on (iii), and a ∈ Cci0+1

β

by the inductive hypothesis on

(i). Hence a∩ [ci0β , α) = a∩ [ci0β , α)∩ [ci0β , ci0+1β ) ∈ Dβ

i0. Thus a is a sum of elements

of Dβi for i+ 1 ≤ i0 and of an element of Dβ

i0, so a ∈ Cβ , as desired in (i).

For (iv), suppose that α < β, and suppose that a ∈ Cβ . Again take i0 suchthat ci0β ≤ α < ci0+1

β .

(1) If i+ 1 ≤ i0, then [ciβ , ci+1β ) ∩ a ∈ Cα.

In fact, clearly [ciβ , ci+1β ) ∩ a ∈ Dβ

i . By (iii) for ci+1β we have [ciβ , c

i+1β ) ∈ Cci+1

β, so

Dβi ⊆ Cci+1

β, and hence [ciβ , c

i+1β )∩ a ∈ Cci+1

β⊆ Cα by the inductive hypothesis on

(i) for α. Thus (1) holds.

(2) [ci0β , α) ∩ a ∈ Cα.

For, again clearly [ci0β , ci0+1β )∩ a ∈ Dβ

i0, so as above, [ci0β , ci0+1

β )∩ a ∈ Cci0+1

β. Then

by the inductive hypothesis on (iv) for ci0+1β ,

[ci0β , α) ∩ a = [ci0β , ci0+1β ) ∩ a ∩ α ∈ Cα,

giving (2). Now by (1) and (2), a∩α =⋃

i+1≤i0(a∩ [ciβ , ci+1

β ))∪ ([ci0β , α)∩a) ∈ Cα.

This finishes the proof of (i)–(iv). For (v), from (i) it is clear that Eα ⊆ Eβ .Now for each a ∈ Eβ , let

g(a) =

{a ∩ α if a ∈ Cβ ,

(ω1\α) ∪ a otherwise.

Now g maps into Eα. For, if a ∈ Cβ , then g(a) = a ∩ α ∈ Cα ⊆ Eα by (iv). Ifω1\a ∈ Cβ , then

ω1\g(a) = ω1\((ω1\α) ∪ a) = ω1 ∩ α ∩ (ω1\a) = α ∩ (ω1\a),

and this is in Cα by (iv) again.

We check that g preserves ∪. Suppose that a, b ∈ Eβ . If a, b ∈ Cβ , thena ∪ b ∈ Cβ , and g(a ∪ b) = (a ∪ b) ∩ α = (a ∩ α) ∪ (b ∩ α) = g(a) ∪ g(b). If a ∈ Cβ

and b /∈ Cβ , then ω1\b ∈ Cβ ⊆ β, hence ω1\β ⊆ b, and so a ∪ b /∈ Cβ ; it followsthat

g(a) ∪ g(b) = (a ∩ α) ∪ (ω1\α) ∪ b

= (ω1\α) ∪ a ∪ b

= g(a ∪ b).

Similarly if a /∈ Cβ and b ∈ Cβ . Finally, if a, b /∈ Cβ , then

g(a) ∪ g(b) = (ω1\α) ∪ a ∪ b = g(a ∪ b).

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So g preserves ∪. For complement, if a ∈ Cβ , then

g(ω1\a) = (ω1\α) ∪ (ω1\a) = ω1\(a ∩ α) = ω1\g(a),

and if a /∈ Cβ , then

ω1\g(a) = ω1\(ω1\α) ∪ a) = α\a = g(ω1\a).

So g is a homomorphism. If a ∈ Cα, clearly g(a) = a. If ω1\a ∈ Cα, then g(a) =(ω1\α) ∪ a = a since (ω1\α) ⊆ a. Thus (v) holds.

For (vi), note that each Eα is clearly a subalgebra of 〈⋃

α∈LCα〉, and so⋃α∈LEα ⊆ 〈

⋃α∈LCα〉. Since

⋃α∈LEα is a subalgebra of P(ω1) containing⋃

α∈LCα, (vi) follows.

Turning to (vii), for any (a, b) ∈ Cα×A we define g(a, b) = a∪ fα[b]. Clearlyg preserves +. For −, we have

−g(a, b) = −(a ∪ fα[b])

= (α + ω)\(a ∪ fα[b])

= ((α + ω)\a) ∩ ((α+ ω)\fα[b])= ((α\a) ∪ [α, α + ω)) ∩ (α ∪ [α, α+ ω)\fα[b])= (α\a) ∪ ([α, α + ω)\fα[b])= (α\a) ∪ fα[−b]= g(−a,−b)= g(−(a, b)).

Thus g is a homomorphism. Its range clearly contains Cα ∪ {fα[a] : a ∈ A} andis contained in 〈Cα ∪ {fα[a] : a ∈ A}〉. It is clearly one-one. So g is the desiredisomorphism.

Finally, (viii) is immediate from (ii). �Proposition 1.13. Assume the notation above.

(i) If x ∈ Eα, then x ∩ α ∈ Cα.

(ii) Cα is a maximal ideal in Eα.

(iii) For every ideal K in Eα, the set K ∩ Cα is an ideal in Cα.

(iv) For every ideal K in Eα and every x ∈ Eα, let f([x]K) = [x∩α]Cα∩K . Thenf is well defined, and is a homomorphism from Eα/K onto Cα/(K ∩ Cα).Moreover, it is one-one on Cα/K.

(v) For every ideal K in Eα, if Eα/K is infinite, then so is Cα/(K ∩ Cα).

Proof. (i): Assume that x ∈ Eα. If x ∈ Cα, then the conclusion is obvious. Supposethat x /∈ Cα. Then (ω1\x) ∈ Cα, and so the set α\(ω1\x) is also in Cα. This setis equal to x ∩ α. So (i) holds.

(ii): Obviously Cα is closed under ∪. If x ∈ Cα and y ⊆ x, with y ∈ Eα, thenby (i), y ∈ Cα. Finally, Cα is obviously maximal.

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(iii): Obvious from (ii).

(iv): If x ∈ K, then (x∩α) ∈ Cα ∩K by (i). So f is well defined. It is clearlya homomorphism from Eα/K into Cα/(K ∩ Cα). For a ∈ Cα we have f([a]K) =[a]Cα∩K , so f is clearly one-one on Cα/K, and f maps onto Cα/(K ∩ Cα).

(v): Let K be an ideal of Eα, with Eα/K infinite. Now Eα/K is partitionedinto two parts of equal size, Cα/K being one of them. From (iv) it then followsthat Cα/(K ∩ Cα) is infinite. �

Proposition 1.14. Let A, B, c be as above. Suppose that E is an infinite homo-morphic image of

∏Bc A. Then there exist an infinite subalgebra F of E and an

infinite homomorphic image G of F such that G is also a homomorphic image ofA or of B.

Proof. Let J be an ideal of∏B

c A such that∏B

c A/J is isomorphic to E. Choose

ai ∈∏B

c A for each i < ω such that [ai]J �= [aj ]J for all i �= j. Choose α < ω1 suchthat {ai : i < ω} ⊆ Eα. For each x ∈ Eα let f([x]Eα∩J) = [x]J . Clearly f is well

defined, and is an isomorphism from Eα/(Eα ∩ J) into∏B

c A/J . By Proposition1.13(iv), Cα/(Cα∩J) is a homomorphic image of Eα/(Eα∩J), and by Proposition1.13(v) it is infinite. Hence it is enough to prove

(1) For every α ∈ L and every infinite homomorphic image G of Cα there is aninfinite homomorphic image F of G such that F is also a homomorphic image ofA or of B.

We prove (1) by induction on α. Since Cω = A, it is obvious for α = ω. Assume it istrue for Cα. By Lemma 1.12(vii), Cα+ω

∼= Cα×A. If G is an infinite homomorphicimage of Cα+ω , then G is isomorphic to H × K for some homomorphic imagesH,K of Cα, A respectively, by Proposition 1.1. One of H,K is infinite, and so thedesired result follows from the inductive assumption on Cα.

Now suppose that α ∈ L2. By Proposition 1.10 let F be an infinite homo-morphic image of G such that F is a homomorphic image of B or of some Dα

i . Thedesired conclusion has been reached if F is a homomorphic image of B. Supposethat F is a homomorphic image of Dα

i . Obviously Dαi is a homomorphic image of

Cci+1α

. Hence the inductive hypothesis applies to finish the proof. �

One-point gluing

Our next algebraic operation is one-point gluing. Suppose we are given a system〈Ai : i ∈ I〉 of BAs, and a corresponding system 〈Fi : i ∈ I〉 of ultrafilters: Fi is anultrafilter on Ai for each i ∈ I. The one-point gluing of the pair (〈Ai : i ∈ I〉, 〈Fi :i ∈ I〉) is the following subalgebra of the direct product

∏i∈I Ai:{

x ∈∏i∈I

Ai : for all i, j ∈ I(xi ∈ Fi iff xj ∈ Fj)

}.

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In the case of two factors Ai and Aj this amounts to identifying the two pointsFi and Fj in the disjoint union of the Stone spaces; this is a special case of thefollowing theorem.

Theorem 1.15. Let 〈Ai : i ∈ I〉 be a system of BAs, and let 〈Fi : i ∈ I〉 be a systemwith Fi an ultrafilter on Ai for each i ∈ I. Let C =

∏i∈I Ai, and let B be the

one-point gluing of the pair (〈Ai : i ∈ I〉, 〈Fi : i ∈ I〉).Then for each i ∈ I the set F ′

idef= {x ∈ C : xi ∈ Fi} is an ultrafilter on

C. Further, let K = {F ′i : i ∈ I}. Let X be the quotient of Ult(C) obtained by

collapsing K to a point. Then Ult(B) is homeomorphic to X.

Proof. The first assertion of the theorem is obvious. Now let π be the naturalcontinuous mapping of Ult(C) onto X . We now define g from Ult(B) into X bysetting g(F ∩B) = π(F ) for any ultrafilter F on C. To see that g is well defined,suppose that F ∩B = G∩B, where F and G are ultrafilters on C. If both F andG are in K, obviously π(F ) = π(G). Now suppose, say, that F /∈ K. We claimthen that F = G. Suppose to the contrary that F �= G. Choose u ∈ F\G. Nowchoose x ∈ F such that SC(x) ∩ K = 0. Here SC is the Stone map associatedwith C. If (x · u)i ∈ Fi for some i ∈ I, then x · u ∈ F ′

i , and hence SC(x) ∩K �= 0,contradiction. Thus (x · u)i /∈ Fi for all i ∈ I, and consequently x · u ∈ B. Butx · u ∈ F , so x · u ∈ G and so u ∈ G, contradiction.

g is one-one: suppose that F and G are ultrafilters on C and π(F ) = π(G);we want to show that F ∩B = G∩B. We may assume that F,G ∈ K. Let x ∈ B;by symmetry we want to show that if x ∈ F then x ∈ G. So, assume that x ∈ F .Thus F ∈ SC(x) so, since F ∈ K, we can choose i ∈ I so that F ′

i ∈ SC(x). Thusx ∈ F ′

i and so xi ∈ Fi. If x /∈ G, then −x ∈ G, and a similar argument gives−xj ∈ Fj for some j ∈ I. This contradicts the assumption that x ∈ B.

g maps onto X : let F ∈ X . If F /∈ K, then g(F ∩ B) = F . If F is the pointwhich is the collapse of K, then for any i ∈ I we have g(F ′

i ∩B) = F .

g is continuous: suppose that U is open in X , and F ∩B ∈ g−1[U ], where Fis an ultrafilter on C. Thus π(F ) ∈ U , and hence F ∈ π−1[U ]. So, choose x ∈ Cso that F ∈ SC(x) ⊆ π−1[U ].

Case 1. F /∈ K. Choose y ∈ C so that F ∈ SC(y) and SC(y) ∩K = ∅. Thus F ∈SC(x ·y). Now if (x ·y)i ∈ Fi, then x ·y ∈ F ′

i , so y ∈ F ′i ∈ K, contradicting SC(y)∩

K = ∅. This being true for all i, it follows that x · y ∈ B. We claim that F ∩B ∈SB(x·y) ⊆ g−1[U ]. Obviously F∩B ∈ SB(x·y). Suppose that x·y ∈ G∩B, whereGis an ultrafilter on C. ThenG ∈ SC(x), and hence g(G∩B) = π(G) ∈ U , as desired.

Case 2. F ∈ K. Thus K ⊆ π−1[U ]. Choose y ∈ C such that K ⊆ SC(y) ⊆ π−1[U ].Thus y ∈ B. We claim that F ∩B ∈ SB(y) ⊆ g−1[U ]. Obviously F ∩B ∈ SB(y).Now suppose that G∩B ∈ SB(y), where G is an ultrafilter on C. Then G ∈ SC(y),and g(G ∩B) = π(G) ∈ U , as desired.

Finally, it is clear that X is Hausdorff. �

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The Alexsandroff duplicate

Given a BA A, its Alexsandroff duplicate, denoted by Dup(A), is the subalgebraof A×P(Ult(A)) whose set of elements is

{(a,X) : a ∈ A,X ⊆ Ult(A), and S(a)�X is finite}.

(It is easy to check that this is a subalgebra of A×P(Ult(A)).)

We show now that this is equivalent to the usual definition. (See, for exam-ple, Gardner, Pfeffer [84], page 1010.) That definition runs like this. Let X be atopological space. We put a topology on X × 2 as follows: a base consists of allsets

• F × {1}, F any subset of X ;

• (G× 2)\(F × {1}), G open in X , F a finite subset of X .

Theorem 1.16. Let A be a BA. Under the above topology, Ult(A)× 2 is a Booleanspace, and Dup(A) is isomorphic to Clop(Ult(A)× 2).

Proof. Hausdorff: If U and V are ultrafilters on A, then we have disjoint openneighborhoods

{(U, 1)} and {(V, 1)} for the elements (U, 1) and (V, 1) when these are distinct;

{(U, 1)} and (Ult(A)× 2)\{(U, 1)} for the elements (U, 1) and (V, 0).

{(V, 1)} and (Ult(A)× 2)\{(V, 1)} for the elements (V, 1) and (U, 0).

S(a)× 2 and S(−a)× 2 for the elements (U, 0) and (V, 0), for distinct U, V , wherea ∈ U\V .

Now we show that X × 2 is compact. Let O be a cover of Ult(A) × 2 bymembers of the given base. Then {G : (G × 2)\(F × {1}) ∈ O for some openG ⊆ X and some finite F ⊆ X} is an open cover of Ult(A), so we can choose

(G1 × 2)\(F1 × {1}), . . . , (Gm × 2)\(Fm × {1}) ∈ O

such that G1, . . . , Gm is a cover of Ult(A). There are only finitely many elementsof Ult(A)× 2 remaining – some of the elements of (F1 × {1})∪ · · · ∪ (Fm ×{1}) –so O has a finite subcover.

Next we determine the clopen subsets of Ult(A)×2. Clearly each set F ×{1}is clopen, when F is a finite subset of Ult(A). And if G is a clopen subset ofUlt(A) and F is a finite subset of Ult(A), then (G× 2)\(F × {1}) is clopen, sinceits complement is [(Ult(A)\G)× 2]∪ (F ×{1}). These two kinds of clopen subsetsform a base for the topology, so every clopen set is a finite join of these two kinds.

So Ult(A) × 2 is a Boolean space. Now we define a function f which willextend to the desired isomorphism. For any a ∈ A let f(a,S(a)) = S(a) × 2, andfor any ultrafilter F on A let f(0, {F}) = {F}×{1}. So f maps a set of generators

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of Dup(A) onto a set of generators of Clop(Ult(A) × 2). An easy application ofSikorski’s extension criterion shows that f extends to a one-one homomorphism,as desired. We give this argument.

Let a0, . . . , an−1 be distinct elements of A, and F0, . . . , Fm−1 distinct ultra-filters on A, with n,m > 0. With other natural assumptions,

(a0,S(a0))ε(0) · . . . · (an−1,S(an−1))ε(n−1) · (0, {F0})δ(0) · . . . · (0, {Fm−1})δ(m−1) = 0

iff one of the following holds:

(1) There are distinct i, j < m such that δ(i) = δ(j) = 1.

(2) There is exactly one i < m such that δ(i) = 1, and

∏j<n

aε(j)j /∈ Fi.

(3) Each δ(i) = 0 and∏

i<n aε(i)i = 0.

On the other hand,

(S(a0)×2)ε(0)·. . .·(S(an−1)×2)ε(n−1)·({F0}×{1})δ(0)·. . .·({Fm−1}×{1})δ(m−1)= 0

holds iff the same conditions hold. �

We give some more simple facts about the Alexsandroff duplicate. Note that theduplicate is always atomic. The following result is easy.

Proposition 1.17. For any BA A, define g : A → Dup(A) by setting g(a) =(a,S(a)) for any a ∈ A. Then g is an isomorphism from A into Dup(A).

The special nature of the duplicate is brought out by the following simpletheorem. For any BA A, let IAat be the ideal of A generated by its atoms.

Theorem 1.18. Let A be any BA. Then A/IAat∼= Dup(A)/I

Dup(A)at .

Proof. Let π be the natural homomorphism from Dup(A) onto Dup(A)/IDup(A)at .

We show that π ◦ g is a homomorphism from A onto Dup(A)/IDup(A)at with kernel

IAat, where g is as in Proposition 1.17. If (a,X) ∈ Dup(A), then (a,S(a))�(a,X) =

(0,S(a)�X) ∈ IDup(A)at ; thus π◦g maps onto Dup(A)/I

Dup(A)at . Now for any a ∈ A,

(π ◦ g)(a) = 0 iff (a,S(a)) ∈ IDup(A)at iff a ∈ IAat, as desired. �

A corollary of this theorem is that if A is superatomic, then so is Dup(A). (It iseasy to see that for any BA B, B is superatomic iff B/IBat is superatomic.)

The ultrafilters on Dup(A) are characterized as follows.

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Proposition 1.19. Let A be an infinite BA and F ⊆ Dup(A). Then F is an ultra-filter on Dup(A) iff one of the following conditions holds:

(i) There is an ultrafilter G on A such that F = {(a,X) ∈ Dup(A) : (0, {G}) ≤(a,X)}.

(ii) There is an ultrafilter G on A such that F = {(a,X) ∈ Dup(A) : a ∈ G}.

Proof. Clearly each of (i), (ii) implies that F is an ultrafilter on Dup(A). Nowsuppose that F is an ultrafilter on Dup(A). If (0, X) ∈ F for some finite nonemptyX , clearly (i) holds. So suppose that (0, X) /∈ F for every finite nonempty X . LetG = {a ∈ A : (a,S(a)) ∈ F}. Clearly G is an ultrafilter on A. Suppose thata ∈ G and (a,X) /∈ F . So (a,S(a)), (−a,Ult(A)\X) ∈ F , hence (0,S(a)\X) ∈ F ,from which it follows that S(a) ⊆ X . Then (a,S(a)) ≤ (a,X), so (a,X) ∈ F ,contradiction. Hence ∀a ∈ G∀X [S(a)�X finite → (a,X) ∈ F ]. Now suppose that(a,X) ∈ F and a /∈ G. so (−a,S(−a))) ∈ F , hence (0, X\S(a)) ∈ F , henceX ⊆ S(a). Now (a,X) ≤ (a,S(a)), so (a,S(a)) ∈ F , contradiction. �

The exponential

We describe here for Boolean algebras the notion of a hyperspace – the space ofclosed subsets of a topological space with the Vietoris topology. Later we considervarious cardinal functions on the associated Boolean algebra, but we have notattempted to describe all known results about it.

Let X be any topological space. Exp (X) is the collection of all non-emptyclosed subspaces of X . We topologize it by taking the collection of sets of thefollowing form as a base:

V (U1, . . . , Um)def= {F ∈ Exp(X) : F ⊆ U1 ∪ · · · ∪ Um and F ∩ Ui �= 0 for all i},

where U1, . . . , Um are open in X . We call Exp (X) the exponential of X .

Theorem 1.20. Let X be a compact Hausdorff space. Then Exp (X) is also Haus-dorff and compact. Moreover, if X is a Boolean space, then so is Exp (X), and theset

{V (U1, . . . , Um) : each Ui clopen}

is a collection of clopen sets forming a base for the topology on Exp (X).

Proof. Hausdorff: suppose that F and G are distinct non-empty closed sets. Sayx ∈ F\G. Let W and U be disjoint open sets such that x ∈ W and G ⊆ U . ThusG ∈ V (U), F ∈ V (X,W ), and V (U) ∩ V (X,W ) = 0.

Compact: First we note some facts. For each open U in X let TU = {F ∈Exp(X) : F ∩ U �= 0}. This is an open set, since TU = V (X,U). Next,

(1) {V (U) : U open} ∪ {TU : U open} is a subbase for the topology on Exp(X).

For, V (U1, . . . , Um) = V (U1 ∪ · · · ∪ Um) ∩ TU1 ∩ · · · ∩ TUm .

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Now to prove compactness of Exp(X), suppose that O is a cover of Exp(X)by subbase members in (1).

Case 1. {W : TW ∈ O} covers X . Choose W1, . . . ,Wm with each TWi ∈ O suchthat X = W1 ∪ · · · ∪Wm. Then {TW1 , . . . , TWm} covers Exp(X), as desired.

Case 2. {W : TW ∈ O} does not cover X . Let Y =⋃

TW∈O W . So Y is a properopen subset of X , and hence X\Y ∈ Exp(X). Therefore X\Y ∈ V (U) for someV (U) ∈ O. Thus X\Y ⊆ U , so X\U ⊆ Y . Since X\U is compact, there existW1, . . . ,Wm with each TWi ∈ O such that X\U ⊆W1∪· · ·∪Wm. Hence {V (U)}∪{TW1 , . . . , TWm} covers Exp(X), as is easily verified.

Now assume that X is a Boolean space. Then

(2) If U1, . . . , Um are clopen in X , then V (U1, . . . , Um) is clopen in Exp(X).

For, suppose that F ∈ Exp(X)\V (U1, . . . , Um).

Case 1. F �⊆ U1∪· · ·∪Um. Then for some Γ ⊆ {1, . . . ,m} we have F ∈ V (X\(U1∪· · · ∪ Um), 〈Ui〉i∈Γ), and this set is disjoint from V (U1, . . . , Um).

Case 2. F ⊆ U1 ∪ · · · ∪ Um. Then there is an i such that F ∩ Ui = ∅. HenceF ∈ V (X\Ui), and V (X\Ui) is disjoint from V (U1, . . . , Um).

(3) If U1, . . . , Um,W1, . . . ,Wn are open and F ∈ V (U1, . . . , Um)∩V (W1, . . . ,Wn),then F ∈ V (〈Ui ∩Wj : F ∩ Ui ∩Wj �= 0〉).

(4) {V (U1, . . . , Um) : each Ui clopen} is a base for the topology on Exp(X).

To prove (4), assume that F ∈ V (U1, . . . , Um) with each Ui open; we want to findclopen W1, . . . ,Wn such that F ∈ V (W1, . . . ,Wn) ⊆ V (U1, . . . , Um). It suffices by(3) to show that the set

{Exp(X)\V (U1, . . . , Um)} ∪ {V (W1, . . . ,Wn) :

each Wi clopen and F ∈ V (W1, . . . ,Wn)}

has empty intersection. Suppose to the contrary that G is in each member of thisset.

Case 1. G �⊆ U1∪· · ·∪Um. Thus G �⊆ F ; say x ∈ G\F . Let W be a clopen set suchthat x ∈W and W ∩F = 0. Thus F ∈ V (X\W ), so G ∈ V (X\W ), contradiction.

Case 2. G ⊆ U1 ∪ · · · ∪ Um. Since G ∈ Exp(X)\V (U1, . . . , Um), it follows thatG ∩ Ui = 0 for some i. Say x ∈ F ∩ Ui. Let W be clopen with x ∈ W andW ∩ G = 0. Then F ∈ V (W,X\W ) or F ∈ V (W ), so G ∈ V (W,X\W ) orG ∈ V (W ), contradiction. �

For any BA A, we denote by Exp(A) the Boolean algebra Clop(Exp(Ult(A))); thisis called the exponential of A.

The following somewhat technical result will be useful.

Proposition 1.21. For any BA A, Exp (A) is generated by {V (S(a)) : a ∈ A}.

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Proof. We already know from Theorem 1.20 that Exp(A) is generated by the set{V (U1, . . . , Um) : each Ui clopen}. So it suffices to see that each element of thisset is generated by {V (S(a)) : a ∈ A}. This follows from:

V (S(a1), . . . ,S(am)) = V (S(a1+ · · ·+am))\(V (S(−a1))∪· · ·∪V (S(−am))). �

Lemma 1.22.

(i) If x �= y, then V (S(x)) �= V (S(y)).(ii) V (S(a)) = {X ⊆ S(a) : X is closed and nonempty}.(iii) V (S(a0)) ∩ · · · ∩ V (S(am−1)) = V (S(a0 · . . . · am−1)).

(iv) V (S(a),S(−a)) = −(V (S(a)) ∪ V (S(−a))).(v) −V (S(a)) = V (S(a),S(−a)) ∪ V (S(−a)).(vi) The following are equivalent:

(a) V (S(a0)) ∩ · · · ∩ V (S(am−1)) ∩−V (S(b0)) ∩ · · · ∩ −V (S(bn−1)) = ∅.(b) ∃i < n[a0 · . . . · am−1 ≤ bi].

Proof. For (i), say x �≤ y. Let F be an ultrafilter such that x · −y ∈ F . Then{F} ∈ V (S(x))\V (S(y)).

(ii)–(iv) are clear.

For (v), note that V (S(−a)) ⊆ −V (S(a)). Hence

V (S(a),S(−a)) ∪ V (S(−a))= −(V (S(a)) ∪ V (S(−a))) ∪ (V (S(−a)) ∩ −V (S(a))= (−V (S(a)) ∩ −V (S(−a))) ∪ (V (S(−a)) ∩ −V (S(a))= −V (S(a)).

For (vi), first assume (a). We may assume that a0 · . . . ·am−1 �= 0. Then S(a0 · . . . ·am−1) ∈ V (S(a0)) ∩ · · · ∩ V (S(am−1)), and so by (a) there is an i < n such thatS(a0 · . . . · am−1) ∈ V (S(bi)). Hence (b) holds. (b)⇒(a) similarly. �

Proposition 1.23. If A is atomic, then so is Exp(A).

Proof. For each atom a of A, let Fa be the principal ultrafilter determined by a.Suppose that 〈a0, . . . , am−1〉 is a system of distinct atoms of A. Then

V (S(a0), . . . ,S(am−1)) = {{Fa0 , . . . , Fam−1}},

and this is hence an atom of Exp(A). Now take any nonzero x ∈ Exp(A). Toshow that there is an atom below x it suffices to take the case in which x has theform V (S(b0), . . . ,S(bm−1)). Note that each bi is nonzero, since x �= 0, and eachelement of x has nonempty intersection with each S(bi). Let ai be an atom belowbi for each i < m. Then {{Fa0 , . . . , Fam−1}} is the desired atom. �

Proposition 1.24. If A is atomless, then so is Exp(A).

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Proof. Suppose that 0 �= x ∈ Exp(A); we want to find 0 < y < x. We may assumethat x has the form V (S(b0), . . . ,S(bm−1)). Note that each bi is nonzero. And weclearly may assume that all the bi are distinct from one another. Finally, we mayassume that b0 is minimal among them, i.e., if bi ≤ b0 then i = 0. Now choose0 < a0 < b0. We claim that the desired element is

y = V (S(a0),S(b1 · −b0), . . . ,S(bm−1 · −b0)).

Clearly y ⊆ x. Let u = a0 + b1 · −b0 + · · ·+ bm−1 · −b0. Then S(u) is a member ofy, so y �= 0. Let v = b0 ·−a0+ b1 ·−b0+ · · ·+ bm−1 ·−b0. Clearly S(v) is a memberof x. Since clearly v · a0 = 0, it is not a member of y. So y < x. �

From this proposition it follows that the free BA on ω free generators is isomor-phic to its own exponential. Sirota [68] proved that also the free BA on ω1 freegenerators is isomorphic to its own exponential. But Shapiro [76a], [76b] showedthat this does not extend to higher cardinals.

The definition of Exp(A) can be given an equivalent formulation which ismore algebraic. Let Id′(A) be the set of all proper ideals of A. For each a ∈ A let

Xa = {I ∈ Id′(A) : ∃b ∈ I[−b ≤ a]},Ya = {I ∈ Id′(A) : a,−a /∈ I} ∪X−a.

Lemma 1.25. Id′(A)\Xa = Ya.

Proof. First suppose that I ∈ Id′(A)\Xa, and suppose that I /∈ X−a. Thus ∀b ∈I[−b · −a �= 0], so −a /∈ I. Also ∀b ∈ I[−b · a �= 0], so a /∈ I. This proves ⊆.

Second, suppose that I ∈ Ya, but suppose that I ∈ Xa. Choose b ∈ I so that−b ≤ a. Thus −a ∈ I. Hence I ∈ X−a. Choose c ∈ I such that −c ≤ −a. Thena ≤ c, so 1 = a+−a ≤ c+ b and I is not proper, contradiction. �

Let Exp′(A) be the subalgebra of P(Id′(A)) generated by all elements Xa.

Theorem 1.26. Exp(A) ∼= Exp′(A).

Proof. For any a ∈ A let f(V (S(a))) = Xa. f is well defined by Lemma 1.22(i).By Proposition 1.21 it suffices to show that f extends to an isomorphism. Weuse Sikorski’s extension criterion. Suppose that a1, . . . , am, b1, . . . , bn are distinctelements of A with m,n > 0. We want to show that the following two conditionsare equivalent:

(1) V (S(a1) ∩ · · · ∩ V (S(am) ∩ −V (S(b1) ∩ · · · ∩ −V (S(bn) = ∅.

(2) Xa1 ∩ · · · ∩Xam ∩ Yb1 ∩ · · · ∩ Ybn = ∅.

By Lemma 1.22(vi), condition (1) is equivalent to

(3) ∃i < n[a1 · . . . · am ≤ bi].

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Assume (3), and suppose that I is a member of (2). Then for all j = 1, . . . ,m thereis a cj ∈ I such that −cj ≤ aj . Hence −c1 · . . . · −cm ≤ a1 · . . . · am ≤ bi. Moreover,c1 + · · · + cm ∈ I. Hence −bi ∈ I, so I ∈ X−bi since I ∈ Ybi . Choose d ∈ I suchthat −d ≤ −bi. Hence bi ≤ d, so −c1 · . . . ·−cm ·−d = 0, hence c1+ · · ·+cm+d = 1and so I is not proper, contradiction.

Now assume that (2) holds, while ∀i < n[a1 · . . . · am · −bi �= 0]. Let I = A �(−a1+ · · ·+−am). Then for any i = 1, . . . ,m we have −ai ∈ I, and hence I ∈ Xai .Hence by (2), there is an i with 1 ≤ i ≤ m such that I /∈ Ybi . We have −bi �≤−a1 + · · ·+−an, so −bi /∈ I. It follows that bi ∈ I. Hence bi ≤ −a1 + · · ·+−am,so a1 · . . . · am ≤ −bi. It follows that I ∈ X−bi , hence I ∈ Ybi , contradiction. �

To make this discussion of the exponential more concrete, we describe the expo-nential for A the finite-cofinite algebra on an infinite cardinal κ. For each α < κlet Fα be the ultrafilter of all Γ ∈ A such that α ∈ Γ. Let G be the ultrafilter ofall cofinite subsets of κ. Thus Ult(A) = {G} ∪ {Fα : α < κ}. Now {S(a) : a ∈ A}is a basis for Ult(A). Note:

a finite ⇒ S(a) = {Fα : α ∈ a};a cofinite ⇒ S(a) = {G} ∪ {Fα : α ∈ a}.

Thus each Fα is isolated. The open subsets of Ult(A) are

{Fα : α ∈ Γ} for any Γ ⊆ κ;

{G} ∪ {Fα : α ∈ Γ} for any cofinite Γ ⊆ κ.

Hence the closed sets are

yΓdef= {G} ∪ {Fα : α ∈ Γ} for any Γ ⊆ κ,

zΓdef= {Fα : α ∈ Γ} for any finite Γ ⊆ κ.

Hence

Exp(Ult(A)) = {yΓ : Γ ⊆ κ} ∪ {zΓ : 0 �= Γ a finite subset of κ}.

Now we claim:

(1) Each zΓ is isolated, Γ a finite nonempty subset of κ.

In fact, write Γ = {α0, . . . , αm−1}, m > 0. Then {zΓ} = V ({Fα0}, . . . , {Fαm−1}),as desired.

The following two statements are obvious:

(2) If a0, . . . , am−1 ∈ A are all finite and m > 0, then

V (S(a0), . . . ,S(am−1)) = {zΓ : Γ ⊆ a0 ∪ · · · ∪ am−1 and Γ∩ ai �= 0 for all i < m}.

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(3) If a0, . . . , am−1 ∈ A and some ai is cofinite, then

V (S(a0), . . . ,S(am−1)) = {yΓ : Γ ⊆ a0 ∪ · · · ∪ am−1 and Γ ∩ ai �= 0

for all i such that ai is finite}∪ {zΓ : Γ ⊆ a0 ∪ · · · ∪ am−1, Γ finite , Γ ∩ ai �= 0

for all i < m}.

Next,

(4) No yΓ is isolated.

For, suppose that yΓ ∈ V (S(a0), . . . ,S(am−1)), m > 0. Since G ∈ yΓ, some ai iscofinite. By the above, if⋃

{ai : ai finite } ⊆ Δ ⊆ a0 ∪ · · · ∪ am−1,

then yΔ ∈ V (S(a0), . . . ,S(am−1)). Thus V (S(a0), . . . ,S(am−1)) has infinitelymany members, as desired. We also proved:

(5) {yΓ : Γ ⊆ κ}, as a subspace of Ult(A), is closed and has no isolated points.

The following is obvious:

(6) {{zΓ} : Γ finite and non-empty} is the set of all atoms of Exp(A), which isatomic.

Let xα = V (S(κ\{α}),S({α})) for all α < κ.

(7) 〈xα/fin : α < κ〉 is a system of independent elements of Exp(A)/fin.

To show this, suppose that Γ and Δ are finite disjoint subsets of κ; we want toshow that

⋂α∈Γ xα ∩

⋂α∈Δ−xα is not a finite sum of atoms. Note by (3) that

xα = {yΩ : α ∈ Ω}∪ {zΩ : α ∈ Ω and Ω �= {α}, Ω finite}.

It follows that yΓ ∈⋂

α∈Γ xα ∩⋂

α∈Δ−xα. Hence (7) holds.

(8) 〈xα/fin : α < κ〉 generates Exp(A)/fin.To prove this, by Proposition 1.21 it suffices to show that if a is cofinite thenV (S(a))/fin is generated by 〈xα/fin : α < κ〉. So (8) follows from

(9) V (S(a))/fin =⋂

α∈κ\a−xα/fin.

To prove this, first note that if α ∈ κ\a, then

V (S(a)) ∩ xα = ({yΓ : ∅ �= Γ ⊆ a} ∪ {zΓ : 0 �= Γ ⊆ a, Γ finite})∩ ({yΓ : α ∈ Γ} ∪ {zΓ : α ∈ Γ, Γ �= {α}, Γ finite})

= ∅.

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This proves ≤ in (9). For the other direction, first note that

−xα = {yΓ : α /∈ Γ} ∪ {zΓ : (α /∈ Γ or Γ = {α}) and Γ finite}.

Hence⎛⎝ ⋂

α∈κ\a−xα

⎞⎠ \V (S(a)) = ({yΓ : Γ ∩ (κ\a) = 0}

∪ {zΓ : ∀α ∈ κ\a(α /∈ Γ or Γ = {α}) and Γ finite})∩ ({yΓ : Γ �⊆ a} ∪ {zΓ : Γ �⊆ a, Γ finite})

= {z{α} : α ∈ κ\a},

as desired.

Some further properties of the exponential will be developed in the discussionof semigroup algebras in the next chapter.

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2 Special Classes of Boolean Algebras

We discuss several special classes of Boolean algebras not mentioned in the Hand-book.

Semigroup algebras

The notion of a semigroup algebra is due to Heindorf [89]. We give basic definitionsand facts only. A subset H of a BA A is said to be disjunctive if 0 /∈ H , andh, h1, . . . , hn ∈ H and h ≤ h1 + · · ·+ hn (n > 0) imply that h ≤ hi for some i. IfP is any partially ordered set, M ⊆ P , and p ∈ P , we define

M ↑ p = {a ∈M : p ≤ a};M ↓ p = {a ∈M : a ≤ p}.

Proposition 2.1. Let A be a BA and H ⊆ A+. Then H is disjunctive iff for everyM ⊆ H there is a homomorphism f from 〈H〉 into P(M) such that f(h) = M ↓ hfor all h ∈ H.

Proof. ⇒: In order to apply Sikorski’s extension criterion, assume that h1, . . . , hm,k1, . . . , kn ∈ H and h1 · . . . · hm ≤ k1 + · · · + kn; we want to show that (M ↓h1)∩· · ·∩(M ↓ hm) ⊆ (M ↓ k1)∪· · ·∪(M ↓ kn). Let x ∈ (M ↓ h1)∩· · ·∩(M ↓ hm).Then x ≤ h1 · . . . · hm, so x ≤ k1 + · · · + kn. Note that n > 0, since otherwisex = 0, contradicting M ⊆ H ⊆ A+. Hence by disjunctiveness, x ≤ ki for some i;so x ∈ (M ↓ ki), as desired.

⇐: Suppose that h, h1, . . . , hm ∈ H and h ≤ h1 + · · · + hm (m > 0). LetM = {h, h1, . . . , hm}, and take the function f corresponding to M . Then

h ∈ (M ↓ h) = f(h) ⊆ f(h1) ∪ · · · ∪ f(hm) = (M ↓ h1) ∪ · · · ∪ (M ↓ hm),

so h ≤ hi for some i. �

A semigroup is an algebra (H, ·) such that H is a nonempty set and · is an as-sociative binary operation on H . A zero of H is an element z ∈ H such thatz · x = x · z = z for all x ∈ H . An identity of H is an element e ∈ H such thate · x = x = x · e for all x ∈ H . Clearly a zero element, if it exists, is unique, and

, ,OI 10.1007/978-3- - - _ ,


014


Basel 2

353

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36 Chapter 2. Special Classes of Boolean Algebras

similarly for identity elements. Also it is clear that if f is a homomorphism froma semigroup H onto a semigroup K, then f preserves the zero and the identity, ifthey exist.

A BA A is a semigroup algebra if it is generated by a subset H with thefollowing properties:

(1) 0, 1 ∈ H ;

(2) H is closed under the operation · of A;(3) H\{0} is disjunctive.Here are three important examples of semigroup algebras:

A. Tree algebras. Let A = TreeAlg(T ). Without loss of generality T has only oneroot. Set H = {T ↑ t : t ∈ T }∪ {0, 1}. The conditions for a semigroup algebra areeasily verified.

B. Interval algebras. Let A = IntAlg(L), where L is a linear ordering with firstelement 0L. Let H = {[0L, a) : a ∈ L} ∪ {1}. Again the indicated conditions areeasily checked.

C. Free algebras. Let A be freely generated by X , and set H = {x ∈ A : x is afinite product of members of X} ∪ {0, 1}. The indicated conditions clearly hold.

It is also useful to note that if A is a semigroup algebra, then so is Dup(A). Weprove this now. Let H be a generating set for A satisfying conditions (1)–(3). Wedefine

H ′ = {(a, S(a)) : a ∈ H} ∪ {(0, {F}) : F ∈ Ult(A)}.We claim that H ′ shows that Dup(A) is a semigroup algebra. Clearly (0, 0), (1, 1) ∈H ′ and H ′ is closed under ·. To show that H ′ is disjunctive, we consider twopossibilities:

(a, S(a)) ≤ h0 + · · ·+ hn

where a �= 0. Say that hi has the form (bi, S(bi)) for all i ∈ M , and the form(0, {Fi}) for all i ∈ (n+1)\M . Hence a ≤

∑i∈M bi, hence a ≤ bi for some i, hence

(a, S(a)) ≤ hi.

Second possibility:(0, {F}) ≤ h0 + · · ·+ hn

The desired conclusion is clear.

Thus H ′\{(0, 0} is disjunctive, so (1)–(3) hold.

Finally, we need to show that H ′ generates Dup(A). Take any (a,X) ∈Dup(A). If a = 0, then X is a finite set of ultrafilters, and so (a,X) is gen-erated by H ′. Suppose that a �= 0. Then since H generates A, we can write

a =∑

i<m

∏j<n b

ε(i,j)ij with each bij ∈ H . Then

(a, S(a)) =∑i<m

∏j<n

(bij , S(bij)ε(i,j),

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Chapter 2. Special Classes of Boolean Algebras 37

so that (a, S(a)) is generated by H ′. Finally, let F = S(a)�X ; so F is finite, andX = S(a)�F . So

(a,X) = [(a, S(a))\(0, F )] ∪ [(0, F\S(a))],

as desired.

Proposition 2.2. Suppose that A is a semigroup algebra with associated semigroupH, B is a BA, and f is a homomorphism from (H, ·) into the semigroup (B, ·)preserving 0 and 1. Then f has a unique extension to a homomorphism from Ainto B. Moreover, if f is onto, the extension is too. Finally, if B is a semigroupalgebra on (K, ·) and f is an isomorphism from (H, ·) into (K, ·) preserving 0 and1, then the extension is an isomorphism into.

Proof. In order to apply Sikorski’s criterion, let b0, . . . , bm−1, c0, . . . , cn−1 be dis-tinct elements of H and suppose that

b0 · . . . · bm−1 · −c0 · . . . · −cn−1 = 0.

Without loss of generality, m > 0 and each ci is different from 0. If n = 0, then

f(b0) · . . . · f(bm−1) = f(b0 · . . . · bm−1) = f(0) = 0,

as desired. Assume that n > 0. Then b0 ·. . .·bm−1 ≤ c0+· · ·+cn−1, so b0 ·. . .·bm−1 ≤ci for some i; hence b0 · . . . · bm−1 · ci = b0 · . . . · bm−1 and

f(b0) · . . . · f(bm−1) · −f(c0) · . . . · −f(cn−1)

= f(b0 · . . . · bm−1) · −f(c0) · . . . · −f(cn−1)

= f(b0 · . . . · bm−1 · ci) · −f(c0) · . . . · −f(cn−1)

= f(b0) · . . . · f(bm−1) · f(ci) · −f(c0) · . . . · −f(cn−1)

= 0,

as desired. Clearly if f is onto, then the extension is onto.

Assume the hypothesis of “Finally. . .”. Let b0, . . . , bm−1, c0, . . . , cn−1 be dis-tinct elements of H such that

f(b0) · . . . · f(bm−1) · −f(c0) · . . . · −f(cn−1) = 0.

We want to show that b0 · . . . · bm−1 · −c0 · . . . · −cn−1 = 0. Wlog m,n > 0. Thusf(b0 · . . . · bm−1) ∈ K, and f(b0 · . . . · bm−1) ≤ f(c0) + · · ·+ f(cn−1), so there is ani < n such that f(b0 ·. . .·bm−1) ≤ f(ci). Hence f(b0 ·. . .·bm−1) = f(b0 ·. . .·bm−1 ·ci),so b0 · . . . · bm−1 = b0 · . . . · bm−1 · ci since f is one-one, and the desired conclusionfollows. �Corollary 2.3. If A and B are semigroup algebras both with the same associatedsemigroup (H, ·), then there is an isomorphism from A onto B which fixes Hpointwise. �

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Now we indicate the connection of the exponential of a BA with semigroup alge-bras.

Proposition 2.4. For any BA A, Exp(A) is a semigroup algebra on a semigroupisomorphic to (A, ·).

Proof. For any a ∈ A let f(a) = V (S(a)), and let H = f [A]. We want to showthat Exp(A) is a semigroup algebra on H and f is an isomorphism from (A, ·)onto (H,∩) preserving 0 and 1. Clearly it preserves 0 and 1. If a, b ∈ A, then

f(a · b) = V (S(a · b)) = V (S(a) ∩ S(b)) = V (S(a)) ∩ V (S(b)) = f(a) ∩ f(b).

If a �= b, say a �≤ b; then S(a) ∈ V (S(a)) but S(a) /∈ V (S(b)); this shows that fis one-one. So we have checked that f is an isomorphism from (A, ·) onto (H,∩)preserving 0 and 1.

Note that H generates Exp(A) by Proposition 1.21. Finally, the disjunctiveproperty follows like this: suppose that V (S(a)) ⊆ V (S(b1)) ∪ · · · ∪ V (S(bm)),m > 0. Now S(a) ∈ V (S(a)), so S(a) ∈ V (S(bi)) for some i, and hence V (S(a)) ⊆V (S(bi)), as desired. �

The following result will also be useful.

Proposition 2.5. For any BA A, Exp(A) embeds in∏

n≥1 A∗n, where A∗n denotes

the free product of n copies of A.

Proof. We use the notation of the proof of Proposition 2.4. For each n ≥ 1 definegn : H → A∗n as follows:

gn(f(a)) =∏i<n

hi(a),

where hi is the natural embedding of A into the ith factor of A∗n. Clearly gnis a homomorphism from (H, ·) into (A∗n, ·) taking 0 to 0 and 1 to 1. Hence byProposition 2.2 it extends to a homomorphism, still denoted by gn, from Exp(A)into A∗n. For any x ∈ Exp(A) let (k(x))n = gn(x) for all n ≥ 1. Clearly k is ahomomorphism from Exp(A) into

∏n≥1 A

∗n, so it suffices to show that k is one-one. We take an arbitrary non-zero member of Exp(A); we may assume that ithas the form V (S(a0), . . . ,S(am−1)), with each ai �= 0. Then

k(V (S(a0), . . . ,S(am−1)))m

= gm(V (S(a0), . . . ,S(am−1)))

= gm(V (S(a0 + · · ·+ am−1))\(V (S(−a0)) ∪ · · · ∪ V (S(−am−1)))

=∏i<m

hi(a0 + · · ·+ am−1) · −∏i<m

hi(−a0) · . . . · −∏i<m

hi(−am−1)

=∏i<m

hi(a0 + · · ·+ am−1) ·∑i<m

hi(a0) · . . . ·∑i<m

hi(am−1)

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≥∏i<m

hi(a0 + · · ·+ am−1) ·∏i<m

hi(ai)

=∏i<m

hi(ai) �= 0. �

Proposition 2.6. For any BA, there is a homomorphism from Exp(A) onto A.

Proof. By Proposition 2.4, Exp(A) is a semigroup algebra on a semigroup H ,with an isomorphism f from (H, ·) onto (A, ·). Then by Proposition 2.2 there isan extension of f to a homomorphism from Exp(A) onto A. �

Proposition 2.7. If f is a homomorphism from A into B, then there is a homo-morphism g from Exp(A) into Exp(B). Moreover, if f is onto, then g may betaken to be onto.

Proof. By Proposition 2.4, the algebras Exp(A) and Exp(B) are semigroup alge-bras on semigroups H and K isomorphic to (A, ·) and (B, ·) respectively; hence fyields a homomorphism from H into K preserving 0 and 1, and the result followsby Proposition 2.2. The last statement of the proposition is obvious. �

Proposition 2.8. Any semigroup algebra can be isomorphically embedded in itsexponential. Hence for any BA A, the algebra Exp(A) can be isomorphically em-bedded in Exp(Exp(A)).

Proof. Let A be a semigroup algebra. By Proposition 2.4, there is an isomorphismf from (A, ·) onto a semigroup (K, ·) such that Exp(A) is a semigroup on (K, ·). ButA is a semigroup algebra on some semigroup (H, ·), so there is an isomorphism of(H, ·) into (K, ·). By the final part of Proposition 2.2, our proposition follows. �

Pseudo-tree algebras

A pseudo-tree is a partially ordered system (T,≤) such that for each t ∈ T theset T ↓ t is simply ordered. Thus this notion generalizes that of a tree, whereT ↓ t is required to be well ordered. We define Treealg(T ) to be the subalgebra ofP(T ) generated by {T ↑ t : t ∈ T }; such algebras are called pseudo-tree algebras.Pseudo-tree algebras are treated thoroughly in Koppelberg, Monk [92].

Much of the theory of tree algebras described in §16 of the BA Handbook,Vol. 1, carries over to pseudo-tree algebras. For the convenience of the reader, wegive a self-contained treatment of the fundamentals of the theory of pseudo-treealgebras.

Theorem 2.9. For any pseudo-tree T there is a pseudo-tree T ∗ with a smallestelement such that Treealg(T ) ∼= Treealg(T ∗).

Proof. We consider two cases.

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Case 1. There is a finite F ⊆ T such that T =⋃

t∈F (T ↑ t), with the members ofF distinct. Fix x ∈ F . We let T ∗ = T , with the new order given by

(∗) s ≤T∗ t iff s ≤T t or s = x.

Let R = T \{x}. Note that (T ∗ ↑ t) = (T ↑ t) for any t ∈ R. For any t ∈ R letf(T ↑ t) = (T ↑ t). Note that T ∗ ↑ x = T . Thus R generates Treealg(T ∗). Weclaim that f extends to an isomorphism from Treealg(T ∗) into Treealg(T ). Thisis clear by Sikorski’s extension criterion. Since (T ↑ x) = −

∑s∈F\{x}(T ↑ x), f

maps onto Treealg(T ).

Case 2. There is no F as in Case 1. Let x be a new element, not in T , and defineT ∗ = T ∪{x}. We define the order on T ∗ by (∗) again. Since T ∗ ↑ x = T ∗, it followsthat {(T ↑ t) : t ∈ T } generates Treealg(T ∗). For each t ∈ T let f(T ↑ t) = (T ↑ t)We claim that f extends to an isomorphism from Treealg(T ∗) into Treealg(T ). Toprove this, by Sikorski’s extension criterion we need to show that the followingconditions are equivalent:

(∗∗) (T ↑ s0) ∩ · · · ∩ (T ↑ sm−1) ∩ [T ∗\(T ↑ t0)] ∩ · · · ∩ [T ∗\(T ↑ tn−1] = ∅;(∗∗∗) (T ↑ s0) ∩ · · · ∩ (T ↑ sm−1) ∩ [T \(T ↑ t0)] ∩ · · · ∩ [T \(T ↑ tn−1] = ∅.

Since T ⊆ T ∗, it is obvious that (∗∗) implies (∗∗∗). Now suppose that (∗∗∗) holds.By the case assumption, m > 0. Hence x is not a member of the set of (∗∗), andso (∗∗) holds.

Obviously the extension of f maps onto Treealg(T ). �

The following normal form theorem is very useful.

Theorem 2.10. Let T be a pseudo-tree with a least element. Then every elementof Treealg(T ) can be written in the form

⋃t∈M

[(T ↑ t)\

⋃s∈Nt

(T ↑ s)],

where:

(i) M is a finite subset of T .

(ii) ∀t ∈M [Nt is a finite subset of (T ↑ t)\{t} consisting of pairwise incomparableelements].

(iii) For all distinct t, u ∈M , the elements

(T ↑ t)\⋃

s∈Nt

(T ↑ s) and (T ↑ u)\⋃

s∈Nu

(T ↑ s)

are disjoint.

(iv) For all distinct t, u ∈M , one of the following holds:

(a) t and u are incomparable.

(b) There is an s ∈ Nt such that s < u.

(c) There is an s ∈ Nu such that s < t.

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Proof. Let a be any element of Intalg(T ). If a = ∅, we can take M = ∅. Supposethat a �= ∅. Then we can write

a =⋃ε∈Γ

⋂t∈P

(T ↑ t)ε(t),

where P is a finite nonempty subset of T , Γ ⊆ P 2, and⋂

t∈P (T ↑ t)ε(t) �= ∅ for allε ∈ Γ. If u is the least element of T , then (T ↑ u) = T ; hence we may assume thatfor each ε ∈ Γ there is a t ∈ P such that ε(t) = 1. Since (T ↑ t)∩ (T ↑ u) = ∅ for tand u incomparable, it follows that for each ε ∈ Γ, all of the elements t ∈ P withε(t) = 1 are comparable, and hence they can be replaced by a single element. Sowe may write

a =⋃t∈M

[(T ↑ t) ∩

⋂s∈Nt

[T \(T ↑ s)]],

where M is a finite subset of T , for each t ∈ M the set Nt is finite, (T ↑ t) ∩⋂s∈Nt

[T \(T ↑ s)] �= ∅ for each t ∈M , and

(T ↑ t) ∩⋂

s∈Nt

[T \(T ↑ s)] ∩ (T ↑ t′) ∩⋂

s∈Nt′

[T \(T ↑ s)] = ∅

for distinct t, t′ ∈ M . If s ∈ Nt is incomparable with t, then T ↑ t is a subset ofT \(T ↑ s), and s can be omitted from Nt. If s ≤ t, then (T ↑ t) ⊆ (T ↑ s), hence(T ↑ t) ∩ [T \(T ↑ s)] = ∅, contradiction. Hence we may assume that t < s forall s ∈ Nt. If s, w ∈ Nt and s ≤ w, then (T ↑ w) ⊆ (T ↑ s), and hence [T \(T ↑s)] ⊆ [T \(T ↑ w)]. So we may assume that all members of Nt are incomparable.We have now arrived at a representation of a satisfying conditions (i)–(iii) of thetheorem. Take such a representation, satisfying these three conditions, with |M |as small as possible. We claim that then also condition (iv) holds. To prove this,suppose that t and u are distinct members of M and they are comparable; sayt < u. If ∀s ∈ Nt[s �≤ u], then u is a member of both (T ↑ t) ∩

⋂s∈Nt

[T \(T ↑ s)and (T ↑ u) ∩

⋂s∈Nu

[T \(T ↑ s), contradiction. So there is an s ∈ Nt such thats ≤ u. If s = u, then

a =

⎛⎝ ⋃

w∈M\{t,u}

[(T ↑ w) ∩

⋂s∈Nw

[T \(T ↑ s)]]⎞⎠

∪

⎛⎝(T ↑ t) ∩

⋂s∈Nt\{u}

[T \(T ↑ s)] ∩⋂

s∈Nu

[T \(T ↑ s)]

⎞⎠ ,

which contradicts the minimality of |M |. Thus (iv)(b) in the theorem holds. �

The normal form is illustrated on the next page.

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a = [(T ↑ t1)\[(T ↑ s1) ∪ (T ↑ s2)]] ∪ (T ↑ t2) ∪ [(T ↑ t3)\(T ↑ s3)]

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Now we want to give an abstract characterization of pseudo-tree algebras.For this purpose we need some easy propositions. A subset R of a BA A is aramification set provided that any two elements of R are either comparable ordisjoint. Thus R is then a pseudo-tree under the inverse ordering of the BA.

Proposition 2.11. Let (P,≤) be a partially ordered system. Then {P ↑ p : p ∈ P}is a disjunctive set in P(P ).

Proof. Obviously ∅ is not in the indicated set. Now suppose that p, p1, . . . , pn ∈ P ,where n > 0, and assume that P ↑ p ⊆ (P ↑ p1)∪· · ·∪ (P ↑ pn). Then p ∈ (P ↑ p),and hence p ∈ (P ↑ pi) for some i, and hence (P ↑ p) ⊆ (P ↑ pi), as desired. �

Corollary 2.12. Every pseudo-tree algebra is a semigroup algebra. �

Proposition 2.13. Let R be a disjunctive ramification set of non-zero elements andlet X,Y be finite subsets of R. Then

∏X ≤

∑Y iff one of the following three

conditions holds:

(1) X = ∅ and∑

Y = 1;

(2) x · y = 0 for some x, y ∈ X;

(3) x ≤ y for some x ∈ X, y ∈ Y .

Proof. Obviously any of (1)–(3) implies that∏

X ≤∑

Y . Now suppose that∏X ≤

∑Y and (1) and (2) do not hold. Note that if X = ∅ then

∑Y = 1; hence

X �= ∅. From the falsity of (2) it then follows that∏

X ∈ X and Y �= ∅. Thendisjunctiveness yields (3). �

Proposition 2.14. Let R ⊆ A+ be a ramification set, and let S be a subset of Rmaximal among disjunctive subsets of R. Then 〈S〉 = 〈R〉.

Proof. We need only show that R ⊆ 〈S〉; so let r ∈ R\S. Since r /∈ S, the setS ∪ {r} is not disjunctive. There are then two cases:

Case 1. r ≤ s1 + · · · + sn for certain s1, . . . , sn ∈ S (n > 0), but r �≤ si for all i.Let n be minimal such that this can happen. By the minimality, r · si �= 0 for alli, so si ≤ r and hence r = s1 + · · ·+ sn ∈ 〈S〉, as desired.Case 2. s1 ≤ r + s2 + · · · + sn for certain s1, . . . , sn ∈ S (n > 0), but s1 �≤ rand s1 �≤ si for all i > 1. Again, take n minimal for this situation. Since s1 �≤ r,we have n > 1. By the minimality of n and the fact that R is a ramificationset, the elements r, s2, . . . , sn are pairwise disjoint. Also by the minimality of n,s1 · r �= 0 �= s1 · si for all i > 1; hence r ≤ s1 and si ≤ s1 for all i > 1. Hences1 = r + s2 + · · · + sn, and it follows that r = s1 · −(s2 + · · · + sn) ∈ 〈S〉, asdesired. �

With these preliminaries over, we can now give our abstract characterization ofpseudo-tree algebras.

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Theorem 2.15. For any BA A, the following conditions are equivalent:

(i) A is isomorphic to Treealg(T ) for some pseudo-tree T with a smallest ele-ment;

(ii) A is isomorphic to Treealg(T ) for some pseudo-tree T ;

(iii) A is generated by a ramification set;

(iv) A is generated by a ramification set S ⊆ A+ such that 1 ∈ S and S isdisjunctive.

Proof. Obviously (i)⇒(ii), and it is also clear that (ii)⇒(iii). For (iii)⇒(iv), sup-pose that R is a ramification set which generates A. We may assume that 0 /∈ Rand 1 ∈ R. Then by Proposition 2.14 we get a ramification set S as desired in (iv).

Finally, we prove (iv)⇒(i). Clearly S is a pseudo-tree with smallest elementunder the converse of the Boolean ordering; so it suffices to show that A is iso-morphic to Treealg(S). By Proposition 2.1, there is a homomorphism f from Ainto P(S) such that f(s) = S ↑ s for all s ∈ S; here S ↑ s is in the tree sense.Clearly f maps onto Treealg(S). It is also one-one; we see this by using Sikorski’scriterion: assume that

t0 · . . . · tm−1 · −s0 · . . . · −sn−1 �= 0,

where all ti and si are in S and m,n ∈ ω. Since 1 ∈ S, we may assume that m > 0.

Then udef= t0 · . . . · tm−1 is an element of S, and u �≤ si for all i. Hence

u ∈ f(t0) ∩ · · · ∩ f(tm−1) ∩ −f(s0) ∩ · · · ∩ −f(sn−1),

as desired. �Proposition 2.16. If T is a pseudo-tree, then {T ↑ t : t ∈ T } is a disjunctiveramification set in Treealg(T ). �

Sikorski’s extension criterion takes the following form for pseudo-tree algebras withminimum element.

Proposition 2.17. Let T be a pseudo-tree with minimum element r. Suppose thatf maps {T ↑ t : t ∈ T } into a BA A. Then

(i) f extends to a homomorphism iff the following conditions hold:

(a) f(T � r) = 1.

(b) If s and t are incomparable elements of T , then f(T ↑ s) · f(T ↑ t) = 0.

(c) If s ≤ t, then f(T ↑ t) ≤ f(T ↑ s).(ii) f extends to a monomorphism iff (a)–(c) hold, along with:

(d) If w ∈ T , F is a finite subset of T , and ∀x ∈ F [x �≤ w], then f(T ↑w) ·

∏x∈F −f(T ↑ x) �= 0.

Proof. (i): Clearly ⇒ holds. For ⇐, assume that (a)–(c) hold. To apply Sikorski’sextension criterion, suppose that t0, . . . , tm−1, s0, . . . , sn−1 are distinct elements of

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T and

(∗) (T ↑ t0) ∩ · · · ∩ (T ↑ tm−1) ∩ [T \(T ↑ s0)] ∩ · · · ∩ [T \(T ↑ sn−1)] = ∅.

We then want to prove that

(∗∗) f(T ↑ t0) · . . . · f(T ↑ tm−1) · −f(T ↑ s0) · . . . · −f(T ↑ sn−1) = 0.

By Propositions 2.13 and 2.16 we then have three cases.

Case 1. m = 0 and sj = r for some j < n. Then by (a), f(T ↑ sj) = 1, and (∗∗)follows.

Case 2. There are i, j < m such that ti and tj are incomparable. Then by (b),f(T ↑ ti) · f(T ↑ tj) = 0, and (∗∗) follows.Case 3. There are i < m and j < n such that sj ≤ ti. Then by (c), f(T ↑ ti) ≤f(T ↑ sj), and (∗∗) follows.

(ii): ⇒: Suppose that w ∈ T , F is a finite subset of T , and ∀x ∈ F [x �≤ w].Then w ∈ (T � w) ∩

⋂x∈X [T \(T � x)], so (T � w) ∩

⋂x∈X [T \(T � x)] �= ∅, hence,

with f a monomorphism, f(T � w) ·∏

x∈X −f(T � x) �= 0, as desired.

⇐: Assume (a)–(d). It suffices now to show that if (∗) fails, then so does (∗∗).Assume that (∗) fails. Let w be a member of the set in (∗). Thus ti ≤ w for alli < m, and sj �≤ w for all j < n. By (d), f(T ↑ w) ·

∏j<n−f(T � sj) �= 0, and by

(c), f(T ↑ w) ≤ f(T � ti) for all i < m. It follows that f(T ↑ w) ·∏

j<n−f(T � sj)is ≤ the left side of (∗∗), and so (∗∗) fails. �

The topological dual of pseudo-tree algebras is given in the next theorem.

Theorem 2.18. Let T be a pseudo-tree with a smallest element r, and let I bethe set of all nonempty initial chains of T , i.e., the collection of all nonemptychains C under the tree ordering such that if s ≤ t ∈ C then s ∈ C. For eachF ∈ Ult(Treealg(T )) let f(F ) = {t ∈ T : (T ↑ t) ∈ F}. Then(i) f [Ult(A)] = I.

(ii) f is injective.

(iii) I is a closed subset of P(T ).

(iv) f : Ult(Treealg(T ))→P(T ) is continuous.

Proof. (i): If F ∈ Ult(A), then (T ↑ r) = 1 ∈ F , and so r ∈ f(F ). So f(F ) isnonempty. If s ≤ t ∈ f(F ), then (T ↑ t) ∈ F and (T ↑ t) ⊆ (T ↑ s), so (T ↑ s) ∈ Fand hence s ∈ f(F ). Thus f(F ) ∈ I.

Conversely, let C ∈ I. Let P = {T ↑ t : t ∈ C} ∪ {T \(T ↑ t) : t ∈ T \C}.Then P has the f.i.p. For, suppose that

(∗) (T ↑ t0) ∩ · · · ∩ (T ↑ tm−1) ∩ [T \(T ↑ s0)] ∩ · · · ∩ [T \(T ↑ sn−1)] = ∅,

with ti ∈ C and sj /∈ C for each i < m and j < n. We may assume thatm > 0 sinceC �= ∅. Let tk be maximum among all ti’s. Then tk ∈ (T ↑ t0) ∩ · · · ∩ (T ↑ tm−1),

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so by (∗), tk ∈ T ↑ sj for some j < n. Hence sj ≤ tk, so sj ∈ C, contradiction.So P does have f.i.p., and so it is contained in an ultrafilter F . Clearly f(F ) = C.This proves (i).

(ii): Suppose that F,G ∈ Ult(A) with F �= G. Then there is a t ∈ T such thatT ↑ t is in one of F,G but not the other. Say (T ↑ t) ∈ F\G. Then t ∈ f(F )\f(g).So (ii) holds.

(iii): Suppose that X ∈ P(T )\I. We want to find an open set U in P(T )such that X ∈ U and U ∩ I = ∅.Case 1. There are incomparable s, t ∈ X . Let U = {Y ⊆ T : s, t ∈ Y }. Then U isas desired.

Case 2. All members of X are comparable, but there exist s, t ∈ T such thats ≤ t ∈ X and s /∈ X . Let U = {Y ⊆ T : t ∈ Y and s /∈ Y }; U is as desired.

Case 3. X = ∅. Let U = {Y ⊆ T : r /∈ Y }.(iv): We take a basic open set U in P(T ). Say U = {Y ⊆ T : H ⊆ Y and

G∩Y = ∅}, whereH and G are finite subsets of T . Suppose that F ∈ f−1[U ]. Thusf(F ) ∈ U , so that H ⊆ f(F ) and G∩f(F ) = ∅. Thus

⋂t∈H(T ↑ t)∩

⋂s∈G[T \(T ↑

s)] ∈ F , so

F ∈ S(⋂

t∈H

(T ↑ t) ∩⋂s∈G

[T \(T ↑ s)])⊆ U. �

Now we give a theorem which is deeper than the ones presented so far. The hardpart, (ii)⇒(i), is due to Purisch and Nikiel, who proved it topologically. The presentpurely algebraic proof is due to Heindorf [97]. For the easy part we need two simplelemmas.

Lemma 2.19. If T is a pseudo-tree and t ∈ T , then (Treealg(T )) � (T ↑ t) is thesame as Treealg(T ↑ t). �

Lemma 2.20. If T is a finite tree with a least element r, then Treealg(T ) can be em-bedded in an interval algebra. In fact, there is an isomorphism f from Treealg(T )into an interval algebra Intalg(L) with the following properties:

(i) For each t ∈ T , f(T ↑ t) has the form [ct, dt) with ct < dt ≤ ∞.

(ii) If s < t, then cs < ct and dt < ds.

(iii) If s and t are incomparable, then dt < cs or ds < ct.

Proof. We prove this by induction on |T |. It is clear if |T | = 1. Now assumeinductively that |T | > 1. Let s0, . . . , sm−1 be all of the immediate successorsof r. By the inductive hypothesis, for each i < m let fi be an isomorphism ofTreealg(T ↑ si) into an interval algebra Intalg(Li) satisfying the conditions of thelemma. We may assume that Li ∩Lj = ∅ if i �= j. Let t0, . . . , tm be new elements.Let

M =⋃i<m

Li ∪ {ti : i ≤ m}

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with the following ordering:

t0 < L0 < t1 < L1 < · · · < Lm−1 < tm

(understood in the natural way), with each Li being ordered in its original way.Then we define g : {T ↑ u : u ∈ T } → M as follows: g(T ) = [t0,∞) (the 1 ofIntalg(M)), while if si ≤ u, then g(T ↑ u) = fi(T ↑ u). Note that g(T ↑ si) =fi(T ↑ si) = [0Li ,∞i) = [0Li, ti+1). Now we want to check the conditions (a)–(d)of Proposition 2.17 as well as conditions (i)–(iii) of our lemma. Conditions (a) and(i) are clear. Conditions (b) and (c) follow from (iii) and (ii) respectively. So wewant to check conditions (d), (ii), and (iii).

For (iii), suppose that u, v ∈ T and u and v are incomparable. Choose i, j <m such that si ≤ u and sj ≤ v. Then g(T ↑ u) = fi(T ↑ u) = [cu, du) andg(T ↑ v) = fj(T ↑ v) = [cv, dv). If i = j, then du < cv or dv < cu by ourassumption on fi. Suppose i �= j; say i < j. Then du < cv. Thus (iii) holds.

For (ii), suppose that u < v. If u = r, say si ≤ v. Then g(T ↑ u) = [t0,∞)and g(T ↑ v) = fi(T ↑ v) = [cv, dv), and t0 < cv and dv ≤ ti+1 <∞. If u �= r, theconclusion is clear. So (ii) holds.

For (d), suppose that w ∈ T , F is a finite subset of T , and ∀x ∈ F [x �≤ w].

Case 1. w = r. Now clearly for all x ∈ F we have r /∈ g(T ↑ x). Hence w ∈⋂x∈F [M\g(T ↑ x)], as desired.

Case 2. r < w. Say si ≤ w. Now take any x ∈ F . Thus g(T ↑ x) = fi(T ↑ x) =[cx, dx). We claim that cw /∈ g(T ↑ x).Subcase 2.1. w < x. By (ii) for fi, cw < cx, so cw /∈ g(T ↑ x).Subcase 2.2. x and w are incomparable. Then by (b), g(T ↑ w) and g(T ↑ x) aredisjoint. Since cw ∈ g(T ↑ w), it follows that cw /∈ g(T ↑ x).This proves our claim. Hence cw ∈ g(T ↑ w) ∩

⋂x∈F [M\g(T ↑ x)], as desired

for (d). �Theorem 2.21. For any BA A the following are equivalent:

(i) A is isomorphic to a pseudo-tree algebra.

(ii) A can be isomorphically embedded into an interval algebra.

Proof. (i)⇒(ii): Let T be a pseudo-tree; by Theorem 2.9 we may assume that it hasa minimum element 0T . We consider a first-order language with a binary relationsymbol <, an individual constant 0, and for each t ∈ T \{0T} two individualconstants at, bt. Let Σ be a set of first-order sentences expressing that in anymodel A of Σ the following hold:

(A,<) is a linear order with first element 0A.

0A < aAt < bAt for all t ∈ T \{0T}.at < as and bs < bt for 0T < t < s in T .

bt < as or bs < at if s, t ∈ T \{0T} and s, t are incomparable.

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To show that Σ has a model, take any finite subset Σ0 of Σ. Then there is a finitesubset T0 of T such that 0T ∈ T0 and for every t ∈ T , if at or bt occurs in somemember of Σ0, then t ∈ T0. We may assume in fact that if one of at or bt occursin some member of Σ0, the so does the other.

By Lemma 2.20, let f be an isomorphic embedding of T0 into an intervalalgebra Intalg(L) satisfying the additional conditions of the lemma. A model ofΣ0 is formed as follows. The universe is L and the ordering of L is assigned to <.For each t for which at occurs in Σ0 let at have the value ct and bt the value dt.Then the conditions of Lemma 2.20 shows that we have a model of Σ0.

Thus by the compactness theorem we obtain a model A of Σ. Now we definea function f mapping Treealg(T ) into Intalg(A,<). Let f(T ) = A. For t ∈ T \{0T}let f(T ↑ t) = [aAt , b

At ). Now we check the conditions of Proposition 2.17 to see

that f is an isomorphic embedding. In fact, (a)–(c) are clear. Now assume thehypotheses of (d).

Case 1. w = 0T . Then for all x ∈ F we have 0A < aAx , and so 0A ∈ [A\[aAx ,bAx )] =

[A\f(T ↑ x)], as desired.Case 2. w �= 0T . Take any x ∈ F .

Subcase 2.1. w < x. Then aAw < aAx , so aAw /∈ [aAx ,bAx ) = f(T ↑ x).

Subcase 2.2. w and x are incomparable. Then bAw < aAx or bA

x < aAw , so again

aAw /∈ [aAx ,bAx ) = f(T ↑ x).

It follows that f(T ↑ w) ∩⋂

x∈F [A\(T ↑ x)] �= ∅.(ii)⇒(i): Let B be the interval algebra on a linear order L which has first

element 0, and let A be a subalgebra of B; we want to show that A is isomorphicto a pseudo-tree algebra. By Theorem 2.15 it suffices to show that A is generatedby some ramification set.

The proof uses the symmetric difference operation extensively, and we startwith some facts about that operation on elements of B.

(1) If 0 < a0 < a1 < · · · < an−1 ≤ ∞ in L (n > 0), then

[0, a0)�[0, a1)�· · ·�[0, an−1)

=

{[a0, a1) ∪ [a2, a3) ∪ · · · ∪ [an−2, an−1) if n is even,

[0, a0) ∪ [a1, a2) ∪ · · · ∪ [an−2, an−1) if n is odd.

We prove (1) by induction on n. It is clear for n = 1. Assume it for n − 1, withn > 1. Then for n even we have

[0, a0)�[0, a1)�· · ·�[0, an−1)

=

([0, a0)�[0, a1)�· · ·�[0, an−2)

)�[0, an−1)

=

([0, a0) ∪ [a1, a2) ∪ · · · ∪ [an−3, an−2)

)�[0, an−1)

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= [0, an−1)\([0, a0) ∪ [a1, a2) ∪ · · · ∪ [an−3, an−2)

)= [a0, a1) ∪ [a2, a3) ∪ · · · ∪ [an−2, an−1).

For n odd we have

[0, a0)�[0, a1)�· · ·�[0, an−1)

=

([0, a0)�[0, a1)�· · ·�[0, an−2)

)�[0, an−1)

=

([a0, a1) ∪ [a2, a3) ∪ · · · ∪ [an−3, an−2)

)�[0, an−1)

= [0, an−1)\([a0, a1) ∪ [a2, a3) ∪ · · · ∪ [an−3, an−2)

)= [0, a0) ∪ [a1, a2) ∪ · · · ∪ [an−2, an−1).

Thus (1) holds. Now let K = {[0, a) : 0 < a ≤ ∞ in L}. Thus K generates B.

(2) If k0, . . . , kn−1 are distinct elements of K, then k0�k1�· · ·�kn−1 �= ∅.

This is clear from (1).

(3) Every element x of B can be written in the form x = k0�k1�· · ·�kn−1 fordistinct elements k0, . . . , kn−1 of K. This form is unique, in the sense that if alsox = l0�· · ·�lm−1 with distinct l0, . . . , lm−1 ∈ K, then m = n and {ki : i < n} ={li : i < n}.

To prove existence in (3) it suffices to show that the set of elements x which can bewritten in the indicated form contains K and is closed under − and ∩. Of courseeach k ∈ K can be written in the indicated form. For closure under −, supposethat x is written as in (3). Then

−x = −(k0�k1�· · ·�kn−1) = [0,∞)�k0�k1�· · ·�kn−1,

and this is of the correct form if each ki �= [0,∞), while if ki = [0,∞), then −x isthe sum of all the kj such that j �= i.

For closure under ·, suppose that x is as in (1), and y = l0�l1�· · ·�lm−1.Then

x ∩ y = (k0�k1�· · ·�kn−1) ∩ (l0�l1�· · ·�lm−1)

= (k0 ∩ l0)�(k1 ∩ l1)�· · ·�(k0 ∩ lm−1)

�(k1 ∩ l0)�(k1 ∩ l1)�· · ·· · ·�(kn−1 ∩ l0)�· · ·�(kn−1 ∩ lm−1).

Now each element ki ∩ lj is a member of K, or is ∅. One can omit ∅ in the aboveexpression, and also omit two equal terms. So doing all these simplifications, wearrive at the desired expression for x∩y. This finishes the proof of existence in (3).

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For uniqueness, suppose that x is written in two different ways, as in theformulation of (3). Then

0 = x�x = k0�k1�· · · kn−1�l0�l1�· · ·�lm−1;

from (2) it then follows that {ki : i < n} = {li : i < m} and hence also m = n.

By (3), we can associate with each x ∈ B the set τ(x) = {k0, . . . , kn−1} givenin (3).

(4) τ(x�y) = τ(x)�τ(y) for all x, y ∈ B.

In fact, it is clear that k is in the representation according to (3) of x�y iff it isin exactly one of the representations of x and y, and this gives (4).

(5) τ(x ∩ y) ⊆ τ(x) ∪ τ(y).

In fact, the above calculation of the representation of x ∩ y essentially shows this,as each ki ∩ lj is one of ki or lj .

Now we can start on the real proof. Let A = {M ⊆ K : M ∩ τ(x) �= ∅ for allx ∈ A+}. Note that A is nonempty, since for example {[0,∞)} ∈ A . If B is achain of members of A under ⊆, then

⋂B ∈ A . In fact, suppose not: so there

is an x ∈ A+ such that⋂

B ∩ τ(x) = ∅. It follows that B does not have a leastelement. Let 〈Cξ : ξ < κ〉 be a strictly decreasing sequence of members of Bcoinitial in B, κ an infinite cardinal. For each ξ < κ there is a zξ ∈ Cξ ∩ τ(x).Since τ(x) is finite, there is a w such that zξ = w for cofinally many ξ < κ. Butthen w ∈

⋂B ∩ τ(x), contradiction. So, we have checked the hypothesis of Zorn’s

lemma for (A ,⊇). By Zorn’s lemma, let M be a minimal member of A under ⊆.(6) If x, y ∈ A and x �= y, then M ∩ τ(x) �= M ∩ τ(y).

In fact, otherwise M ∩ (τ(x)�τ(y)) = ∅, so by (4) M ∩ τ(x�y) = ∅, and x�y �= ∅since x �= y, contradicting the definition of M .

(7) For each m ∈M there is a unique rm ∈ A such that M ∩ τ(rm) = {m}.In fact, by the minimality of M we have (M\{m})∩ τ(x) = ∅ for some nonemptyx ∈ A. On the other hand, M ∩ τ(x) �= ∅; so M ∩ τ(x) = {m}. Thus we can letrm = x. It is unique by (6).

Let R = {rm : m ∈M}. We claim that R is the desired generating ramifica-tion set for A.

To see that R generates A, take any a ∈ A+. Then M ∩ τ(a) is a nonemptyfinite subset of M ; say M ∩ τ(a) = {m0, . . . ,mn−1} with the mi’s distinct. Nowalso

M ∩ τ(rm0�· · ·�rmn−1) = M ∩ (τ(rm0 )�· · ·�τ(rmn−1)) by (4)

= [M ∩ τ(rm0 )]�· · ·�[M ∩ τ(rmn−1)]

= {m0}� · · ·�{mn−1}= {m0, . . . ,mn−1}.

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Thus M ∩ τ(a) = M ∩ τ(rm0�· · ·�rmn−1), and so by (6), a = rm0�· · ·�rmn−1 .This shows that R generates A.

To see that R is a ramification set, take any two distinct m,n ∈M . By (5),we have

M ∩ τ(rm ∩ rn) ⊆ [M ∩ τ(rm)] ∪ [M ∩ τ(rn)] = {m,n}.Thus it follows that M ∩ τ(rm ∩ rn) ∈ {∅, {m}, {n}, {m,n}}.Case 1.M∩τ(rm∩rn) = ∅. Then by the definition ofM it follows that rm∩rn = ∅,as desired.

Case 2. M ∩ τ(rm ∩ rn) = {m}. Then M ∩ τ(rm ∩ rn) = M ∩ τ(rm), so by (6),rm ∩ rn = rm.

Case 3. M ∩ τ(rm ∩ rn) = {n}. Similar to Case 2.

Case 4. M ∩ τ(rm ∩ rn) = {m,n}. Then

M ∩ τ(rm ∩ rn) = {m,n} = [M ∩ τ(rm)]�[M ∩ τ(rn)]

= M ∩ [τ(rm)�τ(rn)] = M ∩ τ(rm�rn),

and so by (6), rm∩rn = rm�rn, which implies that rm = rn = ∅, contradicting (7).�

Theorem 2.22. If A and B are pseudo-tree algebras, then A×B is isomorphic toa pseudo-tree algebra.

Proof. Say A = Treealg(T1) and B = Treealg(T2), where T1 and T2 are pseudo-trees with least elements t1, t2 respectively, and T1 ∩ T2 = ∅. Then T1 ∪ T2

is a pseudo-tree with the ordering <T1 ∪ <T2 . Moreover, Treealg(T1 ∪ T2) �(Treealg(T1 ∪ T2) ↑ t1) = Treealg(T1), and similarly for t2 and T2.

�Theorem 2.23. Every homomorphic image of a pseudo-tree algebra is isomorphicto a pseudo-tree algebra.

Proof. Let A be a pseudo-tree algebra, and f a homomorphism from A onto somealgebra B. Then by Theorem 2.15, A is generated by a ramification set R. Clearlyf [R] is also a ramification set, and it generates B. Hence by Theorem 2.15 again,B is isomorphic to a pseudo-tree algebra. �

Simple extensions of Boolean algebras

Given BAs A and B, we call B a simple extension of A provided that A is asubalgebra of B and B = 〈A ∪ {x}〉 for some x ∈ B; then we write B = A(x).We recall that each element of A(x) can be written in the form a · x+ b · −x witha, b ∈ A; or in the form c + a · x + b · −x with a, b, c pairwise disjoint elementsof A. We now introduce some important ideals for studying simple extensions. IfA is a subalgebra of B and x ∈ B, we let A � x = {a ∈ A : a ≤ x}. This is

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a slight extension of the usual notion; x is not necessarily in A. Under the sameassumptions we let

SmpAx = 〈(A � x) ∪ (A � −x)〉Id,

the ideal in A generated by (A � x) ∪ (A � −x). The three ideals A � x, A � −x,and SmpAx are important for studying simple extensions.

Proposition 2.24. Let A(x) be a simple extension of A. Then A = A(x) iff x ∈SmpAx .

Proof. If A = A(x), then x ∈ A � x ⊆ SmpAx , as desired. Conversely, suppose thatx ∈ SmpAx . Write x = a + b, with a ∈ A � x and b ∈ A � −x. Clearly b = 0, sox = a ∈ A. �

Proposition 2.25. Let A(x) be a simple extension of A, and let a ∈ A. Then thefollowing conditions are equivalent:

(i) a ∈ SmpAx ;

(ii) a = b+ c for some b ∈ A � x and c ∈ A � −x;(iii) a · x ∈ A;

(iv) a · −x ∈ A;

(v) For all y ∈ A(x), if y ≤ a then y · x ∈ A and y · −x ∈ A, and so y ∈ A andA(x) � a = A � a;

(vi) For all y ∈ A(x), if y ≤ a then y ∈ SmpAx .

Proof. Clearly (i)⇔(ii). Assume (ii). Then a · x = b ∈ A, i.e., (iii) holds. Assume(iii). Then a · −x = a · −(a · x) ∈ A, i.e., (iv) holds. Similarly (iv)⇒(iii). If (iii)holds, then (iv) holds and a = a · x + a · −x, so (ii) holds. (ii)⇒(v): Assume (ii),and suppose that y ∈ A(x) and y ≤ a. Write y = u ·x+ v ·−x with u, v ∈ A. Then

y · x = a · y · x = a · u · x ∈ A

by (iii). Similarly y ·−x ∈ A, so y ∈ A. Clearly (v) implies (vi) and (vi) implies (i).�

Corollary 2.26. If A(x) is a simple extension of A, then SmpAx is an ideal of A(x).

Proposition 2.27. Suppose that A(x) and A(y) are simple extensions of A. Thenthe following conditions are equivalent:

(i) There is a homomorphism from A(x) into A(y) which is the identity on Aand sends x to y.

(ii) A � x ⊆ A � y and A � −x ⊆ A � −y.And also the following two conditions are equivalent:

(iii) There is an isomorphism from A(x) onto A(y) which is the identity on Aand sends x to y.

(iv) A � x = A � y and A � −x = A � −y.

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Proof. (i)⇒(ii): obvious. (ii)⇒(i): by Sikorski’s extension criterion. Since the ho-momorphism of (i) is unique, the equivalence of (iii) and (iv) is clear. �

Proposition 2.28. Let A be a BA, and let I0 and I1 be two ideals of A such thatI0 ∩ I1 = {0}. Then there is a simple extension A(x) of A such that A � x = I0and A � −x = I1.

Proof. Let A(y) be a free extension of A by y, that is, let A(y) be the free productof A with a four-element BA B, where 0 < y < 1 in B. Consider the followingideal K of A(y):

K = 〈{a · −y : a ∈ I0} ∪ {a · y : a ∈ I1}〉Id.

Let f be the natural mapping from A onto A/K. It suffices to prove the followingthings:

(1) A ∩K = {0}.(2) (A/K) � (y/K) = f [I0].

(3) (A/K) � −(y/K) = f [I1].

For (1), if a ∈ A ∩K, then a ≤ b · −y + c · y for some b ∈ I0 and c ∈ I1. An easyargument using freeness then yields a ≤ b · c = 0, as desired.

For (2), first suppose that a ∈ A and a/K ≤ y/K. Thus a · −y ∈ K, soa · −y ≤ b · −y + c · y for some b ∈ I0 and c ∈ I1. An easy argument then yieldsa ≤ b, and hence a ∈ I0, as desired. On the other hand, if a ∈ I0, obviouslya · −y ∈ K, and hence a/K ≤ y/K and a/K ∈ (A/K) � (y/K), as desired.

A similar argument proves (3). �

Proposition 2.29. Let A be a BA, and let I0 and I1 be two ideals of A such thatI0∩ I1 = {0}. Suppose that the simple extension A(x) of A is such that A � x = I0and A � −x = I1.

Then A(x) ∼= (A/I0)× (A/I1).

Proof. For a, b ∈ A define f(a · x + b · −x) = ([b]I0 , [a]I1). Then f is well defined,since from a · x + b · −x = a′ · x + b′ · −x it follows that a�a′ ≤ −x, hence[a]I1 = [a′]I1 ; similarly [b]I0 = [b′]I0 .

Clearly now f is a homomorphism. f is one-one, for if [a]I1 = 0 = [b]I0 , thena ∈ I1 and b ∈ I0, hence a ≤ −x and b ≤ x, so a · x+ b · −x = 0. Finally, clearly fis onto. �

Proposition 2.30. If A is superatomic, then so is A(x).

Proof. This follows from Proposition 2.29, since a product of two superatomic BAsis superatomic. �

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Proposition 2.31. Suppose that A(x) is a simple extension of A, and K is an idealin A(x).

(i) There is an isomorphism f of A/(K∩A) into A(x)/K such that f([a]K∩A) =[a]K for any a ∈ A.

(ii) A(x)/K is a simple extension of rng(f) by the element [x]K .

(iii) A(x)/K is isomorphic to a simple extension of A/(K ∩A).

Proof. (i) is clear. For (ii), note that any element of A(x)/K can be written in theform [a · x+ b · −x]K with a, b ∈ A; and

[a ·x+ b · −x]K = [a]K · [x]K + [b]K · −[x]K = f([a]K∩A) · [x]K + f([b]K∩A) · −[x]K .

(iii) follows from (i) and (ii). �

Minimal extensions of Boolean algebras

We say that B is a minimal extension of a BA A, in symbols A ≤m B, if B is anextension of A and there is no subalgebra C of B such that A ⊂ C ⊂ B. Clearlythen B is a simple extension of A. This notion is studied in Koppelberg [89a].First we want to see what this means in terms of the ideal SmpA

x :

Proposition 2.32. Let A(x) be a simple extension of A. Then the following condi-tions are equivalent:

(a) A ≤m A(x).

(b) SmpAx is either equal to A or is a maximal ideal of A.

(c) A = 〈{a ∈ A : a is comparable with x}〉.(d) There is a G ⊆ A which generates A and consists exclusively of elements

comparable with x.

(e) If y ∈ A(x)\A, then x�y ∈ A.

Proof. Obviously (c)⇔(d). (a)⇒(b): assume that (b) fails. Then there is an ele-ment a ∈ A such that neither a nor −a is in SmpAx . Then, we claim, A ⊂ A(a ·x) ⊂A(x). In fact, a ·x /∈ A by Proposition 2.25. And if x ∈ A(a ·x), then we can writex = b · a · x+ c · −(a ·x) with b, c ∈ A. But then x = b · a · x+ c · −a+ c · −x, hencec · −x = 0 and x = b · a · x + c · −a. Therefore −a · x = c · −a ∈ A, and hence byProposition 2.25, −a ∈ SmpAx , contradiction.

(b)⇒(c): Assume (b). Then SmpAx generates A. Let G = {a ∈ A : a iscomparable with x}. Now A � x ⊆ G, and if a ∈ A � −x then −a ∈ G. It followsthat A = 〈SmpA

x 〉 ⊆ 〈G〉, and hence G generates A.

(d)⇒(b): Clearly G ⊆ SmpAx ∪ {a : −a ∈ SmpAx }, and the latter set is asubalgebra of A; hence it is all of A, which means that (b) holds.

(b)⇒(e): Assume (b), and suppose that y ∈ A(x)\A. Write y = a+b·x+c·−x,where a, b, c are pairwise disjoint elements of A. If b and c are both elements ofSmpAx , then b · x and c · −x are both elements of A by Proposition 2.25, and so

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y ∈ A, contradiction. Assume that b /∈ SmpAx . Hence −b ∈ SmpA

x by (b). Nowy · b = x · b, so x�y ≤ −b, and hence x�y ∈ A by Corollary 2.26. If c /∈ SmpAx , weobtain (−x)�y ∈ A similarly; then note that (−x)�y = −(x�y).

(e)⇒(a): If y ∈ A(x)\A, then x = x�y�y ∈ A(y), so A(y) = A(x). �

Proposition 2.33. If A ≤ B, x ∈ B, and A � x is a maximal ideal in A, thenA ≤m A(x).

Proof. By Proposition 2.32. �

The notion of minimal extension can also be expressed using extensions of ultra-filters. Here the following general result motivates what follows.

Proposition 2.34. If A is a proper subalgebra of B, then there is an ultrafilter Fon A such that F does not generate an ultrafilter on B.

Proof. Let b ∈ B\A. We claim that

(∗) {a ∈ A : −a ≤ b} ∪ {a ∈ A : −a ≤ −b}

has fip. Otherwise, there are a0, a1 ∈ A such that −a0 ≤ b, −a1 ≤ −b, anda0 · a1 = 0. Then −b ≤ a0 ≤ −a1 ≤ −b, so −b = a0 ∈ A, contradiction.

Let F be an ultrafilter on A containing the set (∗). Suppose that F generatesan ultrafilter G on B.

Case 1. b ∈ G. Say a ∈ F , a ≤ b. Then −a ∈ (∗) ⊆ F , contradiction.

Case 2. −b ∈ G. Say a ∈ F , a ≤ −b. Then −a ∈ (∗) ⊆ F , contradiction. �

Given A ≤ B, A �= B, b ∈ B\A, and F an ultrafilter on A, we say that b is minimalfor (A,F ) provided that F is the only ultrafilter on A which does not generate anultrafilter on A(b).

Proposition 2.35. Suppose that A(x) is a minimal extension of A, A �= A(x). Thenx is minimal for (A,−SmpA

x ).

Proof. First letH be the filter on A(x) generated by −SmpAx . Suppose that x ∈ H .Say a ∈ −SmpAx and a ≤ x. Then a ∈ SmpA

x , hence 1 ∈ SmpAx , contradictingProposition 2.24. Suppose −x ∈ H . Say a ∈ −SmpAx and a ≤ −x. Then −a ∈SmpAx , so −a ≤ b+ c with b ≤ x and c ≤ −x. Then x ≤ −a ≤ b+ c with c · x = 0,so x ≤ b ≤ x. This gives x = b ∈ A, contradicting Proposition 2.24 again. Thus−SmpAx does not generate an ultrafilter on A(x).

Now suppose that G is an ultrafilter on A different from −SmpAx . Say a ∈G\ − SmpA

x . Since −SmpAx is an ultrafilter on A, it follows that −a ∈ −SmpAx .Thus a ∈ SmpA

x , and it follows from the definition of SmpAx that x or −x is in thefilter generated by G in A(x), so that filter is an ultrafilter. �Proposition 2.36. Suppose that A ≤ B, A �= B, x ∈ B\A, and F is an ultrafilteron A. Suppose that x is minimal for (A,F ). Then F = −SmpAx .

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Proof. By Proposition 2.24, SmpAx �= A, and so also −SmpAx �= A. Hence it isenough to show that F ⊆ −SmpAx . Suppose that a ∈ F\− SmpA

x . Then −SmpAx ∪{−a} has fip, so it is included in an ultrafilter G. Thus F �= G, so G generates anultrafilter H in A(x). Suppose that x ∈ H . Choose b ∈ G such that b ≤ x. Thusb ∈ SmpAx , so −b ∈ −SmpAx ⊆ G, contradiction. If −x ∈ H , a similar contradictionis obtained. �

Proposition 2.37. If A ≤ B, A �= B, x ∈ B\A, F is an ultrafilter on A, and x isminimal for (A,F ), then x,−x /∈ 〈F 〉fiA(x).

Proof. Otherwise 〈F 〉fiA(x) would be an ultrafilter on A(x), contradicting the defi-

nition of x minimal for (A,F ). �

Proposition 2.38. Suppose that A ≤ B, A �= B. Then the following conditions areequivalent:

(i) B is a minimal extension of A;

(ii) ∃x ∈ B\A[B = A(x) and ∃F ∈ UltA(x is minimal for (A,F ))];

(iii) ∀x ∈ B\A[B = A(x) and ∃F ∈ UltA(x is minimal for (A,F ))];

(iv) ∀x ∈ B\A[B = A(x) and −SmpAx is an ultrafilter on A];

(v) ∃x ∈ B\A[B = A(x) and −SmpAx is an ultrafilter on A].

(vi) ∀x ∈ B\A[B = A(x) and x is minimal for (A,−SmpAx )].

(vii) ∃x ∈ B\A[B = A(x) and x is minimal for (A,−SmpAx )].

Proof. (i)⇒(iv): Let x ∈ B\A. Then B = A(x) since B is a minimal extension ofA. By Proposition 2.32, −SmpAx is an ultrafilter on A.

(iv)⇒(v): obvious.

(v)⇒(i): by Proposition 2.32.

(i)⇒(iii): Propositions 2.32, 2.35.

(iii)⇒(ii): obvious.

(ii)⇒(v): Proposition 2.36.

(i)⇒(vi): Proposition 2.35.

(vi)⇒(vii): obvious.

(vii)⇒(ii): obvious. �

Proposition 2.39. Suppose that A ⊆ B, A �= B, x ∈ B\A, and F is an ultrafilteron A. Then the following conditions are equivalent:

(i) {a ∈ A : a · x ∈ A} = −F ;

(ii) {a ∈ A : a · x /∈ A} = F ;

(iii) x is minimal for (A,F ).

(iv) F = −SmpAx .

Proof. (i)⇒(ii): {a ∈ A : a · x /∈ A} = A\{a ∈ A : a · x ∈ A} = A\(−F ) = F .

(ii)⇒(i): {a ∈ A : a · x ∈ A} = A\{a ∈ A : a · x /∈ A} = A\F = −F .

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(ii)⇒(iii): By Proposition 2.34 it suffices to show that every ultrafilter G �= Fon A generates an ultrafilter on A(x). Let G generate the filter H on A(x). Saya ∈ G\F . Thus a · x ∈ A. If a · x ∈ G, then x ∈ H . If a · x /∈ G, then a · −(a · x) =a · −x ∈ G, so −x ∈ H . So H is an ultrafilter.

(iii)⇒(i): by Proposition 2.36, F = −SmpAx , so (i) follows from Proposition2.25 (the equivalence of (i) and (iii) there).

(iii)⇒(iv): by Proposition 2.36.

(iv)⇒(iii): By Proposition 2.35. �Proposition 2.40. Suppose that A ⊆ B, A �= B, x ∈ B\A, and F is an ultrafilteron A. Then x is minimal for (A,F ) iff −x is minimal for (A,F ).

Proof. Note that SmpAx = SmpA−x. So if x is minimal for (A,F ), then by Propo-

sition 2.36, F = −SmpAx = −SmpA−x. Since A(x) = A(−x), it follows that −x isminimal for (A,F ). The converse is similar. �Proposition 2.41. Suppose that A ≤ C ≤ B, G is an ultrafilter on A, and Ggenerates a filter F on C which is an ultrafilter. Suppose that x ∈ B\C and it isminimal for (A,G). Then it is also minimal for (C,F ).

Proof. We haveG = −SmpAx by Proposition 2.36. We now show that F = −SmpCx .Since C �= C(x), by Proposition 2.24 SmpCx �= C. Hence −SmpCx is a proper filterof C, so it suffices to show that F ⊆ −SmpCx .

So, suppose that a ∈ F . Choose b ∈ G such that b ≤ a. then b ∈ −SmpAx ,so −b ∈ SmpAx . Thus −b · x ∈ A by Proposition 2.25. Then −a ≤ −b, so −a · x =−a · −b · x ∈ C. Thus −a ∈ SmpC

x by Proposition 2.25, so a ∈ −SmpCx . �Proposition 2.42. Let A ≤ B, let f : B → Q be an epimorphism, and set P = f [A].Then A ≤m B implies that P ≤m Q.

If, moreover, ker(f) ⊆ A, then A ≤m B iff P ≤m Q, and A �= B iff P �= Q.

Proof. This is a result of universal-algebraic nonsense: the function assigning toeach subalgebra C of Q the subalgebra f−1[C] of B is one-one, and it maps{C : P ≤ C ≤ Q} into {D : A ≤ D ≤ B}. In case ker(f) ⊆ A, it maps ontothe latter set: the preimage of such a D is f [D], and P ≤ f [D] ≤ Q. For thefinal statement, clearly P �= Q implies that A �= B, even without the assumptionker(f) ⊆ A. Now suppose that A �= B; say b ∈ B\A. If f(b) ∈ P , then there isan a ∈ A such that f(b) = f(a). So f(b�a) = 0, hence b�a ∈ ker(f) ⊆ A andb = b�a�a ∈ A, contradiction. �Proposition 2.43. Assume that A ≤ B ≤M ≥ D. Set P = A∩D and Q = B ∩D.Then A ≤m B implies that P ≤m Q.

Proof. Assume all the hypotheses. We may also assume that P �= Q. Now take anyx ∈ Q\P ; we want to show that Q = P (x). To this end, take any y ∈ Q; we showthat y ∈ P (x). Now x ∈ B since x ∈ Q. Also x /∈ A since x ∈ Q ⊆ D and x /∈ P .It follows that B = A(x). We may assume that y /∈ P ; hence y ∈ B\A. Now by

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Proposition 2.32(e) we get x�y ∈ A. Also, x ∈ D and y ∈ D, so x�y ∈ D; hencex�y ∈ P . It follows that y = (x�y)�x ∈ P (x), as desired. �Proposition 2.44. Suppose that A(x) is a proper minimal extension of A. Then thefollowing conditions are equivalent:

(i) A � x and A � −x are non-principal ideals.

(ii) A is dense in A(x).

Proof. Assume (i). Take any non-zero element y of A(x). Wlog we may assumethat y = a · x for some a ∈ A. By Proposition 2.32 there are two cases.

Case 1. a ∈ SmpAx . Say a = b+c with b ∈ A � x and c ∈ A � −x. Then a ·x = b ∈ A

and there is nothing to prove.

Case 2. −a ∈ SmpAx . Say −a = b + c with b ∈ A � x and c ∈ A � −x. Choose

d ∈ A � x with b < d (which we can do because A � x is non-principal). Then

d · −b · −a = d · −b · (b + c) ≤ d · c = 0

since c ∈ A � −x. Hence 0 �= d · −b ≤ a · x, as desired.Now assume (ii). To show that A � x is non-principal, let a ∈ A � x. Now

x · −a �= 0, so we can choose a non-zero b ∈ A such that b ≤ x · −a. Thena < a+ b ≤ x, as desired. Similarly, A � −x is non-principal. �Proposition 2.45. Suppose that A(x) is a proper minimal extension of A, and A � xis a principal ideal generated by an element a∗. Set y = −a∗ · x. Then:(i) y /∈ A, and hence A(x) = A(y);

(ii) for all a ∈ A, y ≤ a iff −a ∈ SmpAx ;

(iii) y is an atom of A(x);

(iv) if D is dense in A, then 〈D ∪ {y}〉 is dense in A(x).

Proof. (i): Clearly y /∈ A, since otherwise y ≤ a∗, y = 0, x ≤ a∗, and x = a∗ ∈ A.So A(x) = A(y) by minimality.

(ii): First assume that y ≤ a. Thus −a ≤ a∗+−x. Now−a = −a·a∗+−a·−a∗,and −a · −a∗ ≤ −x, proving that −a ∈ SmpA

x .

Second, assume that −a ∈ SmpAx . Say −a = b + c with b ∈ A � x and

c ∈ A � −x. Then −a∗ · x · −a = −a∗ · b = 0, showing that y ≤ a.

(iii): Clearly y �= 0, by (i). Suppose that z ≤ y; we show that z = 0 or z = y.Say z = a ·x+b ·−x with a, b ∈ A. Since y ≤ x, we have b ·−x = 0. By Proposition2.32, either a ∈ SmpAx or −a ∈ SmpA

x . If −a ∈ SmpAx , then y ≤ a by (ii), hencey ≤ x·a = z and so y = z, as desired. Now suppose that a ∈ SmpAx . Write a = c+dwith c ∈ A � x and d ∈ A � −x. Then c ≤ a∗, hence a · x = c · x ≤ a∗ · x, but alsoa · x = z ≤ y = −a∗ · x, so z = a · x = 0, as desired.

(iv): Assume that u is a non-zero element of A(x). If u · y �= 0, then y ≤ uby (iii), as desired. So, assume that u · y = 0. Then we can write u = b · −y withb ∈ A, by (i). Choose d ∈ D+ so that d ≤ b. If d ·−y = 0, then d ≤ y, hence d = y,

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from which it follows that y ∈ A, contradicting (i). So 0 �= d · −y ≤ b · −y = u, asdesired. �

Proposition 2.46. Suppose that B is superatomic and A is a proper subalgebra ofB. Then there is an x ∈ B\A such that A ≤m A(x) ≤ B.

Proof. Assume the hypotheses. We recall the definition of the standard sequence〈Iα : α an ordinal〉 of ideals in B:

I0 = {0};Iα+1 = the ideal generated by {a ∈ B : a/Iα is an atom of B/Iα};

Iγ =⋃α<γ

Iα for γ limit.

Because B is superatomic, there is an ordinal α such that Iα = B; since A is aproper subalgebra of B, there is a smallest ordinal β such that Iβ �⊆ A. Clearly β isa successor ordinal γ+1. Hence there is an atom b/Iγ of B/Iγ such that b /∈ A. For

any x ∈ A/Iγ we have x · (b/Iγ) = 0 or (b/Iγ) ≤ x, so that SmpA/Iγb/Iγ

is a maximal

ideal of A/Iγ . It then follows by Proposition 2.32 that A/Iγ ≤m (A/Iγ)(b/Iγ). Notethat A(b)/Iγ = (A/Iγ)(b/Iγ). Let f : A(b) → A(b)/Iγ be the natural mapping.Then ker(f) = Iγ . Hence by Proposition 2.42 we get A ≤m A(b). �

Minimally generated Boolean algebras

Let A and B be BAs. A representing chain for B over A is a sequence 〈Cα : α < ρ〉of BAs with the following properties:

(1) α < β < ρ implies that Cα ≤ Cβ .

(2) If λ is a limit ordinal less than ρ, then Cλ =⋃

α<λ Cα.

(3) C0 = A.

(4)⋃

α<ρCα = B.

B is minimally generated over A if there is a representing chain for B over Asuch that Cα ≤m Cα+1 whenever α + 1 < ρ. And B is minimally generated ifit is minimally generated over 2. We write A ≤mg B to abbreviate that B isminimally generated over A. Finally, if A ≤mg B, then len(B : A) is the smallestordinal ρ demonstrating the minimal generation of B over A, and if B is minimallygenerated, then len(B) = len(B : 2). The notion of a minimally generated BA isdue to S. Koppelberg [89a].

Proposition 2.47.

(i) Suppose that A ≤ B and f is an epimorphism from B onto Q. Let P = f [A].Then:

(a) if A ≤mg B, then P ≤mg Q and len(Q : P ) ≤ len(B : A);

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(b) if A includes the kernel of f , then A ≤mg B iff P ≤mg Q, and if oneand hence both of these holds then len(Q : P ) = len(B : A).

(ii) A homomorphic image Q of a minimally generated BA B is minimally gen-erated. Moreover, len(Q) ≤ len(B).

Proof. By Proposition 2.42. �

Proposition 2.48.

(i) Suppose that A ≤ B ≤M ≥ D and A ≤mg B. Set P = A∩D and Q = B∩D.Then P ≤mg Q, and len(Q : P ) ≤ len(B : A).

(ii) Every subalgebra D of a minimally generated BA B is minimally generated.Moreover, len(D) ≤ len(B).

(iii) If A ≤mg B and A ≤ C ≤ B, then A ≤mg C.

Proof. (i) and (ii) are clear from Proposition 2.43. For (iii), let D = C in (i). �

Proposition 2.49. Suppose that A is an atomless subalgebra of B and there is anelement u ∈ B which is independent over A, i.e., a ·u �= 0 �= a ·−u for all a ∈ A+.Then B is not minimally generated over A.

Proof. Otherwise we would haveA ≤mg A(u) by Proposition 2.48(iii). Hence thereis an x ∈ A(u) such that x /∈ A and A ≤m A(x). Say x = a + b · u + c · −u witha, b, c pairwise disjoint elements of A. From the independence of u over A it thenfollows that A � x = A � a, a principal ideal in A. In fact, if v ∈ A and v ≤ x, thenv ≤ a+ b+ c; if v · b �= 0, then v · b · −u �= 0. But v · b ≤ x · b = b · u, contradiction.So v · b = 0, and similarly v · c = 0. Similarly, A � −x is a principal ideal in A. SoSmpAx is a principal ideal in A. By Proposition 2.32, SmpA

x is a maximal ideal; sothis gives us an atom of A, contradiction. �

Proposition 2.50. If A and B are minimally generated, then so is A×B.

Proof. Let 〈Cα : α < ρ〉 be a representing sequence for A’s minimal generation(thus with C0 = 2 and Cα ≤m Cα+1 for all α + 1 < ρ), and let 〈Dα : α < σ〉 besimilarly chosen for B. The desired sequence for A× B is 〈G〉�〈Eα : α < ρ+ σ〉,where G is the two-element subalgebra {(0, 0), (1, 1)} of A× B and for α < ρ weset

Eα = {(a, b) : a ∈ Cα and b ∈ {0, 1}},

and for α < σ we set

Eρ+α = {(a, b) : a ∈ A and b ∈ Dα}.

We show that this works. Note that E0 = {(a, b) : a ∈ {0, 1} and b ∈ {0, 1}}, soE0 = G((0, 1)) and there is no algebra strictly between G and E0. Now take anyα with α + 1 < ρ. Let Cα+1 = Cα(u), with no algebra properly in between. Weclaim that Eα+1 = Eα((u, 1)) with no algebra properly in between. For, take any

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(x, y) ∈ Eα+1. Thus x ∈ Cα+1 and y ∈ {0, 1}. Hence we can write x = a ·u+ b ·−uwith a, b ∈ Cα. Then

(x, 1) = (a, 1) · (u, 1) + (b, 1) · (−u, 0) and

(x, 0) = (a, 0) · (u, 1) + (b, 0) · (−u, 0).So Eα+1 = Eα((u, 1)). Now suppose that Eα ⊆ F ⊆ Eα+1. Then Cα ⊆ pr0[F ] ⊆Cα+1, which gives two possibilities.

Case 1. Cα = pr0[F ]. We claim then that Eα = F . For, suppose that (u, v) ∈ F .Then u ∈ Cα and v ∈ {0, 1}, as desired.Case 2. Cα+1 = pr0[F ]. Similarly.

Now consider Eρ. If ρ = ξ + 1, then A = Cξ, Eξ = A × {0, 1}, and Eρ = Eξ. If ρis a limit ordinal, then Eρ = {(a, b) : a ∈ A and b ∈ {0, 1}} =

⋃ξ<ρEξ.

Next, take α < σ with α+1 < σ. LetDα+1 = Dα(u) with no algebra properlyin between. We claim that Eρ+α+1 = Eρ+α((1, u)) with no algebra properly inbetween. For, take any (x, y) ∈ Eρ+α+1. Thus y ∈ Dα+1 and x ∈ A. Hence we canwrite y = a · u+ b · −u with a, b ∈ Dα. Then for any x ∈ A we have

(x, y) = (x, a) · (1, u) + (0, b) · (0,−u).So Eρ+α+1 = Eρ+α((1, u)). Now suppose that Eρ+α ⊆ F ⊆ Eρ+α+1. Hence Dα ⊆pr1[F ] ⊆ Dα+1. This gives two cases.

Case 1. Dα = pr1[F ]. Then we claim that Eρ+α = F . For, suppose that (u, v) ∈ F .Then v ∈ Dα, so (u, v) ∈ Eα.

Case 2. Dα+1 = pr1[F ]. Similarly.

Finally, clearly⋃

α<ρ+σ Eα = A×B. �Proposition 2.51. If 〈Ai : i ∈ I〉 is a system of minimally generated BAs, then sois∏w

i∈I Ai.

Proof. Wlog each Ai is nontrivial, and I is an infinite cardinal κ. For all β < κlet 〈Cβα : 0 < α < ρβ〉 be a representing sequence for Aβ (for technical reasonsstarting at 1 rather than 0) such that Cβα ≤m Cβ,α+1 and Cβα �= Cβ,α+1 ifα + 1 < ρβ . Thus Cβ1 = 2, and each ρβ is at least 2. Let σ =

∑β<κ ρβ . For

each ξ < σ we define a subalgebra Eξ of∏w

β<κAβ . Choose δ < κ such that

μdef=∑

β<δ ρβ ≤ ξ <∑

β≤δ ρβ ; say ξ = μ+ ε with ε < ρδ. If ε = 0 we set

Eξ =

{x ∈

w∏β<κ

Aβ : ∀θ ∈ [δ, κ)[xθ = 1] or ∀θ ∈ [δ, κ)[xθ = 0]

}.

If ε �= 0 we set

Eξ =

{x ∈

w∏β<κ

Aβ : xδ ∈ Cδε and ∀θ ∈ (δ, κ)[xθ = 1] or ∀θ ∈ (δ, κ)[xθ = 0]

}.

Then 〈Eξ : ξ < σ〉 is as desired.

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To prove this, first note that for ξ = 0 we have δ = 0 and so E0 is thetwo-element subalgebra of

∏wi∈I Ai. Clearly each Eξ is a subalgebra of

∏wβ<κAβ .

Now suppose that ξ < η; we show that Eξ ⊆ Eη. Say δ < κ, μdef=∑

β<δ ρβ ≤ξ <

∑β≤δ ρβ , and ξ = μ + ε with ε < ρδ, and δ′ < κ, μ′ def

=∑

β<δ′ ρβ ≤ η <∑β≤δ′ ρβ , and η = μ′ + ε′ with ε′ < ρδ′ . Let x ∈ Eξ.

Case 1. δ = δ′. So μ = μ′ and ε < ε′.

Subcase 1.1. ε = 0. Then ∀θ ∈ [δ, κ)[xθ = 1] or ∀θ ∈ [δ, κ)[xθ = 0]. Hence xδ = 1or xδ = 0, so that xδ ∈ Cδ′ε′ , and ∀θ ∈ (δ′, κ)[xθ = 1] or ∀θ ∈ (δ′, κ)[xθ = 0]. Thusx ∈ Eη.

Subcase 1.2. ε > 0. Then xδ ∈ Cδε and [∀θ ∈ (δ, κ)[xθ = 1] or ∀θ ∈ (δ, κ)[xθ = 0].Hence, since Cδε = Cδ′ε ⊆ Cδ′ε′ , we have xδ ∈ Cδ′ε′ . Thus x ∈ Eη.

Case 2. δ < δ′.

Subcase 2.1. ε′ = 0. Clearly ∀θ ∈ [δ′, κ)[xθ = 1] or ∀θ ∈ [δ′, κ)[xθ = 0]. Thusx ∈ Eη.

Subcase 2.2. ε′ > 0. Clearly xδ′ = 0 or xδ′ = 1, so xδ′ ∈ Cδ′ε′ . Also clearly∀θ ∈ (δ′, κ)[xθ = 1] or ∀θ ∈ (δ′, κ)[xθ = 0]. Thus x ∈ Eη.

Next we show that for ξ limit the set Eξ is the union of previous Eη’s. For, letx ∈ Eξ. Assuming notation as in the definition of Eξ, we take two cases.

Case 1. ε = 0. So ξ = μ is a limit ordinal. This gives two subcases.

Subcase 1.1. For all θ ∈ [δ, κ) we have xθ = 1. Since [δ, κ) is infinite, we must have{ν < κ : xν �= 1} finite. Now we consider two subcases.

Subsubcase 1.1.1. δ is a successor ordinal δ′+1. Now ξ = μ =∑

β<δ′ ρβ+ρδ′ ,so ρδ′ is a limit ordinal. Now xδ′ ∈ Aδ′ , so there is a σ < ρδ′ such that xδ′ ∈ Cδ′σand 0 < σ. Let ξ′ =

∑β<δ′ ρβ + σ. Then x ∈ Eξ′ and ξ′ < ξ, as desired.

Subsubcase 1.1.2. δ is a limit ordinal. Then by the fact that {ν < κ : xν �= 1}is finite, there is a δ′ < δ such that x ∈ Eξ′ , with ξ′ =

∑β<δ′ ρβ . Note that ξ

′ < ξ.

Subcase 1.2. For all θ ∈ [δ, κ) we have xθ = 0. This is similar to Subcase 1.1.

Case 2. ε is a limit ordinal. Since xδ ∈ Cδε, it follows that there is an η < ε suchthat xδ ∈ Cδη and η �= 0. Hence x ∈ Eμ+η and μ+ η < ξ, as desired.

Next, we need to show that Eξ ≤m Eξ+1 for all ξ < σ. Let ξ, δ, μ, ε be as in thedefinition. Take any y ∈ Eξ+1\Eξ; we want to show that Eξ+1 = Eξ(y). To thisend, take any z ∈ Eξ+1\Eξ; we want to show that z ∈ Eξ(y).

Case 1. ε = 0. Note that Cδ1 = {0, 1} and ξ + 1 = μ+ 1. We have some subcases.

Subcase 1.1. zθ = 1 = yθ for all θ ∈ (δ, κ). Since y, z /∈ Eξ, it follows thatzδ = 0 = yδ. So y�z ∈ Eξ, and hence z ∈ Eξ(y).

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Subcase 1.2. zθ = 1 for all θ ∈ (δ, κ), and yθ = 0 for all τ ∈ (δ, κ). Since y, z /∈ Eξ,it follows that zδ = 0 and yδ = 1. Then (y�z)θ = 1 for all θ ∈ [δ, κ), so againy�z ∈ Eξ.

The other subcases are similar.

Case 2. ε > 0. Since yδ, zδ ∈ Cδ,ε+1, it follows that there exist a, b ∈ Cδε such thatzδ = a · yδ + b · −yδ. Now define u, v, w ∈

∏wα<κ Aα as follows: for any θ < κ,

uθ =

{0 if θ �= δ,

a otherwise;

vθ =

{0 if θ �= δ,

b otherwise;

wθ =

{zθ if θ �= δ,

0 otherwise.

Then u, v, w ∈ Eξ and z = u · y + v · −y + w.

Finally, we show that⋃

ξ<σ Eξ =∏w

β<κAβ . Take any x ∈∏w

β<κ Aβ . Chooseδ < κ such that ∀θ ∈ (δ, κ)[xθ = 1] or ∀θ ∈ (δ, κ)[xθ = 0]. Let μ =

∑β<δ ρβ . Now

xδ ∈ Aδ, so there is an ε < ρδ such that 0 < ε and xδ ∈ Cδε. Then with ξ = μ+ εwe have x ∈ Eξ. �

Proposition 2.52. Every interval algebra is minimally generated.

Proof. Let A be an interval algebra. Then it is generated by a chain C. EnumerateC: C = {cα : α < ρ}. For each α < 1 + ρ let Bα = 〈{cβ : β < α}〉. Then byProposition 2.32 we have Bα ≤m Bα+1 whenever α+1 ≤ 1+ ρ, so this shows theminimal generation of A. �

Corollary 2.53. Every pseudo-tree algebra is minimally generated.

Proof. By Propositions 2.52, 2.48(ii), and 2.21. �

Proposition 2.54. Every superatomic BA is minimally generated.

Proof. This follows from Proposition 2.46 by an easy transfinite construction. �

Theorem 2.55. A is superatomic iff B ≤mg A for every B ≤ A.

Proof. ⇒ follows from Proposition 2.46 by an easy transfinite construction. Nowsuppose that A is not superatomic. Then A has an atomless subalgebra, andhence has a denumerable atomless subalgebra B. B is freely generated by someset X . Take any x ∈ X . Then x is independent over 〈X\{x}〉. It follows then fromProposition 2.49 that A is not minimally generated over 〈X\{x}〉. �

Proposition 2.56. A(x) is minimally generated over A iff A/SmpAx is superatomic.

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Proof. First suppose that A/SmpAx is superatomic. Note that SmpAx is an ideal

of both A and A(x), by Corollary 2.26. By Proposition 2.31(iii), A(x)/SmpAx is

isomorphic to a simple extension of A/SmpAx , and so it is superatomic. By Theorem

2.55, A/SmpAx ≤mg A(x)/SmpAx . Hence by Proposition 2.47, A ≤mg A(x).

Conversely, suppose that A/SmpAx is not superatomic. Then there is an idealK on A such that SmpA

x ⊆ K and A/K is atomless. Let L = 〈K〉idA(x). Let

h :A(x)/SmpAx →A(x)/L be the natural homomorphism, taking y/SmpAx to y/L.

(1) x/L is independent over h[A/SmpAx ].

In fact, take any a ∈ A such that h(a/SmpAx ) �= 0. Thus a /∈ L. If (x/L)·(a/L) = 0,then x · a ∈ L, hence there is a b ∈ K such that x · a ≤ b. This gives a · −b ≤ −x,so that a · −b ∈ SmpA

x ⊆ K. Now a ≤ b+ a · −b ∈ K, so a ∈ K, contradiction. So(x/L) · (a/L) �= 0. Similarly, −(x/L) · (a/L) �= 0. This proves (1).

It now follows from Proposition 2.49 that h[A/SmpAx ] �≤mg A(x)/L. Let l :

A(x) → A(x)/L be the natural homomorphism. Then l[A] = h[A/SmpAx ]. sol[A] �≤mg A(x)/L. Then by Proposition 2.47, A �≤mg A(x). �

Proposition 2.57. The free BA A on ω1 free generators is not minimally generated.

Proof. Suppose it is, and let 〈Bα : α < σ〉 be a representing chain which demon-strates this. Wlog σ = len(A), and Bα ⊂ Bα+1 whenever α + 1 < σ. Clearlyσ ≥ ω1.

We claim that σ = ω1. Otherwise Bω1 is a subalgebra of A. Now |Bω1 | = ω1,so by Theorem 9.16 of the Handbook, it has a subalgebra C isomorphic to A.Hence by Proposition 2.48,

len(A) > ω1 ≥ len(Bω1) ≥ len(C) = len(A),

contradiction.

Let {xα : α < ω1} be the set of free generators of A, and for each α < ω1

let Cα = 〈{xβ : β < α}〉. This gives another representing chain for A. Hence

Kdef= {α < ω1 : Bα = Cα} is a club in ω1. Take any infinite member α of K.

Now xα is independent over Bα and A is minimally generated over Bα, whichcontradicts Proposition 2.49. �

Proposition 2.58. If an infinite BA A satisfies any of the following conditions thenit is not minimally generated:

(1) A is complete;

(2) A is σ-complete;

(3) A is ω1-saturated (in the sense of model theory);

(4) A has the countable separation property;

(5) A is the product of infinitely many non-trivial algebras.

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Proof. Each of (1), (2), (3) implies (4), and (5) implies that A has an infinite CSPsubalgebra. Hence it suffices to show that if A is CSP then it is not minimallygenerated (using Proposition 2.48(ii)). Now A has P(ω) as a homomorphic image,and P(ω) has an independent set of size ω1, so the same is true of A, and theconclusion follows from Proposition 2.57 and Proposition 2.48(ii). �Theorem 2.59. For every minimally generated BA B there is a dense subalgebraA of B such that A is isomorphic to a tree algebra and B is minimally generatedover A.

Proof. Fix a subset X of B generating B and a well-ordering <X of X such thatif we let Bx = 〈{y : y <X x}〉 then the chain 〈Bx : x ∈ X〉 demonstrates theminimal generation of B; we assume that 1 ∈ X , and, moreover, that x /∈ Bx forall x ∈ X\{1}. In particular, 0 /∈ X . Define S = {x ∈ X : x �= 1, Bx is dense inBx(x)}, and set T = X\S. Then:

(∗) If x ∈ T \{1}, then there is an ultrafilter F on Bx and an element y ∈ Bx(x)such that Bx(x) = Bx(y) and ∀a ∈ Bx(y ≤ a iff a ∈ F ).

In fact, by Proposition 2.44, one of Bx � x and Bx � −x is a principal ideal. Thus(∗) follows from Proposition 2.45.

By (∗), wlog we may assume:

(∗∗) If x ∈ T \{1} then there is an ultrafilter F on Bx such that ∀a ∈ Bx(x ≤ a iffa ∈ F ).

Next we claim that T is a tree under the inverse of the Boolean ordering. Infact, suppose x, y, z ∈ T and x < y, x < z; we want to show that y and z arecomparable. Say y <X z. Then y ∈ Bz , so by (∗∗), z ≤ y or z ≤ −y; and z ≤ −y isruled out since 0 �= x ≤ y · z. Thus z and y are comparable. Moreover, if x, y ∈ Tand x < y, then y <X x; otherwise x <X y, hence x ∈ By, and so by this sameproperty, y ≤ x or y ≤ −x, both of which are false. So, T is a tree.

Also note that for u, v ∈ T we have that u and v are incomparable iff u·v = 0.In fact, assume that u and v are incomparable; say u <X v. Then by (∗∗) it followsthat u /∈ Fv and hence −u ∈ Fv and v ≤ −u, as desired.

Next, T is disjunctive. For, assume that t, t1, . . . , tn ∈ T , where n > 0, andt ≤ t1 + · · ·+ tn, but t �≤ ti for all i. Now for all i, either t and ti are comparable,or they are disjoint, by the tree property and the previous paragraph. So wemay assume that ti < t for each i. Similarly, we may assume that the ti’s arepairwise disjoint. So, t = t1 + · · ·+ tn. If t is <X -maximum in {t, t1, . . . , tn}, thent ∈ Bt, contradiction. Otherwise some ti is <X -maximum in {t, t1, . . . , tn}, andti = t ·

∑j �=i −tj, hence ti ∈ Bti , contradiction. So, T is disjunctive.

Now by the proof of Theorem 2.15(iv) ⇒ (i) it follows that Adef= 〈T 〉 is

isomorphic to TreealgT .

We claim that A is dense in B. To prove this, we show by <X -induction onx ∈ X that 〈{y ∈ T : y <X x}〉 is dense in Bx for all x ∈ X . For x the smallest

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element of X the algebra Bx has only two elements, and the conclusion is obvious.Now suppose that the statement holds for x ∈ X , and x′ ∈ X is the immediatesuccessor of x under <X . Now Bx′ = Bx(x). We consider two cases.

Case 1. x ∈ T . Let F be as in (∗∗). By the inductive hypothesis, the set Ddef=

〈{y ∈ T : y <X x}〉 is dense in Bx. We claim that 〈D ∪ {x}〉 is dense in Bx(x).Suppose that 0 �= u ∈ Bx(x). Write u = x ·a+−x · b with a, b ∈ Bx. First supposethat x ·a �= 0. If −a ∈ F , then x ≤ −a by (∗∗), contradiction. So a ∈ F , and hencex ≤ a and so x · a = x and the desired conclusion holds. Suppose that −x · b �= 0.Choose d ∈ D+ such that d ≤ b. Suppose that d · −x = 0.

Subcase 1.1. d ∈ F . Then x ≤ d by (∗∗), so x = d ∈ D, contradiction.

Subcase 1.2. −d ∈ F . Then x ≤ −d by (∗∗), so x · d = 0. Now d ≤ x, so d = 0,contradiction.

Thus d · −x �= 0 and the desired conclusion again holds.

Case 2. x ∈ S. Then Bx is dense in Bx(x) and the desired conclusion is obvious.

Next, for x limit under <X , the induction hypothesis clearly implies the desiredconclusion. Hence our inductive statement holds, and then it is clear that A isdense in B.

If T = X , then A = B and we are through. So assume that S �= 0. We definea new well-ordering� on X by putting T before S: for x, y ∈ X , we define x� yiff (x, y ∈ T and x <X y) or (x, y ∈ S and x <X y) or (x ∈ T and y ∈ S). For eachx ∈ S let Cx = 〈{y ∈ X : y � x}〉. We claim that 〈{Cx : x ∈ S}〉 demonstratesthat B is minimally generated over A. Since Cs = A for s the least member of Sunder �, by Corollary 2.53 all we really need to prove is that Cx ≤m Cx(x) forall x ∈ S. To do this, it suffices to take any y � x and show that y · x ∈ Cx ory · −x ∈ Cx. In fact, if we can do this, then by Proposition 2.25, y or −y will bein SmpCx

x , and since this will be true for each generator of Cx, it will follow thatSmpCx

x is either all of Cx or is a maximal ideal in Cx, so that Cx ≤m Cx(x) byProposition 2.32.

If y <X x, then y ∈ Bx ≤m Bx(x), and so by Propositions 2.25 and 2.32, oneof y · x or y · −x is in Bx. Now Bx ⊆ Cx since x ∈ S, so we obtain the desiredconclusion. Assume, on the other hand, that x <X y. This can only happen ify ∈ T . Hence by (∗∗) applied to By, either y ≤ x and hence y · x = y ∈ Cx, ory ≤ −x and y · x = 0 ∈ Cx, as desired. �

Two BAs A and B are co-complete iff they have isomorphic completions.

Corollary 2.60. Any minimally generated BA A is co-complete with an intervalalgebra.

Proof. By Theorem 2.59, A has a dense subalgebra B isomorphic to a tree algebra.From Theorem 2.21 we know that B can be embedded in an interval algebra, solet f be an isomorphism from B into an interval algebra C. Let I be an ideal of C

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maximal with respect to the property that f [B]∩ I = {0}. Then if g : C → C/I isthe natural homomorphism, g ◦ f is still an embedding, and C/I is isomorphic toan interval algebra. (See the Handbook, Proposition 15.9.) Now g[f [B]] is densein C/I. For, suppose that c ∈ C and c/I �= 0. Then 〈I ∪ {c}〉id, and so there isa b ∈ B such that 0 �= f(b) ≤ d + c for some d ∈ I. Then 0 �= g(f(b)) ≤ c/I, asdesired. Therefore we may assume in the original situation that f [B] is dense inC. Now by the uniqueness of completions (see the remark following 4.14 in theHandbook), it follows that B = A is isomorphic to C. �

Our next result on minimal generation requires several elementary but interestinglemmas.

Lemma 2.61. If I is an ideal in A and C is a subalgebra of A, then C/(I ∩ C) ∼=〈I ∪ C〉/I.

Proof. For any c ∈ C let f([c]I∩C) = [c]I . Clearly f is well defined, and is in facta monomorphism. Now note that

〈I ∪ C〉 ={a+

∑i<m

(ci · −di) : a ∈ I,m ∈ ω, ci ∈ C, di ∈ I

}.

Given an element of 〈I∪C〉 of the form indicated, f([∑

i<m ci]I∩C) = [a+∑

i<m(ci·−di)]I is clear. Hence f is the desired isomorphism. �

For the next lemmas, note that if I is an ideal in A, then 〈I〉 = (I ∪ −I), where−I = {−a : a ∈ I}.Lemma 2.62. Let I be an ideal of A and C be a subalgebra of A. Let f : A→ A/I

be the natural homomorphism, taking any a ∈ A to [a]I . Then gdef= 〈f [B] : 〈I〉 ≤

B ≤ 〈I ∪C〉〉 is an isomorphism from ({B : 〈I〉 ≤ B ≤ 〈I ∪C〉,≤) onto ({D : D ≤〈I ∪ C〉/I},≤).

Proof. Clearly g maps ({B : 〈I〉 ≤ B ≤ 〈I ∪C〉,≤) into ({D : D ≤ 〈I ∪C〉/I},≤),and 〈I〉 ≤ B ≤ B′ ≤ 〈I ∪ C〉 implies that g(B) ≤ g(B′). Now g is one-one.For, suppose that 〈I〉 ≤ B,B′ ≤ 〈I ∪ C〉, B �= B′, and g(B) = g(B′). Say bysymmetry that b ∈ B\B′. Then [b]I ∈ g(B) = g(B′), so there is a b′ ∈ B′ suchthat [b]I = [b′]I . Thus b�b′ ∈ I ⊆ B′. Hence b = (b�b′)�b′ ∈ B′, contradiction.So g is one-one.

Next, if D ≤ 〈I ∪ C〉/I, then 〈I〉 ≤ f−1[D] ≤ 〈I ∪ C〉, and g(f−1[D]) =f [f−1[D]] = D. So g is a surjection.

Finally, if g(B) ≤ g(B′) ≤ 〈I ∪ C〉/I, then

B = g−1(g(B)) = f−1[g(B)] ≤ f−1[g(B′)] = g−1(g(B′)) = B′. �

Proposition 2.63. If I is an ideal in A, C is a subalgebra of A, and C is minimallygenerated, then 〈I〉 ≤mg 〈I ∪ C〉.

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Proof. Note that C/(I ∩C) is minimally generated, by Proposition 2.47(ii); henceby Proposition 2.61, so is 〈I ∪ C〉/I. Now we apply Proposition 2.47(i) to 〈I〉,〈I ∪C〉, 〈I ∪C〉/I, f in place of A,B,Q, f , where f is the natural homomorphismfrom 〈I ∪C〉 onto 〈I ∪ C〉/I; the conclusion of our proposition follows. �Lemma 2.64. Let A be a BA and a ∈ A with a �= 1. Then 〈A � a〉 ∼= (A � a)× 2.

Proof. We define f : 〈A � a〉 → (A � a)× 2 by setting, for any x ∈ 〈A � a〉,

f(x) =

{(x, 0) if x ∈ (A � a),(x · a, 1) if x ∈ −(A � a).

Note that f is well defined because a �= 1.

Now f preserves +. In fact,

if x, y ∈ (A � a), then x+ y ∈ (A � a), andf(x+ y) = (x+ y, 0) = (x, 0) + (y, 0) = f(x) + f(y);

if x ∈ (A � a) and y ∈ −(A � a), then x+ y ∈ −(A � a), andf(x+ y) = ((x + y) · a, 1) = (x, 0) + (y · a, 1) = f(x) + f(y);

if x /∈ (A � a) and y ∈ (A � a), then x+ y ∈ −(A � a), andf(x+ y) = ((x + y) · a, 1) = (x · a, 1) + (y, 0) = f(x) + f(y);

if x, y ∈ −(A � a), then x+ y ∈ −(A � a), andf(x+ y) = ((x + y) · a, 1) = (x · a, 1) + (y · a, 1) = f(x) + f(y).

Similarly, f preserves −:if x ∈ (A � a), then −x /∈ (A � a), and

f(−x) = (−x · a, 1) = −(x, 0) = −f(x);if x /∈ (A � a), then −x ∈ (A � a), and

f(−x) = (−x, 0) = −(x · a, 1) = −f(x).

So f is a homomorphism.

Clearly f is one-one. Finally, f maps onto, since if x ∈ (A � a), then f(x) =(x, 0), and f(x+−a) = ((x +−a) · a, 1) = (x, 1). �Proposition 2.65. If I is an ideal on A and A � a is minimally generated for everya ∈ I, then the subalgebra 〈I〉 of A is also minimally generated.

Proof. Let 〈aα : α < κ〉 enumerate I, κ an infinite cardinal. For each α < κ letJα be the ideal generated by {aβ : β < α}. Since I =

⋃α<κ Jα, we also have

〈I〉 =⋃

α<κ〈Jα〉. Hence it suffices to show that 〈Jα〉 ≤m 〈Jα+1〉 for every α < κ.We may assume that aα /∈ 〈Jα〉.

Let C = 〈A � aα〉. Now C is minimally generated by Lemma 2.64 andProposition 2.50; and so C/(Jα∩C) is minimally generated by Proposition 2.47(ii).Then by Lemma 2.61, 〈Ja ∪ C〉/Jα is minimally generated. It follows by Lemma2.63 that 〈Jα〉 ≤mg 〈Jα ∪C〉. Clearly 〈Jα ∪C〉 = 〈Jα+1〉, so 〈Jα〉 ≤m 〈Jα+1〉. �

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In conclusion of our treatment of minimally generated BAs we mention the fol-lowing problem from Koppelberg, Monk [92].

Problem 1. Is there in ZFC an infinite minimally generated BA with no countablyinfinite homomorphic image?

The problem is formulated in this way since Koppelberg [88] showed that♦ impliesthat there is a BA of this sort.

Tail algebras

For any partial order P , let the tail algebra of P be the subalgebra Tailalg(P ) ofP(P ) generated by {P ↑ p : p ∈ P}. Thus these algebras generalize tree algebrasand pseudo-tree algebras. The notion is due to Gary Brenner. It is studied inKoppelberg, Monk [92], the main results being due to Koppelberg and Blass.

Theorem 2.66. Every semigroup algebra is isomorphic to a tail algebra.

Proof. Let A be a semigroup algebra, and choose a generating set H for A such

that 0, 1 ∈ H , H is closed under ·, and Pdef= H\{0} is disjunctive. Let f be the

homomorphism from A onto Tailalg (P−1) given by Proposition 2.1: f(p) = P ↓ pfor any p ∈ P . We show that f is one-one, which will finish the proof. By Sikorski’scriterion, we have to show that f(p1) ∩ · · · ∩ f(pn) ⊆ f(q1) ∪ · · · ∪ f(qm) (wherepi, qj ∈ P ) implies p1 · · · · · pn ≤ q1 + · · · + qm. Without loss of generality, p =p1 · · · · · pn is nonzero and hence is in P . Now p ∈ f(p1)∩ · · · ∩ f(pn). So p ∈ f(qj)for some j, p ≤ qj , and p ≤ q1 + · · ·+ qm, as desired. �

We also need a set-theoretic lemma:

Lemma 2.67. Let P be an infinite partially ordered set. Then: either P has astrictly ascending chain of type ω, or P has a strictly descending chain of type ω,or P is well founded (with, say, Pα as its αth level) and there is some n ∈ ω suchthat Pn is infinite.

Proof. Assume P has no descending chain of type ω (so P is well founded) andno infinite level Pn (n ∈ ω). For each n ∈ ω, let

Tn = {(p0, . . . , pn) : pi ∈ Pi for all i ≤ n, and p0 < · · · < pn}.

So T =⋃

n∈ω Tn is a tree in which every level is finite and non-empty. But thenT has an infinite branch, which yields an increasing chain of type ω in P . �

An algebra which is generated by a disjunctive set is called disjunctively generated.Clearly tail algebras are disjunctively generated.

Theorem 2.68. Every infinite disjunctively generated algebra has a countably infi-nite homomorphic image.

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Proof. Say A = 〈P 〉, where P is an infinite disjunctive subset of A. We applyLemma 2.67 to P−1, and have three cases.

Case 1. There is in P an ascending sequence (pn : n ∈ ω). Let then M = {pn : n ∈ω}, and consider the homomorphism fM given by Proposition 2.1. Then fM mapseach p ∈ P to an initial segment of M , and since fM (pn) = {p0, . . . , pn} and Pgenerates A, it follows that the image of A under fM is the finite-cofinite algebraon M , a countable algebra.

Case 2. There is in P a descending sequence of type ω. This is similar to Case 1,again considering M = {pn : n ∈ ω}.Case 3. P−1 is well founded, and for some (minimal) n ∈ ω, Pn is infinite. ConsiderM = Pn and f = fM as in Proposition 2.1. Note that

1. f(p) = ∅ if p ∈ Pα, α > n

2. f(p) = {p} for p ∈ Pn

3. {f(p) : p ∈ Pk, k < n} is finite.

It follows that the image of A under f is superatomic, since its quotient under theideal generated by the atoms is finite. It is well known, and easy to check, thatevery superatomic algebra has a countable homomorphic image, giving the desiredresult. �

Corollary 2.69. No infinite Boolean algebra having the countable separation prop-erty is disjunctively generated. �

Theorem 2.70. Every BA can be embedded into a tail algebra.

Proof. This is trivial for a finite Boolean algebra B with, say, n atoms – just takea tree with n roots and no other points. So let B be an infinite Boolean algebra;we may assume that it is the algebra of clopen subsets of some Boolean space X .

For each b ∈ B, take two new points pb, qb such that the points pb, qb (b ∈ B),are pairwise distinct and not in X . Then put

U = {pb, qb : b ∈ B}, P = U ∪X

and define a partial order on P by setting pb < x and qb < x for all x ∈ b. Thus, forb ∈ B, P ↑ pb = {pb}∪b, P ↑ qb = {qb}∪b and b = (P ↑ pb)∩(P ↑ qb) ∈ Tailalg(P ).

We define a map e from B into the power set algebra of P by fixing a non-isolated point x∗ of X and putting e(b) = b if x∗ /∈ b and e(b) = U ∪ b if x∗ ∈ b.It is easily checked that e embeds B into the power set algebra of P and thate(b) ∈ Tailalg(P ) if x∗ /∈ b; hence e is an embedding from B into Tailalg(P ). �

The following problem is mentioned in Koppelberg, Monk [92].

Problem 2. Is every disjunctively generated BA isomorphic to a tail algebra?

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Initial chain algebras

Let T be a pseudo-tree. The initial chain algebra of T , denoted by Init(T ), is thesubalgebra of P(T ) generated by {T ↓ t : t ∈ T }. A thorough treatment of thesealgebras is found in Baur [00] and Baur, Heindorf [97]. Here we state some of thefacts about these algebras and give just a few proofs.

Lemma 2.71. Let A be a BA generated by a set H with the following properties:

(i) 0 /∈ H;

(ii) H ∪ {0} is closed under ·;(iii) H ↓ h is linearly ordered, for all h ∈ H.

Then H is disjunctive.

Proof. Suppose that h, h1, . . . , hn ∈ H , n > 0, and h ≤ h1 + · · · + hn. We mayassume that h ·hi �= 0 for all i. Now h = h ·h1+ · · ·+h ·hn, and h ·hi ∈ (H ↓ h) foreach i, so by (iii) there is an i such that h·hi ≤ h·hj for all j. Hence h = h·hi ≤ hi,as desired. �

Lemma 2.72. Let A be a BA generated by a set H with the following properties:

(i) 0 /∈ H;

(ii) H ∪ {0} is closed under ·;(iii) H ↓ h is linearly ordered, for all h ∈ H;

(iv) for every nonzero a ∈ A there is an h ∈ H such that a · h �= 0.

Then H is a pseudo-tree under the Boolean ordering, and A is isomorphic toInit(H).

If we replace (iii) by

(iii′) H ↓ h is well ordered, for all h ∈ H,

Then we can conclude that H is a tree under the Boolean ordering.

Proof. ObviouslyH is a pseudo-tree under the Boolean ordering, and also the finalstatement of the Proposition holds. By Lemma 2.71 and Proposition 2.1, there isa homomorphism f from A into P(H) such that f(h) = H ↓ h for all h ∈ H .Thus f maps onto Init(H). We need to show that f is one-one. Assume that

(H ↓ h1) ∩ · · · ∩ (H ↓ hm) ∩ [H\(H ↓ k1)] ∩ · · · ∩ [H\(H ↓ kn)] = 0,

but h1 · . . . · hm · −k1 · . . . · −kn �= 0. By (iv) choose h ∈ H such that h · h1 · . . . ·hm · −k1 · . . . · −kn �= 0. Now

h · h1 · . . . · hm ∈ (H ↓ h1) ∩ · · · ∩ (H ↓ hn),

so h·h1·. . .·hm ∈ (H ↓ ki) for some i. Thus h·h1·. . .·hm·−ki = 0, contradiction. �

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Theorem 2.73. For any BA A the following three conditions are equivalent:

(i) A is isomorphic to the initial chain algebra on some pseudo-tree T ;

(ii) A has a set of generators H such that the conditions of Lemma 2.71 hold.

(iii) A has a set of generators H such that the conditions of Lemma 2.72 hold.

Proof. (i)⇒(ii): let U = [T ]<ω\{∅}, and let

H =

{⋂t∈F

(T ↓ t) : F ∈ U

}\{0}.

Clearly the conditions (i) and (ii) of Lemma 2.71 hold. Condition (iii) requiresmore work. For each F ∈ U , let vF =

⋂t∈F (T ↓ t). Suppose that F,G,H ∈ U

and vG, vH ⊆ vF ; we want to show that vG and vH are comparable (under ⊆).Suppose that vG �⊆ vH and vH �⊆ vG. Say s ∈ vG\vH and t ∈ vH\vG. Now sinceF is nonempty, choose u ∈ F . Since vG, vH ⊆ vF , we have s, t ≤ u; so s and t arecomparable. By symmetry, say s ≤ t. For any w ∈ H we have t ≤ w, and hences ≤ w. So s ∈ vH , contradiction. This verifies (iii) of Lemma 2.71, so (ii) of thepresent theorem holds.

(ii)⇒(iii): Suppose that (iv) of Lemma 2.72 fails to hold. Then there is somemonomial x over H such that x · h = 0 for all h ∈ H , with x �= 0. Write x =h0 · . . . ·hm−1 · −k0 · . . . · −kn−1 with all hi, kj ∈ H . Then m = 0 since x ·h = 0 forall h ∈ H . Let H ′ = H ∪ {x}. Clearly H ′ satisfies the conditions of Lemma 2.72.

(iii)⇒(i): by Lemma 2.72. �

Theorem 2.74. For any BA A the following three conditions are equivalent:

(i) A is isomorphic to the initial chain algebra on some tree;

(ii) A has a set of generators H such that the conditions of Lemma 2.71 hold,with (iii) replaced by the condition that H ↓ h is well ordered for all h ∈ H.

(iii) A has a set of generators H such that the conditions of Lemma 2.72 hold,with the same replacement as in (ii).

Proof. We just need to add to the proof of (i)⇒(ii) in the proof of Theorem 2.73.Suppose that vf0 ⊃ vf1 ⊃ · · · . Choose ti ∈ vFi\vFi+1 for all i ∈ ω. Then forany i ∈ ω, ti, ti+1 ∈ Fi, and hence ti and ti+1 are comparable. If ti < ti+1, thenti ∈ vFi+1 , contradiction. So ti+1 < ti. Thus we have a strictly decreasing sequenceof elements of H , contradiction. �

We say that a pseudo-tree T is well met iff for any two x, y ∈ T , if z ≤ x, y forsome z ∈ T , then x and y have a glb in T . Note that for T a tree, this just meansthat for any two distinct x, y ∈ T of the same limit level, the sets T ↓ x and T ↓ yare different.

Corollary 2.75. Every initial chain algebra on a pseudo-tree is isomorphic to aninitial chain algebra on a well-met pseudo-tree.

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Proof. This is clear from the proof of Theorem 2.73. �Theorem 2.76. Let T be a pseudo-tree and A = Init(T ). Then every homomorphicimage of A is isomorphic to an initial chain algebra of a pseudo-tree.

Similarly with “pseudo-tree” replaced by “tree”.

Proof. We may assume that T is well met by Corollary 2.75. Suppose that I is anideal on A. Denote members of A/I by [a]. Then with T ′ = {[h] : h ∈ T }, the firstpart is clear by Theorem 2.73(ii).

For the second part, suppose that [h0] > [h1] > · · · with h0, h1, . . . membersof T . For each i, let ki be the glb of h0, . . . , hi. Then each ki is in T , and k0 >k1 > · · · , contradicting the assumption in the second part that T is a tree. �

The following rather obvious normal form theorem for elements is sometimes use-ful.

Theorem 2.77. f T is a well-met pseudo-tree and x ∈ Init(T ), then x is a finitesum of elements of the following forms:

(i) T ↓ s.(ii) (T ↓ s)\(T ↓ t).(iii) T \

⋃s∈F (T ↓ s), with F a finite subset of T .

Proof. Obviously x is a finite sum of elements of the following form:

(T ↓ s0) ∩ · · · ∩ (T ↓ sm−1) ∩ [T \(T ↓ t0)] ∩ · · · ∩ [T \(T ↓ tn−1)].

If m = 0, we have condition (iii). If m > 0, by the well-met condition we mayassume that m = 1. Then n = 0 gives condition (i). If n > 0, then note that

(T ↓ s0)∩[T \(T ↓ t0)] ∩ · · · ∩ [T \(T ↓ tn−1)]

= (T ↓ s0) ∩ [T \[(T ↓ s0) ∩ (T ↓ t0)]] ∩ · · · ∩ [T \[(T ↓ s0) ∩ (T ↓ tn−1)]];

the elements (T ↓ s0)∩(T ↓ t0), . . . , (T ↓ s0)∩(T ↓ tn−1) are all comparable, by thewell-met condition, since they are all below T ↓ s0. Thus this gives condition (ii).

�Theorem 2.78. Every initial chain algebra on a tree is superatomic.

Proof. By Theorem 2.76 it suffices to show that Init(T ) is always atomic for T atree. Note that if t ∈ T is not at a limit level, then {t} ∈ A. We may assume thatT is well met. Let x be a non-zero element of Init(T ). By Theorem 2.77 we mayassume that x has one of the three forms mentioned there:

Case 1. x = (T ↓ t) for some t. Let s be the root such that s ≤ t. Then {s} is thedesired atom below x.

Case 2. x = (T ↓ t)\(T ↓ s) for some s, t. If (T ↓ t) ∩ (T ↓ s) = 0, then (T ↓t)\(T ↓ s) = (T ↓ t) and we are back in Case 1. So let (T ↓ t) ∩ (T ↓ s) = (T ↓ r).Choose u at a successor level, r < u ≤ t. Then {u} is the desired atom below x.

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Case 3. x = T \⋃

s∈F (T ↓ s) for some finite subset F of T . Since x �= 0, chooset ∈ x. Then (T ↓ t)\

⋃s∈F (T ↓ s) reduces to Case 1 or Case 2, as desired. �

Theorem 2.79. Every initial chain algebra on a pseudo-tree is a semigroup algebra.

Proof. This is immediate from Lemma 2.71 and Theorem 2.73. �

For any BA A we define

IAint = {a ∈ A : A � a is isomorphic to an interval algebra}.

Corollary 2.80. IAint is an ideal of A.

Proof. If b ≤ a ∈ IAint, then A � b is a homomorphic image of A � a, and henceA � b is isomorphic to an interval algebra by the Handbook Proposition 15.9. Ifa, b ∈ IAint, then (A � (a+ b)) = (A � a)× (A � (b ·−a)), so A � (a+ b) is isomorphicto an interval algebra by the Handbook Proposition 15.9 and 15.11. �Theorem 2.81. If A is an initial chain algebra on a pseudo-tree, then IAint = A orIAint is a maximal ideal of A.

Proof. By Theorem 2.73 there is a setH generatingA which satisfies the conditionsof Lemma 2.72. For each h ∈ H let f : A → (A � h) be the natural mapping:f(a) = a · h for all a ∈ A. So f is a surjective homomorphism. Since H generatesA, f [H ] generates A � h. But f [H ] = (H ↓ h) is a chain by Lemma 2.71(iii). ThusA � h is isomorphic to an interval algebra. This is true for any h ∈ H , and Hgenerates A. The conclusion of our theorem follows. �Theorem 2.82. Every initial chain algebra is minimally generated.

Proof. This is immediate from Theorem 2.81, Proposition 2.52, and Proposition2.65. �

Kinds of subalgebras

We now indicate the relationships between the kinds of subalgebras described inChapter 0. See Monk [10].

Proposition 2.83. A ≤free B implies that A ≤proj B.

Proof. Assume that A ≤free B; say B = A⊕D, with D free. Let q = e = IdB. �Proposition 2.84. A ≤proj B implies that A ≤rc B.

Proof. Suppose that b ∈ B; we want to find the least u ∈ A such that b ≤ u.Assume the notation in the definition of A ≤proj B. Write e(b) =

∑i<m(ai · ci)

with each ai ∈ A and each ci ∈ C. Thus

b = q(e(b)) =∑i<m

q(ai) · q(ci) =∑i<m

(ai · q(ci)) ≤∑i<m

ai ∈ A.

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Suppose that b ≤ d ∈ A. Then e(b) ≤ e(d) = d, and so∑

i<m(ai · −d · ci) = 0, andhence

∑i<m ai ≤ d. �

Proposition 2.85. A ≤rc B implies that A ≤reg B.

Proof. Suppose that X ⊆ A and∑A

X exists. Let b ∈ B be an upper bound forX ;

we want to show that∑A

X ≤ b. Let c ∈ A be maximum such that c ≤ b. Now if

x ∈ X , then x ≤ b; hence x ≤ c. It follows that∑A X ≤ c, hence

∑A X ≤ b. �

The following two propositions are clear.

Proposition 2.86. A ≤m B implies that A ≤s B. �

Proposition 2.87. A ≤m B implies that A ≤mg B. �

Proposition 2.88. A ≤π B implies that A ≤reg B.

Proof. See 4.17 in the Handbook. �

Proposition 2.89. A ≤rc B implies that A ≤σ B. �

Proposition 2.90. A ≤free B implies that A ≤u B.

Proof. Suppose that A ≤free B; say B = A⊕C with C free. Let F be an ultrafilteron A. Let c ∈ C\{0, 1}. Clearly F ∪ {c} and F ∪ {−c} both have fip. �

Proposition 2.91. If A ≤u B, A is infinite, and A �= B, then not(A ≤m B).

Proof. This is immediate from Proposition 2.35. �

Example 2.92. There exist BAs A,B such that A ≤u B, A infinite, A �= B, andA ≤mg B.

Let C = finco(ω). Let f({i}) = {2i, 2i + 1} for all i ∈ ω, and extend f to anisomorphism from C into C. Let A = f [C]. Then A ≤mg C by Proposition 2.55.Now the principal ultrafilters on A have the form {u ∈ A : {2i, 2i+ 1} ⊆ u} forsome i ∈ ω, and each of these has two extensions to an ultrafilter on C. A has onemore ultrafilter, namely {u ∈ A : u∩F = ∅ for some finite subset F of ω such that∀i ∈ ω[2i ∈ F iff 2i+ 1 ∈ F ]}. This ultrafilter extends only to the ultrafilter on Cconsisting of cofinite subsets of ω. So it suffices to define a minimal extension B ofC in which this ultrafilter has two extensions. Let I0 be the ideal of C consisting offinite sets all of whose members are even, and let I1 be the ideal of C consisting offinite sets all of whose members are odd. Then I0 ∩ I1 = {∅}. By Proposition 2.28let B = C(x) be such that C � x = I0 and C〈−x = I1. Now SmpCx = 〈I0 ∪ I1〉id isthe maximal ideal of all finite subset of ω, so C ≤m C(x). The ultrafilter in C ofcofinite subsets of ω extends to two different ultrafilters on C(x); one with x as amember, and one with −x as a member.

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These propositions can be summarized in the diagram below.

To see that no further relations exist, we need several examples. The diagram canbe used to check that these examples take care of all possibilities; see also thetable below. For the first example we need two lemmas of independent interest.

Proposition 2.93. Suppose that A ≤proj B, with C, q, e as in the definition. Alsosuppose that C ≤ D with D free. Let h : D → C be a homomorphism such thath � C = IdC , and let q′ : A ⊕ D → B be such that q′(a) = a for all a ∈ A andq′(d) = q(h(d)) for all d ∈ D. Then D, q′, e also witness that A ≤proj B. Namely,q′ ◦ e = IdB and e � A = q′ � A = IdA.

Proof. This is immediate, upon observing that q′ � (A⊕ C) = q. �

Lemma 2.94. If A1 ≤proj B1 and A2 ≤proj B2, then A1 ×A2 ≤proj B1 ×B2.

Proof. We first give a different proof for the following fact, Example 11.6(c) in theHandbook.

(1) If B,C,D are BAs, then (B × C)⊕D ∼= (B ⊕D)× (C ⊕D).

In fact, by the basic property of free products there is a homomorphism h of(B × C)⊕D into (B ⊕D)× (C ⊕D) such that h(b, c) = (b, c) for any b ∈ B andc ∈ C, and h(d) = (d, d) for any d ∈ D. If h((b, c) · d) = (0, 0), then b · d = 0 andc · d = 0, and so by the free product property, d = 0 or b = c = 0. This shows thath is one-one. Finally, h maps onto since for any b ∈ B, c ∈ C, and d ∈ D we haveh((b, 0) · d) = (b · d, 0), h((0, c) · d) = (0, c · d), and h(1, 0) = (1, 0).

Turning to the proof of our lemma, by hypothesis we get Ci, qi, ei for i ∈ {1, 2}such that Ci is free, qi : Ai ⊕Ci → Bi is a homomorphism, ei : Bi → Ai ⊕Ci is ahomomorphism, qi ◦ ei = IdBi , and ei � Ai = qi � Ai = IdAi . By Proposition 2.93we may assume that C1 = C2 = C, say. Let h : (A1×A2)⊕C → (A1⊕C)×(A2⊕C)be the isomorphism given in the proof of (1). Define e′ : B1×B2 → (A1×A2)⊕Cby setting e′(b1, b2) = h−1(e1(b1), e2(b2)). Define q′ : (A1 × A2) ⊕ C → B1 × B2

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by setting q′(h−1(u, v)) = (q1(u), q2(v)). Clearly e′ and q′ are homomorphisms. Ifb1 ∈ B1 and b2 ∈ B2, then

q′(e′(b1, b2)) = q′(h−1(e1(b1), e2(b2))) = (q1(e1(b1)), q2(e2(b2))) = (b1, b2).

For a1 ∈ A1 and a2 ∈ A2 we have

q′(a1, a2) = q′(h−1(h(a1, a2))) = q′(h−1(a1, a2))

= (q1(a1), q2(a2)) = (a1, a2)

and

e′(a1, a2) = h−1(e1(a1), e2(a2)) = h−1(a1, a2) = (a1, a2). �

Example 2.95. Let the free generators of Fr(ω1) be 〈xα : α < ω1〉. Let A =Fr(ω)× Fr(ω) and B = Fr(ω)× Fr(ω1). Now obviously Fr(ω) ≤free Fr(ω1), and soalso Fr(ω) ≤proj Fr(ω1). Hence A ≤proj B by Lemma 2.94.

We claim that not(A ≤u B). Let F be any ultrafilter on A such that (1, 0) ∈F . Suppose that G and H are ultrafilters on B such that F ⊆ G,H ; we want toshow that G = H . Let (a, b) be any element of G. Since (1, 0) ∈ F ⊆ G, we have(a, 0) ∈ G. If (a, 0) /∈ F , then (−a, 1) ∈ F , hence (−a, 0) = (−a, 1) · (1, 0) ∈ F , andso (−a, 0) ∈ G, contradiction. So (a, 0) ∈ F . Hence (a, 0) ∈ H and consequently(a, b) ∈ H . This shows that G ⊆ H ; so G = H .

It is also clear that not(A ≤π B) and not(A ≤s B).

Finally, we claim that not(A ≤mg B). For, suppose to the contrary thatA ≤mg B. Let I be the ideal of B generated by the set

{(1, 0)} ∪ {(0, xn) : n ∈ ω}.

Then B/I ∼= Fr(ω1). In fact, let f be the homomorphism of Fr(ω1) into B/I suchthat f(xα) = [(0, xω+α)]I for every α < ω1. To show that f is a surjection, considerthe generating set

{(xn, 0) : n ∈ ω} ∪ {(1, 0)} ∪ {(0, xα) : α < ω1}

of B. Under the natural homomorphism from B onto B/I, this set goes over intothe generating set {[(0, xω+α)]I : α < ω1 of B/I, showing that f is a surjection.To see that f is an injection, it suffices to take any y ∈ Fr(ω1) of the form y =∏

α∈F xε(α)α (F ⊆ ω1 finite, ε ∈ F 2) and show that f(y) �= 0. In fact, clearly

f(y) = [∏

α∈F (0, xω+α)ε(α)]I , and clearly this last element is nonzero. This proves

that B/I ∼= Fr(ω1).

It is also clear that A/(I ∩ A) is the two element BA. Now since A ≤mg B,by Proposition 2.47 we it follows that B/I is minimally generated. So Fr(ω1) isminimally generated, contradicting Proposition 2.57.

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Example 2.96. Let A = Fr(ω) and B = A ⊕ Finco(ω1). Here Finco(ω1) is thealgebra of finite and cofinite subsets of ω1. So A ≤rc B by the Handbook 11.8.Suppose that A ≤proj B, with notation as in the definition. Then A ⊕ C is free,hence has ccc, but B is isomorphically embeddable in C, contradiction.

We have A ≤u B. In fact, if F is any ultrafilter on A, then the two sets{a · {0} : a ∈ F} and {a · {1} : a ∈ F} both have fip, and this gives distinctextensions of F to an ultrafilter on B.

Also notice that A �≤mg B. In fact, suppose that A ≤mg B. Let x = {i}.Then by Proposition 2.48(iii), A ≤mg A(x). This contradicts Proposition 2.49.

Example 2.97. Let A = Finco(ω1) and B = P(ω1). Thus A ≤π B. Let b = {2α :α < ω1}. Then

〈A � b〉id = [b]<ω.

Clearly this ideal is not countably generated. So not(A ≤σ B). Also, not(A ≤mg

B). In fact, otherwise with the ideal I of finite sets we would get by Proposition2.47 that B/I is minimally generated, contradicting Proposition 2.58.

Example 2.98. Let A = Fr(ω) and B = Fr(ω) ⊕ Fr(ω1). Note that not(A ≤π B).Also, not(A ≤mg B); see the end of Example 2.96.

Example 2.99. Let A′ = Fr(ω), let f be the Stone isomorphism of A′ into Bdef=

P(Ult(A′)) (f(a) = {F ∈ Ult(A′) : a ∈ F} for any a ∈ A′), and let A be therange of f . Let 〈xi : i ∈ ω〉 be a system of free generators of A′. We claim that∑A′

i∈ω xi = 1. In fact, if a is a nonzero element of A′, write a =∑

ε∈Γ

∏i∈M x

ε(i)i ,

where M is a finite subset of ω and Γ ⊆ M2. Then a · xj �= 0 for any j ∈ ω\M .This proves the claim.

It follows that∑A

i∈ω f(xi) = 1. However,⋃

i∈ω f(xi) =∑B

i∈ω f(xi) �= 1, since

for example the ultrafilter generated by {−xi : i ∈ ω} is not in∑B

i∈ω f(xi). Hencenot(A ≤reg B).

We claim that A ≤u B. For, suppose that F ∈ Ult(A). Then there is anultrafilter F ′ on A′ such that f [F ′] = F . For each a ∈ F ′ let G′

a be an ultrafilteron A′ such that a ∈ G′

a and G′a �= F ′. This is possible since A′ is atomless. Let

X = {G′a : a ∈ F ′}. Thus X ⊆ Ult(A′), and so X ∈ B. Then F ∪ {X} has the

fip. In fact, if a ∈ F , then f−1(a) ∈ F ′, hence f−1(a) ∈ G′f−1(a), hence G′

f−1(a) ∈f(f−1(a)) = a, so that G′

f−1(a) ∈ a ∩X . Let H be an ultrafilter on B containing

F ∪ {X}. Thus H extends F . But also the ultrafilter Ldef= {K ∈ B : F ′ ∈ K}

extends F . In fact, if a ∈ F , then f−1(a) ∈ F ′, hence F ′ ∈ f(f−1(a)) = a, asdesired. Now X ∈ H\L, so H �= L. This shows that A ≤u B.

Finally, A �≤mg B. For, suppose that A ≤mg B. Let I be the ideal in Agenerated by {xi : i ∈ ω}. Then f [I] is a maximal ideal in A, so |A/I| = 2.Let J be the ideal in B generated by f [I]. Then J = {X ∈ B : there is a finiteF ⊆ ω such that X ⊆ f(

∑i∈F xi)}. We claim that B/J is infinite. For, let

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〈Mi : i ∈ ω〉 be a partition of ω into infinite sets. For each j ∈ ω let Gj be theultrafilter on A′ such thatxj ∈ Gj and −xk ∈ Gj for all k �= j. For each i ∈ ω letXi = {Gj : j ∈ Mi}. Thus Xi ∩Xk = ∅ for i �= k. We claim that Xi /∈ J . In fact,otherwise there is a finite F ⊆ ω such that Xi ⊆ f(

∑k∈F xk). Take Gl ∈ Xi such

that l /∈ F . Then Gl ∈ f(∑

k∈F xk), so that∑

k∈F xk ∈ Gl. But −xk ∈ Gl for allk ∈ F , contradiction. It follows that B/J is infinite. By Proposition 2.47, B/J isminimally generated. This contradicts Proposition 2.58.

Example 2.100. Let A = Fr(ω), let I be a maximal ideal in A, and let J = {0}. ByProposition 2.28 there is a simple extension B

def= A(x) of A such that A � x = I

and A � −x = J . By Proposition 2.32, A ≤m B. We have∑A I = 1 while∑B

I = x; so not(A ≤reg B). Since A � (−x) = {0}, it follows that A �≤π B.

Example 2.101. Let A = Finco(ω1) and B = A⊕ Fr(ω). Clearly A ≤u B. For the

element bdef=∑

α∈ω1{2α}, the ideal A � b is not countably generated. A �≤mg B by

the argument at the end of Example 2.96. A �≤s B since every simple extension ofA is superatomic by Proposition 2.30.

Example 2.102. Let A = Finco(ω1), I = 〈{{2α} : α < ω1}〉, and J = 〈{{2α+ 1} :α < ω1}〉. Let B = A(x) with B � x = I and B � (−x) = J . Then A ≤m B andnot(A ≤σ B), since A � x is not countably generated. A �≤π B by Proposition2.44. Clearly A ≤reg B.

The properties of these examples are summarized in the following table.

Example ≤reg ≤σ ≤π ≤rc ≤proj ≤u ≤free ≤s ≤mg ≤m

2.95 Y Y N Y Y N N N N N

2.96 Y Y N Y N Y N N N N

2.97 Y N Y N N N N N N N

2.98 Y Y N Y Y Y Y N N N

2.99 N Y N N N Y N N N N

2.100 N Y N N N N N Y Y Y

2.101 Y N N N N Y N N N N

2.102 Y N Y N N N N Y Y Y

There are 10 subalgebra notions considered, and there are 13 implications be-tween them, given in the diagram following Proposition 2.93. None of the other77 implications hold. This is shown by the above examples, or in some cases isobvious:

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≤free and �≤π 2.98

≤free and �≤m 2.98

≤free and �≤mg 2.98

≤free and �≤s 2.98

≤proj and �≤free 2.95

≤proj and �≤u 2.95

≤proj and �≤π 2.95

≤proj and �≤m 2.95

≤proj and �≤s 2.95

≤proj and �≤mg 2.95

≤u and �≤free 2.96

≤u and �≤proj 2.96

≤u and �≤rc 2.99

≤u and �≤σ 2.101

≤u and �≤reg 2.99

≤u and �≤π 2.98

≤u and �≤m 2.98

≤u and �≤s 2.98

≤u and �≤mg 2.98

≤rc and �≤free 2.95

≤rc and �≤proj 2.96

≤rc and �≤u 2.95

≤rc and �≤π 2.96

≤rc and �≤m 2.95

≤rc and �≤s 2.95

≤rc and �≤mg 2.95

≤π and �≤free 2.97

≤π and �≤proj 2.97

≤π and �≤u 2.97

≤π and �≤rc 2.97

≤π and �≤σ 2.97

≤π and �≤m 2.97

≤π and �≤s 2.97

≤π and �≤mg 2.97

≤reg and �≤u 2.95

≤reg and �≤σ 2.97

≤reg and �≤m 2.95

≤reg and �≤s 2.95

≤reg and �≤mg 2.95

≤reg and �≤free 2.95

≤reg and �≤proj 2.96

≤reg and �≤rc 2.97

≤reg and �≤π 2.96

≤ σ and �≤u 2.95

≤ σ and �≤reg 2.99

≤ σ and �≤π 2.95

≤ σ and �≤m 2.95

≤ σ and �≤s 2.95

≤ σ and �≤mg 2.95

≤ σ and �≤free 2.95

≤ σ and �≤proj 2.96

≤ σ and �≤rc 2.99

≤m and �≤free 2.100

≤m and �≤proj 2.100

≤m and �≤u 2.100

≤m and �≤rc 2.100

≤m and �≤π 2.100

≤m and �≤reg 2.100

≤m and �≤σ 2.102

≤s and �≤free 2.100

≤s and �≤proj 2.100

≤s and �≤u 2.100

≤s and �≤rc 2.100

≤s and �≤π 2.100

≤s and �≤reg 2.100

≤s and �≤σ 2.102

≤s and �≤m obv.

≤s and �≤mg obv.

≤mg and �≤free 2.100

≤mg and �≤proj 2.100

≤mg and �≤u 2.100

≤mg and �≤rc 2.100

≤mg and �≤π 2.100

≤mg and �≤reg 2.100

≤mg and �≤σ 2.102

≤mg and �≤m obv.

≤mg and �≤s obv.

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Other classes of Boolean algebras

We briefly mention some other classes of Boolean algebras.

⊥-free Boolean algebras. A BA A is ⊥-freely generated by a subset D iff D gen-erates A, and if f : D → B is a mapping from D into any BA B such that∀a, b ∈ D(a · b = 0 ⇒ f(a) · f(b) = 0), then f extends to a homomorphism fromA into B. This notion was introduced and studied in Heindorf [94]. Corey Brunshas studied cardinal functions on these algebras, and has generalized the notion;see Bruns [13].

Moderate Boolean algebras. A subset X of a BA A is moderate iff 0 /∈ X and forevery a ∈ A, the set {x ∈ X : a · x �= 0 �= −a · x} is finite. A BA A is moderate iffall of its ideals are generated by moderate sets. This notion was introduced andstudied in Heindorf [92]; he also proved some facts about cardinal functions onthis class.

Retractive Boolean algebras. A BA A is retractive iff for every epimorphism f :A→ B there is a homomorphism g : B → A such that f ◦ g is the identity on B.See Heindorf [86].

A theorem of M. Rubin is that every subalgebra of an interval algebra isretractive; see the Handbook. Thus by Theorem 2.13, every pseudo-tree algebra isretractive.

Projective Boolean algebras. A BA A is projective iff for all homomorphisms f :A → B and g : C → B with g onto, there is a homorphism h : A → C such thatf = g ◦ h. See Koppelberg [89b] for an extensive treatment of this notion.

Weakly projective Boolean algebras. A BA A is weakly projective iff A has a denseprojective subalgebra.

Cohen algebras. A Cohen algebra is a BA which is co-complete with a product offree BAs.

Co-interval algebras. A BA A is a co-interval algebra iff it is co-complete with aninterval algebra.

Rc-filtered Boolean algebras. A BA A is rc-filtered iff for all homomorphismsf : A → B and g : C → B with g surjective there is an order-preserving maph : A → C which also preserves disjointness and is such that g ◦ h = f . SeeHeindorf, Shapiro [94].

σ-filtered Boolean algebras. A BA A is σ-filtered iff there is a function I : A →[A]≤ω such that for all a, b ∈ A, a ≤ b implies that there is an x ∈ I(a)∩ I(b) suchthat a ≤ x ≤ b. See Heindorf, Shapiro [94].

Lindelof Boolean algebras. A BA A is Lindelof iff every ideal in A is countablygenerated. This notion and the next one were introduced and studied in Hein-dorf [85].

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Paracompact Boolean algebras. A BA A is paracompact iff every ideal in A isgenerated by a set of pairwise disjoint elements.

Model-theoretic classes of Boolean algebras. Many classes of BAs turn out to bemodel-theoretically definable in the following sense. Suppose that L is a first-orderlanguage. Expand L to LP by adding a unary relation symbol P . A sentence ϕof LP is universal over L iff it has the form ∀xϕ with ϕ a Boolean combinationof formulas of L and atomic formulas of the form Pτ , τ a term of L . Supposethat Σ is a set of formulas of LP , each universal over L . Given an L -structureA, we say that U ⊆ A is a Σ-subset of A iff (A,U) |= Σ. We let X(A,Σ) be the setof all Σ-subsets of A. Thus X(A,Σ) is a subset of P(A), and under the Tychonofftopology on P(A) (homeomorphic to A2), it is closed. We let B(Σ, A) be the BAof clopen subsets of X(A,Σ). For any theory T in L we then define the Booleanalgebra model class of T under Σ to be

{B(A,Σ) : A |= T }.

See Koppelberg [93] for this notion.

Free poset algebras. Given a poset P , the free poset algebra on P is the BAgenerated by P freely – subject only to the conditions p ≤ q when this holds in P .(The rigorous definition is the quotient of the free BA on P divided by the idealgenerated by {p · −q : p ≤ q}.) See Bonnet, Rubin [04].

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3 Cellularity

A BA A is said to satisfy the κ-chain condition (κ-cc) if every disjoint subsetof A has power < κ. Thus for κ non-limit, this is the same as saying that thecellularity of A is < κ. Of most interest is the ω1-chain condition, called ccc forshort (countable chain condition). We shall return to it below.

The attainment problem for cellularity is covered by two classical theorems ofErdos and Tarski: see the Handbook Theorem 3.10 and Example 11.14. Cellularityis attained for any singular cardinal, while for every weakly inaccessible cardinalthere are examples of BAs with cellularity not attained.

If A is a subalgebra of B, then obviously c(A) ≤ c(B) and the difference canbe arbitrarily large. We now consider the special kinds of subalgebras defined inChapter 0; see also Chapter 2 for relationships between them.

• With A ≤rc B the difference can be large; for example, Fr(ω) ≤rc Fr(ω) ⊕Finco(κ); see the Handbook, 11.8. By the known relationships, this exampleapplies also to ≤reg and ≤σ.

• With A ≤π B it is clear that c(A) = c(B).

• With A ≤s B the difference can be large. Namely, let κ be any infinitecardinal, and let A be the free BA on the system 〈xα : α < κ〉 of freegenerators. Let I be the ideal in A generated by all elements xα · xβ withα < β < κ. Clearly xα /∈ I for all α < κ. Let J = {0}. Then there isa simple extension B = A(y) of A such that I = {b ∈ B : b ≤ y} andJ = {b ∈ B : b ≤ −y}. (See Proposition 2.28.) Then 〈xα · −y : α < κ〉 is asystem of pairwise disjoint nonzero elements in B.

• With A ≤m B and A infinite we have c(A) = c(B). In fact, let B = A(x). IfA � x and A � −x are both nonprincipal, then c(A) = c(B) by Proposition2.44. If one of these ideals is principal, then by Proposition 2.45 we mayassume that x is an atom of B. In this case, A is isomorphic to B � −x, andthen an easy argument gives c(A) = c(B) again.

• With A ≤mg B, since every superatomic BA is minimally generated, it isclear that the difference here can be large.

• With A ≤free B, we claim that c(A) = c(B). Write B = A⊕F with F a freealgebra, and suppose that 〈uα : α < c(A)+〉 is a system of pairwise disjointnon-zero elements of B; we want to get a contradiction. We may assume

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014


Basel 24

83

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84 Chapter 3. Cellularity

that each uα has the form bα · cα with bα ∈ A and cα ∈ F . By 9.16 of theHandbook, we may assume that either the cα’s are all equal, or they forman independent set. Then 〈bα : α < c(A)+〉 is a system of pairwise disjointnonzero elements of A, contradiction.

• With A ≤σ B, the difference in cellularities can be large, by the example for≤rc.

• With A ≤proj B, since B embeds in a free extension of A, it follows from theremark about ≤free that c(A) = c(B).

• With A ≤u B, we can get an example with A ccc and B of arbitrarily largecellularity. Namely, let κ be an infinite cardinal. Take A countable, let A(x)be the free extension of A by x (the free product of A with the four elementalgebra {0, x,−x, 1}), and let B = A(x)⊕ Finco(κ).

If B is a homomorphic image of A, then cellularity can change either way from Ato B. For example, if A is a free BA, then it has cellularity ω, while a homomorphicimage of A can have very large cellularity. On the other hand, given any infiniteBA A, it has a homomorphic image of cellularity ω: take a denumerable subalgebraB of A, and by Sikorski’s theorem extend the identity mapping from B into thecompletion B of B to a homomorphism of A into B.

By an easy argument, c(∏

i∈I Ai

)= |I|+ supi∈Ic(Ai), if all the Ai are non-

trivial and either I is infinite or some Ai is infinite. The same computation holdsfor weak products.

Now we turn to chain conditions in free products, where there has been a lotof work done. We formulate some results here using the function c′ instead of c.Recall that c′(A) is the least cardinal greater than |X | for every disjoint subset Xof A. Thus c′(A) = (c(A))+ if c(A) is attained, while c′(A) = c(A) otherwise. Ifc′(A) = c(A), then c′(A) is a regular limit cardinal.

The most general question for free products can be formulated as follows.

Problem 3. For which infinite cardinals κ, λ are there BAs A, B such that c′(A) =κ, c′(B) = λ, while c′(A⊕B) > max(κ, λ)?

See Problems 4, 5 below for more specific forms of this question. Before listingsome specific results relevant to Problem 3 we make some general remarks.

In considering cellularity, it is frequently useful to work with partially or-dered sets rather than BAs. If P is a partially ordered set, then one can take thecompletion of P ; this is a complete BA P such that there is a function e : P → Psatisfying some natural conditions; see the Handbook, page 55. Elements p, q ∈ Pare compatible iff there is an r ∈ P with r ≤ p, q. Then p and q are compatible iffe(p) · e(q) �= 0. We say that P has the κ-cc iff every compatible family of pairwiseincompatible elements of P has size < κ. Then the countable chain condition, ccc,is by definition the ω1-cc. Clearly P has the κ-cc iff every pairwise disjoint subsetof P has size < κ.

Given partially ordered sets P,Q, we make P × Q into a partially orderedset by defining (p1, q1) ≤ (p2, q2) iff p1 ≤ p2 and q1 ≤ q2.

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Chapter 3. Cellularity 85

Lemma 3.1. Let P and Q be partially ordered sets. Then

P ×Q ∼= P ⊕Q.

Proof. Let eP be the embedding of P into P ; and let eQ be the embedding of Q

into Q. We now define e : P × Q → P ⊕Q by setting e(p, q) = eP (p) · eQ(q) forany p ∈ P , q ∈ Q. The following conditions are then clear:

(1) If (p, q) ≤ (p′, q′), then e(p, q) ≤ e(p′, q′).

(2) If (p, q) and (p′, q′) are incompatible (i.e., there is no (r, s) ∈ P ×Q such that(r, s) ≤ (p, q), (p′, q′), then e(p, q) · e(p′, q′) = 0.

(3) e[P ×Q] is dense in P ⊕Q.

Hence the lemma follows from Theorem 4.14 of the Handbook. �

The following simple fact justifies consideration of c(A ⊕ A) instead of the moregeneral c(A⊕B).

Lemma 3.2. The following conditions are equivalent, for any infinite cardinal κ:

(i) There are infinite BAs A and B such that c′(A), c′(B) ≤ κ and c′(A⊕B) > κ.

(ii) There is an infinite BA A such that c′(A) ≤ κ and c′(A⊕A) > κ.

In fact, (ii) obviously implies (i). Now assume that (i) holds. Then by the Hand-book, Example 11.6(c),

(A×B)⊕ (A×B) ∼= (A⊕A)× (A⊕B)× (B ⊕A)× (B ⊕B).

By the results on products, c′(A × B) ≤ κ, while c′(A ⊕ B) > κ, so c((A × B) ⊕(A×B)) > κ. �

Now we list some facts relevant to Problem 3, giving proofs for some of them.

(C1) c′(A⊕B) ≤ (2c(A)·c(B))+.

To see this, suppose that 〈xα : α < (2cA·cB)+〉 is a system of disjoint elements ofA⊕B. Without loss of generality we may assume that for each α < (2cA·cB)+, theelement xα has the form aα × bα, where aα ∈ A and bα ∈ B (we use × to makeclear that the indicated product of elements is in the algebra A ⊕ B). Thus fordistinct α, β we have aα ·aβ = 0 or bα ·bβ = 0, and hence the Erdos–Rado partitiontheorem (2κ)+ → (κ+)2κ implies that there is a subset Y of (2cA·cB)+ of power(c(A) ·c(B))+ such that either aα ·aβ = 0 for all distinct α, β ∈ Y or bα ·bβ = 0 forall distinct α, β ∈ Y , which is impossible. (For this partition relation, see Erdos,Hajnal, Mate, Rado [84], pp. 98–100.

(C2) Under GCH, c′(A⊕B) ≤ max((c(A))++, (c(B))++).

This follows from (C1).

(C3) If c′(A) ≤ (2κ)+ and c′(B) ≤ κ+, then c′(A⊕B) ≤ (2κ)+.

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This holds by the partition theorem (2κ)+ → ((2κ)+, κ+)2 (see the above book,Corollary 17.5).

(C4) Under GCH, if c′(A) ≤ κ++ and c′(B) ≤ κ+, then c′(A⊕B) ≤ κ++.

This follows from (C3).

(C5) If κ is strong limit, c′(A) ≤ κ+, and c′(B) ≤ cf(κ), then c′(A⊕B) ≤ κ+. Thisfollows from Corollary 17.2 in the above book, upon noting that Lκ+(κ+) = cf(κ).

The correspondence between partially ordered sets and BAs is used in the proofof the following result.

(C6) Let M be a countable transitive model of ZFC, κ an uncountable regular car-dinal in M , and form M [G] by Cohen forcing using the set of finite functions froma subset of κ into 2. Then in M [G] there is a ccc BA A such that c′(A⊕A) ≥ κ+.

This fact is due to Fleissner [78]; see also Kunen [80], Exercise (C6) on page 292.We give the argument here.

Lemma 3.3. Let κ be any infinite cardinal. For each subset C of [κ]2 we let P (C)be the partially ordered set consisting of

{s ⊆ κ : [s]2 ⊆ C and s is finite},

with the partial ordering ⊇. Then(i) Elements s1, s2 of P (C) are compatible iff [s1 ∪ s2]

2 ⊆ C.

(ii) For any C ⊆ [κ]2, the partially ordered set P (C)×P (κ\C) has a family of κpairwise incompatible elements.

Proof. (i) is clear.

For (ii), we let the family be {({α}, {α}) : α < κ}. Suppose that α < β < κand ({α}, {α}) and ({β}, {β}) are compatible. By (i) it follows that {α, β} ∈C ∩ (κ\C), contradiction. �Theorem 3.4. Let M be a countable transitive model of ZFC, κ an uncountableregular cardinal in M , and form M [G] by Cohen forcing using the set of finite func-tions from a subset of κ into 2. Then cardinals are preserved in going from M toM [G], and in M [G] 2ω ≥ κ and there are ccc BAs A,B such that c′(A⊕B) ≥ κ+.

Proof. We use the framework of Kunen [80]. Let R be the partially ordered set ofall finite functions ⊆ κ× 2 under ⊇, and let S be the partially ordered set of allfinite functions ⊆ [κ]2×2, under the relation ⊇. If f is a bijection from κ onto [κ]2,then R is isomorphic to S via p !→ p ◦ f−1. Hence it suffices to take the genericextension using the partially ordered set S.

Let G be M generic over S. Clearly⋃G is a function mapping [κ]2 into 2.

We defineC =

{{α, β} :

(⋃G)({α, β}) = 1

}.

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Thus C ⊆ [κ]2. By Lemma 3.3, P (C)×P (κ\X) has an incompatible family of sizeκ. Let A = P (C) and B = P (κ\C). Thus by Lemma 3.1 we have c′(A⊕B) ≥ κ+.

Note that

P (C)={s⊆κ :s is finite and for all distinct α,β∈s

[(⋃G)({α,β})=1

]};

P (κ\C)={s⊆κ :s is finite and for all distinct α,β∈s

[(⋃G)({α,β})=0

]}.

Thus by symmetry it suffices to show that P (C) is ccc. Suppose not. Let f be abijection from ω1 onto an antichain in P (C). We may assume that all members ofrng(f) have the same finite size m. Clearly m �= 0. Thus

(1) If α ∈ ω1, then [f(α)]2 ⊆ C.

(2) If α, β ∈ ω1 and α �= β, then there exist ξ ∈ f(α) and η ∈ f(β) such that{ξ, η} /∈ C.

Say f = τG. Then there is a p ∈ G such that

(3) p �τ is a function ∧ dmn(τ) = ω1 ∧ τ is one-one

∧ ∀α ∈ ω1∀ξ, η ∈ τ(α)[ξ �= η → ∃q ∈ Γ[op(up(ξ, η), 1) ∈ q]]

∧ ∀α, β ∈ ω1[α �= β → ∃ξ ∈ τ(α)∃η ∈ τ(β)∃q ∈ Γ[op(up(ξ, η), 0) ∈ q]].

For up and op, see Kunen VII2.16; for Γ, see Kunen VII2.12. Now fix α ∈ ω1.From (3) it follows that

p � ∃ξ ∈ κ[ξ is the first element of τ(α)].

Hence there exist an ξα0 ∈ κ and a p0α ≤ p with p0α ∈ G such that

p0α � ξα0 is the first element of τ(α).

Continuing by induction, we obtain ξα0 , . . . , ξαm−1 and pm−1

α ≤ · · · ≤ p0α ≤ p suchthat each pαi ∈ G and

piα � ξαi is the (i+ 1)-st element of τ(α)

for i < m. We write pα in place of pm−1α . Let sα = {ξα0 , . . . , ξαm−1}. Then clearly

(4) pα � τ(α) = sα.

(5) [sα]2 ⊆ C.

In fact, suppose that ξ, η ∈ sα and ξ �= η. Then by the second line of (3), pα � ∃q ∈Γ[op(up(ξ, η), 1) ∈ q]. Since pα ∈ G, we get q ∈ ΓG = G such that q({ξ, η}) = 1;thus {ξ, η} ∈ C.

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Clearly

(6) For all {ξ, η} ∈ [κ]2, the set {q : {ξ, η} ∈ dmn(q)} is dense in S.

By (6), we may assume that for all α < ω1, [sα]2 ⊆ dmn(pα).

(7) If ξ, η ∈ sα and ξ �= η, then pα({ξ, η}) = 1.

This is clear from (5) and (6), since pα ∈ G.

Now let M be an uncountable subset of ω1 such that 〈sα : α ∈ M〉 is aΔ-system, say with root r. Fix α ∈M . We claim now that the set

(8) {q : ∃β ∈M∀ξ ∈ sα\r∀η ∈ sβ\r[{ξ, η} ∈ dmn(q) and q({ξ, η}) = 1]}

is dense in S. For, suppose that t is any member of S. Then the set

{β ∈M : ∃ξ ∈ sα\r∃η ∈ sβ\r[{ξ, η} ∈ dmn(t)}

is finite. Choose β ∈M not in this set, and let q extend t by letting q({ξ, η}) = 1whenever ξ ∈ sα\r and η ∈ sβ\r. Thus q is in the set (8). So that set is dense.

Choose q ∈ G which is in the set (8). A common extension of pα and qcontradicts the third line of (3). �

We return to our general list of results about cellularity of free products.

(C7) Under CH there are ccc partial orders P,Q such that P ×Q is not ccc.

This result is due to Laver; see Kunen [80], Exercise VIII(C8).

The following result is complementary to (C6) and (C7):

(C8) Under MA + ¬CH, the free product of two ccc BAs is again ccc.

We go through a proof of this.

Lemma 3.5 (MA + ¬CH). Suppose that 〈xα : α < ω1〉 is a system of elements ina ccc BA A. Then there is an uncountable S ⊆ ω1 such that 〈xα : α ∈ S〉 has thefinite intersection property.

Proof. We may assume that A is complete. For each α < ω1 let yα =∑

γ>α xγ .Then, we claim,

(∗) There is an α < ω1 such that for all β > α we have yβ = yα.

Otherwise, since clearly α < β → yα ≥ yβ , we easily get an increasing sequence〈β(ξ) : ξ < ω1〉 of ordinals less than ω1 such that yβ(ξ) > yβ(η) whenever ξ < η <ω1. But then 〈yβ(ξ) · −yβ(ξ+1)〉 is a disjoint family of power ω1, contradiction.

Thus (∗) holds, and we fix an α as indicated there. The partial ordering Pthat we want to apply Martin’s axiom to is {x ∈ A : 0 �= x ≤ yα} under ≤. It is accc partial ordering since A is a ccc BA. Now for the dense sets. For each β < ω1 let

Dβ = {p ∈ P : there is a γ > β such that p ≤ xγ}.

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To see that Dβ is dense in P , let p ∈ P be arbitrary. Choose δ ∈ ω1 with δ > α, β.Then yα = yδ, so from 0 �= p ≤ yα we infer that there is a γ > δ such thatp · xγ �= 0. Thus p · xγ is the desired element of Dβ which is ≤ p.

Now let G be a filter on P intersecting each dense set Dβ for β < ω1, by

MA + ¬CH. Then it is easy to see that Sdef= {xγ : γ < ω1, and p ≤ xγ for some

p ∈ G} is the set desired in the lemma. �

Theorem 3.6 (MA + ¬CH). The free product of ccc BAs A and B is again ccc.

Proof. Let 〈xα : α < ω1〉 be a disjoint system of elements of A⊕B. Without lossof generality we may assume that each xα has the form aα × bα where aα ∈ Aand bα ∈ B. By the lemma, let S be an uncountable subset of ω1 such that〈aα : α ∈ S〉 has the finite intersection property. But then, by the free productproperty, 〈bα : α ∈ S〉 is a disjoint system in B, contradiction. �

Now we turn to higher cellularity.

(C9) For μ>ℵ1 regular, there is a BA A such that c′(A)=μ+ while c′(A⊕A)>μ+.

This is proved in Shelah [91]. We give the construction here. It depends upon acombinatorial result that should be useful in other situations. The proof of thisresult requires considerable notation, and for the reader’s convenience we indicatein boxes the most important notations when they are introduced. We define ageneral combinatorial principle; see Shelah [94], Appendix 1, for more informationon this principle.

Pr1(λ, μ, κ, θ)means that λ is an infinite cardinal, λ ≥ μ ≥ κ+ θ, and

∃c : [λ]2 → κ∀ξ ∈ θ\1∀a ∈ μ([λ]ξ)∀γ < κ[∀α, β < μ[α �= β → aα ∩ aβ = ∅]→ ∃α, β ∈ μ[α �= β and ∀δ ∈ aα∀ε ∈ aβ [c({δ, ε}) = γ]]].

Pr1(λ, μ, κ, θ)

For any infinite cardinal λ, a strict club system for λ is a system 〈eα : α < λ〉satisfying the following conditions:

(a) ∀α < λ[eα is club in α of order type cf(α)].

(b) C0 = ∅ and ∀α < λ[Cα+1 = {α}](c) For every limit ordinal α < λ and every β ∈ eα, if β is not a supremum ofelements of eα, then β is not a limit ordinal.

A stationary subset S of a regular cardinal λ is non-reflecting iff S ∩ α is non-stationary in α for every α < λ of cofinality greater than ω. For example, if μ isregular, then {α < λ : cf(α) = μ} is clearly a non-reflecting stationary subset ofμ+. This is used to derive (C9) from the more general result.

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Theorem 3.7. Suppose that λ is a regular cardinal > ω2, and S is a non-reflectingstationary subset of λ such that cf(α) > ω1 for every α ∈ S, with every memberof S a limit ordinal. Then Pr1(λ, λ, ω1, ω) holds.

λ S

Proof. (1) There is a strict club system 〈eα : α < λ〉 for λ such that eα ∩ S = ∅for every limit ordinal α < λ.

In fact, for each ordinal α < λ of uncountable cofinality, let Dα be club in α withDα ∩ S = ∅, and let e′α be a cofinal subset of Dα with order type cf(α). Replaceeach member β of e′α which is not a supremum of other members of e′α by β + 1,and let eα be the result. For α < λ of cofinality ω, let eα be the range of a strictlyincreasing sequence of successor ordinals with supremum α. Further, let e0 = ∅and eα = {β} for α = β + 1. Then (1) holds.

eα

The proof uses Todorcevic walks in an essential way. γ is the function defined forall pairs (β, α) such that α < β < λ by

γ(β, α) γ(β, α) = min(eβ\α).

Thus α ≤ γ(β, α) < β. Now we define

γl(β, α)

γ0(β, α) = β;

γl+1(β, α) =

{γ(γl(β, α), α) if γl(β, α) > α

undefined, otherwise.

This gives a strictly decreasing sequence of ordinals, and so there is a greatestk(β, α) < ω such that γk(β,α) is defined; then γk(β,α) = α.

k(β, α)

Clearly k(β, α) ≥ 1. We define

ρ(β, α) ρ(β, α) = 〈γ0(β, α), . . . , γk(β,α)(β, α)〉.

Now let 〈Sα : α < ω1〉 be a partition of S into stationary subsets.

Sα

We define H : λ→ ω1 by setting, for any α ∈ λ,

H H(α) =

{β if α ∈ Sβ with β < ω1,

0 if α /∈ S.

Thus H−1[{β}] = Sβ for all β < ω1.

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Let 〈Rα : α < ω1〉 be a partition of ω1 into stationary subsets.

Rα

For the next definitions, we omit the arguments α, β from γp, k, etc., if thesearguments should be clear.

For α < β < λ define

w1(β, α) w1(β, α) =

{p ∈

(k

2, k

]: ∀q <

k

2[H(γp) > H(γq)]

},

and

p1(β, α) p1(β, α) =

{min(w1) if w1 �= ∅,0 otherwise.

Thus p1(β, α) ≤ k(β, α). Next,

w2(β, α) w2(β, α) =

{q <

k

2: ∀p ∈

(k

2, k

][p /∈ w1 → H(γq) > H(γp)]

}.

Let p2(β, α) be the s ∈ ω1 such that min{H(γq) : q ∈ w2} ∈ Rs if w2 �= ∅, andp2(β, α) = 0 otherwise.

p2(β, α)

Finally, we set

c({α, β}) c({α, β}) ={γ if p1 − p2 ≥ 0 and γp1−p2(β, α) ∈ Sγ ,

0 otherwise,

where α < β < λ.

The rest of the proof is to check that c is as desired. Before doing this we makesome claims.

If s = 〈si : i < m〉 and t = 〈ti : i < n〉, with m,n ≥ 2, and sm−1 = t0, thens�t is the sequence 〈s0, . . . , sm−1, t1, . . . , tn−1〉.

Claim 1. Suppose that δ ∈ S and δ < β < λ. Then:

(i) If l < k(β, δ)− 1, then eγl(β,δ) ∩ δ is bounded below δ.

(ii) Define χ(β, δ) = sup{sup{eγl(β,δ)∩δ} : l < k(β, δ)−1}}, whenever δ < β < λ.

χ(β, δ)

Thus χ(β, δ) < δ by (i). Suppose that α ∈ (χ(β, δ), δ). Then

(a) If l < k(β, δ)− 1, then eγl(β,δ)\δ = eγl(β,δ)\α.(b) If l < k(β, δ), then γl(β, δ) = γl(β, α).

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(c) γk(β,δ)−1 = δ + 1 = γk(β,α)−1.

(d) ρ(β, α) = ρ(β, δ)�ρ(δ, α).

Proof. (i): The assumption l < k(β, δ)− 1 implies that

δ < γl+1(β, δ) = γ(γl(β, δ), δ) = min(γl(β, δ)\δ),

and hence δ /∈ eγl(β,δ). Hence either γl(β, δ) is a successor ordinal α + 1, henceeγl(β,δ) = {α}, so that eγl(β,δ) ∩ δ ⊆ {α}, and hence eγl(β,δ) ∩ δ is bounded belowδ, or it is a limit ordinal, and so eγl(β,δ) ∩ δ is bounded below δ because eγl(β,δ) isclub.

(ii)(a): Clearly eγl(β,δ) ∩ δ ⊆ α, so eγl(β,δ)\α ⊆ eγl(β,δ)\δ ⊆ eγl(β,δ)\α, giving(ii)(a).

(ii)(b): We prove this by induction on l. First, γ0(β, δ) = β = γ0(β, α). Nowassume that γl(β, δ) = γl(β, α), with l < k(β, δ)− 1. Then

γl+1(β, δ) = min(eγl(β,δ)\δ)= min(eγl(β,δ)\α) by (ii)(a)

= min(eγl(β,α)\α) induction hypothesis

= γl+1(β, α).

(ii)(c): Note that δ < γk(β,δ)−1(β, δ). If γk(β,δ)−1(β, δ) is a limit ordinal, theneγk(β,δ)−1(β,δ) is unbounded in γk(β,δ)−1(β, δ), contradicting

δ = min(eγk(β,δ)−1(β,δ)\δ).

It follows that γk(β,δ)−1(β, δ) is a successor ordinal, and since

δ = min(eγk(β,δ)−1(β,δ)\δ),

we must have γk(β,δ)−1(β, δ) = δ + 1. Hence also γk(β,α)−1 = δ + 1 by (ii)(b).

(ii)(d): Immediate from (ii)(c).

This finishes the proof of Claim 1. �

For any A ⊆ λ we let A′ = {α < λ : α is limit and α = sup(A ∩ α)}.

A′

Clearly if A is unbounded, then A′ is a club in λ. Note that all members of A′ arelimit ordinals.

Claim 2. If A,B ∈ [λ]λ and l ∈ ω, then there exist α ∈ A and β ∈ B such thatα < β and k(β, α) > l.

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Proof. Let C0 = A′. If Ci has been defined and is a club in λ, note that S ∩ Ci isunbounded. Let Ci+1 = (S ∩ Ci)

′; it is a club. We now define γl, γl−1, . . . , γ0 andχl, χl−1, . . . , χ0 by recursion. Let γl ∈ Cl∩S. Then choose β ∈ B such that β > γl,and let χl = χ(β, γl). Now suppose that γi+1 and χi+1 have been defined so thatγi+1 ∈ Ci+1 ∩ S and χi+1 < γi+1. Choose γi ∈ S ∩Ci such that χi+1 < γi < γi+1.Then χ(γi+1, γi), χi+1 < γi. Choose χi so that χ(γi+1, γi), χi+1 < χi < γi. Thisfinishes the construction.

Note that k(β, γl) ≥ 1. For i < l we have χ(β, γi+1) = χi+1 < γi < γi+1, andhence by Claim 1, ρ(β, γi) extends ρ(β, γi+1), and so k(β, γi) ≥ k(β, γi+1) + 1. Itfollows that k(β, γi) ≥ 1+ l− i for each i ≤ l. Hence k(β, γ0) ≥ 1+ l. Now χ0 < γ0.Clearly Ci ⊆ A′ for all i ∈ ω; so γ0 ∈ A′. Choose α ∈ A such that χ0 < α < γ0.Since χ0 = χ(β, γ0), it follows by Claim 1 that ρ(β, α) extends ρ(β, γ0), and hencek(β, α) ≥ 2 + l.

This ends the proof of Claim 2. �

Now we define ρH(β, α) = 〈H(γl(β, α)) : l ≤ k(β, α)〉.

ρH(β, α)

If σ ∈ <ωω1 and i < ω1, then σi is the sequence of the same length as σ such that

σi σi(l) =

{σ(l) if σ(l) < i,

ω1 otherwise.

If T ⊆ λ, δ < λ, and R is a stationary subset of ω1, then we define

U(δ, T,R) = {η ∈ <ω(ω1 + 1)\<ωω1 : ∀i < ω1∃β ∈ T \(δ + 1)

[ρH(β, δ)i = η and min{(ρH(β, δ))(l) : ηi(l) = ω1} ∈ R]}.

U(δ, T,R)

For χ < λ, U(δ, T,R, χ) is defined similarly, with the added stipulation thatχ(β, δ) < χ.

U(δ, T,R, χ)

Claim 3. Suppose that R is a stationary subset of ω1, and 〈βi : i ∈ R〉 and〈δi : i ∈ R〉 are systems of ordinals less than λ such that βi > δi for all i ∈ R.Then there exist k ∈ ω, η ∈ <ω(ω1 +1), and a stationary subset R′ ⊆ R such thatthe following conditions hold:

(i) ∀i ∈ R′[| rng(ρH(βi, δi))| = k].

(ii) ∀i, j ∈ R′∀l < k[H(γl(βi, δi)) < i↔ H(γl(βj , δj)) < j].

(iii) ∀i ∈ R′[ρH(βi, δi)i = η].

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Proof.

R =⋃k∈ω

{i ∈ R : | rng(ρH(βi, δi))| = k}.

Hence there is a k ∈ ω such that R1def= {i ∈ R : | rng(ρH(βi, δi))| = k} is

stationary. Also, for each i ∈ R1 define fi ∈ k2 by setting

fi(l) =

{1 if H(γl(βi, δi)) < i,

0 otherwise.

Then R1 =⋃

s∈k2{i ∈ R1 : fi = s}. So there exist a stationary subset R2 of R

and a function s ∈ k2 such that ∀i ∈ R2[fi = s]. Thus

∀i, j ∈ R2∀l < k[H(γl(βi, δi)) < i↔ H(γl(βj , δi)) < j].

Now let B = {l < k : H(γl(βi, δi)) < i}, for any (hence all) i ∈ R2. Say B ={l1, . . . , lm}. Now ∀i ∈ R2[H(γl1(βi, δi)) < i]. So by Fodor’s theorem there exist astationary set R3,1 ⊆ R2 and an ordinal μ1 such that H(γl1(βi, δi)) = μ1 for alli ∈ R3,1. Continuing, we end up with a stationary set R3,m ⊆ R and a functionη ∈ <ω(ω1 + 1) such that ∀i ∈ R3,m[ρH(βi, δi)

i = η].

This ends the proof of Claim 3. �Claim 4. If T ∈ [λ]λ and R is a stationary subset of ω1, then there is a δ(T ) < λsuch that

δ(T )∀δ ∈ [δ(T ), λ)[U(δ, T,R) �= ∅] and

∀δ ∈ [δ(T ), λ)[cf(δ) > ω1 → ∃χ < δ[U(δ, T,R, χ) �= ∅]].

Proof. For each i < ω1 let Ai = {δ < λ : ∀β ∈ T ∩ (δ, λ)[i /∈ rng(ρH(β, δ))]}. �

Subclaim. ∀i < ω1[|Ai| < λ].

Proof of the Subclaim. Suppose that i < ω1 and |Ai| = λ. Pick δ ∈ Si ∩ A′i, and

then pick β ∈ T such that δ < β. Choose γ ∈ Ai ∩ (χ(β, δ), δ). Then δ is an entryof ρ(β, γ) by Claim 1, and so i = H(δ) ∈ rng(ρH(β, γ)), contradiction. This provesthe subclaim.

By the subclaim, there is an ordinal δ(T ) < λ such that⋃

i<ω1Ai ⊆ δ(T ).

Now suppose that δ ∈ [δ(T ), λ).

(2) ∀i < ω1∃βi ∈ (δ, λ) ∩ T [i ∈ rng(ρH(βi, δ))].

This is immediate from the definition of Ai. Now we apply Claim 3 and get k ∈ ω,η ∈ <ω(ω1 + 1), and a stationary subset R′ of R such that

(3) ∀i ∈ R′[| rng(ρH(βi, δ)| = k].

(4) ∀i, j ∈ R′∀l < k[H(γl(βi, δ)) < i↔ H(γl(βj , δ)) < j].

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(5) ∀i ∈ R′[ρH(βi, δ)i = η].

Since i ∈ rng(ρH(βi, δ)), we have η /∈ ωω1. If i ∈ R′, say that i = H(γl(βi, δ)). Thusη(l) = (ρH(βi, δ)

i)(l) = ω1, and (ρH(βi, δ))(l) = H(γl(βi, δ) = i. If m < k andη(m) = ω1, then (ρH(βi, δ))(m) ≥ i. Thus min{(ρH(βi, δ))(m) : η(m) = ω1} = i ∈R′ ⊆ R. So η ∈ U(δ, T,R), as desired.

Now suppose that cf(δ) > ω1. Then {χ(βi, δ) : i ∈ R′} is bounded below δ,say by θ, Hence η ∈ U(δ, T,R, θ).

This finishes the proof of Claim 4.

Now suppose that T ⊆ λ and δ < λ. We define

L(δ, T ) = {ρ ∈ <ω(ω1 + 1)\<ωω1 : there is a stationary R′ ⊆ ω1 such that ∀i ∈R′∀α < δ∃β ∈ T ∩ (α, δ)[ρH(δ, β)i = ρ]}.

L(δ, T )

For T ∈ [λ]λ we define C(T ) =⋂

i<ω1(Si ∩ T ′)′. Note that this is club in λ.

C(T )

Claim 5. If T ∈ [λ]λ, δ ∈ C(T ), and cf(δ) ≥ ω1, then L(δ, T ) �= ∅.

Proof. Case 1. cf(δ) = ω1. Let 〈δi : i < ω1〉 be strictly increasing with supremumδ. Since δ ∈ C(T ), we can choose αi ∈ Si∩T ′∩(δi, δ) for each i < ω1. Then for eachi < ω1 choose βi ∈ T ∩(δi, αi) with χ(δ, αi) < βi. Then i = H(αi) ∈ rng(ρH(δ, βi))by claim 1. We now apply claim 3 and get k, η, and a stationary subset R′ of ω1

such that the following condition holds:

(6) ∀i ∈ R′[ρH(δ, βi)i = η].

Since i ∈ rng(ρH(δ, βi)) we have η /∈ <ωω1. Hence η ∈ L(δ, T ), as desired.

Case 2. cf(δ) > ω1. Let 〈δα : α < cf(δ)〉 be strictly increasing with supremum δ.Since δ ∈ C(T ), for each α < cf(δ) and each i < ω1 we can choose γiα ∈ Si ∩ T ′

such that δα < γiα < δ. By Claim 1 we have χ(δ, γiα) < γiα, so we can chooseβiα ∈ T ∩ (χ(δ, γiα), γiα). Now by Claim 1 applied to γiα < δ < λ, the ordinal γiαis a term of ρ(δ, βiα); hence i = H(γiα) ∈ rng(ρH(δ, βiα)). Now if we apply Claim 3to δ and βiα for i < ω1 we get a stationary subset R′ of ω1 and an ηα ∈ <ω(ω1+1)such that ∀i ∈ R′[ρH(δ, βiα)

i = ηα]. Since i = H(γiα) ∈ rng(ρH(δ, βiα)), we haveηα /∈ <ωω1. Since ω1 < cf(δ), there is an η ∈ <ω(ω1+1) such that ηα = η for cf(δ)many α. Hence η ∈ L(δ, T ), as desired.

This finishes the proof of Claim 5. �

Now we begin the actual checking that c, defined above, works. Thus assume thatn ∈ ω\1, γ < ω1, and a ∈ λ([λ]n) is a system of pairwise disjoint sets.

n γ a

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(7) We may assume that α < min(aα), for all α < λ.

In fact, define 〈αη : η < λ〉 by recursion. If αξ has been defined for all ξ < η, thenthe set {αξ : ξ < η} ∪ {β < λ : min(aβ) ≤ η} has size less than λ. Let αη be anyordinal less than λ which is not in this set. Let a′η = aαη for all η < λ. These setsare clearly pairwise disjoint, and η < min(aαη ) = min(a′η). So (7) holds.

Write aα = {ζ0α, . . . , ζn−1α } for each α < λ, with ζ0α < · · · < ζn−1

α .

ζlα

Thus

(8) ∀α < λ∀l < n[α < ζlα].

Now define δ0 = 0, δα+1 =(supβ≤δα ζn−1

β

)+1, and δλ =

⋃α<λ δα for λ limit. Let

C = {δλ : λ limit}.δα C

(9) C is club, and ∀α, δ[α < δ ∈ C → ζn−1α < δ].

In fact, assume that α < δ ∈ C. Say δ = δλ, λ limit. Choose β < λ such thatα < δβ. Then ζn−1

α ≤ δβ+1 < δλ, as desired.

Now if δ ∈ S ∩ C, then

δ =⋃s∈U

{α < δ : 〈ρH(δ, ζlα) : l < n〉 = s},

where U = n (<ωω1). Since cf(δ) > ω1, there is an s ∈ U such that the indicatedset is cofinal in δ. Thus there is a system 〈νδl : l < n〉 such that

(10) sup{α < δ : ∀l < n[ρH(δ, ζlα) = νδl ]} = δ.

NowS ∩C =

⋃t∈U

{δ ∈ S ∩ C : 〈νδl : l < n〉 = t}.

Hence there is a t ∈ U such that T1def= {δ ∈ S ∩C : 〈νδl : l < n〉 = t} is stationary.

We can write t = 〈νl : l < n〉. So (10) becomes

(11) ∀δ ∈ T1[sup{α < δ : ∀l < n[ρH(δ, ζlα) = νl]} = δ.

T1 νl

Next, by Claim 5,

S ∩ C(T1) ∩ C =⋃ρ∈W

{δ ∈ S ∩ C(T1) ∩C : ρ ∈ L(δ, T1)},

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where W = <ω(ω1+1)\<ωω1. Hence there is a stationary T2 ⊆ S ∩C(T1)∩C anda τ ∈ <ω(ω1 + 1)\<ωω1 such that τ ∈ L(δ, T1) for all δ ∈ T2. Let l

∗ = min{l ∈ ω :τ(l) = ω1}.

T2 τ l∗

Next, by Claim 5 again we have

Sγ ∩ C(T2) =⋃ρ∈W

{δ ∈ Sγ ∩ C(T2) : ρ ∈ L(δ, T2)},

with W = <ω(ω1 + 1)\<ωω1, and so there exist a stationary T3 ⊆ Sγ ∩C(T2) anda ρ ∈ <ω(ω1 + 1)\<ωω1 such that ρ ∈ L(δ, T2) for all δ ∈ T3.

T3 ρ

Now

S =⋃σ∈X

{δ ∈ S : ρH(ζ0δ , δ) = σ},

where X = <ωω1. Hence there exist a stationary subset S′ ⊆ S and a σ ∈ X suchthat the indicated set is stationary. Continuing, we arrive at a stationary subsetT 1 ⊆ S and a system 〈νl : l < n〉 such that

(12) ∀δ ∈ T 1∀l < n[ρH(ζlδ, δ) = νl].

T 1 νl

Now for each δ ∈ T 1 we have χ(ζ0δ , δ) < δ. Hence there exist a stationary subsetT 10 of T 1 and an ordinal θ0 such that χ(ζ0δ , δ) = θ0 for all δ ∈ T 1

0 . Continuing,we arrive at a stationary subset T 2 ⊆ T 1 and a function θ : n → λ such thatχ(ζlδ, δ) = θ(l) for all δ ∈ T 2 and all l < n. Letting χ2 = (supl<n θ(l))+ 1, we thenhave

T 2 χ2

(13) ∀δ ∈ T 2∀l < n[χ(ζlδ, δ) < χ2].

(14) There are δ(T 2) < λ, η, χ3 ∈ (χ2, λ), and a stationary T 3 ⊆ (S ∩ C)\δ(T 2)such that ∀δ ∈ T 3[η ∈ U(δ, T 2, Rl∗+|ρ|, χ3)].

η χ3 T 3

In fact, by Claim 4 there is a δ(T 2) < λ such that ∀δ ∈ [δ(T 2), λ)[cf(δ) > ω1

implies that ∃χ < δ[U(δ, T 2, Rl∗+|ρ|, χ) �= ∅]. Hence

∀δ ∈ (S ∩ C)\δ(T 2)∃χ < δ[U(δ, T 2, Rl∗+|ρ|, χ) �= ∅],

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so by Fodor’s theorem there exist a stationary T ′ ⊆ (S ∩ C)\δ(T 2) and χ3 < λsuch that ∀δ ∈ T ′[U(δ, T 2, Rl∗+|ρ|, χ3) �= ∅]. Now

T ′ =⋃η∈V

{δ ∈ T ′ : η ∈ U(δ, T 2, Rl∗+|ρ|, χ3),

where V = <ω(ω1+1)\<ωω1. Hence there exist a stationary T 3 ⊆ T ′ and an η ∈ Vsuch that the condition in (14) holds.

Now we apply Claim 2 with A = T3\(χ3 + 1) and B = T 3 to obtain β3 ∈T3\(χ3 + 1) and β3 ∈ T 3 such that β3 < β3 and

β3 β3

(15) k(β3, β3) > max{|νl| : l < n}+ |τ | + |ρ|+ |η|+max{|νl| : l < n}.

Now ρ ∈ L(β3, T2), so for all α < β3 there are uncountably many i < ω1 such that∃β ∈ T2 ∩ (α, β3)[ρH(β3, β)

i = ρ]. We apply this to α = max(χ3, χ(β3, β3)) andwith i0 < ω1 greater than all countable ordinals in ρH(β3, β3), η, 〈νl : l < n〉, 〈νl :l < n〉 to get a β2 ∈ T2 such that χ3, χ(β3, β3) < β2 < β3 and ρH(β3, β2)

i0 = ρ.

i0 β2

Since χ3 < β2, χ(β3, β3) < β2, and χ(β3, β2) < β2, choose χ2 ∈ (χ3, β2) such that

χ(β3, β3) < χ2 and χ(β3, β2) < χ2.

χ2

Now let R′ be a stationary subset of ω1 given by the definition of L(β2, T1).

R′

Fix i1 ∈ R′ greater than i0 and each ordinal in ρH(β3, β2).

i1

Now β3 ∈ T 3, so η ∈ U(β3, T 2, Rl∗+|ρ|, χ3). Applying this to i1, we get β ∈T 2\(β3 + 1) such that ρH(β, β3)i1 = η and χ(β, β3) < χ3.

β

Since β ∈ T 2 ⊆ T 1, we have by (13)

(16) ∀l < n[ρH(ζlβ , β) = νl and χ(ζlβ , β) < χ2 < χ3].

Now since β2 ∈ T2, we have τ ∈ L(β2, T1). Applying the definition of L(β2, T1) toχ2 < β2 we see that for ∀i ∈ R′∃θ ∈ T1 ∩ (χ2, β2)[ρH(β2, θ)

i = τ ]. We apply thisto an i2 ∈ R′ such that i2 is greater than i1 and all ordinals in ρH(β, β3).

i2

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This gives β1 ∈ T1 ∩ (χ2, β2) such that ρH(β2, β1)i2 = τ .

β1

Now χ(β2, β1), χ2 < β1. Fix χ1 < β1 with χ(β2, β1), χ2 < χ1.

χ1

Since β1 ∈ T1, by (11) there is an α ∈ (χ1, β1) such that ∀l < n[ρH(β1, ζlα) = νl].

α

Now note that

(17) α < β1 < β2 < β3 < β.

(18) If l,m < n, then ρ(ζlβ , ζmα ) = ρ(ζlβ , β)

�ρ(β, ζmα ).

For, β ∈ T 2 ⊆ T 1 ⊆ S, β < ζlβ by (8), and χ(ζlβ , β) < χ2 < χ3 < χ2 < χ1 < α, sothe conclusion follows by Claim 1.

(19) If m < n, then ρ(β, ζmα ) = ρ(β, β3)�ρ(β3, ζmα ).

In fact, we apply Claim 1 with δ, β, α replaced by β3, β, ζmα : we have β3 < β < λby the definition of β; by the definitions of β, χ2, χ1, and α we have χ(β, β3) <χ3 < χ2 < χ1 < α, and then (8) gives χ(β, β3) < ζmα ; by (17) and the definitionof β3 we have α < β3 ∈ C, and then (9) gives ζmα < β3.

(20) If m < n, then ρ(β3, ζmα ) = ρ(β3, β3)�ρ(β3, ζ

mα ).

For, β3 < β3. Also, α < β3 ∈ T2 ⊆ C, so by (9), ζmα < β3. Finally, since χ(β3, β3) <

χ2 < χ1 < α < ζmα by (8), Claim 1 applies again, with β3, β3, ζmα in place of δ, β, α.

(21) If m < n, then ρ(β3, ζmα ) = ρ(β3, β2)

�ρ(β2, ζmα ).

For, β2 < β3. Also, α < β2 ∈ T2 ⊆ C, so ζmα < β2 by (9). Also, χ(β3, β2) < χ2 <χ1 < α < ζmα , so Claim 1 again applies with δ, β, α replaced by β2, β3, ζ

mα .

(22) If m < n, then ρ(β2, ζmα ) = ρ(β2, β1)

�ρ(β1, ζmα ).

For, β1 < β2. Also, α < β1 ∈ T1 ⊆ C, so by (9), ζmα < β1. Since χ(β2, β1) < χ1 <α < ζmα , Claim 1 applies again, with δ, β, α replaced by β1, β2, ζ

mα .

Putting (18)–(22) together, we get the following two statements; for alll,m < n,

(23) ρ(ζlβ , ζmα ) = ρ(ζlβ , β)

�ρ(β, β3)�ρ(β3, β3)�ρ(β3, β2)

�

ρ(β2, β1)�ρ(β1, ζ

mα );

(24) ρH(ζlβ , ζmα ) = ρH(ζlβ , β)

�ρH(β, β3)�ρH(β3, β3)�ρH(β3, β2)

�

ρH(β2, β1)�ρH(β1, ζ

mα ).

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Now note that the lengths of the six components of this long sequence are, respec-tively, |νl|, |η|, s, |ρ|, |τ |, and |νm|, where s > max{|νl : l < n} + |τ | + |ρ|+ |η|+max{|νl| : l < n}. Let k be the length of the long sequence. Then for any p, ifp < k

2 then p comes before the component ρ(β3, β2).

Now

(25) All entries of ρH(ζlβ , β) are less than i0.

This holds by (12), since β ∈ T 2 ⊆ T 1.

(26) Some entries of ρH(β, β3) are ≥ i1, but all entries are < i2.

This holds since ρH(β, β3)i1 = η /∈ <ωω1.

(27) All entries of ρH(β3, β3) are less than i0.

(28) Some entries of ρH(β3, β2) are ≥ i0, but all entries are less than i1.

This holds since ρH(β3, β2)i0 = ρ /∈ <ωω1.

(29) Some entries of ρH(β2, β1) are ≥ i2, while other entries are less than i1.

This holds by the definition of i2 and the fact that i1 ∈ R′.

(30) All entries of ρH(β1, ζmα ) are less than i0.

It now follows that w1(ζlβ , ζ

mα ) consists of all p in the ρ(β2, β1) portion with

value ≥ i2. In particular, w1(ζlβ , ζ

mα ) �= ∅. Hence p1(ζ

lβ , ζ

mα ) = |ρ(ζlβ , β3)| + |ρ| +

l∗. Now w2(ζlβ , ζ

mα ) consists of those q in the ρH(β, β3) portion whose value is

≥ i1. Now ρH(β, β3)i1 = η, β3 ∈ T 3, and η ∈ U(β3, T 2, Rl∗+|ρ|, χ3). Thus

min{(ρH(β, β3))(l) : ηi1 (l) = ω1} ∈ Rl∗+|ρ|, i.e., p2(ζlβ , ζmα ) = l∗ + |ρ|. So γp1−p2 =

γs = β3 ∈ Sγ , so c(ζlβ , ζmα ) = γ, as desired. �

Theorem 3.8. Suppose that λ is a regular cardinal, and S is a non-reflecting sta-tionary subset of λ such that cf(α) > ω1 for every α ∈ S, with every member of Sa limit ordinal. Then Pr1(λ, λ, 2, ω) holds.

Proof. Let c be obtained by Theorem 3.7. Define

c′(β, α) ={0 if c(β, α) = 0,

1 otherwise.

Clearly c′ is as desired. �Theorem 3.9. Suppose that μ > ℵ1 is a regular cardinal. Then Pr1(μ

+, μ+, 2, ω)holds.

Proof. See the remark before Theorem 3.7. �Theorem 3.10. For any regular cardinal λ, Pr1(λ, λ, 2, ω) implies that there existBAs B0, B1 such that B0 and B1 satisfy the λ-cc but B0 ⊕B1 does not.

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Proof. Choose c by the definition of Pr1(λ, λ, 2, ω).

Let A = Fr(〈xα : α < λ〉), and for each ε ∈ 2 let

Iε = 〈{xα · xβ : α �= β and c(α, β) = ε}〉id and Bε = A/Iε.

(1) [xα]Iε �= 0.

In fact, otherwise we get

xα ≤ xβ(0) · xγ(0) + · · ·+ xβ(m−1) · xγ(m−1)

with β(i) �= γ(i) for all i < m. Mapping xα to 1 and all other generators to 0,then extending to a homomorphism from A into 2, gives a contradiction.

(2) For F and G disjoint finite subsets of λ we have∏α∈F

[xα]Iε ·∏α∈G

−[xα]Iε = 0 iff there are distinct α, β ∈ F such that c(α, β) = ε.

In fact, ⇐ is clear. Now suppose that∏

α∈F [xα]Iε ·∏

α∈G−[xα]Iε = 0 whilec(α, β) = 1− ε for all distinct α, β ∈ F . It follows that∏

α∈F

xα ·∏α∈G

−xα ≤ xβ(0) · xγ(0) + · · ·+ xβ(m−1) · xγ(m−1)

for some β(0), γ(0), . . . , β(m−1), γ(m−1), where β(i) �= γ(i) and c(β(i), γ(i)) = εfor each i < m. Mapping xα to 1 for each α ∈ F and all other generators to 0,and extending to a homomorphism from A into 2, we get a contradiction. So (2)holds.

Now clearly 〈[xα]I0 · [xα]I1 : α < λ〉 is a system of pairwise disjoint nonzeroelements of B0 ⊕B1, so B0 ⊕B1 does not have the λ-cc.

Now fix ε ∈ 2. We want to get a contradiction from assuming that 〈bζ : ζ < λ〉is a system of pairwise disjoint nonzero elements of Bε. Wlog each bζ has the form

bζ =∏

α∈Mζ

[xα]δ(α,ζ)

with Mζ a finite subset of λ and δ(α, ζ) ∈ 2 for all α ∈Mζ . Wlog 〈Mζ : ζ < λ〉 isa Δ-system with kernel N . Wlog we can write each bζ as follows:

bζ =∏α∈N

[xα]δ(α) ·

∏α∈Pζ

[xα] ·∏

α∈Qζ

−[xα],

with N ∩ Pζ = ∅ = N ∩ Qζ, Pζ ∩ Qζ = ∅, and Pζ , Qζ finite. Since bζ �= 0, wehave ∀α, β ∈ Pζ [α �= β → c(α, β) = 1 − ε. Now wlog |Pζ | = n for all ζ < λ.Now by the condition Pr1(λ, λ, 2, ω) it follows that there are ξ < η < λ such thatc(α, β) = 1 − ε for all α ∈ Pξ and β ∈ Pη. Now by (2) it follows that bξ · bη �= 0,contradiction. �

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Corollary 3.11. For any regular cardinal μ > ℵ1, there are BAs B0, B1 such thatB0 and B1 satisfy the μ+-cc but B0 ⊕B1 does not.

Proof. By Theorems 3.9 and 3.10. �

Now we continue with our list of results about free products.

(C10) There is a BA A such that c′(A) = ℵ2 while c′(A⊕A) > ℵ2.

This is proved in Shelah [97].

Note that the results given so far answer the productivity question exceptfor cellularity a limit cardinal.

(C11) For singular λ, there is a BA A such that c′(A) = λ+ while c′(A⊕A) > λ+.

For this, see Shelah [94], Conclusion 4.1 in Chapter II, and Appendix 1, Diagram1.8. We give this proof here. It involves a part of pcf theory, and also uses an inter-esting elementary substructure argument which is an instance of a proof methodwidely used at present.

It is useful to recall an argument proving the downward Lowenheim–Skolemtheorem.We suppose thatH is a first-order structure which has among its relationsa well-order < of H itself. Then with each formula ∃xϕ(x, y) of the given languagewe associate itsSkolem function f defined as follows:

f∃xϕ(x,y)(a) = the <-least element b such that H |= ϕ[b, a],

if there is such an element,

= a0, otherwise.

Here we assume that a = 〈a0, a1, . . . , an−1〉 is a sequence of elements of H , with npositive, y being a sequence of distinct variables of length n. Note that any Skolemfunction is definable in the given language.

The following is a version of the well-known lemma of Tarski concerningelementary substructures. A first-order language is relational iff it does not haveany operation symbols.

Theorem 3.12. Suppose that H is a first-order structure over a relational languagewhich has among its relations a well-order < of H itself. Suppose also that X isa nonempty subset of H. Let Y be the union of the ranges of all of the Skolemfunctions over H, each one restricted to X. Let the relations of Y be the ones ofH restricted to Y . Then Y is an elementary substructure of H.

Proof. We prove by induction on the formula ψ that for any a ⊆ Y , H |= ψ[a] iffY |= ψ[a]. Atomic ψ works by the definition of the relations on Y . The inductivesteps using sentential connectives are straightforward. If Y |= ∃xϕ(x, a), then theinductive step easily applies. Suppose that H |= ∃xϕ(x, a). Let χ be the following

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formula, where a has length n, and ai = fi(bi), each fi a Skolem function andeach bi ⊆ X :

∃z0, . . . , zn−1

(∧i<n

(zi = fi(yi)) ∧ ϕ(x, z)

).

Then with g the Skolem function associated with ∃xχ we have

H |= ϕ(g(b0, . . . , bn−1), a),

and the inductive hypothesis gives the desired conclusion. �

The restriction to a relational language in this theorem is made just to shortenthe proof, since we will apply the theorem only to such a language.

Lemma 3.13. Let L be a relational language extending the language of set the-ory, with in particular an additional binary relation symbol <. Suppose that χ isan uncountable regular cardinal, M is an elementary substructure of H(χ), andθ ∈ M is a cardinal. Assume that < well-orders H(χ). Let N be the elementarysubstructure of H(χ) determined by Theorem 3.12, using the set X = M ∪θ. Thenfor any regular cardinal σ ∈M\θ+ we have

sup(M ∩ σ) = sup(N ∩ σ).

Proof. Since M ⊆ N , we have sup(M ∩ σ) ⊆ sup(N ∩ σ). Now suppose thatα ∈ N ∩ σ; we want to find β ∈M ∩ α such that α ≤ β. By Theorem 3.12 we canwrite α = f(b, c), with f a Skolem function, b ⊆ θ, and c ⊆ M\θ. Say that n isthe length of b. We define a function F with domain nθ by setting

F (d) =

{f(d, c) if this is an ordinal less than σ,

0 otherwise.

Clearly F is elementarily definable with parameters from M . Let β = sup(rng(F )).Then β ∈ M by the assumption that M is an elementary substructure of H(χ).Since α ∈ rng(F ), we have α ≤ β. Also, rng(F ) is a set of at most θ ordinals lessthan σ, so β < σ by regularity. �

Theorem 3.14. If μ is a singular cardinal, then Pr1(μ+, μ+, 2, ω) holds.

Proof. By Theorems 2.23 and 2.26 of Abraham, Magidor [10] let 〈λα : α < cf(μ)〉and 〈fα : α < μ+〉 be such that the following conditions hold:

(1) 〈λα < α < cf(μ)〉 is a strictly increasing sequence of regular cardinals withsupremum μ such that cf(μ) < λ0.

(2) For f, g ∈∏

α<cf(μ) λα define f <cf(μ) g iff ∃β < cf(μ)∀α ∈ (β, cf(μ))[f(α) <

g(α)]. Then ∀α, β ∈ cf(μ)[α < β → fα <cf(μ) fβ].

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(3) ∀g ∈∏

α<cf(μ) λα∃α < μ+[g <cf(μ) fα].

Now we define, for α < β < μ+, d(α, β) to be the maximum i < cf(μ) suchthat fβ(i) < fα(i) if there is such an i, and 0 otherwise. Let h : cf(μ) → 2 besuch that h−1[{0}] and h−1[{1}] have size cf(μ). Define c : [μ+]2 → 2 by settingc({α, β}) = h(d(α, β)) for α < β.

The rest of the proof is to show that c works. Suppose that n ∈ ω\1 and

a ∈ μ+

([μ+]n), with the aα’s pairwise disjoint, and ϕ < 2.

(4) We may assume that ∀α < μ+[α < min(aα] and ∀α, β < μ+[α < β →max(aα) < min(aβ)].

To prove this, we define by recursion a sequence 〈γβ : β < μ+〉 of ordinals lessthan μ+. Suppose that γα has been defined for all α < β. Then for each α < β,the set {δ < μ+ : min(aδ) ≤ max(aγα)} has at most max(aγα) + 1 < μ+ elements.Also, {γ < μ+ : min aγ ≤ β} has size less than μ+. Hence we can choose γβ ∈ μ+

greater than

sup

⎛⎝⋃

α<β

{δ < μ+ : min(aδ) ≤ max(aγα)}

⎞⎠ ∪ sup({γ < μ+ : min aγ ≤ β}).

Let a′β = aγβfor all β < μ+. If α < β < μ+, then max(a′α) = max(aγα) <

min(aγβ) = min(a′β). If β < μ+, then β < min aγβ

= min a′β. Thus (4) holds.Now for β < μ+ we define for i < cf(μ)

f infβ (i) = min{fα(i) : α ∈ aβ};

f supβ (i) = max{fα(i) : α ∈ aβ}.

(5) ∀β ∈ cf(μ)∃γ ∈ cf(μ)∀δ ∈ (γ, cf(μ))[f infβ (δ) = fmin(aβ)(δ)].

In fact, choose γ ∈ cf(μ) so that ∀ε ∈ aβ\{minaβ}∀δ ∈ (γ, cf(μ))[fmin(aβ)(δ) <fε(δ)]. Clearly this proves (5).

(6) ∃α < cf(δ)∀i ∈ (α, cf(δ))[λi = sup{f infβ (i) : β < μ+}].

In fact, suppose not. Then there is an M ∈ [cf(μ)]cf(μ) such that for all i ∈ Mwe have sup{f inf

β (i) : β < μ+} < λi. Let g(i) = sup{f infβ (i) : β < μ+} + 1 for all

i ∈M , and g(i) = 0 otherwise. Choose β < μ+ such that g <cf(μ) fβ. Then by (4),

also g <cf(μ) fminaβ. By (5), choose γ < cf(μ) such that ∀δ ∈ (γ, cf(μ))[f inf

β (δ) =fminaβ

(δ)]. We may also assume that ∀δ ∈ (γ, cf(μ))[g(δ) < fminaβ(δ)]. Choose

δ ∈M ∩ (γ, cf(μ)). Then g(δ) < fmin(aβ)(δ) = f infβ (δ) < g(δ), contradiction. So (6)

holds.

Now let M be an elementary submodel of H(χ) with |M | = cf(μ), cf(μ) ⊆M , and μ, μ+, f, a ∈ M . Note that if i < cf(μ), then sup(M ∩ λi) < λi, since

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|M | = cf(μ) < λi and λi is regular. Hence if we define, for each i < cf(μ),

ChM (i) = sup(M ∩ λi),

we get an element of∏

i<cf(μ) λi. Thus we can choose α < μ+ such that ChM <cf(μ)

fα; hence by (4) also ChM <cf(μ) fmin(aα).

(7) There is an i∗ < cf(μ) such that the following conditions hold:

(a) h(i∗) = ϕ.

(b) λi∗ = sup{f infβ (i∗) : β < μ+}.

(c) ∀i ∈ [i∗, cf(μ))[ChM (i) < f infα (i)].

This is clear from (6). By (b), choose β < μ+ such that f supα (i∗) < f inf

β (i∗).Let N be the elementary substructure of H(χ) obtained from Lemma 3.13

using the set X = M ∪ λi∗ , with |N | = λi∗ . By Lemma 3.13,

(8) sup(M ∩ σ) = sup(N ∩ σ) if σ is a regular cardinal greater than λi∗ .

Thus

(9) ChM � [i∗ + 1, cf(μ)) = ChN � [i∗ + 1, cf(μ)).

Let δ = f infβ (i∗). Then δ ∈ λi∗ ⊆ N , i∗ < cf(μ) ⊆ M ⊆ N , and μ+ ∈ M ⊆ N .

Moreover, ∃ε < μ+[f infε (i∗) = δ]. Hence by elementarity, there is an ε ∈ N with

ε < μ+ such that f infε (i∗) = δ. So f sup

α (i∗) < f infε (i∗). But for any i ∈ (i∗, cf(μ))

we havef supε (i) < ChN (i) = ChM (i) < f inf

α (i).

Hence if γ ∈ aα and δ ∈ aε, then d(γ, δ) = i∗, and so c(γ, δ) = h(d(γ, δ)) = ϕ. �Theorem 3.15. For any singular λ, there is a BA A such that c′(A) = λ+ whilec′(A⊕A) > λ+. �

Now we have discussed c′(A⊕A) in all cases except that with c′(A) a regular limitcardinal. Here we have, first of all:

(C12) If κ is weakly compact and c′(A) = κ, then c′(A⊕A) = κ.

In fact, assume the hypothesis, and suppose that 〈xα : α < κ〉 is a system ofpairwise disjoint nonzero elements of A ⊕ A. Wlog we may assume that each xα

has the form aα · bα, with aα in the first copy of A and bα in the second copy.Then define f : [κ]2 → 2 by setting, for any distinct α, β < κ,

f({α, β}) ={0 if aα · aβ = 0,

1 if bα · bβ = 0.

Then a homogeneous set of size κ gives a system of κ nonzero pairwise disjointelements of A, contradiction.

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On the other hand, under V = L for every regular limit cardinal λ which isnot weakly compact, there is a non-reflecting stationary subset S of λ such thatevery member of S is a limit ordinal of cofinality > ω1; see Devlin [84], Theorem1.2′ on page 304. Hence by Theorem 3.8 V = L implies the existence of a BA Awith c′(A) = λ while c′(A⊕A) > λ.

This completes our exposition of results concerning the cellularity of free productsof two Boolean algebras. The following problems remain open; they are versionsof Problem 1 in Monk [96].

Problem 4. Can one construct BAs A,B such that c′(B) < c′(A) < c′(A ⊕ B)with c′(B) regular limit?

Problem 5. Can one prove in ZFC that if κ is inaccessible but not weakly compact,then there a BA A such that c′(A) = κ while c′(A⊕A) > κ?

Another important and quite elementary fact about free products is that

c(⊕i∈IAi) = sup{c(⊕i∈FAi) : F ∈ [I]<ω}.

In fact, ≥ is clear. Now let κ = sup{c(⊕i∈FAi) : F ∈ [I]<ω}, and suppose thatX is a disjoint subset of ⊕i∈IAi of size κ+. For each x ∈ X choose a finiteF (x) ⊆ I such that x ∈ ⊕i∈F (x)Ai. We may assume that each x ∈ X has theform x =

∏i∈F (x) y

xi , where y

xi ∈ Ai for each i ∈ F (x). Without loss of generality,

〈F (x) : x ∈ X〉 forms a Δ-system, say with kernel G. But then, by the freeproduct property, 〈

∏i∈G yxi : x ∈ X〉 is a disjoint system of elements of ⊕i∈GAi,

contradiction.

Next, we consider amalgamated free products where the following fact is basic:

c(A⊕C B) ≤ 2c(A)·c(B)·|C|.

To prove this, let κ = c(A) · c(B) · |C|, and suppose that 〈cα : α < (2κ)+〉 is adisjoint system in A⊕C B. We may assume that each cα is non-zero, and has theform aα · bα, with aα ∈ A and bα ∈ B. Thus for all distinct α, β < (2κ)+ there is ac ∈ C such that aα · aβ ≤ c and bα · bβ ≤ −c. Hence by the Erdos–Rado theorem

there is a Γ ∈ [(2κ)+]κ+

and a c ∈ C such that aα · aβ ≤ c and bα · bβ ≤ −c forall distinct α, β ∈ Γ. Thus (aα · −c) · (aβ · −c) = 0 and (bα · c) · (bβ · c) = 0 for alldistinct α, β ∈ (2κ)+. Since cA < κ+, it follows that there is a Δ ∈ [Γ]κ such thataα · −c = 0 for all α ∈ Γ\Δ; and there is a Θ ∈ [Γ\Δ]κ such that bα · c = 0 for allα ∈ (Γ\Δ)\Θ. But then for any α ∈ (Γ\Δ)\Θ we have aα · bα = 0, contradiction.

The above inequality is best-possible, in a sense. To see this, consider P(ω) ⊕C

P(ω), where C is the BA of finite and cofinite subsets of ω. Let 〈Γα : α < 2ω〉 bea system of infinite almost disjoint subsets of ω; and also assume that each Γα isnot cofinite. For each α < 2ω let yα be the element Γα · (ω\Γα) of P(ω)⊕C P(ω).

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These elements are clearly non-zero. For distinct α, β < 2ω let F = Γα ∩Γβ . ThenΓα∩Γβ = F and (ω\Γα)∩(ω\Γβ) ⊆ (ω\F ), which shows that the system is disjoint.This demonstrates equality above, with A = B = P(ω) and C = Finco(ω).

For free amalgamated products with infinitely many factors we have

c(⊕C

i∈IAi

)≤ 2|C| · 2supi∈Ic(Ai).

To prove this, let κ be the cardinal on the right, and suppose that 〈yα : α < κ+〉is a disjoint system of elements of ⊕C

i∈I(Ai). We may assume that each yα has theform

yα =∏i∈Fα

aαi ,

where Fα is a finite subset of I and aαi ∈ Ai for all i ∈ Fα. We may assume, infact, that the Fα’s form a Δ-system, say with kernel G; and that they all have thesame size. Thus by a change of notation we may write

yα =∏j<m

aαiαj ·∏j<n

aαkj,

where Fα\G = {iαj : j < m} and G = {kj : j < n}. For distinct α, β < κ+ therethen exist cj ∈ C for j < m, dj ∈ C for j < m, and ej ∈ C for j < n such that

aαiαj ≤ cj for all j < m, aβiβj≤ dj for all j < m, and aαkj

·aβkj≤ ej for all j < n, such

that ∏j<m

cj ·∏j<m

dj ·∏j<n

ej = 0.

Using the Erdos–Rado theorem again, we get Γ ∈ [κ+]λ and c, d ∈ mC, e ∈ nCsuch that the above holds for all distinct α, β ∈ Γ, where λ = (|C| · supi∈Ic(Ai))

+.If j < n and α, β are distinct members of Γ, then aαkj

· −ej · aβ · −ej = 0, so since

c(Akj ) < κ+ there is a Δj ∈ Γκ such that aαkj· −ej = 0 for all α ∈ Γ\Δj . Choose

α ∈ Γ\⋃

j<n Δj . Then yα ≤∏

j<m cj ·∏

j<m dj ·∏

j<n ej = 0, contradiction.

Since P(ω)⊕C P(ω) can be considered as a subalgebra of ⊕Ci∈ωP(ω), with

C as in the example for the free product of two factors, it follows that the aboveinequality is again best possible.

The behaviour of cellularity under unions of well-ordered chains is clear onthe basis of cardinal arithmetic. We restrict ourselves, without loss of generality,to well-ordered chains of regular type. Actually, we can formulate a more generalfact about increasing chains of BAs; this fact will apply to several of our cardinalfunctions, namely to the ordinary sup-functions (see the introduction).

Theorem 3.16. Let κ and λ be infinite cardinals, with λ regular. Suppose that kis an ordinary sup-function with respect to P . Then the following conditions areequivalent:

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(i) cf(κ) = λ.

(ii) There is a strictly increasing sequence 〈Aα : α < λ〉 of BAs each satisfyingthe κ− k-chain condition such that

⋃α<λ Aα does not satisfy this condition.

Proof. (i)⇒(ii): Assume (i). Let 〈μξ : ξ < λ〉 be a strictly increasing sequence ofordinals with sup κ (maybe κ is a successor cardinal, so that we cannot take the μξ

to be cardinals). Let A be a BA of size κ with a set X ∈ P (A) such that |X | = κ.Write A = {aα : α < κ}. For each ξ < λ let Bξ = 〈{aα : α < μξ}〉. Thus Bξ ⊆ Bη

if ξ < η, and |Bξ| < κ for all ξ < λ. Hence a strictly increasing subsequence is asdesired (since λ is regular).

(ii)⇒(i). Assume that (ii) holds but (i) fails. Let X be a subset of⋃

α<λ Aα

of power κ which is in P(⋃

α<λ Aα

). If λ < cf(κ), then the facts that X =⋃

α<λ(X ∩ Aα), |X | = κ, and |X ∩Aα| < κ for all α < λ, give a contradiction.

So, assume that cf(κ) < λ. Now for all α < λ there is a β > α such thatX∩Aα ⊂ X∩Aβ, since otherwise some Aα would contain X . It follows that λ ≤ κ,and so κ is singular in the case we are considering. Let 〈μα : α < cf(κ)〉 be a strictlyincreasing sequence of cardinals with sup κ. Since supα<λ|X ∩ Aα| = κ, for eachα < cf(κ) choose ν(α) < λ such that |X ∩ Aν(α)| ≥ μα. Let ρ = supα<cf(κ)ν(α).Then ρ < λ since cf(κ) < λ and λ is regular. But |X ∩Aρ| = κ, contradiction. �

With regard to Theorem 3.16, see also Theorem 3.57.

Now we turn to ultraproducts. Many of the things which we mention hold forother cardinal functions as well. First we consider countably complete ultrafilters;the main result we want to give here is that if F is a countably complete ultrafilteron an infinite set I and Ai is a ccc BA for each i ∈ I, then

∏i∈I Ai/F also satisfies

ccc. An analogous statement holds for many of our other functions. The resultfollows from the following standard facts. If F is countably complete and non-principal, then there is an uncountable measurable cardinal, and |I| is at least asbig as the first such – call it κ. (See Comfort, Negrepontis [74], p. 196.) Also, Fis κ-complete. To see this, suppose not, and let λ be the least cardinal such thatF is not λ-complete. Thus ω1 < λ ≤ κ. Then there exist a cardinal μ < λ anddisjoint aα ⊆ I for α < μ such that I\aα ∈ F for all α < μ, while

⋃α<μ aα ∈ F .

Let G = {S ⊆ μ :⋃

α∈S aα /∈ F}. Then it is easy to check that G is a σ-completenon-principal maximal ideal on μ, which is a contradiction, since μ is less than κ.

Now we can give the simple BA argument from these set-theoretical facts.Suppose

∏i∈I Ai/F does not satisfy ccc. Let 〈[aα] : α < ω1〉 be a system of non-

zero disjoint elements of the product; [x] denotes the equivalence class of x under

F . Since F is ω2-complete, the sets Jαβdef= {i ∈ I : (aα)i · (aβ)i = 0} for α �= β

and the sets Kαdef= {i ∈ I : (aα)i �= 0} have a non-zero intersection, since that

intersection is in F . But this is obviously a contradiction.

Thus countably complete ultrafilters tend to preserve chain conditions; weskip trying to give a more general version of the above argument.

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Problem 6. Describe the possibilities for chain conditions in ultraproducts withrespect to countably complete ultrafilters.

Next, if F is a countably incomplete ultrafilter on I and each algebra Ai is infinite,then

∏i∈I Ai/F never has ccc. This follows from the fact that the ultraproduct is

ω1-saturated in the model-theoretic sense; see Chang, Keisler [73], p. 305.

Now we present some results of Douglas Peterson; see Peterson [97]. Theydepend on some well-known notions and results. An ultrafilter F on an infinite setI is regular if there is a system 〈ai : i ∈ I〉 of elements of F such that

⋂j∈J aj = 0

for every infinite subset J of I. The following concept is useful. Let F be anultrafilter on I, and let 〈αi : i ∈ I〉 be a system of ordinals. We define the essentialsupremum of 〈αi : i ∈ I〉 over F to be

ess.supFi∈Iαi = min

{supi∈b

αi : b ∈ F

}.

Keisler, Prikry [74] show that if F is a regular ultrafilter on an infinite set I and〈κi : i ∈ I〉 is a system of infinite cardinals, then |

∏i∈I κi/F | = (ess.supFi∈Iκi)

|I|.Given an infinite cardinal κ and an ultrafilter F on some set I, we call F

κ-descendingly incomplete provided that there is a system 〈aα : α < κ〉 of elementsof F such that aα ⊇ aβ whenever α < β < κ, and

⋂α<κ aα = 0. We need the

following well-known fact:

(∗) If F is a regular ultrafilter on an infinite set I and κ is an infinite cardinal suchthat κ ≤ |I|, then F is κ-descendingly incomplete.

To prove this fact, let 〈aα : α < |I|〉 be a system of elements showing the regularityof F . We may assume that κ ⊆ I. For each α < κ, let bα =

⋃κ>β>α aβ . Thus

〈bα : α < κ〉 is descending. Now⋂α<κ

bα =⋂α<κ

⋃κ>γ>α

aγ =⋃f∈F

⋂α<κ

af(α),

where F =∏

α<κ(α, κ). Given f ∈ F , we have

⋂α<κ

af(α) ⊆⋂

n∈ω\1afn(0) = ∅,

as desired.

Now we begin Peterson’s results, with two useful lemmas.

Lemma 3.17. Suppose that κ is an uncountable limit cardinal, I is a set such thatcf(κ) ≤ |I|, and F is a cf(κ)-descendingly incomplete ultrafilter on I. Then thereis a sequence 〈λi : i ∈ I〉 of infinite cardinals such that λi < κ for all i ∈ I andess.supFi∈Iλi = κ.

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Proof. By the cf(κ)-descending incompleteness of F let 〈aα : α < cf(κ)〉 be asystem of elements of F such that aα ⊇ aβ whenever α < β < cf(κ), and⋂

α<cf(κ) aα = 0. We may assume that a0 = I and aλ =⋂

α<λ aα for λ limit.

Let 〈μδ : δ < cf(κ)〉 be a strictly increasing continuous sequence of infinite car-dinals with supremum κ. Now we define the sequence 〈λi : i ∈ I〉. Let i ∈ I.Then there is a unique γ < cf(κ) such that i ∈ aγ\aγ+1, and we define λi = μγ .Now to show that ess.supFi∈Iλi = κ, take a ∈ F and δ < cf(κ); we shall showthat sup{λi : i ∈ a} ≥ μδ. For any i ∈ aδ we have λi ≥ μδ. Now a ∩ aδ �= 0 sincea∩aδ ∈ F , and if we choose i ∈ a∩aδ we have μδ ≤ λi, and so μδ ≤ sup{λi : i ∈ a},as desired. �

Lemma 3.18. Suppose that F is a regular ultrafilter on an infinite set I, κ is alimit cardinal, 〈κi : i ∈ I〉 is a system of infinite cardinals, κ = ess.supFi∈Iκi,cf(κ) ≤ |I|, and ω < κi < κ for all i ∈ I. Then there is a system 〈λi : i ∈ I〉 ofinfinite cardinals such that λi < κi for all i ∈ I and ess.supFi∈Iλi = κ.

Proof. Let 〈aα : α < |I|〉 be a system of elements of F showing the regularity ofF , with a0 = I. Let 〈δα : α < cf(κ)〉 be a strictly increasing continuous sequenceof cardinals with supremum κ such that δ0 = ω. Let G = {i ∈ I : κi = δα for somelimit α, or for α = 0}, and for each i ∈ G let α(i) be such that κi = δα(i). Setaiξ =

⋃ξ≤α<α(i) aα for each i ∈ G and ξ < α(i). Then we have:

(1)⋂

ξ<α(i) aiξ = ∅ for each i ∈ G.

(2) If i, j ∈ G are such that α(i) < α(j), then for each ξ < α(i) we have aiξ ⊆ ajξ.

(3) If i ∈ G and ξ is a limit ordinal < α(i), then⋂

γ<ξ aiγ = aiξ.

(1) is proved like (∗) above. (2) is clear. For (3), suppose the hypotheses of (3)hold. Note that aiξ ⊆ aiγ for all γ < ξ, hence aiξ ⊆

⋂γ<ξ a

iγ .

Suppose that k ∈(⋂

γ<ξ aiγ

)\aiξ. Then k /∈

⋃ξ≤α<α(i) aα, so for each γ < ξ,

k ∈⋃

γ≤α<α(i) aα\⋃

ξ≤α<α(i) aα, hence k is in some aα with γ ≤ α < ξ. Thisclearly implies that k is in infinitely many aα’s, contradiction.

Now let i ∈ I be arbitrary. We will define λi by cases.

Case 1. i /∈ G. There is an α < cf(κ) such that δα ≤ κi < δα+1. If α is a successorordinal β + 1, let λi = δβ. Otherwise α is a limit ordinal or 0, and by i /∈ G wehave δα < κi < δα+1, and we let λi = δα. Under either possibility we then haveλi < κi.

Case 2. i ∈ G. Since a0 = I, we have i ∈ ai0. Let ξ = sup{η < α(i) : i ∈ aiη}. Thenby (1), ξ < α(i), and by (3), i ∈ aiξ\aiξ+1; let λi = δξ. Thus λi = δξ < δα(i) = κi.

In order to show that ess.supFi∈Iλi = κ, suppose that a ∈ F and ω ≤ ρ < κ;we show that sup{λi : i ∈ a} ≥ ρ. Choose α < cf(κ) such that δα ≤ ρ < δα+1. Weconsider two cases.

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Case 1. G /∈ F . Let a′ = {i ∈ I : κi > δα+2}. Then a′ ∈ F since ess.supFi∈Iκi = κ.If i ∈ a′\G, then λi ≥ δα+1 > ρ. Since I\G ∈ F , there is a j ∈ (a ∩ a′)\G. Thensup{λi : i ∈ a} ≥ λj > ρ.

Case 2. G ∈ F . Since ess.supi∈Iκi = κ, for each γ < κ we have Mγdef= {i ∈

G : δα(i) > γ} ∈ F . Hence choose k ∈ G such that δα(k) > δα+1. Then choose

j ∈ Mδα(k)∩ akα+1 ∩ a. Then j ∈ ajα+1 since ajα+1 ⊇ akα+1 by (2). Choose ε

such that j ∈ ajε\ajε+1. Then ε ≥ α + 1, and so λj = δε ≥ δα+1 > ρ. Hence

sup{λi : i ∈ a} ≥ λj > ρ. �

We need three more simple results.

Theorem 3.19. If F is a regular ultrafilter over a set I then there is a system〈ni : i ∈ I〉 of natural numbers such that

∣∣∏i∈I ni/F

∣∣ = 2|I|.

Proof. Let 〈ai : i ∈ I〉 be a system showing that F is regular. For each i ∈ I letMi = {j ∈ I : i ∈ aj}. Thus |Mi| < ω. We will show that 2|I| ≤

∣∣∏i∈I

Mi2/F∣∣,

proving the theorem. For each g ∈ I2 define g′ ∈∏

i∈IMi2 by g′(i) = g � Mi. If

g, h ∈ I2 and g �= h, pick j ∈ I such that g(j) �= h(j); then for any i ∈ aj we havej ∈ Mi, and hence g′(i) �= h′(i). This shows that aj ⊆ {i ∈ I : g′(i) �= h′(i)}, andhence g′/F �= h′/F . �

Recall that if k is a cardinal function defined by supremums with respect to afunction P , then

k′(A) = min{κ : |X | < κ for all X ∈ P (A)}.

Proposition 3.20. If k is an ultra-sup function, 〈Ai : i ∈ I〉 is a system of infiniteBAs, with I infinite, F is an ultrafilter on I, and κi < k′(Ai) for all i ∈ I. thenk(∏

i∈I Ai/F)≥∣∣∏

i∈I κi/F∣∣. �

Corollary 3.21. If k is an ultra-sup function, 〈Ai : i ∈ I〉 is a system of BAs withI infinite, and F is an ultrafilter on I, then k

(∏i∈I Ai/F

)≥ ess.supFi∈Ik(Ai).

Proof. Let λ = ess.supFi∈Ik(Ai). If λ is a successor cardinal, then we may assumethat k(Ai) = λ for all i ∈ I, and we are done by Proposition 3.20. If λ is a limitcardinal, then for each regular cardinal γ < λ we have k

(∏i∈I Ai/F

)≥ |Iγ/F | ≥

γ, and the result follows. �

Now we are ready for the main result of Peterson:

Theorem 3.22. Let k be an ultra-sup function with respect to P such that if Ais an infinite BA then P (A) contains arbitrarily large finite sets. Suppose that〈Ai : i ∈ I〉 is a system of BAs with I infinite and F is a regular ultrafilter on I.Then k

(∏i∈I Ai/F

)≥∣∣∏

i∈I k(Ai)/F∣∣.

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Proof. Let λ = ess.supFi∈Ik(Ai), and recall that λ|I| =∣∣∏

i∈I k(Ai)/F∣∣ (the result

of Keisler and Prikry mentioned above). We now consider several cases.

Case 1. λ = ω. Then λ|I| = 2|I|, and by Theorem 3.19, k(∏

i∈I Ai/F)≥ 2|I|.

Case 2. ω < λ ≤ |I|. Then k(∏

i∈I Ai/F)≥ |Iω/F | = 2|I| = λ|I|.

Case 3. cf(λ) ≤ |I| < λ, and {i ∈ I : k(Ai) = λ} ∈ F . Then we may assume thatk(Ai) = λ for all i ∈ I. By Lemma 3.17 and (∗) before it, let 〈κi : i ∈ I〉 be asystem of infinite cardinals such that κi < λ for all i ∈ I and ess.supFi∈Iκi = λ.Then by Proposition 3.20, k

(∏i∈I Ai/F

)≥∣∣∏


Case 4. cf(λ) ≤ |I| < λ, and {i ∈ I : k(Ai) < λ} ∈ F . This is like Case 3, exceptLemma 3.18 is used.

Case 5. |I| < cf(λ). Then λ|I| = sup{κ|I| : κ < λ}. If ω ≤ κ < λ, thenk(∏

i∈I Ai/F)≥ |Iκ/F | = κ|I|. Hence k

(∏i∈I Ai/F

)≥ λ|I|. �

Recall that c is an ultra-sup function. Thus Theorem 3.22 gives a lower bound forc(∏

i∈I Ai/F), at least for regular F . The following simple result gives an upper

bound.

Theorem 3.23. Let 〈Ai : i ∈ I〉 is a system of infinite BAs, with I infinite, andsuppose that F is a uniform ultrafilter on I. Let κ = max(|I|, ess.supFi∈Ic(Ai)).Then c

(∏i∈I Ai/F

)≤ 2κ.

Proof. Let λ = ess.supFi∈Ic(Ai). We may assume that c(Ai) ≤ λ for all i ∈ I.In order to get a contradiction, suppose that 〈fα/F : α < (2κ)+〉 is a system ofdisjoint elements. We may assume that fα(i) �= 0 for all i ∈ I and α < (2κ)+.Thus [(2κ)+]2 =

⋃i∈I{{α, β} : fα(i) · fβ(i) = 0}, so by the Erdos–Rado theorem

(2κ)+ → (κ+)2κ we get a homogeneous set which gives a contradiction. �

Concerning the inequality in Theorem 3.22, an example with strict inequality isgiven in Shelah [90], in fact, with |

∏i∈ω c(Ai)/F | < c

(∏i∈ω Ai/F

)for any uniform

ultrafilter F on ω. We give a form of this result, using Theorem 3.7.

Theorem 3.24. (Shelah). Let μ be a regular cardinal greater than ℵ1. Then thereis a system 〈Bn : n ∈ ω〉 of BAs each satisfying the μ+-cc such that for anynon-principal ultrafilter D on ω, the ultraproduct

∏n∈ω Bn/D does not satisfy the

μ+-cc.

Proof. By Theorem 3.7, Pr1(μ+, μ+, ω1, ω) holds. Let c : [μ+] → ω1 be as in the

definition of Pr1(μ+, μ+, ω1, ω).

Temporarily fix n ∈ ω. Let Cn be freely generated by 〈xnα : α < μ+〉. Let In

be the ideal of Cn generated by the set

{xnα · xn

β : α < β < μ+ and c({α, β}) < n}.

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Let ynα = xnα/In for each α < μ+. Set Bn = Cn/In. Then for α < β < μ+ we

have {n ∈ ω : ynα · ynβ = 0} ⊇ {n ∈ ω : c({α, β}) < n}; the latter set is cofinite, so(ynα/D) · (ynβ/D) = 0. We claim that ynα �= 0 for all n ∈ ω. Otherwise we would get

xnα ≤ xn

γ1· xn

δ1 + · · ·+ xnγm· xn

δm

with γi �= δi for all i. Mapping xnα to 1 and all other generators to 0 then extending

to a homomorphism, we get a contradiction.

To show that each Bn satisfies μ+-cc, assume that 〈bα : α < μ+〉 ∈ μ+

Cn issuch that bα · bβ ∈ In for all distinct α, β < μ+, while each bα /∈ In; we want toget a contradiction. Without loss of generality, we may assume that each bα hasthe following form:

bα =∏

β∈Fα

(xnβ)

εα(β),

where Fα is a finite subset of μ+ and εα ∈ Fα2. Without loss of generality wemay assume that: 〈Fα : α < μ+〉 forms a Δ-system, say with kernel G; εα � G isthe same for all α < μ+; and |Fα\G| = |Fβ\G| for all α, β < μ+. Cutting downfurther, we may assume that Fα\G < Fβ\G if α < β (that is, γ < δ if γ ∈ Fα\Gand δ ∈ Fβ\G). Now by the definition of Pr1(μ

+, μ+, ω1, ω), choose α < β < μ+

so that c({ε, ζ}) = n for all ε ∈ Fα\G and ζ ∈ Fβ\G. Now we can write

(1) bα · bβ ≤ xnγ1· xn

δ1 + · · ·+ xnγm· xn

δm

with γi < δi and c({γi, δi}) < n; moreover, we assume that m is minimal so thatsuch an inequality holds. It follows that γi, δi ∈ Fα ∪ Fβ for all i. If γi ∈ Fα, thenεα(γi) = 1, since otherwise the summand xn

γi· xn

δicould be dropped. Similarly

γi ∈ Fβ implies that εβ(γi) = 1, and similarly for the δi’s. Now it follows, sincebα /∈ In, that we cannot have γ1, δ1 ∈ Fα. Similarly, γ1, δ1 /∈ Fβ . It follows, then,that γ1 ∈ Fα\G and δ1 ∈ Fβ\G. But then c({γ1, δ1}) = n, contradiction. �

Corollary 3.25. For any infinite cardinal κ there is a system 〈Bn : n ∈ ω〉such that for any non-principal ideal D on ω, |

∏n∈ω c(Bn)/D| ≤ (2κ)+ while

c(∏

n∈ω Bn/D) ≥ (2κ)++.

Proof. Apply Theorem 3.24 to μ = (2κ)+. �

According to a result of Donder [88], it is consistent that the lower bound inTheorem 3.22 always holds (since his result says that it is consistent that everyuniform ultrafilter is regular). It is also consistent that the bound fails (for cellu-larity); see Magidor, Shelah [98]. The example they give uses some large cardinals.This answers Problem 2 in Monk [96].

Now we consider the upper bound given in Theorem 3.23. By the following theo-rem, that bound can be attained. This theorem is a consequence of Lemma 1.5.3in McKenzie, Monk [82].

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Theorem 3.26. Suppose that κ is an infinite cardinal and 〈Aα : α < κ〉 is a systemof infinite BAs. Then there is a non-principal filter F on κ such that for everynon-principal ultrafilter G ⊇ F , the BA

∏α<κ Aα/G has a chain of order type 2κ,

and hence has a pairwise disjoint set of size 2κ.

Proof. By Comfort, Negrepontis [74] Corollary 3.17, there is an S ⊆ κω with thefollowing properties:

(1) |S| = 2κ.

(2) For every finite M ⊆ S and every a ∈ Mω there is an α < κ such thatf(α) = a(f) for every f ∈M .

Now for each α < κ let 〈xαi : i ∈ ω〉 be a system of elements of Aα such that0 < xα0 < xα1 < · · · . Let 〈fξ : ξ < 2κ〉 be a one-one enumeration of S. For eachξ < 2κ we define gξ ∈

∏α<κ Aα by setting gξ(α) = xαfξ(α). Now for ξ < η < 2κ

we defineJξη = {α < κ : gξ(α) < gη(α)}.

We claim:

(3) {Jξη : ξ < η < 2κ} has fip.

For, suppose that N is a finite set of pairs (ξ, η) with ξ < η < 2κ. Let M be theset of all fξ such that ξ is the first or second coordinate of some member of N . Wecan write M = {fξ(i) : i < m} for some natural number m, such that ξ(i) < ξ(j)if i < j < m. Then let a(fξ(i)) = i for all i < m. By (2) choose α < κ such thatfξ(i)(α) = a(fξ(i)) for every i < m. Then for i < j < m,

gξ(i)(α) = xαfξ(i)(α) = xαa(fξ(i)) = xαi < xαj = xαa(fξ(j)) = xαfξ(j)(α) = gξ(j)(α),

and hence α ∈⋂

(ξ,η)∈N Jξη. This proves (3).

So, let F be the filter generated by all sets Jξη for ξ < η < 2κ. Clearly thisgives the desired conclusion. �

There are trivial examples where the bound in Theorem 3.23 is not attained. Forexample, one can take I = ω and Ai = Finco(2ω) for every i ∈ ω. Thus c(Ai) = 2ω

for each i ∈ ω, while for any ultrafilter F on ω, the ultraproduct∏

i∈ω Ai/F hassize 2ω, and hence has cellularity 2ω.

We turn to the examination of cellularity under other operations on Boolean al-gebras.

We describe the situation with subdirect products. Suppose that B is a sub-direct product of BAs 〈Ai : i ∈ I〉; what is the cellularity of B in terms of thecellularity of the Ai’s? Well, since a direct product is a special case of a subdirectproduct, we have the upper bound c(B) ≤ supi∈Ic(Ai)∪|I|. The lower bound ω isobvious. And that lower bound can be attained, even if the algebras Ai have highcellularity. In fact, consider the following example. Let κ be any infinite cardinal,

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let A be the free BA on κ free generators, and let B be the algebra of finite andcofinite subsets of κ. We show that A is isomorphic to a subdirect product ofcopies of B. To do this, it suffices to take any non-zero element a ∈ A and finda homomorphism of A onto B which takes −a to 0. In fact, A � a is still free onκ free generators, and so there is a homomorphism of it onto B. So our desiredhomomorphism is obtained as follows:

A→ (A � −a)× (A � a)→ A � a→ B.

For moderate products we have c(∏B

i∈I Ai

)= |I| + supi∈Ic(Ai). In fact, ≥ is

clear. Now suppose that κ = |I| + supi∈Ic(Ai) and 〈h(bα, Fα, aα) : α < κ+〉 is a

system of disjoint elements of∏B

i∈I Ai. Then 〈bα : α < κ+ is a system of disjointelements of B, and B ⊆ P(I), so fewer than κ+ of the bα’s are nonzero. So wemay assume that bα = ∅ for all α < κ+. Now let 〈Fα : α ∈ M〉 be a Δ-system,say with kernel G, with |M | = κ+. Then 〈Fα\G : α ∈ M〉 is a system of disjointsubsets of I, so at most κ of these sets are nonempty. Hence we may assume thatFα\G = ∅ for all α ∈ M . Now 〈aα : α ∈ M〉 is a system of disjoint nonemptyelements of

∏i∈G Ai, and |M | = κ+, contradiction.

Next, let B be obtained from algebras 〈Ai : i ∈ I〉 by one-point gluing,as described in Chapter 1. With respect to cellularity, B behaves like the fulldirect product: If B is infinite and all algebras Ai have at least four elements,then c(B) = |I| + supi∈Ic(Ai). In fact, ≤ holds since B ≤

∏i∈I Ai. Now suppose

that i ∈ I and X ⊆ Ai is disjoint. At most one member of X is in Fi, so wemay assume that all members of X are not in Fi. For each x ∈ X let fx be themember of

∏j∈I Aj such that fx(i) = x and fx(j) = 0 for j �= i. Then fx ∈ B,

and 〈fx : x ∈ X〉 is a disjoint system in B. Hence c(Ai) ≤ c(B). Next, let i ∈ I.Choose x ∈ Ai such that x /∈ Fi; this is possible since |Ai| ≥ 4. Let gi(i) = x andgi(j) = 0 for j �= i. Then 〈gi : i ∈ I〉 is a system of nonzero elements of B, and so|I| ≤ c(B).

Our next algebraic operation is the Alexsandroff duplicate.

Note that |Dup(A)| = |Ult(A)|, and Dup(A) is atomic with |Ult(A)| atoms.Hence c(Dup(A)) = |Ult(A)|.

We consider now the exponential of a given BA A. The main result here isdue to Fedorchuk and Todorcevic [97]: the cellularity of Exp(A) is equal to thecellularity of the ωth free power of A. The proof depends on the following lemma.

Let n be a positive integer. A system 〈〈a0ξ , . . . , an−1ξ 〉 : ξ ∈ I〉 of n-tuples of

members of A+ refines another sequence 〈〈b0ξ , . . . , bn−1ξ 〉 : ξ ∈ I〉 iff aiξ ≤ biξ for all

i < n and ξ ∈ I.

Lemma 3.27. Let n be a positive integer, τ an uncountable regular cardinal, and Aa BA. Suppose that 〈〈a0ξ , . . . , an−1

ξ 〉 : ξ ∈ τ〉 is a system of elements of A+. Then

there exist an I ∈ [τ ]τ , a refinement 〈〈c0ξ , . . . , cn−1ξ 〉 : ξ ∈ I〉 of 〈〈a0ξ, . . . , an−1

ξ 〉 : ξ ∈I〉, a system 〈b0, . . . , bk−1〉 of pairwise disjoint elements of A+, with k a positive

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integer, and a partition 〈M0, . . . ,Mk−1〉 of n into nonempty sets, such that thefollowing conditions hold:

(i) For each j < k, ciξ ≤ bj for all i ∈M j and all ξ ∈ I.

(ii) If j < k, |M j| > 1, and ξ, η ∈ I with ξ < η, then either (∑

i∈Mj ciξ) ·(∑

i∈Mj ciη) = 0 or there is an l ∈M j such that∑

i∈Mj ciη ≤ clξ.

Proof. Induction on n. For n = 1 we can take I = τ , ciξ = aiξ for all i < n and

ξ < τ , k = 1 with b0 = 1 and M0 = n. Now assume that n > 1 and the lemmaholds for positive integers less than n.

Case 1. There exist a u ∈ A+ and an i < n such that the set

J iu

def= {ξ < τ : aiξ · u �= 0 �= aiξ · −u}

has size τ .

Subcase 1.1. There exist an I ⊆ J iu and a j < n such that |I| = τ and ajξ · u = 0

for all ξ ∈ I. Let N = {k < n : |{ξ ∈ J iu : akξ · u �= 0}| < τ}, and let I =

{J iu\⋃

k∈N{ξ ∈ J iu : akξ · u �= 0}. So |I| = τ , and aiξ · u = 0 for all k ∈ N and all

x ∈ I. Note that j ∈ N and i /∈ N . Let M = n\N . So akξ ·u �= 0 for all k ∈M and

ξ ∈ I. We apply the induction hypothesis to 〈〈akξ · u : k ∈ M〉 : ξ ∈ I〉 to obtain

a K ∈ [I]τ , a refinement 〈〈dkξ : k ∈ M〉 : ξ ∈ K〉 of 〈〈akξ · u : k ∈ M〉 : ξ ∈ K〉, asystem 〈bk : k < l〉 of pairwise disjoint elements of A+, with l a positive integer,and a partition 〈P k : k < l〉 of M into nonempty sets, such that the followingconditions hold:

(1) For each k < l, diξ ≤ bk for all i ∈ P k and all ξ ∈ I.

(2) If j < l, |P j | > 1, and ξ, η ∈ I with ξ < η, then either (∑

i∈P j diξ)·(∑

i∈P j diη) =

0 or there is an s ∈ P j such that∑

i∈P j diη ≤ dsξ.

Then apply the induction hypothesis to 〈〈akξ : k ∈ N〉 : ξ ∈ K〉 to obtain an

L ∈ [K]τ , a refinement 〈〈ekξ : k ∈ N〉 : ξ ∈ L〉 of 〈〈akξ : k ∈ N〉 : ξ ∈ L〉, a system

〈fk : k < m〉 of pairwise disjoint elements of A+, with m a positive integer, and apartition 〈Qk : k < m〉 of N into nonempty sets, such that the following conditionshold:

(3) For each k < m, eiξ ≤ fk for all i ∈ Qk and all ξ ∈ L.

(4) If j < m, |Qj| > 1, and ξ, η ∈ L with ξ < η, then either (∑

i∈Qj eiξ) ·(∑

i∈Qj eiη) = 0 or there is an s ∈ Qj such that∑

i∈Qj eiη ≤ esξ.

Then 〈〈dkξ : k ∈ M〉 ∪ 〈ekξ : k ∈ N〉 : ξ ∈ L〉 is a refinement of 〈〈a0ξ, . . . , an−1ξ 〉 :

ξ ∈ I〉, 〈b0 · u, . . . , bl−1 · u, f0 · −u, . . . , fm−1 · −u〉 is a system of pairwise disjointelements of A+, and 〈P k : k < l〉 ∪ 〈Qk : k < m〉 is a partition of n into nonemptysets, such that the conditions of the lemma hold.

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Subcase 1.2. Subcase 1.1 does not hold. Thus

(5) ∀I ∈ [J iu]

τ∀j < n∃ξ ∈ I[ajξ · u �= 0].

Let J iu = {αν : ν < τ} without repetitions. Then by (5), ∀ρ < τ∃ξ ∈ {αν : ρ ≤

ν < τ}[a0ξ · u �= 0]. Hence there is an I0 ∈ [J iu]

τ such that for all ξ ∈ I0 we have

a0ξ · u �= 0. Similarly we find I0 ⊇ I1 ⊇ · · · ⊇ In−1 = I such that I ∈ [J iu]

τ and

∀ξ ∈ I∀j < n[ajξ · u �= 0]. We apply the induction hypothesis to 〈〈ajξ · u : j <

n, j �= i〉 : ξ ∈ I〉 to obtain a K ∈ [I]τ , a refinement 〈〈djξ : j < n, j �= i〉 : ξ ∈ K〉of 〈〈ajξ · u : j < n, j �= i〉 : ξ ∈ K〉, a system 〈bk : k < l〉 of pairwise disjoint

elements of A+, with l a positive integer, and a partition 〈P k : k < l〉 of n\{i}into nonempty sets, such that the following conditions hold:

(6) For each k < l, diξ ≤ bk for all i ∈ P k and all ξ ∈ I.

(7) If j < l, |P j | > 1, and ξ, η ∈ I with ξ < η, then either (∑

k∈P j diξ) ·(∑

k∈P j diη) = 0 or there is an s ∈ P j such that∑

k∈P j dkη ≤ dsξ.

We may assume that bk ≤ u for all k < l. Now 〈〈djξ : j < n, j �= i〉�〈aiξ · −u〉 : ξ ∈K〉 is a refinement of 〈〈ajξ · u : j < n〉 : ξ ∈ K〉, 〈bk : k < l〉�〈−u〉 is a system of

pairwise disjoint elements of A+, 〈P k : k < l〉�{i}〉 is a partition of n, and theconditions of the lemma hold.

Case 2. Case 1 fails. We apply this to b = ajη. Thus for all i < n the set {ξ < τ :

aiξ · ajη �= 0 �= aiξ · −ajη} has size less than τ ; so the same is true of

⋃i,j<n

{ξ < τ : aiξ · ajη �= 0 �= aiξ · −ajη}.

Let Aη be the complement of this set. Thus

Aη =⋂

i,j<n

{ξ < τ : aiξ · ajη = 0 or aiξ · −ajη = 0}

= {ξ < τ : ∀i, j < n[aiξ · ajη = 0 or aiξ · −ajη = 0]}.

Let B = �η<τAη. So B is club, and

B = {ξ < τ : ∀η < ξ[ξ ∈ Aη]}= {ξ < τ : ∀η < ξ∀i, j < n[[aiξ · ajη = 0 or aiξ · −ajη = 0]}.

Subcase 2.1. There is a u ∈ A+ such that |C| = τ , where

C = {ξ ∈ B : ∃i, j < n[i �= j and aiξ · u �= 0 �= ajξ · −u]}.

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NowC =

⋃i,j<ni�=j

{ξ ∈ C : aiξ · u �= 0 �= ajξ · −u},

so there are distinct i, j < n such that Ddef= {ξ ∈ C : aiξ ·u �= 0 �= ajξ · −u} has size

τ . Now let

E0 =

{D\{ξ ∈ D : a0ξ · u �= 0} if |{ξ ∈ D : a0ξ · u �= 0}| < τ ,

{ξ ∈ D : a0ξ · u �= 0} otherwise.

Thus either a0ξ · u = 0 for all ξ ∈ E0, or a0ξ · u �= 0 for all ξ ∈ E0; and |E0| = τ .

Continuing analogously, we arrive at En−1 such that for all k < n, either akξ ·u = 0

for all ξ ∈ En−1, or akξ · u �= 0 for all ξ ∈ En−1; and |En−1| = τ . Workinganalogously with −u, we end up with F ⊆ En−1 such that |F | = τ and for allk < n, either akξ · −u = 0 for all ξ ∈ F , or akξ · −u �= 0 for all F . Now for k < nand ξ ∈ F we define

ckξ =

{akξ · u if ∀η ∈ F [akη · u �= 0] and k �= j,

akξ · −u otherwise.

Let M = {k < n : ∀η ∈ F [akη ·u �= 0] and k �= j} and N = n\M . So M �= ∅ �= N , sowe can apply the induction hypothesis to M and N ; putting the results togetheras above, we again get the conditions of the lemma.

Subcase 2.2. Subcase 2.1 fails. Thus

(8) ∀u ∈ A+|{ξ ∈ B : ∀i, j < n[i �= j → aiξ · u = 0 or ajξ · −u = 0}| = τ.

Now we claim

(9) |{ξ ∈ B : ∃i < n[aiξ · u �= 0]→ ∀j < n[ajξ · u �= 0]}| = τ.

In fact, take ξ in the set of (8), and suppose that i < n and aiξ · u �= 0. Then by

the condition of (8), ajξ · −u = 0 for all j �= i, hence ajξ · u = ajξ �= 0. So (9) holds.Similarly,

(10) |{ξ ∈ B : ∃i < n[aiξ · −u �= 0]→ ∀j < n[ajξ · −u �= 0]}| = τ.

Suppose that ξ, η ∈ B and η < ξ. We consider two possibilities.

Subsubcase 2.2.1. ∃i, j < n[aiη · ajξ �= 0]. Then by (9) for η with ajξ for

u, ∀k < n[akη · ajξ �= 0]. Then by the definition of B, akη ≤ ajξ for all k < n,

Subsubcase 2.2.2. ∀i, j < n[aiη · ajξ = 0].

These two subsubcases give the desired conclusion. �

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Theorem 3.28. For any infinite BA A, c(Exp(A)) = c(A∗ω), where A∗ω is the ωthfree power of A.

Proof. By Proposition 2.5 and the remark following Problem 5 we have

c(Exp(A)) ≤ c(A∗ω).

For the other direction it suffices by that same remark to show that c(A∗n) ≤c(Exp(A)) for any positive integer n. We may assume that n > 1. Suppose that〈eξ : ξ < τ〉 is a system of pairwise disjoint nonzero elements of A∗n. We mayassume that each eξ has the form a0ξ · . . . · an−1

ξ , where the ai come from differentfree factors of A∗n. Now we apply Lemma 3.27 to get a set I ∈ [τ ]τ , a refinement〈〈c0ξ, . . . , cn−1

ξ 〉 : ξ ∈ I〉 of 〈〈a0ξ , . . . , an−1ξ 〉 : ξ ∈ I〉, a system 〈b0, . . . , bk−1〉 of

pairwise disjoint elements of (A∗n)+, with k a positive integer, and a partition〈M0, . . . ,Mk−1〉 of n into nonempty sets, such that the following conditions hold:

(1) For each j < k, ciξ ≤ bj for all i ∈M j and all ξ ∈ I.

(2) If j < k, |M j | > 1, and ξ, η ∈ I with ξ < η, then either (∑

i∈Mj ciξ) ·(∑

i∈Mj ciη) = 0 or there is an l ∈M j such that∑

i∈Mj ciη ≤ clξ.

Now for each ξ ∈ I we let dξ be the following element of Exp(A):

V (S(c0ξ + · · ·+ cn−1ξ )) ∩

⋂{−V (S(−bj)) : |M j | = 1}∩⋂

{−V (S(−bj + ciξ)) : |M j| > 1, i ∈M j}.

First we check that each dξ is nonzero. Suppose that dξ = 0. Then by Lemma1.22(vi) there are two cases:

Case 1. c0ξ + · · · + cn−1ξ ≤ −bj for some j with |M j | = 1. Say M j = {i}. then

(c0ξ + · · ·+ cn−1ξ ) ∩ bj = ciξ, contradiction.

Case 2. c0ξ + · · · + cn−1ξ ≤ −bj + ciξ for some j with |M j | > 1 and some i ∈ M j .

Then (c0ξ + · · ·+ cn−1) ∩ bj ≤ ciξ, a contradiction since |M j | > 1.

Now suppose that ξ < η; we show that dξ ∩ dη = 0.

Case 1. There is a j such that |M j| > 1 and(∑

i∈Mj ciξ

)∩(∑

i∈Mj ciη)= 0. Then(∑

i<n

ciξ

)∩(∑

i<n

ciη

)∩ bj =

(∑i∈Mj

ciξ

)∩(∑

i∈Mj

ciη

)= 0,

hence dξ ∩ dη = 0 by Lemma 1.22(vi).

Case 2. There is a j such that |M j | > 1 and(∑

i∈Mj ciη)≤ clξ for some l ∈ M j .

Then (∑i<n

ciξ

)∩(∑

i<n

ciη

)∩ bj =

(∑i∈Mj

ciξ

)∩(∑

i∈Mj

ciη

)≤ clξ,

hence dξ ∩ dη = 0 by Lemma 1.22(vi).

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Case 3. |M j | = 1 for all j. Write M j = {ij} for all j. Choose j, k such that

cijξ ·cikη = 0. By the free product property we must have ij = ik, hence j = k. Then

(∑i<n

ciξ

)∩(∑

i<n

ciη

)∩ bj = c

ijξ · cijη = 0,

and again Lemma 1.22(vi) applies. �

Now we proceed to discuss the derived functions associated with cellularity. Firstwe show that cH+ is the same as spread. For this, it is convenient to have anequivalent definition of spread. A subset X of a BA A is ideal independent ifx /∈ 〈X\{x}〉Id for every x ∈ X ; recall that 〈Y 〉Id denotes the ideal generated byY , for any Y ⊆ A.

Theorem 3.29. For any infinite BA A, s(A) = sup{|X | : X is an ideal independentsubset of A}.

Proof. First suppose that D is a discrete subspace of Ult(A). For each F ∈ D,let aF ∈ A be such that S(aF ) ∩ D = {F}. Then 〈aF : F ∈ D〉 is one-oneand {aF : F ∈ D} is ideal independent. In fact, suppose that F,G0, . . . , Gn−1

are distinct members of D such that aF ≤ aG0 + · · · + aGn−1 . Then S(aF ) ⊆S(aG0) ∪ · · · ∪ S(aGn−1), and so F ∈ S(aG0) ∪ · · · ∪ S(aGn−1), which is clearlyimpossible.

Conversely, suppose that X is an ideal independent subset of A. Then foreach x ∈ X , {x} ∪ {−y : y ∈ X\{x}} has the finite intersection property, and sois included in an ultrafilter Fx. Let D = {Fx : x ∈ X}. Then S(x) ∩D = {Fx} foreach x ∈ X , so D is discrete and |D| = |X |, as desired. �

The proof of Theorem 3.29 shows that s(A) is attained in the discrete subspacesense iff it is attained in the ideal independence sense.

Theorem 3.30. For any infinite BA A, cH+(A) is equal to s(A), the spread of A.

Proof. First let f be a homomorphism from A onto a BA B, and let X be adisjoint subset of B+. We show that |X | ≤ s(A); this will show that cH+(A) ≤s(A). For each x ∈ X choose ax ∈ X such that f(ax) = x. Then 〈ax : x ∈X〉 is one-one and {ax : x ∈ X} is ideal independent. In fact, suppose thatx, y(0), . . . , y(n − 1) are distinct elements of X , and ax ≤ ay(0) + · · · + ay(n−1).Applying the homomorphism f to this inequality we get x ≤ y(0)+ · · ·+ y(n− 1).Since the elements x, y(0), . . . , y(n− 1) are pairwise disjoint, this is impossible.

For the converse, suppose that X is an ideal independent subset of A; wewant to find a homomorphic image B of A having a disjoint subset of size |X |.Let I = 〈{x · y : x, y ∈ X, x �= y}〉Id. It suffices now to show that [x] �= 0 for eachx ∈ X . ([u] is the equivalence class of u under the equivalence relation naturallyassociated with the ideal I). Suppose that [x] = 0. Then x is in the ideal I, andhence there exist elements y0, z0, . . . , yn−1, zn−1 of X such that yi �= zi for all

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i < n, and x ≤ y0 · z0 + · · ·+ yn−1 · zn−1. Without loss of generality, x �= yi for alli < n. But then x ≤ y0+· · ·+yn−1, contradicting the ideal independence ofX . �

For later purposes it is convenient to note the following corollary to the proof ofthe previous two theorems.

Corollary 3.31. cH+(A) and s(A) have the same attainment properties, in the sensethat s(A) is attained (in either the discrete subspace or ideal independence sense)iff there exist a homomorphic image B of A and a disjoint subset X of B suchthat |X | = cH+(A). �

Note in this corollary that attainment of cH+(A) involves two sups, while attain-ment of s(A) involves only one. Thus if s(A) is not attained, there are still twopossibilities according to Corollary 3.31: there can exist a homomorphic image Bof A with s(A) = c(B) but c(B) is not attained, or there is no homomorphic imageB of A with sA = c(B). Both possibilities are consistent with ZFC; we shall returnto this shortly and indicate the examples.

It is easy to see that cH−(A) = ω for any infinite BA A: let B be a de-numerable subalgebra of A, and extend the identity homomorphism h of B intoB to a homomorphism from A into B; the image of A under h is a ccc BA.(We are using here Sikorski’s extension theorem; recall that B is the comple-tion of B.) It is obvious that cS+(A) = c(A) and cS−(A) = ω for any infiniteBA A. ch+(A) is equal to s(A), since a disjoint family of open subsets of a sub-space Y of Ult(A) gives a discrete subset of Ult(A) of the same size, so thatch+(A) ≤ s(A) = cH+(A) ≤ ch+(A). It is obvious that ch−(A) = ω, and an easyargument gives that dcS+(A) = c(A) = dcS−(A).

The function cmm(A) has been studied for the algebra P(ω)/fin, using thenotation a. For this reason, we use a(A) rather than cmm(A). Thus for any BA A,

a(A) = min{|P | : P is an infinite partition of A};aspect(A) = {|P | : P is an infinite partition of A}.

For general BAs, a(A) is discussed in Monk [96], [96a], [01], [01a], and [02], and inMckenzie, Monk [04].

We consider what happens to a in subalgebras. It is possible to have A ≤ Bwith aspect(A) ∩ aspect(B) = ∅. Namely let A be arbitrary and extend A to anatomless BA with high saturation. One can have A a subalgebra of B with a(A)much smaller than a(B); for example, with B κ-saturated and A countable. Onthe other hand, if B is κ-saturated and C = B×D with D countable, then B canbe considered as a subalgebra of C, and a(C) = ω < κ ≤ a(B). That a(C) = ωis clear, but is stated and proved in Proposition 3.36 below. We now consider ourspecial kinds of subalgebras with respect to a.

Proposition 3.32. If A ≤reg B, then aspect(A) ⊆ aspect(B) and a(B) ≤ a(A). Here≤reg can be replaced by ≤free, ≤proj, ≤rc, or ≤π. �

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That the inequality here cannot in general be replaced by equality follows fromthe next proposition.

Proposition 3.33. Suppose that A ≤free B with B = A⊕C, C an infinite free BA.Then ω ∈ aspect(B). In particular, a(B) = ω.

Proof. Since C satisfies ccc. �

Applying Proposition 3.33 with a(A) > ω gives a(B) < a(A). Here we have A ≤free

B, and hence also A ≤proj B, A ≤u B, A ≤rc B, A ≤σ B, and A ≤reg B. For anexample with A ≤π B and a(B) < a(A) one can take any A with a(A) > ω andlet B be the completion of A.

If C is finite in Proposition 3.33, then equality does hold; this follows from Propo-sition 3.37 below.

Proposition 3.34. If A ≤m B, then aspect(A) ⊆ aspect(B), and a(B) ≤ a(A).

Proof. We assume thatA �= B. LetX be a partition of unity inA, with |X | = a(A).Write B = A(x).

Case 1. A � x and A � −x are nonprincipal. Then by Proposition 2.44, A is densein A(x) and our result follows from Proposition 3.32.

Case 2. (By symmetry) A � x is principal. By Proposition 2.45 we may assumethat x is an atom in A(x). Now the mapping a !→ a ·−x is clearly a homomorphismfrom A onto A � −x. It is one-one since a ·−x = 0 implies that a ≤ x, hence a = 0since x is an atom in A(x) and x /∈ A. Thus A(x) is isomorphic to A× 2. It is easyto check that aspect(A) = aspect(A× 2). �

On the other hand, there are BAs A,B with A ≤mg B, a(A) = ω, and a(B) > ω.In fact, let B = Finco(ω1), let A be a countably infinite subalgebra of B, andapply Theorem 2.55.

Proposition 3.35. Suppose that a(A) > ω and A ≤σ B. Then aspect(A) ⊆ aspect(B)and a(B) ≤ a(A).

Proof. Assume the hypotheses, and suppose that X is an uncountable partitionof unity in A. Suppose that b ∈ B, b < 1, and ∀a ∈ X [a ≤ b]; we want to get acontradiction. Now A � b is a countably generated ideal; say that it is generatedby the countable set Y . Note that

∑F ≤ b < 1 for every finite F ⊆ Y . We

may assume that Y is closed under +. For each a ∈ X choose ya ∈ Y such thata ≤ ya. Then

∑a∈X ya = 1; since {ya : a ∈ X} is countable, this gives a countable

partition of unity in A, contradiction. �

One can have A ≤σ B with a(A) = ω < a(B). For example, let B = finco(ω1),and let A be the subalgebra of B generated by {{n} : n ∈ ω}.

There are BAs A,B with A ≤u B and a(A) < a(B). In fact, one can take Ato be a denumerable atomless BA. Any ultrafilter on A has two extensions to an

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ultrafilter on A(x), the free extension of A by adding one element x. A(x) is stilldenumerable and atomless, and can be extended to an ω1-saturated BA B.

There exist BAs A,B such that A ≤m B and a(B) < a(A). In fact, letA = Finco(κ), where κ is an uncountable cardinal. Thus a(A) = κ. Let

I = 〈{{n} : n ∈ ω}〉Id,J = 〈{{α} : ω ≤ α < κ}〉Id.

Let B = A(x) be the simple extension of A such that B � x = I and B � −x = J .Clearly I ∪ J generates a maximal ideal, so we have A ≤m B. We claim that

{{n} : n ∈ ω} ∪ {−x}

is a partition of unity in B. Clearly it is a collection of pairwise disjoint elements.Suppose that a ·x+b ·−x is disjoint from each member of this set. Then b ·−x = 0.For each n ∈ ω, a · {n} · x = 0; since {n} ≤ x, this implies that a is a finite subsetof κ\ω. Hence a ∈ J , and so a ≤ −x. Thus a · x = 0. So we have shown thata · x+ b · −x = 0, proving the claim.

Problem 7. Does A ≤s B imply that a(B) ≤ a(A)?

Proposition 3.36. Let A and B be infinite BAs.

(i) aspect(A×B) = aspect(A) ∪ aspect(B).

(ii) a(A×B) = min(a(A), a(B)).

Proof. Clearly (ii) follows from (i). For (i), suppose that X is an infinite partitionof unity of A; then

{(x, 0) : x ∈ X} ∪ {(0, 1)}

is an infinite partition of unity in A × B. A similar argument works for B, so ⊇holds. Now suppose that X is an infinite partition of unity in A×B. Let

Y = {a ∈ A+ : ∃b ∈ B[(a, b) ∈ X ]},Z = {b ∈ B+ : ∃a ∈ A[(a, b) ∈ X ]}.

Then Y is a partition of unity in A and Z is a partition of unity in B. At leastone of them is of size |X |, and this gives ⊆. �

Proposition 3.37. If A is infinite and B is finite with at least 4 elements, thenaspect(A) = aspect(A⊕B).

Proof. By the Handbook 11.6(d), A ⊕ B is isomorphic to nA for some positiveinteger n, so our conclusion is immediate from Proposition 3.36(i). �

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Proposition 3.38. Suppose that 〈Ai : i ∈ I〉 is a system of infinite BAs, with Iinfinite. Then

aspect

(∏w

i∈IAi

)= {|I|} ∪

⋃i∈I

aspect(Ai) (i)

∪{κ : κ > |I|, κ is singular, and

∃J ⊆ I∃λ ∈∏

j∈Jaspect(Aj)

(J �= ∅ and κ = supj∈J λj

)}.

(ii) a(∏w

i∈I Ai) = min(|I|,mini∈I a(Ai)).

Proof. For brevity let B =∏w

i∈I Ai.

Clearly (i) implies (ii). For (i), by Proposition 3.36 it is clear that⋃

i∈I aspect(Ai) ⊆aspect(B), and an easy argument shows that |I| ∈ aspect(B). Now suppose that Jis a nonempty subset of I, λ ∈

∏j∈J aspect(Aj), and |I| < κ = supj∈J λj , with κ

singular. For each j ∈ J let Xj be a partition of unity in Aj such that |Xj | = λj .Now let K = {(j, x) : j ∈ J, x ∈ Xj} ∪ {(i, 0) : i ∈ I\J}. For each j ∈ J andx ∈ Xj , let y(j,x) be the member of

∏i∈I Ai which takes the value x at j and the

value 0 elsewhere. For each j ∈ I\J let y(j,0) be the member of∏

i∈I Ai whichtakes the value 1 at j and the value 0 elsewhere. Clearly 〈yk : k ∈ K〉 is a partitionof unity of size κ. This finishes the proof of ⊇.

For ⊆, suppose that |X | /∈ {|I|}∪⋃

i∈I aspect(Ai), with X an infinite partitionof unity of B.

Case 1. There is an x ∈ X such that Jdef= {i ∈ I : xi �= 1} is finite. If y ∈ X\{x}

and i ∈ I\J , then yi = 0, since x · y = 0. In particular, y � J �= z � J for alldistinct y, z ∈ X . If w is a nonzero element of

∏i∈J Ai, extend it to an element w′

of∏

i∈I AI by letting w′i = 0 for all i ∈ I\J . Choose y ∈ X such that w′ · y �= 0.

Then w · (y � Y ) �= 0. Thus {y � J : y ∈ X}\{0} is an infinite partition of unity in∏i∈J Ai, and so by Proposition 3.36, |X | ∈

⋃i∈J aspect(Ai), contradiction.

Case 2. For every x ∈ X the set {i ∈ I : xi �= 0} is finite. Since∑

X = 1, it isclear that |I| ≤ |X |, so |I| < |X |. Let J = {i ∈ I : {xi : x ∈ X} is infinite}. Foreach i ∈ I let Ki = {x ∈ X : xi �= 0}. Then X =

⋃i∈I Ki. If J = ∅, then each Ki

is finite, and so clearly |X | = |I|, contradiction. So J �= ∅. For each j ∈ J , the set{xj : x ∈ X}\{∅} is clearly an infinite partition of Aj . Let λj be the cardinalityif this set. Clearly λj = |Kj| for each j ∈ J . Since |X | /∈ aspect(Aj), we also haveλj < |X |. Moreover, |J | ≤ |I| < |X |. Now

|X | =

∣∣∣∣∣∣⋃

i∈I\JKi ∪

⋃j∈J

Kj

∣∣∣∣∣∣ ≤∑

i∈I\J|Ki|+

∑j∈J

|Kj |

≤ ω · |I\J |+∑j∈J

λj ≤ |X |,

so |X | = supj∈J λj and |X | is singular, as desired. �

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Kevin Selker has generalized Proposition 3.38 to moderate products. The gener-alization is somewhat complicated, but it has as a corollary the following simplegeneralization of Proposition 3.38(ii):

For a moderate product, a(∏B

i∈I Ai) = min(a(B),mini∈I a(Ai)).

Proposition 3.39. Let 〈Ai : i ∈ I〉 be a system of infinite BAs, with I infinite.Then:

aspect

(∏i∈I

Ai

)= [ω, |I|] ∪

⋃i∈I

aspect(Ai) (i)


∃J ⊆ I∃λ ∈∏j∈J

aspect(Aj)

[J �= ∅ and κ = sup

j∈Jλj

]}.

(ii) a(∏

i∈I Ai

)= ω.

Proof. Let B =∏

i∈I Ai. (ii) obviously follows from (i). For ⊇ in (i), first supposethat κ ∈ [ω, |I|]. Write I =

⋃α<κ Jα with each Jα �= ∅, and the Jα’s pairwise

disjoint. For each α < κ let fα be the member of B which takes the value 1 on Jαand the value 0 elsewhere. Clearly this gives a partition of B of size κ.

Each aspect(Ai) is a subset of aspect(B) by Proposition 3.36.

For κ, J, λ as indicated in the statement, one can follow the procedure de-scribed in the proof of Proposition 3.38. So ⊇ holds.

For ⊆ the proof follows that of Proposition 3.38. In fact, one can skip Case1, and start Case 2 with the condition |I| < |X |. �

Proposition 3.40. Let A and B be infinite BAs. Then

(i) aspect(A) ∪ aspect(B) ⊆ aspect(A⊕B).

(ii) a(A⊕B) ≤ min(a(A), a(B)) for any infinite BAs A,B. �

We will see in Corollary 4.50 that if a(A), a(B) > ω, then also a(A⊕B) > ω. Thefollowing more general question is open.

Problem 8. Is it true that for all infinite BAs A,B one has

a(A⊕B) = min(a(A), a(B))?

Note that a consequence of Corollary 3.11 is that there are BAs A, B such thataspect(A) ∪ aspect(B) ⊂ aspect(A⊕B).

Proposition 3.41. If 〈Ai : i ∈ I〉 is an infinite system of infinite BAs, thena(⊕i∈IAi) = ω.

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Proof. Let f : ω → I be an injection. For each i ∈ ω let ai be a member of Af(i)

such that 0 < ai < 1. Now for each j ∈ ω let bj = aj ·∏

i<j −ai. Thus 〈bj : j ∈ ω〉is a system of pairwise disjoint elements of ⊕i∈IAi. Clearly

∑i∈ω ai = 1, and∑

j∈ω bj =∑

i∈ω ai = 1, so we have an infinite partition of size ω. �

Proposition 3.42. If K is a set of infinite cardinals with a largest element, thenthere is an atomic BA A such that aspect(A) = K.

Proof. Let κ be the largest element of K. Clearly |K| ≤ κ. Let f be a func-tion mapping κ onto K, and for each α < κ let Bα = Finco(f(α)). Note thataspect(Bα) = {f(α)}. Let A =

∏wα<κ Bα. Then A is as desired, by Proposition

3.38. �

Note that if A is atomic, then aspect(A) has a largest element, namely the numberof atoms of A. Also, if sup(K) is singular, then any BA A with aspect(A) = Kmust have sup(K) ∈ aspect(A), by the Erdos–Tarski theorem.

We would now like to generalize Theorem 3.42 to the case of two BAs A ≤ B. Weadapt some arguments from Monk [07].

Lemma 3.43. If A is an infinite BA, then there is a BA B such that A ≤ B andno infinite partition in A remains a partition in B.

Proof. We take B = P(Ult(A)) and the Stone isomorphism S from A into B.Suppose that X is an infinite partition in A. Then by compactness,

⋃S[X ] �=

Ult(A). �

Lemma 3.44. If A is an infinite BA and κ is a regular cardinal, then there is aBA B such that A ≤ B and every partition in B has size at least κ.

Proof. We construct 〈Cα : α < κ〉 by recursion. Let C0 = A. Having defined Cα,let Cα+1 be obtained by Lemma 3.43: Cα ≤ Cα+1 and no infinite partition in Cα

is a partition in Cα+1. For α limit < κ, let Cα =⋃

β<αCβ . Now let B =⋃

α<κ Cα.Suppose that X is a partition in B of size less than κ. Then X ⊆ Cα for someα < κ, and hence X is not a partition in Cα+1, hence also not a partition in B,contradiction. �

Proposition 3.45. Suppose that K is a set of infinite cardinals with a largest mem-ber. Also suppose that κ is a regular cardinal > sup(K), and L is a set of cardinalsless than κ such that L has a largest element.

Then there exist BAs A,B such that A ≤ B, aspect(A) = K, and aspect(B)∩κ = L.

Proof. By Proposition 3.42, let A be a BA with aspect(A) = K. Then by Lemma3.44 let C be a BA such that A ≤ C and every partition of C has size at leastκ. Suppose that L has a largest element λ. Clearly |L| ≤ λ. Let f map λ onto L.Then set B = C ×

∏wα<λ Finco(f(α)). B is as desired, by Proposition 3.38. �

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These results on a leave several problems open in addition to Problem 8 above.

Problem 9. Suppose that K is a set of infinite cardinals without greatest element,with sup(K) inaccessible. Is there a BA A such that aspect(A) = K?

Problem 10. Describe in terms not involving BAs the sets K,L such that thereexist BAs A,B with A ≤reg B, aspect(A) = K, and aspect(B) = L.

Problem 11. Describe aspect(⊕i∈IAi) in terms of the sets aspect(⊕i∈FAi) for F afinite subset of I.

Problem 12. Describe the behaviour of a(A) under ultraproducts.

In connection with this problem, recall that if 〈Ai : i ∈ I〉 is a system of infiniteBAs and F is a countably incomplete ultrafilter on I, then

∏i∈I Ai/F is ω1-

saturated in the model-theoretic sense; see Chang, Keisler [73], Theorem 6.1.1. Inparticular, a(

∏i∈I Ai/F ) ≥ ω1 under these assumptions.

The homomorphic spectrum of cellularity is interesting. First, we can easilysee that [ω, s(A)) ⊆ cHsA ⊆ [ω, s(A)] (for cardinals κ < λ, [κ, λ) denotes the setof all cardinals μ such that κ ≤ μ < λ; similarly for [κ, λ]). This follows fromthe fact already proved that s(A) = cH+(A): given a homomorphic image B of Aand a disjoint subset X of B, one can use Sikorski’s extension theorem to get ahomomorphic image C of B such that c(C) = |X |.

It is more difficult to decide whether s(A) ∈ cHsA. This amounts to thefollowing question: is there always a homomorphic image B of A such that c(B) =s(A)? In case s(A) is attained, this is true by Corollary 3.31. For s(A) not attained,there are three consistency results which clarify things here and with respect tothe question raised above after Corollary 3.31. First, example 11.14 of the BAhandbook shows that for each weakly inaccessible cardinal κ there is a BA A suchthat |A| = c(A) = s(A) = κ and c(A) is not attained but s(A) is attained (asis easily checked). Second, the interval algebra of a κ-Suslin line, for κ stronglyinaccessible but not weakly compact, gives an example of a BA A such that |A| =c(A) = s(A), with neither c(A) nor s(A) attained; see Juhasz [75], example 6.6(V=L or something beyond ZFC is needed for the existence of a κ-Suslin line).Third, an example of Todorcevic [86], Theorem 12, shows that it is consistent tohave a BA A in which s(A) is not attained, while there is no homomorphic imageB of A with c(B)=s(A). This example involves some interesting ideas, and weshall now give it. It depends on the following lemma about the real numbers.

Lemma 3.46. There exist disjoint subsets E0 and E1 of [0, 1] which are of cardi-nality 2ω, are dense in [0, 1], and satisfy the following two conditions:

(i) For any κ < 2ω there is a strictly increasing function from some subset ofE0 of size κ into E1.

(ii) There is no strictly monotone function from a subset of E0 of size 2ω into E1.

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Proof. The idea of the proof is to construct E0 and E1 in steps, “killing” all ofthe possible big strictly monotone functions as we go along. The very first thingto do is to see that we can list out in a sequence of length 2ω all of the functionsto be “killed”.

For the empty set ∅ we let sup ∅ = 0, inf ∅ = 1. For any subset W of [0,1] welet cl(W ) be its topological closure in [0,1], and we let C1(W ) = {f : f : W →[0, 1], and f is strictly monotone}. For W ⊆ [0, 1] and f ∈ C1(W ) (say f strictlyincreasing) we define fcl : cl(W )→ [0, 1] by

fcl(x) =

⎧⎨⎩f(x) if x ∈ W ,

sup{f(y) : x > y ∈ W} if x /∈ W and x = sup{y ∈ W : y < x},inf{f(y) : x < y ∈ W} if x /∈ W and x �= sup{y ∈ W : y < x}.

(A similar definition is given if f is strictly decreasing.) Note that if x ∈ cl(W )\Wthen x = sup{y ∈ W : y < x} or x = inf{y ∈ W : x < y}. Now fcl is increasing.For, suppose that x, x′ ∈ cl(W ) and x < x′. If x, x′ ∈ W , then fcl(x) = f(x) <f(x′) = fcl(x

′). Suppose that x /∈ W and x′ ∈ W . If x = sup{y ∈ W : y < x},then f(y) < f(x′) for all y ∈W with y < x, and so fcl(x) ≤ f(x′) = fcl(x

′). If x �=sup{y ∈W : y < x}, then x = inf{y ∈ W : x < y}, hence fcl(x) ≤ f(x′) = fcl(x

′).Other possibilities for x and x′ are treated similarly. Now if x, y ∈ cl(W )\W ,x < y, and fcl(x) = fcl(y), then x = sup{z ∈W : z < x}, y = inf{z ∈W : y < z},and |(x, y)∩W | ≤ 1. For, if x �= sup{z ∈W : z < x}, then x = inf{z ∈ W : x < z},and so there are u, v ∈W with x 1 a contradiction is easily reached.

Next, note that for each z ∈ [0, 1] the set f−1cl [{z}] has at most three elements.

For, if it has four or more, at most one of them is in W ; so this gives three elementsw < x < y of cl(W )\W all with the same value under fcl. If x = sup{u ∈ W :u < x}, this gives infinitely many elements of W between w and x, contradiction;similarly if x = inf{u ∈W : x < u.

For any W ⊆ [0, 1] let C2(W ) be the set of all functions f : W → I such that

(1) f is monotone.

(2) f−1[{y}] is finite for all y ∈ [0, 1],

(3) |{x ∈ W : f(x) �= x}| = 2ω.

Thus by the above, fcl ∈ C2(cl(W )) whenever f ∈ C1(W ), |W | = 2ω, and f(x) �= xfor all x ∈ W .

Now let W be a closed subset of [0, 1]. Choose a countable dense subset F1 ofW (pick wrs ∈ (r, s)∩W for each pair r < s of rationals such that (r, s)∩W �= 0,and let F1 be the set of all such elements wrs). Now take any member f of C2(W ).Let

F f2 = {x ∈ W : sup{f(y) : x ≥ y ∈ F1} < inf{f(y) : x ≤ y ∈ F1}}.

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If x ∈ F f2 , then x /∈ F1. Now for each x ∈ F f

2 let Ux = (sup{f(y) : x ≥ y ∈F1}, inf{f(y) : x ≤ y ∈ F1}). This is a nonempty open interval, and if x, y ∈ F f

2

with x < y we can choose z ∈ F1 such that x < z < y, and then inf{f(u) : x ≤u ∈ F1} ≤ f(z) ≤ sup{f(u) : y ≥ u ∈ F1}, showing that Ux ∩ Uy = ∅. so that

Ux ∩ Uy = 0 for x �= y. Therefore, F f2 is countable.

Now if f, g ∈ C2(W ), F f2 = F g

2 , and f � (F1 ∪ F f2 ) = g � (F1 ∪ F g

2 ), then

f = g. For, suppose that x ∈W\(F1 ∪ F f2 ). Then

f(x) = sup{f(y) : x ≥ y ∈ F1} = sup{g(y) : x ≥ y ∈ F1} = g(x).

From these considerations it follows that |C2(W )| ≤ 2ω. Also recall that there arejust 2ω closed sets, since every closed set is the closure of a countable dense subset.Hence the set

C =⋃{C2(F ) : F ⊆ [0, 1], F closed, |F | = 2ω}

has cardinality ≤ 2ω. Let 〈fα : α < 2ω〉 be an enumeration of C. Let h be a strictlydecreasing function from R onto (0,1); thus h−1 is also strictly decreasing. (Forexample, let h(x) = 1/(ex + 1) for all x ∈ R.) Moreover, fix a well-ordering of R.

Now we construct by induction pairwise disjoint subsets Aα of [0, 1] for α <2ω. At the end we will let E0 be the union of the Aα with even α and E1 be theunion of the rest. We will carry along the inductive hypothesis that |α| ≤ |Aα| ≤|α| + ω. Let A0 and A1 be denumerable disjoint subsets of [0,1] which are densein [0,1].

Now suppose that Aα has been constructed for all α < β, where β ≥ 2.Let Bβ =

⋃α<β Aα and B∗

β = Bβ ∪⋃

α<β fα[Bβ] ∪⋃

α<β f−1α [Bβ ]. Note by our

assumptions that |β| ≤ |B∗β | ≤ |β|+ ω. For every real number r, let

Cr = h[{r + h−1(b) : b ∈ B∗β}].

We claim that there is an r ∈ R such that Cr ∩ B∗β = 0. Suppose not. For every

r ∈ R choose br ∈ Cr ∩ B∗β. Since |B∗

β| < 2ω, there exist a set S ⊆ R and anelement c ∈ B∗

β such that |S| > B∗β and br = c for all r ∈ S. Say c = h(d) with

d ∈ {r + h−1(x) : x ∈ B∗β} for all r ∈ S. Thus h(d − r) ∈ B∗

β for all r ∈ S. Sothere exist distinct r, s ∈ S such that h(d− r) = h(d−s). This contradicts h beingone-one.

Finally, let r be the least real number (in the well-ordering fixed above) suchthat Cr∩B∗

β = 0, and let Aβ = Cr. Clearly the inductive hypothesis remains true.This finishes the construction of the sets Aα, α < 2ω.

Let E0 =⋃

α evenAα and E1 =⋃

α oddAα. So E0 and E1 are disjoint subsetsof [0,1], and both of them are dense in [0,1]. Since all of the sets Aα are non-empty,it is clear that both E0 and E1 are of power 2ω.

Now suppose that g is a strictly monotone function from a subset of E0 ofpower 2ω into E1. Say gcl = fα. Now |

⋃β≤αAα| < 2ω, so choose y ∈ rng(g) such

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that y ∈ Aγ for some γ > α. Say g(x) = y with x ∈ Aδ. Now δ is even and γ isodd. If δ < γ, then y ∈ B∗

γ , so y ∈ Aγ is a contradiction. If γ < δ, then x ∈ B∗δ , so

x ∈ Aδ is a contradiction. Thus (ii) of the lemma has been verified.

If ω ≤ κ < 2ω, choose β < 2ω odd with β > κ. Say Aβ = Cr, as in thedefinition. Now b !→ h(r + h−1b) is an increasing mapping from B∗

β into Aβ , andE0 ∩B∗

β has at least κ elements. This verifies (i). �

The example also depends upon the following lemma, which will also be usefullater on.

Lemma 3.47. Let A be the interval algebra on R. Then there does not exist in A astrictly increasing sequence 〈Iα : α < ω1〉 of ideals.

Proof. Suppose that there is such a sequence. For each α < ω1 define r ≡α s iffr, s ∈ R and either r = s or else if, say, r < s, then [r, s) ∈ Iα. Then ≡α is anequivalence relation on R and the equivalence classes are intervals. For each r ∈ R

the left endpoints of the intervals [r]α are decreasing for increasing α, and the rightendpoints, increasing ([r]α denotes the equivalence class of r under the equivalencerelation ≡α). Since there is no strictly monotone sequence of real numbers of typeω1, there is an ordinal βr < ω1 such that both the left and right endpoints of [r]αare constant for α > βr. Let γ = sup{βr : r rational}. Then all of the equivalenceclasses are constant for α > γ, contradiction. �Corollary 3.48. Let A be a subalgebra of the interval algebra on R. Then A doesnot have an uncountable ideal independent subset.

Proof. Suppose that X is an uncountable ideal independent subset of A. Let 〈aα :α < ω1〉 be a one-one enumeration of some elements of X . For each α < ω1 letIα = 〈{aβ : β < α}〉Id. Clearly then 〈Iα : α < ω1〉 is a strictly increasing sequenceof ideals in B, contradicting 3.47. �

Finally, we are ready for the example. The main content of the example is fromTodorcevic [86], Theorem 12, as we mentioned.

Theorem 3.49. There is a BA A of power 2ω such that:

(i) Ult(A) has, for each κ < 2ω, a discrete subspace of power κ, and A has anatomic homomorphic image B with κ atoms;

(ii) Ult(A) has no discrete subspace of power 2ω;

(iii) If B is any homomorphic image of A, then there is a dense subset X of Bsuch that we can write X = W ∪

⋃i∈ω Zi with W the set of all atoms of B

and for each i ∈ ω, the set Zi has the finite intersection property.

Proof. Let E0 and E1 be as in Lemma 3.46. Without loss of generality, 0, 1 /∈E0 ∪ E1. For i < 2 let Ki be the linearly ordered set obtained from [0,1] byreplacing each element r ∈ Ei by two new points r− < r+. Taking the ordertopology on Ki, we obtain a Boolean space. In fact, Ki is homeomorphic to theStone space of the interval algebra on Ei ∪ {0}. Namely, the following function

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f from Ult(Intalg(Ei ∪ {0})) into Ki is the desired homeomorphism. Take anyF ∈ Ult(Intalg(Ei ∪ {0})). Let r = inf{a ∈ Ei : [0, a) ∈ F}; so r ∈ [0, 1]. If r ∈ Ei

and [0, r) ∈ F , let f(F ) = r−; if r ∈ Ei and [0, r) /∈ F , let f(F ) = r+; and ifr /∈ Ei let f(F ) = r. Clearly f is one-one and maps onto Ki. To show that itis continuous, we first note that the following clopen subsets of Ki constitute asubbase for its topology:

(∗) {[0, s+) : s ∈ Ei} ∪ {(r−, 1] : r ∈ Ei}.

In fact, it suffices to show that each nonempty open interval of Ki is a union offinite intersections of elements of (∗). Let N = {r− : r ∈ Ei} ∪ {r+ : r ∈ Ei}. Weconsider several cases:

[0, s+) is in (∗);

[0, s−) =⋃

s>t∈Ei

[0, t+);

[0, b) =⋃

b>s∈Ei

[0, s+) for b /∈ N ;

(r+, 1] =⋃

r<s<1,s∈Ei

(s−, 1];

(r−, 1] is in (∗);

(b, 1] =⋃

b<s<1,s∈Ei

(s−, 1] forb /∈ N ;

(r−, s−) =⋃

r<t<s,t∈Ei

([0, t+) ∩ (r−, 1]);

(r−, s+) = (r−, 1] ∩ [0, s+);

(r+, s−) =⋃

r<t<u<s,t,u∈Ei

((t+, 1] ∩ [0, u+);

(r+, s+) =⋃

r<t<s,t∈Ei

((t−, 1] ∩ [0, s+));

(a, r−) =⋃

a<t<u<rt,u∈Ei

([0, u+) ∩ (t−, 1]) for a /∈ N ;

(a, r+) =⋃

a<t<r,t∈Ei

((t−, 1] ∩ [0, r+)) for a /∈ N ;

(r−, a) =⋃

r<t<a,t∈Ei

((r−, 1] ∩ [0, t+)) for a /∈ N ;

(r+, a) =⋃

r<t<u<at,u∈Ei

([0, u+) ∩ (t−, 1]) for a /∈ N ;

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(a, b) =⋃

a<t<u<bt,u∈Ei

([0, u+) ∩ (t−, 1]) for a, b /∈ N.

Next we claim thatf−1[[0, s+)] = {F : [0, s) ∈ F};f−1[(r−, 1)] = {F : [r, 1] ∈ F}.

To prove these equations, let F be any ultrafilter on Intalg(Ei ∪ {0}). Definet = inf{a ∈ Ei : [0, a) ∈ F}. We now consider three cases.

Case 1. t ∈ Ei and [0, t) ∈ F . Hence f(F ) = t−, and

F ∈ f−1[[0, s+)] iff t− ∈ [0, s+)

iff t ≤ s

iff [0, s) ∈ F ;

F ∈ f−1[(s−, 1]] iff t− ∈ (s−, 1]iff s < t

iff [s, 1) ∈ F.

Case 2. t ∈ Ei and [0, t) /∈ F . Hence f(F ) = t+, and

F ∈ f−1[[0, s+)] iff t+ ∈ [0, s+)

iff t < s

iff [0, s) ∈ F ;

F ∈ f−1[(s−, 1]] iff t+ ∈ (s−, 1]iff s ≤ t

iff [s, 1) ∈ F.

Case 3. t /∈ Ei. Hence f(F ) = t, and

F ∈ f−1[[0, s+)] iff t ∈ [0, s+)

iff t < s

iff [0, s) ∈ F ;

F ∈ f−1[(s−, 1]] iff t ∈ (s−, 1]iff s < t

iff [s, 1) ∈ F.

This completes the proof that f is a homeomorphism from Ult(Intalg(Ei ∪ {0}))onto Ki.

By Corollary 3.48, neither K0 nor K1 has an uncountable discrete subspace.Also, K0×K1 is a Boolean space, and we let A be the BA of closed-open subsetsof it.

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First we check that for any κ < 2ω, K0 ×K1 has a discrete subset of powerκ. Let f be a strictly increasing function from a subset of E0 of power κ into E1.

Then we claim that Ddef= {(r−, (f(r))+) : r ∈ dmn(f)} is discrete. To show this,

for each r ∈ dmn(f) let ar = [0, r+) × ((f(r))−, 1]. Suppose (s−, (f(s))+) ∈ arand s �= r. Thus s− < r− and (f(r))+ < (f(s))+, contradiction.

From the proofs of 3.29 and 3.30 it now follows that A has a homomorphicimage C which has a disjoint subset of power κ. By an easy application of theSikorski extension theorem, A has an atomic homomorphic image B with κ atoms.

Next we prove (ii). Suppose that D is a discrete subspace of K0 × K1 ofsize 2ω. Now K1 has no uncountable discrete subspace, so for each x ∈ dmn(D),the set {y : (x, y) ∈ D} is countable. It follows that we may assume that D is afunction. Similarly, we may assume that D is one-one.

For (r, s) ∈ D let ars and brs be open intervals in K0 and K1 respectivelysuch that (ars × brs) ∩ D = {(r, s)}. Let F0 and F1 be countable dense subsetsof K0 and K1 respectively (in the sense that if a < b in K0 and (a, b) �= 0then there is a c ∈ F0 such that a < c < b; similarly for K1). Suppose thatdmn(D)\({r− : r ∈ E0}∪ {r+ : r ∈ E0}) has power 2ω. Then we may successivelyassume that dmn(D) ∩ ({r− : r ∈ E0} ∪ {r+ : r ∈ E0})=0, that each ars is anopen interval with endpoints in F0, and that all of the ars are equal, which impliesthat D has only one element, contradiction.

Thus we may assume that dmn(D) ⊆ {r− : r ∈ E0} ∪ {r+ : r ∈ E0}, andsimilarly for rng(D). Hence we may assume that there are ε, δ ∈ {−,+} suchthat dmn(D) ⊆ {rε : r ∈ E0} and rng(D) ⊆ {rδ : r ∈ E1}. Further, we mayassume that there exist qi ∈ Fi, i = 0, 1, such that q0 ∈ aεδ and q1 ∈ bεδ for each(rε, sδ) ∈ D. Finally, we may assume that one of the following four cases holds:

(1) ε = −, δ = −, and for each (r−, s−) ∈ D, the right endpoint of ar−s− is r+,and the right endpoint of br−s− is s+.

(2) ε = −, δ = +, and for each (r−, s+) ∈ D, the right endpoint of ar−s+ is r+,and the left endpoint of br−s+ is s−.

(3) ε = +, δ = −, and for each (r+, s−) ∈ D, the left endpoint of ar+s− is r−, andthe right endpoint of br+s− is s+.

(4) ε = +, δ = +, and for each (r+, s+) ∈ D, the left endpoint of ar+s+ is r−, andthe left endpoint of br+s+ is s−.

Next,

(5) If (1) or (2) holds, (r−, sδ), (u−, vδ) ∈ D, and r < u, then r− ∈ au−vδ .

In fact, since q0 ∈ ar−sδ and ar−sδ has right endpoint r+, we have q0 ≤ r− < u−,and then q0 ∈ au−vδ implies that r− ∈ au−vδ .

By symmetry we have

(6) If (1) or (3) holds (rε, s−), (uε, v−) ∈ D, and s < v, then s− ∈ buεv− .

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(7) If (3) or (4) holds, (r+, sδ), (u+, vδ) ∈ D, and r < u, then u+ ∈ ar+sδ .

In fact, since q0 ∈ au+vδ and au+vδ has left endpoint u−, we have r+ < u+ ≤ q0,and then q0 ∈ ar+sδ implies that u+ ∈ ar+sδ .

Again by symmetry we have

(8) If (2) or (4) holds, (rε, s+), (uε, v+) ∈ D, and s < v, then v+ ∈ brεs+ .

Now we obtain a contradiction, as follows. If (1) holds, then the mapping r !→ s iff(r−, s−) ∈ D is strictly decreasing, as otherwise we would get (r−, s−), (u−, v−) ∈D with r < s and u < v, and then (5) and (6) would give (r−, s−) ∈ (au−v− ×bu−v−), contradiction. But then Lemma 3.46(ii) is contradicted. Similarly, (4) givesa strictly decreasing function, while (2) and (3) give strictly increasing functions.

Now we turn to the last part of the theorem. Suppose that B is a homomorphicimage of A. Let Fi be a countable dense subset of Ei for i = 0, 1. Let E+

i = {r+ :r ∈ Ei}, E−

i = {r− : r ∈ Ei} for i = 0, 1. Now we are going to define some subsetsX ...

... of B indexed by various objects in countable sets; each subset will satisfy thefinite intersection property, and this will be obvious in each case. What is not soobvious is what these sets are good for. We show after defining them that theirunion with the set of atoms of B is dense in B, which is the desired conclusionof the theorem. It is convenient to work with the dual of B, which is some closedsubspace Y of K0 ×K1.

Suppose that p, q ∈ F0, r, s ∈ F1, p < q, r < s, and ([p+, q−]× [r+, s−])∩Y �=0; then we set

X1pqrs = {([p+, q−]× [r+, s−]) ∩ Y }.

Next, suppose that q ∈ F0, r, s ∈ F1, and r < s. Then we set

X2qrs = {([x, q−]× [r+, s−]) ∩ Y : x ∈ E+

0 , x < q−, and

∃y(r+ < y < s− and (x, y) ∈ Y )}.

The next three sets are similar to X2qrs. Suppose that p ∈ F0, r, s ∈ F1, and r < s.

SetX3

prs = {([p+, x]× [r+, s−]) ∩ Y : x ∈ E−0 , p+ < x,

and ∃y(r+ < y < s− and (x, y) ∈ Y )}.Suppose that p, q ∈ F0, s ∈ F1, and p < q. Set

X4pqs = {([p+, q−]× [y, s−]) ∩ Y : y ∈ E+

1 , y < s−,

and ∃x(p+ < x < q− and (x, y) ∈ Y )}.

Suppose that p, q ∈ F0, r ∈ F1, and p < q. Set

X5pqr = {([p+, q−]× [r+, y]) ∩ Y : y ∈ E−

1 , r+ < y,

and ∃x(p+ < x < q− and (x, y) ∈ Y )}.

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Now suppose that p, q ∈ F0, r, s ∈ F1, p < q, and r < s. Set

X6pqrs = {([x, q−]× [y, s−]) ∩ Y : x ∈ E+

0 , y ∈ E+1 , x < p−,

y < r−, and ([p+, q−]× [r+, s−]) ∩ Y �= 0}.

The next three sets are similar to X6pqrs. For each of them we suppose that p, q ∈

F0, r, s ∈ F1, p < q, and r < s.

X7pqrs = {([x, q−]× [r+, y]) ∩ Y : x ∈ E+

0 , y ∈ E−1 , x < p−,

s+ < y, and ([p+, q−]× [r+, s−]) ∩ Y �= 0}.

X8pqrs = {([p+, x]× [y, s−]) ∩ Y : x ∈ E−

0 , y ∈ E+1 , q+ < x,

y < r−, and ([p+, q−]× [r+, s−]) ∩ Y �= 0}.

X9pqrs = {([p+, x]× [r+, y]) ∩ Y : x ∈ E−

0 , y ∈ E−1 , q+ < x,

s+ < y, and ([p+, q−]× [r+, s−]) ∩ Y �= 0}.Next, if p ∈ F0 and r, s ∈ F1 with r < s, we set

X10prs ={([p+, x]× [r+, y]) ∩ Y : x ∈ E−

0 , y ∈ E−1 , s+ < y, p < x,

and there is a v such that (x, v) ∈ Y and r+ < v < s−}.

The other sets are similar to this one; with obvious assumptions,

X11prs ={([p+, x]× [y, s−]) ∩ Y : x ∈ E−

0 , y ∈ E+1 , y < r−, p < x,


X12qrs ={([x, q−]× [r+, y]) ∩ Y : x ∈ E+

0 , y ∈ E−1 , s+ < y, x < q,


X13qrs ={([x, q−]× [y, s−]) ∩ Y : x ∈ E+

0 , y ∈ E+1 , y < r−, x < q,


X14pqs ={([x, q−]× [y, s−]) ∩ Y : x ∈ E+

0 , y ∈ E+1 , x < p−, y < s,

and there is a u such that (u, y) ∈ Y and p+ < u < q−}.

X15pqr ={([x, q−]× [r+, y]) ∩ Y : x ∈ E+

0 , y ∈ E−1 , x < r−, r < y,


X16pqs ={([p+, x]× [y, s−]) ∩ Y : x ∈ E−

0 , y ∈ E+1 , q+ < x, y < s,


X17pqr ={([p+, x]× [r+, y]) ∩ Y : x ∈ E−

0 , y ∈ E−1 , q+ < x, r < y,


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Now we show that the union of these sets with the set of atoms of B is dense in B.Suppose that U is a non-zero element of B; we may assume that U has the form((a, b)× (c, d)) ∩ Y , and that it is not ≥ any atom of B. Fix an element (x, y) ofU . We consider various possibilities.

Case 1. x /∈ E−0 ∪ E+

0 and y /∈ E−1 ∪ E+

1 . Then clearly there exist p, q, r, s suchthat (x, y) ∈ Z ⊆ U with Z ∈ X1

pqrs.

Case 2. x ∈ E−0 and y /∈ E−

1 ∪E+1 . There are p, r, s such that (x, y) ∈ Z ⊆ U with

Z ∈ X3prs.

Case 3. x ∈ E+0 and y /∈ E−

1 ∪E+1 . There are q, r, s such that (x, y) ∈ Z ⊆ U with

Z ∈ X2qrs.

Case 4. x /∈ E−0 ∪ E+

0 . Similar to above cases, using X1..., X

4..., or X

5....

Case 5. x ∈ E−0 and y ∈ E−

1 . Then it is easy to find p ∈ F0, r ∈ F1 so that(x, y) ∈ [p+, x]× [r+, y] ∩ Y ⊆ U . Now there are two subcases.

Subcase 5.1. There is a (u, v) ∈ Y such that p+ < u < x and r+ < v < y. Then itis easy to find q, s so that (x, y) ∈ Z ⊆ U and Z ∈ X9

pqrs.

Subcase 5.2. Otherwise, since we are assuming that U is not ≥ any atom of B,either there is a v such that (x, v) ∈ Y and r+ < v < y, or there is a u suchthat (u, y) ∈ Y and p+ < u < x. In the first instance there is an s such that(x, y) ∈ Z ⊆ U with Z ∈ X10

prs. In the second instance we use X17.

Case 6. x ∈ E−0 and y ∈ E+

1 . This is like Case 5. We use X8, X16, and X11.

Case 7. x ∈ E+0 and y ∈ E−

1 . This is like Case 5. We use X7, X15, and X12.

Case 8. x ∈ E+0 and y ∈ E+

1 . This is like Case 5. We use X6, X14, and X13. �Corollary 3.50. Assume that 2ω is a limit cardinal. Then there is a BA A of power2ω with spread 2ω not attained, such that A has no homomorphic image B suchthat c(B) = s(A).

Proof. The first part of the conclusion follows immediately from the theorem. Nowsuppose that B is a homomorphic image of A such that c(B) = s(A). Since s(A) isnot attained, it follows from Corollary 3.31 that c(B) is not attained. Now let X ,W , etc., be as in (iii) of the theorem. Then |W | < 2ω since c(B) is not attained.Let Z be a disjoint subset of B of power |W |+. For each z ∈ Z choose xz ∈ Xsuch that xz ≤ z, and let X ′ = {xz : z ∈ Z}. Then there has to exist an i ∈ ωsuch that |X ′ ∩ Zi| > 2, which is a contradiction, since X ′ is disjoint and Zi hasthe finite intersection property. �

Returning to the program described in the introduction, we note that it is obviousthat cSs(A) = [ω, c(A)]. The caliber notion associated with cellularity has beenworked on a lot. There are several variants of this notion. For a survey of resultsand problems, see Comfort, Negrepontis [82].

We shall compare c with other cardinal functions one-by-one in the discussionof those functions.

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We turn to the relation cSr; see the end of the introduction. We do not havea purely cardinal number characterization of this relation; this is Problem 3 inMonk [96].

Problem 13. Give a purely cardinal number characterization of cSr.

Some restrictions to put on cSr are given in the following simple theorem:

Theorem 3.51. For any infinite BA A the following conditions hold:

(i) If (κ, λ) ∈ cSr(A), then κ ≤ λ ≤ |A| and κ ≤ c(A).

(ii) For each κ ∈ [ω, c(A)] we have (κ, κ) ∈ cSr(A).

(iii) If (κ, λ) ∈ cSr(A) and κ ≤ μ ≤ λ, then (κ, μ) ∈ cSr(A).

(iv) If (λ, (2κ)+) ∈ cSr(A) for some λ ≤ κ, then (ω, (2κ)+) ∈ cSr(A).

(v) (c(A), |A|) ∈ cSr(A).

(vi) If ω ≤ λ ≤ |A| then (κ, λ) ∈ cSr(A) for some κ. �

The proof of this theorem is easy; for (iv), use Theorem 10.1 of Part I of theHandbook. Another general theorem about cSr is as follows; this is Theorem 14 ofMonk, Nyikos [97].

Theorem 3.52. For every infinite cardinal κ, and every BA A, if c(A) ≥ κ++ and(κ, κ++) ∈ cSr(A), then (κ+, κ++) ∈ cSr(A).

Proof. Suppose not. Let B be a subalgebra of size κ++ with cellularity κ.

(1) There is an a ∈ A such that B � a has cellularity κ++.

(Recall here that B � a = {b · a : b ∈ B}; we do not assume that a ∈ B.) To prove(1), let X ⊆ A be pairwise disjoint of size κ+. Then 〈B ∪ X〉 is of size κ++ andhas cellularity greater than κ, so by supposition its cellularity is κ++; let Y be apairwise disjoint subset of size κ++. We may assume that each element y ∈ Y hasthe form y = by · ay with by ∈ B and ay ∈ 〈X〉. Since |X | < κ++, we may in factsuppose that each ay is equal to some element a, as desired in (1).

Choose such an a, and let X ∈ [B]κ++

be such that 〈x ·a : x ∈ X〉 is a systemof nonzero pairwise disjoint elements. Let Y be a subset of X of size κ+, and let

C = 〈{x · a : x ∈ Y } ∪ {x · −a : x ∈ X\Y }〉.

Now define x ≡ y iff x, y ∈ X\Y and x · −a = y · −a. Then

(2) Every ≡-class has size at most κ.

For, suppose that |x/ ≡ | > κ. For any y ∈ (x/ ≡)\{x} we have

y · −x = y · −x · a+ y · −x · −a= y · a · −(x · a) + x · −x · −a= y · a.

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This means that B has a pairwise disjoint subset of size greater than κ, contra-diction. So (2) holds.

From (2) it follows that |C| = κ++. Since clearly c(C) ≥ κ+, by suppositionwe must have c(C) = κ++. Now each element of C has the form

∑i<m ui · vi with

each ui ∈ 〈{x · a : x ∈ Y }〉 and each vi ∈ 〈{x · −a : x ∈ X\Y }〉. Hence we mayassume that in a disjoint subset of C of size κ++ each element has the form u · v,with u ∈ 〈{x · a : x ∈ Y }〉 and v ∈ 〈{x · −a : x ∈ X\Y }〉. Hence clearly there is ad ∈ 〈{x · a : x ∈ Y }〉 and a

Z ∈ [〈{x · −a : x ∈ X\Y }〉]κ++

such that 〈z · d : z ∈ Z〉 is a system of nonzero pairwise disjoint elements. We mayassume that each z ∈ Z has the form

xz,0 · −a · . . . · xz,m−1 · −a· (−yz,0 + a) · . . . · (−yz,n−1 + a),

where each xz,i and yz,j is in X\Y , and m and n do not depend on z.

Now since 〈{x · a : x ∈ Y }〉 is isomorphic to Finco(κ+), there are two cases.

Case 1. d =∑

x∈F (x · a) for some finite F ⊆ Y . In this case we have m = 0, andthen each z · d is just equal to d, contradiction.

Case 2. d = −∑

x∈F (x · a) for some finite F ⊆ Y . Thus d = −a+ a · −∑

x∈F x.Note that

(3) a · −∑

x∈F x �= 0.

In fact, choose y ∈ Y \F . Then

a · y ·∑x∈F

x = a · y ·∑x∈F

a · x = 0;

hence 0 �= a · y ≤ a · −∑

x∈F x �= 0, proving (3).

If m = 0, then each z · d is ≥ a · −∑

x∈F x, so the elements z · d are notdisjoint by (3), contradiction. Thus m > 0. Hence z · d = z ≤ −a for each z ∈ Z.For each z ∈ Z write ez = xz,0 · . . . · xz,m−1 and cz = ez · −yz,0 · . . . · −yz,n−1. Soz = cz · −a for each z ∈ Z. Note that if z, w ∈ Z and ez �= ew then ez · ew · a = 0.Define z ∼= w iff z, w ∈ Z and ez = ew. If z �∼= w, then ez �= ew, hence z �= w and

cz · cw = cz · cw · a+ cz · cw · −a = z · w = 0.

Since cz ∈ B for each z ∈ Z, it follows that there are at most κ ∼=-classes. So,some ∼=-class has κ++ members. Thus we may assume that all of the ez’s are thesame. Thus for any z ∈ Z we have

z = x0 · . . . · xm−1 · −yz,0 · . . . · −yz,n−1 · −a,cz = x0 · . . . · xm−1 · −yz,0 · . . . · −yz,n−1.

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Note that

cz · a = (x0 · a) · . . . · (xm−1 · a) · −(yz,0 · a) · . . . · −(yz,n−1 · a).

It follows that either cz · a = 0, or else m = 1 and cz · a = x0 · a. Thus we mayassume that either cz · a = 0 for all z ∈ Z, or cz · a = x0 · a for all z ∈ Z. Hencecz · a · −(cw · a) = 0 for all distinct z, w ∈ Z. So if z �= w, then

cz · −cw = cz · −cw · a+ cz · −cw · −a= cz · a · −(cw · a) + cz · −a · −(cw · −a)= z · −w = z.

So if we fix w ∈ Z, then 〈cz ·−cw : z ∈ Z\{w}〉 is a system of κ++ nonzero pairwisedisjoint elements of B, a contradiction. �

To understand more about the possibilities for the relation cSr(A), consider thefollowing examples. If κ is an infinite cardinal and A is the finite-cofinite algebra onκ, then cSr(A) = {(λ, λ) : λ ∈ [ω, κ]}. If A is the free algebra on κ free generators,then cSr(A) = {(ω, λ) : λ ∈ [ω, κ]}. If A is an infinite interval algebra and weassume GCH, then cSr(A) does not have any gaps of size 2 or greater. That is, if(κ, λ) ∈ cSr(A), then λ = κ or λ = κ+. This is seen by using Theorem 10.1 of theHandbook again: such a gap would imply the existence in A of an uncountableindependent subset, which does not exist in an interval algebra. Without CH therecan be gaps; for example with Intalg(R).

There are two deeper results:

(1) Todorcevic in [87a] shows that it is consistent (namely, it follows from V=L)to have for each regular non-weakly compact cardinal κ a κ-cc interval algebraA of size κ such that any subalgebra or homomorphic image B of A of size < κhas a disjoint family of size |B|. His proof is based on the following result fromTodorcevic [81] (Theorem 4.9 there):

(∗) Assume V = L. Let κ be a regular uncountable non-weakly compact cardinal.Then there exists a κ-Suslin tree with no λ-Aronszajn nor ν-Cantor subtrees, forevery λ �= κ and every ν.

Here “subtree” merely means a subset considered as a tree under the inducedordering; it is not assumed to be closed downwards.

Applying this result to subalgebras and to non-limit cardinals, this means inour terminology that it is consistent to have an algebra A with cSr(A) = {(λ, λ) :λ ∈ [ω, κ]} ∪ {(κ, κ+)}.

(2) In models of Kunen [78] and Foreman, Laver [88], every ω2-cc algebra of sizeω2 contains an ω1-cc subalgebra of size ω1. Thus in these models certain relationscSr are ruled out; cf. (1).

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Nowwe survey what we know about cSr for BAs of size≤ ω2. Note that there are sixpairs here: (ω, ω), (ω, ω1), (ω, ω2), (ω1, ω1), (ω1, ω2), and (ω2, ω2). The range of anyrelation cSr(A) consists of all infinite cardinals≤ |A|. We go through all possibilitiesby the number of pairs, and for equal numbers of pairs, lexicographically, skippingthose possibilities ruled out by simple considerations.

(S1) cSr(A) = {(ω, ω), (ω, ω1)} for A = Fr(ω1).

(S2) cSr(A) = {(ω, ω), (ω1, ω1)} for A = Finco(ω1).

(S3) cSr(A) = {(ω, ω), (ω, ω1), (ω, ω2)} for A = Fr(ω2).

(S4) cSr(A) = {(ω, ω), (ω, ω1), (ω1, ω1)} for Fr(ω1)× Finco(ω1).

(S5) cSr(A) = {(ω, ω), (ω1, ω1), (ω1, ω2)} for the algebra A of (1). Note that in themodels mentioned in (2), such a value for cSr is not possible.

(S6) cSr(A) = {(ω, ω), (ω1, ω1), (ω2, ω2)} for A = Finco(ω2).

(S7) cSr(A) = {(ω, ω), (ω, ω1), (ω1, ω1), (ω1, ω2)} for the following BA A, assumingCH. Let

L = ω12\{f ∈ ω12 : ∃α < ω1(fα = 0 and ∀β > α(fβ = 1))},M = {f ∈ ω12 : ∃α < ω1(fα = 1 and ∀β > α(fβ = 0))}.

We take the lexicographic order on L and M . Clearly M is dense in L, and |M | =ω1 by CH. Let L′ be a subset of L of size ω2 which contains M . Then A

def=

Intalg (L′) is as desired. For, by the denseness of M it has ω2-cc, and it clearly hasdepth ω1, and hence cellularity ω1; so (ω1, ω2) ∈ cSr(A). We have (ω, ω2) /∈ cSr(A)by Theorem 10.1 of Part I of the Handbook. Obviously (ω1, ω1) ∈ cSr(A). Theordered set L′′ constructed from ω2 similarly to L′ from ω12 has size ω1 and adense subset of size ω. Then Intalg (L′′) is isomorphic to a subalgebra of Intalg (L′)by Remark 15.2 of the BA Handbook, so (ω, ω1) ∈ cSr(A). It is a problem to getan example as in (S7) without CH. This is Problem 5(i) in Monk [96].

Problem 14. Can one prove in ZFC that there is a BA A with

cSr(A) = {(ω, ω), (ω, ω1), (ω1, ω1), (ω1, ω2)}?

(S8) cSr(A) = {(ω, ω), (ω, ω1), (ω1, ω1), (ω2, ω2)} for A = Finco(ω2) × Fr(ω1). Toprove this, it suffices to show that any subalgebra B of A of size ω2 has cellularity

ω2. Now Cdef= {x ∈ Finco(ω2) : ∃y(x, y) ∈ B} is a subalgebra of Finco(ω2) of size

ω2, and hence there is a system 〈cα : α < ω2〉 of nonzero disjoint elements of C.Say (cα, dα) ∈ B for all α < ω2. Now there are only ω1 possibilities for the dα’s, sowlog we may assume that they are all equal. Then (−c0,−d0) · (cα, dα) = (cα, 0)for all α > 0, giving a disjoint family of size ω2.

(S9) cSr(A) = {(ω, ω), (ω1, ω1), (ω1, ω2), (ω2, ω2)} for A = B×Finco(ω2), where Bis the algebra of (1), assuming V = L. For, (ω, ω2) /∈ cSr(A) by CH and Theorem

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10.1 of the Handbook, Part I. So we just need to show that (ω, ω1) /∈ cSr(A).Suppose that D is a subalgebra of A of size ω1. Let E = {b ∈ B : (b, c) ∈ D forsome c} and F = {c ∈ Finco(ω2) : (b, c) ∈ D for some b}.Case 1. |E| = ω1. By the basic property of B, let 〈eα : α < ω1〉 be a system ofnonzero disjoint elements of E. Say (eα, fα) ∈ D for all α < ω1. If some fβ iscofinite, replace 〈(eα, fα) : α < ω1〉 by 〈(eα, fα · −fβ) : α < ω1, α �= β〉; so wlog allfα are finite. Wlog the fα’s form a Δ-system, say with kernel g. Pick distinct β, γ <ω1. Note that for α �= β, γ we have (eα, fα\g) = (eα, fα) · −[(eβ, fβ) ∩ (eγ , fγ)];hence

〈(eα, fα\g) : α < ω1, α �= β, γ〉

is a system of disjoint, nonzero elements of D, as desired.

Case 2. |F | = ω1 and |E| < ω1. We may assume that each member of F is finite.Let F = {cα : α < ω1}, and choose bα ∈ B so that (bα, cα) ∈ D. We may assumethat all bα’s are the same. Let 〈cα : α ∈ M〉 be a Δ-system, with |M | = ω1. Fixα ∈M . Then 〈(bβ ·−bα, cβ ·−cα) : β ∈M\{α}〉 is a system of ω1 disjoint elements.

Note that in the models of (2), an algebra A of the sort just described is notpossible.

(S10) cSr(A) = {(ω, ω), (ω, ω1), (ω1, ω1), (ω, ω2), (ω1, ω2)} for A = Finco(ω1) ×Fr(ω2).

(S11) cSr(A) = {(ω, ω), (ω, ω1), (ω1, ω1), (ω, ω2), (ω2, ω2)} is ruled out by Theorem3.52. This answers Problem 4 in Monk [96].

(S12) cSr(A) = {(ω, ω), (ω, ω1), (ω1, ω1), (ω1, ω2), (ω2, ω2)} for B = A × Fincoω2,with A as in (S7), assuming CH; the argument for this is easy.

We do not know whether CH is really needed here; this is Problem 5(ii) inMonk [96].


cSr(A) = {(ω, ω), (ω, ω1), (ω1, ω1), (ω1, ω2), (ω2, ω2)}?

(S13) cSr(A) = {(ω, ω), (ω, ω1), (ω1, ω1), (ω, ω2), (ω1, ω2), (ω2, ω2)}, where A is thealgebra B × Finco(ω2), B the subalgebra of P(ω1) generated by the singletonsand a set of ω2 independent elements.

We turn to the relation cHr(A).

Problem 16. Give a cardinal number characterization of cHr(A).

This is Problem 6 in Monk [96].

We begin the discussion of cHr(A) with a general theorem. Part (v) of this theoremis due to Piotr Koszmider.

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(i) If (κ, λ) ∈ cHr(A), then κ ≤ λ ≤ |A| and κ ≤ s(A).

(ii) For each κ ∈ [ω, s(A)) there is a λ ≤ 2κ such that (κ, λ) ∈ cHr(A).

(iii) If (λ, (2κ)+) ∈ cHr(A) for some λ ≤ κ, then (ω, (2κ)+) ∈ cHr(A).

(iv) (c(A), |A|) ∈ cHr(A).

(v) If (κ′, λ′) ∈ cHr(A), where κ′ is a successor cardinal or a singular cardinaland κ′ < cf(|A|), then there is a κ′′ ≥ κ′ such that (κ′′, |A|) ∈ cHr(A).

Proof. Only (ii) and (v) need need proofs. For (ii), let κ ∈ [ω, s(A)). Take a homo-morphic image B of A such that c(B) > κ; let C be a subalgebra of B generatedby a disjoint set of power κ, and extend the identity on C to a homomorphismfrom B onto a subalgebra D of C; then D is as desired.

Now we prove (v). For brevity let λ = |A|. There is nothing to prove ifκ′ = λ, so assume that κ′ < λ. Let f be a homomorphism from A onto a BA Bwith |B| = λ′ and c(B) = κ′. By the Erdos–Tarski theorem, there is a system of〈bξ : ξ < κ′〉 of nonzero disjoint elements in B. For each ξ < κ′ choose aξ ∈ A suchthat f(aξ) = bξ. We now consider two cases.

Case 1. |A � aξ| < λ for all ξ < κ′. Let J be the ideal in A generated by {aξ · aη :ξ < η < κ′}. Then

(∗) |J | < λ.

In fact, a ∈ J if and only if there is a finite set Γ of ordered pairs (ξ, η) withξ < η < κ′ such that

a ≤∑

(ξ,η)∈Γ

(aξ · aη),

and the number of such sets Γ is κ′. Take any such set Γ. Write Γ = {(ξi, ηi) : i <n}. Define ci = aξi · −

∑j<i aξj for each i < n. Then

∑i<n ci =

∑i<n aξi , and

hence ∣∣∣∣∣{a ∈ A : a ≤

∑i<n

(aξi · aηi)

}∣∣∣∣∣ ≤∣∣∣∣∣∏i<n

(A � ci)∣∣∣∣∣ < λ,

which proves (∗), since κ′ < λ.

It is also clear that aξ /∈ J for all ξ < κ′. It follows that in A/J there is asystem of κ′ nonzero disjoint elements, and |A/J | = λ, as desired.

Case 2. There is a ξ0 < κ′ such that |A � aξ0 | = λ. Then if we take the homomor-phism

A ∼= (A � aξ0)× (A � −aξ0)→ (A � aξ0)× (B � −bξ0)

determined by the identity and f � (A � −aξ0), we get a homomorphism from Aonto an algebra C of size λ and with κ′ disjoint elements. �

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We now list some additional facts concerning cHr.

(1) (Theorem 4 of Monk, Nyikos [97]) Suppose that ω ≤ ρ ≤ κ. Let A =〈[κ]≤ρ〉P(κ). Then cHr(A) = S ∪ T ∪ U , where

S = {(μ, ν) : ω ≤ μ ≤ ν ≤ 2ρ, νω = ν};T = {(μ, μρ) : 2ρ < μ ≤ κ};U = {(μ, κρ) : 2ρ < μ, μρ = κρ, κ < μ}.

Note that S, T, U are all needed here. This can be seen as follows. For ρ = ω andκ = (2ω)+ we have (ω, 2ω) ∈ S\(T ∪ U). For ρ = ω and κ = (2ω)++ we have((2ω)+, (2ω)+) ∈ T \(S ∪ U). Finally, for ρ = ω1 and κ = ω1 we have (2ω1 , 2ω1) ∈U\(S ∪ T ).

(2) As a corollary of (1), assume CH, and let A = 〈[ω2]≤ω〉P(ω2). Then

cHr(A) = {(ω, ω1), (ω1, ω1), (ω2, ω2)}.

(3) (Part of Theorem 7 of Monk, Nyikos [97]) Suppose that λ is uncountable andstrongly inaccessible, and λ ≤ κ. Let A = 〈[κ]<λ〉P(κ). Then cHr(A) = S ∪ T ∪U ∪ {(λ, λ)}, where

S = {(μ, ν) : ω ≤ μ ≤ ν < λ, νω = ν};T = {(μ, μ<λ) : λ < μ ≤ κ};U = {(μ, κ<λ) : λ < μ, μ<λ = κ<λ, κ < μ}.

The sets here all needed. In fact, (ω, 2ω) ∈ S\(T ∪ U ∪ {(λ, λ)}). For κ = λ+ wehave (λ+, λ+) ∈ T \(S ∪ U ∪ {(λ, λ)}. For U , let κ = λ+ω and μ = λ+ω+1.

(4) (Theorem 10 of Monk, Nyikos [97]) Assume GCH, and suppose that A ∈[[κ+]κ

+

]κ++

, with any two distinct members of A having intersection of size atmost κ. Let A be the κ+-complete subalgebra of P(κ+) generated by A ∪ {{α} :α < κ+}. then

cHr(A) = {(μ, ν) : ω ≤ μ ≤ ν ≤ κ+, cf(ν) > ω} ∪ {(κ+, κ++), (κ++, κ++)}.

(5) (A corollary of (4)) Under GCH there is a BA A such that

cHr(A) = {(ω, ω1), (ω1, ω1), (ω1, ω2), (ω2, ω2)}.

(6) (A result of Nyikos, based upon a consistency result of Laver; see Monk, Nyikos[97], Theorems 11, 12) It is consistent to have a BA A such that

cHr(A) = {(ω, ω), (ω1, ω1), (ω, ω2), (ω2, ω2)}.

(7) (A result of Nyikos; see Monk, Nyikos [97], Theorem 13) If (κ+, κ++) ∈ cHr(A)and (κ, κ++) /∈ cHr(A), then (κ+, κ+) ∈ cHr(A).

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(8) By Corollary 3.50, κ = s(A) is not in general possible in Theorem 3.53(ii).

(9) If A is complete and (κ, λ) ∈ cHr(A), then λω = λ.

(10) If A is the finite-cofinite algebra on an infinite cardinal κ, then cHr(A) ={(λ, λ) : ω ≤ λ ≤ κ}.

(11) If A is the free BA on κ free generators, κ infinite, then cHr(A) = {(λ, μ) :ω ≤ λ ≤ μ ≤ κ}.

(12) If A an infinite interval algebra and GCH is assumed, then there do not existcardinals κ, λ such that κ++ ≤ λ, (κ, μ), (λ, μ) ∈ cHr(A), while (ρ, μ) /∈ cHr(A) forall ρ ∈ (κ, λ)card.

(13) The algebra A of Todorcevic [87a] (assuming V = L) has cHr(A) = {(λ, λ) :λ ∈ [ω, κ]} ∪ {(κ, κ+)}. See (1) in the discussion of cSr.

(14) Under CH, cHr(P(ω)) = {(ω, ω1), (ω1, ω1)}.

(15) If 2ω = ω2 then cHr(P(ω)) = {(ω, ω2), (ω1, ω2), (ω2, ω2)}.

(16) (See Koppelberg [77]) Assuming MA, if A is an infinite BA with |A| < 2ω,then A has a countable homomorphic image. This is proved in Theorem 9.9.

(17) A special case of a result of Just, Koszmider [91] is that it is consistentto have 2ω = ω2 with an algebra A having homomorphic cellularity relation{(ω, ω1), (ω1, ω1)}.

(18) In Juhasz [92] it is shown that if κ > ω and |A| ≥ κ, then A has a homomorphicimage of size λ for some λ with κ ≤ λ ≤ 2<κ. In particular, if CH holds and|A| = ω2, then A has a homomorphic image of size ω1.

(19) Fedorchuk [75] constructed, assuming ♦, a BA A such that cHrA = {(ω, ω1)}.

(20) Koszmider [99] has modified Fedorchuk’s construction as follows. If M is ac.t.m. of ZFC and λ is a cardinal of M of uncountable cofinality with λ > ω1, thenthere is a generic extension M [G] of M preserving cardinalities and cofinalitiessuch that in M [G] we have 2ω = λ and there are BAs A, B, C, and D with thefollowing properties:

(a) hd(A) = ω, |A| = ω1, and A does not have a countably infinite homo-morphic image. (Theorem 5.5 of Koszmider [99])

(b) hd(B) = ω, |A| = λ, and all homomorphic images of A have size λ.(Theorem 5.5 of Koszmider [99])

(c) cHr(C) = {(ω, λ), (ω1, λ)}. (Theorem 6.11 of Koszmider [99])

(d) cHr(D) = {(ω, ω1), (ω1, ω1)}. (Theorem 6.11 of Koszmider [99])

Later we show that c(D) ≤ hd(D) for any BA D.

Using Proposition 1.1 we then get

(e) cHr(A×B) = {(ω, ω1), (ω, λ)}.

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(f) cHr(C ×D) = {(ω, ω1), (ω1, ω1), (ω, λ), (ω1, λ)}.(g) cHr(A× C) = {(ω, ω1), (ω, λ), (ω1, λ)}.

The following interesting result of Koszmider was used to construct the algebra Cabove.

Proposition 3.54. Suppose that cHr(A) = {(ω, λ)} with λ ≥ ω2. Then there is aBA E such that cHr(E) = DepthHr(E) = {(ω, λ), (ω1, λ)}.

Proof. By Sikorski’s extension theorem we may assume that Finco(ω) ≤ A ≤P(ω). Let E =

∏Ac A, using the notation concerning moderate products in Chap-

ter 1. It is easy to see that |E| = λ. Hence by Proposition 1.10, every homomorphicimage of E has size λ. By Lemma 1.12(i), Depth(E) ≥ ω1. So it remains only toprove that E does not have an ideal L such that E/L has a disjoint set of sizegreater than ω1. Suppose that L is such an ideal, with X a subset of E such that〈[x]L : s ∈ X〉 is a one-one system of nonzero pairwise disjoint elements of E/L,with |X | = ω2. By Lemma 1.11, let Y ∈ [X ]ω2 and n ∈ ω be such that 〈Y 〉E canbe isomorphically embedded in nA. By Sikorski’s extension theorem, 〈Y 〉/L canbe isomorphically embedded in a homomorphic image F of nA. By Proposition1.1, F is isomorphic to a finite product H of homomorphic images of A. Hencec(H) ≥ ω2, contradiction. �

Now we consider small cardinals, like we did for cSr. There are many more problemshere. The problems are of two sorts: cases in which we know that the existence ofthe appropriate BA is consistent but have no construction in ZFC, and cases inwhich we know that the existence of the appropriate BA is inconsistent, but haveno proof of non-existence in ZFC. Even if we know both the consistency of theexistence of a BA with a specified cHr and the consistency of the nonexistence ofsuch, there may be problems remaining, but we do not formulate them.

Possibilities for cHr. For the convenience of the reader we mention all of the 63a priori possibilities. As a guide through these, we arrange the six possible pairslexicographically and go through them in order of the number of pairs.

(H1) {(ω, ω)}. Any countably infinite BA works.

(H2) {(ω, ω1)}. The Fedorchuk example (19) gives this, assuming ♦. If MA+2ω >ω1, it is ruled out by Koppelberg (16).

(H3) {(ω, ω2)}. This is impossible under CH, by Juhasz (18). If MA + 2ω > ω2,it is ruled out by Koppelberg (16). A consistent example is given by Koszmider’salgebra B in (20).

(H4) {(ω1, ω1)}. Any BA has a homomorphic image of countable cellularity, sothis relation is impossible.

(H5) {(ω1, ω2)}. Impossible; see (H4).

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(H6) {(ω2, ω2)}. Impossible; see (H4).

(H7) {(ω, ω), (ω, ω1)}. A subalgebra of Intalg(R) of size ω1 gives an example. Infact let L be an unbounded subset of R of size ω1. Let A be the subalgebra ofIntalg(R) generated by {(−∞, r) : r ∈ L}. For each r ∈ L let f((−∞, r)) = [0, n),with n ∈ ω minimum such that r < n. Then f extends to a homomorphism ofIntalg(L) into an infinite subalgebra of Intalg(ω); see the Handbook, 15.2.

(H8) {(ω, ω), (ω, ω2)}. This is impossible under CH, by Juhasz (18). Assuming2ω = ω2, the algebra Intalg(R) works; see Theorem 9.13.

(H9) {ω, ω), (ω1, ω1)}. Finco(ω1) works.

(H10) {(ω, ω), (ω1, ω2)}. This is impossible, by the result of Nyikos (7).

(H11) {(ω, ω), (ω2, ω2)}. Any BA of cellularity ω2 has a homomorphic image ofcellularity ω1, so this relation is impossible.

(H12) {(ω, ω1), (ω, ω2)}. Not possible under CH, by Theorem 10.1 of the Hand-book. If MA+2ω > ω1, it is ruled out by Koppelberg’s Theorem (16). A consistentexample is given by A×B, where A and B are Koszmider’s algebras; see (20)(e).

(H13) {(ω, ω1), (ω1, ω1)}. Under CH, P(ω) works; and also the Ju example (17)works. Koppelberg’s Theorem (16) indicates that it is not possible to have suchan example in ZFC.

(H14) {(ω, ω1), (ω1, ω2)}. This is ruled out by the result of Nyikos (7).

(H15) {(ω, ω1), (ω2, ω2)}. Not possible: see (H11).

(H16) {(ω, ω2), (ω1, ω1)}. This is not possible, since the homomorphic image ofsize ω1 should have a homomorphic image which is ccc.

(H17) {(ω, ω2), (ω1, ω2)}. This is impossible under CH, by Juhasz (18). If MA +2ω > ω2, it is ruled out by Koppelberg’s Theorem (16). A consistent example isgiven by Koszmider’s algebra C; see (20).

(H18) {(ω, ω2), (ω2, ω2)}. Not possible; see (H11).

(H19) {(ω1, ω1), (ω1, ω2)}. Not possible; see (H4).



(H22) {(ω, ω), (ω, ω1), (ω, ω2)}. Not possible under CH, since then there is an un-countable independent set. Under ¬CH, a product of certain interval algebrasworks. Namely, with L0, L1, L2 subsets of R of sizes ω, ω1, ω2 respectively, eachcontaining Q, Intalg(L0)× Intalg(L1)× Intalg(L2) works.

(H23) {(ω, ω), (ω, ω1), (ω1, ω1)}. A free algebra of size ω1 works.

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(H24) {(ω, ω), (ω, ω1), (ω1, ω2)}. This is not possible, by the result of Nyikos (7).

(H25) {(ω, ω), (ω, ω1), (ω2, ω2)}. Not possible; see (H11).

(H26) {(ω, ω), (ω, ω2), (ω1, ω1)}. Ruled out by Theorem 3.53(v).

(H27) {(ω, ω), (ω, ω2), (ω1, ω2)}. This is impossible under CH, by Juhasz (18). Aconsistent example is given by C ×Finco(ω), where C is Koszmider’s algebra; see(20).

(H28) {(ω, ω), (ω, ω2), (ω2, ω2)}. Not possible; see (H11).

(H29) {(ω, ω), (ω1, ω1), (ω1, ω2)}. Under V=L this is possible, by the result ofTodorcevic (13). This is not possible in the models of Kunen [78] and of Foremanand Laver; see (2) in the discussion of cSr. Namely, suppose that cHrA is theindicated relation in one of the indicated models A. Let B be an ω1-cc subalgebraof A of size ω1. By the Sikorski extension theorem, there is a homomorphism fromA onto some BA C such that B ≤ C ≤ B. Thus C has ccc and size ω1 or ω2,contradiction.

(H30) {(ω, ω), (ω1, ω1), (ω2, ω2)}. Finco(ω2) works.

(H31) {(ω, ω), (ω1, ω2), (ω2, ω2)}. This is ruled out by the result of Nyikos (7).

(H32) {(ω, ω1), (ω, ω2), (ω1, ω1)}. This is ruled out by Theorem 3.53(v).

(H33) {(ω, ω1), (ω, ω2), (ω1, ω2)}. This is impossible under CH, by Juhasz (18). Aconsistent example is given by A× C, both Koszmider’s algebras; see (20).

(H34) {(ω, ω1), (ω, ω2), (ω2, ω2)}. This is not possible; see (H11).

(H35) {(ω, ω1), (ω1, ω1), (ω1, ω2)}. If MA + 2ω > ω2, this is ruled out by Koppel-berg’s Theorem (16). It is open whether an example is consistent; this is Problem8(ii) in Monk [96].

Problem 17. Is it consistent that cHr(A) = {(ω, ω1), (ω1, ω1), (ω1, ω2)} for someBA A?

(H36) {(ω, ω1), (ω1, ω1), (ω2, ω2)}. If MA+2ω > ω2, this is ruled out by Koppel-berg’s theorem (16). Under CH, an example exists; see (2). This solves Problem8(i) in Monk [96].

(H37) {(ω, ω1), (ω1, ω2), (ω2, ω2)}. This is ruled out by the result of Nyikos (7).

(H38) {(ω, ω2), (ω1, ω1), (ω1, ω2)}. The homomorphic image of size ω1 and cellu-larity ω1 must have a homomorphic image of cellularity ω, contradiction.

(H39) {(ω, ω2), (ω1, ω1), (ω2, ω2)}. Impossible; see (H38).

(H40) {(ω, ω2), (ω1, ω2), (ω2, ω2)}. This is impossible under CH, by Juhasz (18).Assuming 2ω = ω2, P(ω) has this relation. If MA + 2ω > ω2, it is ruled out byKoppelberg’s Theorem (16).

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(H41) {(ω1, ω1), (ω1, ω2), (ω2, ω2)}. Not possible; see (H4).

(H42) {(ω, ω), (ω, ω1), (ω, ω2), (ω1, ω1)}. This is ruled out by Theorem 3.53(v).

(H43) {(ω, ω), (ω, ω1), (ω, ω2), (ω1, ω2)}. Not possible under CH, since then theremust be an independent subset of size ω2, and one of the pairs must be (ω2, ω2).A consistent example with this relation is C×E, where C is Koszmider’s exampleand E is a subalgebra of Intalg(R) of size ω1.

(H44) {(ω, ω), (ω, ω1), (ω, ω2), (ω2, ω2)}. Not possible: see (H11).

(H45) {(ω, ω), (ω, ω1), (ω1, ω1), (ω1, ω2)}. Under GCH the following algebra works:the standard interval algebra constructed from ω12; see (S7). It is open whetherthere is an example in ZFC; this is Problem 7(i) in Monk [96].

Problem 18. Can one prove in ZFC that there is a BA A such that

cHr(A) = {(ω, ω), (ω, ω1), (ω1, ω1), (ω1, ω2)}?

(H46) {(ω, ω), (ω, ω1), (ω1, ω1), (ω2, ω2)}. Let B be a subalgebra of Intalg(R) ofsize ω1, and set A = B × Finco(ω2). Proposition 1.1 implies that cHr(A) is asdesired.

(H47) {(ω, ω), (ω, ω1), (ω1, ω2), (ω2, ω2)}. By the result of Nyikos (7) this is impos-sible.

(H48) {(ω, ω), (ω, ω2), (ω1, ω1), (ω1, ω2)}. Not possible under CH, since then theremust be an independent subset of size ω2, and one of the pairs must be (ω2, ω2). Aconsistent example is provided by C×Finco(ω1), where C is Koszmider’s algebra;see (20).

(H49) {(ω, ω), (ω, ω2), (ω1, ω1), (ω2, ω2)}. Not possible under GCH, since thenthere must be an independent subset of size ω2, and one of the pairs must be(ω1, ω2). The Nyikos, Laver example (6) gives a consistent example.

(H50) {(ω, ω), (ω, ω2), (ω1, ω2), (ω2, ω2)}. This is impossible under CH, by Juhasz.Assuming that 2ω = ω2, the algebra Finco(ω)×P(ω) gives an example.

(H51) {(ω, ω), (ω1, ω1), (ω1, ω2), (ω2, ω2)}. This is possible under V=L, namelywith the algebra Finco(ω2)×A, with A the algebra of Todorcevic (13) with κ = ω1.

This relation is not possible in the models of Kunen and of Foreman andLaver mentioned in (2), in the discussion of cSr. In fact, in those models, if A is aBA of size ω2 with cellularity ω1, then A has a subalgebra of size ω1 which is ccc.Hence by Sikorski’s extension theorem A has an uncountable homomorphic imagewhich is ccc, so that the relation (H51) is not possible.

(H52) {(ω, ω1), (ω, ω2), (ω1, ω1), (ω1, ω2)}. Not possible under CH, since then thereis an independent subset of size ω2, and one of the ordered pairs must be (ω2, ω2).

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Also ruled out by Koppelberg’s theorem (16) if MA + 2ω > ω2. A consistentexample is provided by C ×D, where C and D are Koszmider’s algebras (20).

(H53) {(ω, ω1), (ω, ω2), (ω1, ω1), (ω2, ω2)}. Not possible under GCH, since thenthere is an independent set of size ω2, and one of the ordered pairs must be(ω1, ω2). Also ruled out by Koppelberg’s theorem (16) if MA + 2ω > ω2. Theconsistency of the existence of such an algebra is open; this is problem 8(iv) inMonk [96]:

Problem 19. Is it consistent that cHr(A) = {(ω, ω1), (ω, ω2), (ω1, ω1), (ω2, ω2)} forsome BA A?

(H54) {(ω, ω1), (ω, ω2), (ω1, ω2), (ω2, ω2)}. Not possible under CH since then thereis an independent set of size ω2, and hence (ω1, ω1) would have to be present.If MA + 2ω > ω2, this is ruled out by Koppelberg’s Theorem (16). A consistentexample is given by A×P(ω), where A is Koszmider’s algebra (20).

(H55) {(ω, ω1), (ω1, ω1), (ω1, ω2), (ω2, ω2)}. If MA + 2ω > ω2, this is ruled out byKoppelberg’s Theorem (16). Under GCH there is such an algebra; see (5). Thisanswers Problem 8(iii) in Monk [96].

(H56) {(ω, ω2), (ω1, ω1), (ω1, ω2), (ω2, ω2)}. This is impossible; a BA of size ω1 hasa ccc homomorphic image.

(H57) {(ω, ω), (ω, ω1), (ω, ω2), (ω1, ω1), (ω1, ω2)}. This is not possible under CH,since then there must be an independent subset of size ω2, and one of the pairsmust be (ω2, ω2). Assuming that 2ω > ω1, we can take the standard linear orderwhich is a subset of ω2, take a subset L of size ω2, containing a dense subset ofsize ω, and let B = Intalg(L) and A = B × Finco(ω1).

(H58) {(ω, ω), (ω, ω1), (ω, ω2), (ω1, ω1), (ω2, ω2)}. This is not possible under CH,since then there must be an independent subset of size ω2, and one of the pairsmust be (ω1, ω2). A consistent example is given by A×B, where A is the algebraof Laver and Nyikos (6) and B is a subalgebra of Intalg(R) of size ω1.

(H59) {(ω, ω), (ω, ω1), (ω, ω2), (ω1, ω2), (ω2, ω2)}. Assuming 2ω = ω2, one can takeA×P(ω), where A is the interval algebra on a subset of R of size ω1. Under CH,no algebra with this cHr exists; for suppose that A is such an algebra. Then Ahas a ccc homomorphic image B of size ω2. Then B has an independent set ofsize ω2, and it follows that P(ω) is a homomorphic image of B. In turn, P(ω)has a homomorphic image of size and cellularity ω1. Thus (ω1, ω1) ∈ cHr(A),contradiction.

(H60) {(ω, ω), (ω, ω1), (ω1, ω1), (ω1, ω2), (ω2, ω2)}. This is possible under GCH:take

Intalg(ω12)× Intalg(R)× Finco(ω2),

where ω12 has the lexicographic order. It is open to give an example in ZFC; thisis Problem 7(ii) in Monk [96].

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Problem 20. Can one construct in ZFC a BA A such that

cHr(A) = {(ω, ω), (ω, ω1), (ω1, ω1), (ω1, ω2), (ω2, ω2)}?

(H61) {(ω, ω), (ω, ω2), (ω1, ω1), (ω1, ω2), (ω2, ω2)}. Not possible under CH, sincethen there must be an independent subset of size ω2, and one of the pairs mustbe (ω, ω1). Assuming 2ω = ω2, P(ω)× Finco(ω2) works.

(H62) {(ω, ω1), (ω, ω2), (ω1, ω1), (ω1, ω2), (ω2, ω2)}. P(ω1) works, assuming GCH.Ruled out by Koppelberg’s theorem (16) if MA + 2ω > ω2.

(H63) {(ω, ω), (ω, ω1), (ω, ω2), (ω1, ω1), (ω1, ω2), (ω2, ω2)}. Many examples.

The following problem is related to some of the above problems, but is simpler tostate.

Problem 21. Can one prove in ZFC that there is a BA A of size ω2 such that(ω1, ω2) ∈ cHr(A) but (ω, ω2) /∈ cHr(A)?

To conclude this chapter, we consider cellularity for special classes of BAs. Foran atomic BA A, c(A) coincides with the number of atoms of A. Also note thatsome of the free product questions are trivial for atomic algebras; in particular,c(A⊕B) = max{cA, cB} if A and B are atomic.

Recall from Propositions 3.42 our partial characterization of the sets aspect.Atomic BAs were used there. It is natural to ask for a similar result with atomlessBAs. We now give such a theorem.

Lemma 3.55. For any regular cardinal κ there is an atomless interval algebra Asuch that aspect(A) = {κ}.

Proof. The case κ = ω is taken care of by Intalg(Q), so assume that κ is uncount-able.

By Hausdorff [1908] there is a dense linear order L such that every element ofL has character (κ, κ∗), and L has coinitiality and cofinality κ. We claim that wemay assume that |L| = κ. To see this we define by recursion a sequence 〈Mi : i ∈ ω〉of subsets of L. Let M0 be a subset of L whose order type is κ∗ + κ, consistingof a coinitial subset of type κ∗ and a cofinal subset of type κ. Having definedM2i+1 ⊆ L of size κ, let 〈(aα, bα) : α < κ〉 list all pairs (c, d) with c, d ∈ M2i+1

and c < d. For each α < κ choose cα ∈ L such that aα < cα < bα, and letM2i+2 = M2i+1 ∪ {cα : α < κ}. Now list out the elements of M2i+2: 〈dα : α < κ〉.For each α < κ let 〈eαξ : ξ < κ〉 be strictly increasing in L with supremum dα,and let 〈fαξ : ξ < κ〉 be strictly decreasing in L with infimum dα. Let

M2i+3 = M2i+2 ∪ {eαξ : α, ξ < κ} ∪ {fαξ : α, ξ < κ}.

Thus |M2i+3| = κ. Let N =⋃

i∈ω Mi. Clearly |N | = κ, N is dense, and eachelement of N has character (κ, κ∗).

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This proves our initial claim in this proof. Now let A = Intalg(L). Now Ldoes not have a first or last element, since all its elements have character (κ, κ∗).So in forming Intalg(L) we introduce a first element −∞ to L. External to L wealso have a largest element ∞. Since L has a subset of order type κ, clearly A hasa partition of size κ, and on cardinality grounds it does not have any partitionsof larger size. Suppose now that X is an infinite partition of A of size less than κ;we want to get a contradiction. We may assume that each element of X has theform [a, b) with a < b.

(1) There is an element of X of the form [−∞, b).

Suppose not. Let Y = {a : [a, b) ∈ X}. Since |X | < κ, it follows that there is ac ∈ L with −∞ < c and c < a for all a ∈ Y . Then [−∞, c) ∩ [a, b) = ∅ for all[a, b) ∈ X , contradiction.

So let b be such that [−∞, b) ∈ X . By an argument similar to that proving (1),there is an element of X of the form [b, c) with b < c. In fact, by induction we getelements [di, di+1) ∈ X with d0 = b, for each i ∈ ω. Let Y = {[e, f) ∈ X : di < efor all i ∈ ω}. Since ω, |X | < κ, it follows that there are elements u < v in L suchthat di < u for all i ∈ ω and v < e for all [e, f) ∈ Y . Then [u, v)∩ [e, f) = ∅ for all[e, f) ∈ X , contradiction. �

Now if we apply Proposition 3.38 as in the proof of Proposition 3.42 we obtainthe following:

Proposition 3.56. If K is a set of regular cardinals with a largest element, thenthere is an atomless BA A such that aspect(A) = K.

There is an interesting result which comes up in considering cellularity and unionsfor complete BAs; this result is evidently due to Solovay, Tennenbaum [71]:

Theorem 3.57. Let κ and λ be uncountable regular cardinals, and suppose that〈Aα : α < λ〉 is an increasing sequence of complete BAs satisfying the κ-chaincondition, such that Aα is a complete subalgebra of Aβ for α < β < λ, and forγ limit < λ,

⋃α<γ Aα is dense in Aγ . Then

⋃α<λ Aα also satisfies the κ-chain

condition.

Proof. By Theorem 3.16 we may assume that κ = λ. Let B =⋃

α<κ Aα. For eachα < κ we define cα mapping B into Aα by setting

cα(x) =

Aα∏x≤a∈Aα

a.

Now, in order to get a contradiction, assume that X is a disjoint subset of B ofsize ≥ κ. We may assume that X is maximal disjoint. Temporarily fix α < κ.Now

∑BX = 1, and hence

∑B{cα(x) : x ∈ X} = 1. Hence clearly∑Aα{cα(x) :

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x ∈ X} = 1 too. Since Aα satisfies the κ-chain condition, choose Xα ⊆ X of size< κ such that

(1)∑Aα{cα(x) : x ∈ Xα} = 1.

(See Lemma 3.12 of the Handbook.) Choose βα < κ such that Xα ⊆ Aβα ; theordinal βα exists since |Xα| < κ and κ is regular.

Now unfix α. Define δ0 = 0 and if δn has been defined, let δn+1 be an ordinalless than κ such that δn < δn+1 and βα < δn+1 for all α ≤ δn. Finally, letγ = supn∈ω δn. Thus if α < γ, then there is an n ∈ ω such that α < δn, and soβα < δn+1 < γ. Thus

(2) γ is a limit ordinal, and for any α < γ we have βα < γ.

We shall now prove that X ⊆ Aγ (contradiction!).

Let x ∈ X be arbitrary. Since⋃

α<γ Aα is dense in Aγ , choose a non-zerob ∈

⋃α<γ Aα such that b ≤ cγ(x). Say b ∈ Aα with α < γ. By (1), choose a ∈ Xα

such that cα(a) · b �= 0. If b · a = 0, then a ≤ −b and hence cαa ≤ −b and socαa · b = 0, contradiction. Thus b · a �= 0, and so cγ(x) · a �= 0. It follows thatx · a �= 0, by the same argument as above. But both x and a are in X , so x = a.Thus x ∈ Xα ⊆ Aγ , as desired. �

Small cellularity is rather trivial for complete BAs: a(A) = ω for A complete, and

aspect(A) =

{[ω, c(A)] if c(A) is attained,

[ω, c(A)) otherwise.

There is a large literature on cellularity for BAs of the form P(κ)/I; for a start,see Baumgartner, J., Taylor, A., Wagon, S. [82]. Usually BA terminology is notused in such investigations; saturation of ideals is the term used.

Note that c(P(ω)/fin) = 2ω, and this value is attained; see the Handbook,Example 5.28. The function a for this algebra has been carefully studied. The mainresults are that ω1 ≤ a(P(ω)/fin), Martin’s axiom implies that a(P(ω)/fin) = 2ω,and it is relatively consistent that a(P(ω)/fin) < 2ω. The first two of these resultswill be proved at the end of Chapter 4.

If L is a linearly ordered set with first element, then clearly c(Intalg(L)) isthe supremum of the cardinalities of sets of pairwise disjoint open intervals in L.

Theorem 3.58. If A is an interval algebra, then |A| ≤ 2c(A).

Proof. Let A = Intalg(L), and suppose that |A| > 2c(A). Let ≺ be a well-order ofL. Then

[L]2 = {{a, b} : a, b ∈ L, a <L b, a ≺ b} ∪ {{a, b} : a, b ∈ L, a <L b, b ≺ a}.

Applying the Erdos–Rado theorem we get a subset of L of size (c(A))+ which iswell ordered or inversely well ordered. This yields a pairwise disjoint set of thatsize also, contradiction. �

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An interval algebra A such that a(A) < c(A) is provided by Intalg(κ×Q), whereκ is an uncountable cardinal and the product × uses the lexicographic order.

Clearly a(A) ≥ κ if A = Intalg(L) with L κ-saturated. This gives rise to thefollowing problem.

Problem 22. Are the following conditions equivalent, for any linearly ordered set L?

(i) a(Intalg(L)) ≥ κ.

(ii) L is κ-saturated.

Another related problem, somewhat vague, is as follows.

Problem 23. For each infinite cardinal κ, give a criterion, purely in terms of thelinear order L, for there to exist a partition of size κ in Intalg(L).

The cellularity of tree algebras has been described in Brenner [82]:

Theorem 3.59. For A = Treealg(T ), T a tree, c(A) is the maximum of |{t ∈T : t has finitely many immediate successors}| and sup{|X | : X is a collection ofpairwise incomparable elements of T }.

Proof. If t has finitely many immediate successors, then {t} ∈ A. And if s andt are incomparable, then (T ↑ s) ∩ (T ↑ t) = 0. Hence ≥ is clear. Now supposethat X is a collection of pairwise disjoint elements of A; we want to show that|X | is ≤ the indicated maximum. Without loss of generality we may assume thateach element x ∈ X has the form (T ↑ tx)\

⋃s∈Fx

(T ↑ s), where Fx is a finiteset of s > tx. And we may assume that if tx has only finitely many immediatesuccessors, then x = {tx}. Write X = X0 ∪X1, where X0 is the set of singletonsin X and X1 = X\X0. Thus if x ∈ X1, then tx has infinitely many immediatesuccessors. Therefore, if x, y ∈ X1, x �= y, then either tx and ty are incomparable,or s ≤ ty for some s ∈ Fx, or s ≤ tx for some s ∈ Fy . For each x ∈ X1 let ux be animmediate successor of tx such that ux �≤ s for all s ∈ Fx. Then it is easy to checkthat if x and y are distinct elements of X1, then ux and uy are incomparable. Thisproves that |X1| is ≤ the sup mentioned in the theorem. �

This characterization does not work for pseudo-tree algebras: for example, if L isa dense linear order of size ω1 with an increasing subset of order type ω1, thenc(Treealg(L)) = ω1 (recall that for L a linear order, Treealg(L) = Intalg(L)).

There are problems for tree algebras similar to Problems 10 and 24:

Problem 24. If K is a nonempty set of infinite cardinals, is there a tree algebra Asuch that aspect(A) = K?

Problem 25. For each infinite cardinal κ, give a criterion, purely in terms of thetree T , for there to exist a partition of size κ in Treealg(T ).

A characterization of cellularity for pseudo-tree algebras was obtained by Brown[06], answering Problem 9 in Monk [96]. To describe it, we need some notation.

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Let T be a pseudo-tree. A fan element of T is an element t such that there is afinite set F of pairwise incomparable elements of T greater than t, with |F | ≥ 2,such that every element greater than t is comparable with exactly one s ∈ F . Apure chain of T is a chain C of size at least 2 such that for all a, b ∈ C with a < band for all c > a, c is comparable with b. Then the characterization is as follows:

The cellularity of a pseudo-tree algebra is equal to the maximum of the followingfour cardinals:

(i) the supremum of the cellularities of Intalg(L) for chains L ⊆ T .

(ii) the supremum of the cardinalities of sets of pairwise incomparable elementsof T .

(iii) the number of fan elements of T .

(iv) the number of maximal pure chains of T .

Moreover, there are examples showing that none of these four conditions is redun-dant; two of the examples involve consistency results, and such consistency resultsare really necessary.

Similarly to the above, we have the following problems.

Problem 26. If K is a nonempty set of infinite cardinals, is there a pseudo-treealgebra A such that aspect(A) = K?

Problem 27. For each infinite cardinal κ, give a criterion, purely in terms of thepseudo-tree T , for there to exist a partition of size κ in Treealg(t).

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4 Depth

Recall that Depth(A) is the supremum of cardinalities of subsets of A which arewell ordered by the Boolean ordering. There are two main references for resultsabout this notion: McKenzie, Monk [82] and (implicitly) Gratzer, Lakser [69].

If A is a subalgebra of B, then clearly Depth(A) ≤ Depth(B). The differencecan be arbitrarily large: consider B = A ⊕ C. This same example shows thatthe difference can be arbitrarily large for A ≤reg B, A ≤rc B or A ≤σ B; seeProposition 11.8 in the Handbook, and the diagram concerning special subalgebrasin Chapter 2.

Concerning A ≤π B, note that if B is the interval algebra on κ and A ≤π Bis the finite-cofinite algebra on κ, then Depth(B) = κ, while Depth(A) = ω. It ispossible to get atomless examples here. Namely, let B = κFr(ω) and let A be theweak power which is a subalgebra of B. It is easy to see, and is proved below, thatDepth(B) = κ, while Depth(A) = ω.

For A ≤s B, we can use the same idea as for cellularity to obtain a big difference.

For ≤m we have the following result.

Proposition 4.1. If A ≤m B, then Depth(A) = Depth(B).

Proof. Let B = A(x). Since Depth(B) = max(Depth(B � x),Depth(B � −x))(proved below), by symmetry it suffices to show that Depth(B � x) ≤ Depth(A).So, suppose that 〈aα · x : α < κ〉 is strictly increasing, where each aα ∈ A. Nowwe consider two cases.

Case 1. |{α < κ : aα ∈ SmpAx }| = κ. Then we may assume that aα ∈ SmpAx for

all α < κ. Write aα = bα + cα with bα ≤ x and cα ≤ −x. Then aα · x = bα, so〈bα : α < κ〉 is a strictly increasing sequence of element os A.

Case 2. |{α < κ : aα ∈ SmpAx }| < κ. Then we may assume that −aα ∈ SmpAx forall α < κ. For each α < κ write −aα = uα + vα with uα, vα ∈ A and uα ≤ x,vα ≤ −x. Then −(aα ·x) = −aα +−x = uα+−x, and it follows that 〈uα : α < κ〉is strictly decreasing. �

Since every interval algebra is minimally generated (Proposition 2.52), it is clearthat one can have A ≤mg B with Depth(A) much less than Depth(B).

, ,OI 10.1007/978-3- - - _ ,


014


Basel 2

1555

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156 Chapter 4. Depth

Clearly Depth(A) = Depth(B) if A ≤free B. Hence easily also Depth(A) =Depth(B) if A ≤proj B.

Depth can increase arbitrarily when A ≤u B; see the argument for cellularity.

If A is a homomorphic image of B, then depth can change in either way in goingfrom B to A; see the argument for cellularity.

Now we turn to products. Some of the results which we shall present depend onthe following simple lemma.

Lemma 4.2. Let A and B be BAs, and let X be a chain in A × B of infinitecardinality κ. Then the projections of X are chains, and at least one of them hascardinality κ. Furthermore, if X has order type κ, then X has a subset of ordertype κ on which one of the two projections is one-one.

Proof. For any z ∈ X write z = (z0, z1). For i = 0, 1 write z ≡i w iff z, w ∈ X andzi = wi. Now note that

{{x} : x ∈ X} ⊆ {a ∩ b : a ∈ X/ ≡0, b ∈ X/ ≡1}\{0};

hence one of the two equivalence relations ≡0,≡1 has κ equivalence classes, andthe lemma follows. �Theorem 4.3. Depth(

∏i∈I Ai) = max(|I|, supi∈IDepth(Ai)) if each Ai is nontriv-

ial and either I is infinite or some Ai is infinite.

Proof. Clearly≥ holds. Suppose = fails to hold, and let f be an order isomorphismof κ+ into

∏i∈I Ai, where κ = max(|I|, supi∈IDepth(Ai)). For each i ∈ I there is

an ordinal αi < κ+ such that (f(α))i = (f(β))i for all β > αi. Let γ = supi∈Iαi.Then for all δ > γ we have f(δ) = f(γ), contradiction. �Theorem 4.4. Let κ = supi∈IDepth(Ai), and suppose that κ is regular. Then thefollowing conditions are equivalent:

(i) Depth(∏

i∈I Ai) is not attained.

(ii) |I| < κ, and for all i ∈ I, Ai has no chain of order type κ. �

The proof of this theorem is very similar to that of Theorem 4.3. The case ofsingular cardinals is a little more involved:

Theorem 4.5. Let κ = supi∈I Depth(Ai), and suppose that κ is singular. Then thefollowing conditions are equivalent:

(i) Depth(∏

i∈I Ai) is not attained.

(ii) These four conditions hold:

(a) |I| < κ.

(b) For all i ∈ I, Ai has no chain of type κ.

(c) |{i ∈ I : Depth(Ai) = κ}| < cf(κ).

(d) sup{Depth(Ai) : i ∈ I, Depth(Ai) < κ} < κ.

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Chapter 4. Depth 157

Proof. Let 〈μα : α < cf(κ)〉 be a strictly increasing continuous sequence of cardi-nals with supremum κ, with μ0 = 0.

(i)⇒(ii): (a) and (b) are clear. Suppose that (c) fails to hold; we show that(i) fails. Let i be a one-one function from cf(κ) into {i ∈ I : Depth(Ai) = κ}.For each α < cf(κ) let 〈aiβ : μα ≤ β < μα+1〉 be a strictly increasing sequence ofelements of Aiα . Now we define a sequence 〈xβ : β < κ〉 of elements of

∏i∈I Ai.

For each β < κ choose α < cf(κ) so that μα ≤ β < μα+1, and for any j ∈ I set

xβ(j) =

⎧⎨⎩1 if j = iγ for some γ < α;

aiβ if j = iα;

0 otherwise.

Clearly this sequence is as desired.

Next we show that if (d) fails then (i) fails. By induction we can define iαfor α < cf(κ) so that

supβ<α

(Depth(Aiβ ) ∪ μβ) < Depth(Aiα) < κ,

and then we can proceed as for (c).

(ii)⇒(i): Assume (ii), and suppose that 〈xα : α < κ〉 is strictly increasing in∏i∈I Ai. Define

Ji = {α < κ : xα(i) < xα+1(i)} for i ∈ I;

K = {i ∈ I : Depth(Ai) = κ};λ = sup{Depth(Ai) : i ∈ I, Depth(Ai) < κ}.

Then by the above assumptions we have λ < κ, |Ji| ≤ λ for all i ∈ I\K, |K| <cf(κ), and |Ji| < κ for all i ∈ K. It follows that |

⋃i∈I Ji| < κ. But for any

α ∈ κ\⋃

i∈I Ji we have xα = xα+1, contradiction. �

The above theorems completely describe the depth of products. The case of weakproducts is even simpler:

Theorem 4.6. Let κ = supi∈I Depth(Ai), and suppose that cf(κ) > ω. Then thefollowing conditions are equivalent:

(i)∏w

i∈I Ai has no chain of order type κ.

(ii) For all i ∈ I, Ai has no chain of order type κ.

Proof. (i)⇒(ii) is clear. (ii)⇒(i): Suppose that 〈xα : α < κ〉 is strictly increasingin∏w

i∈I Ai. For any y ∈∏w

i∈I Ai let S(y) = {i ∈ I : yi �= 0}.Case 1. S(xα) is finite for all α < κ. Since cf(κ) > ω, it follows that there is anα < κ such that S(xα) = S(xβ) whenever α < β < κ. But then Lemma 4.2 easilygives a contradiction.

Case 2. Otherwise we may assume that {i ∈ I : xα(i) �= 1} is finite for all α < κ,and a contradiction is reached as in Case 1. �

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Corollary 4.7. Depth(∏w

i∈I Ai) = supi∈IDepth(Ai). �Now we shall show that Depth(A) is attained if Depth(A) is a successor cardinalor a cardinal of cofinality ω; otherwise, there are counterexamples.

Theorem 4.8. If cf(Depth(A)) = ω, then Depth(A) is attained.

Proof. Let κ = Depth(A). We may assume that κ is an uncountable limit cardinal.Let 〈λi : i < ω〉 be a strictly increasing sequence of cardinals with supremum κ,and with λ0 = 0 and λ1 infinite. Now we call an element a of A an ∞-element ifλi is embeddable in A � a for all i < ω. We claim

(∗) If a is an ∞-element, and a = b+ c with b · c = 0, then b is an ∞-element or cis an ∞-element.

In fact, by Lemma 4.2, for each i < ω, λi is embeddable in A � b or A � c, so (∗)follows.

Using (∗), we construct a sequence 〈ai : i < ω〉 of elements of A by induction.

Suppose that aj has been constructed for all j < i so that bdef=∏

j<i−aj is an∞-element. Let 〈c(α) : α < λi+1〉 be an isomorphic embedding of λi+1 into b.By (∗), one of the elements c(λi) and b · −c(λi) is an ∞-element, while clearly λi

is embeddable in both of these elements. So we can choose ai ≤ b so that λi isembeddable in ai, and

∏j≤i−aj is an ∞-element. This finishes the construction.

For each i < ω let 〈biα : α < λi〉 be an embedding of λi into ai. Note thatai · aj = 0 for i < j < ω. Hence the following sequence 〈dα : α < κ〉 is clearly thedesired embedding of κ into A. Given α < κ, there is a unique i < ω such thatλi ≤ α < λi+1. We let dα = a0 + · · ·+ ai + bi+1,α. �

Theorem 4.6 enables us to easily show that Theorem 4.8 is best possible: if κ isa limit cardinal with cf(κ) > ω, then it is easy to construct a weak product Bsuch that Depth(B) = κ but depth is not attained in B. Namely suppose that〈λξ : ξ < cf(κ)〉 is strictly increasing with supremum κ. For each ξ < cf(κ) letAξ = Intalg(λξ). Then

∏wξ<cf(κ) Aξ is as desired.

Depth in free products is discussed thoroughly in McKenzie, Monk [82]. The mainresults of that paper are as follows.

• If cf(κ) > ω, A has no chain of order type cf(κ), and B has no chain of ordertype κ, then A⊕B has no chain of order type κ.

• If Depth(B) = κ and A has a chain of order type cf(κ), then A ⊕ B has achain of order type κ.

• If κ is regular and uncountable, and for every i ∈ I the BA Ai has no chainof order type κ, then also ⊕i∈IAi has no chain of order type κ.

• Depth (⊕i∈IAi) = supi∈I Depth(Ai) for any system 〈Ai : i ∈ I〉 of infiniteBAs.

We give a proof of a special case of the first of these results, as it will be neededlater.

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Theorem 4.9. If κ is an uncountable regular cardinal and neither A nor B has achain of order type κ, then A⊕B also does not have such a chain.

Proof. An element x of A⊕B is of length n if we can write x =∑

i<n ai · bi where∀i < n[0 �= ai ∈ A and 0 �= bi ∈ B], and bi ·bj = 0 for distinct i, j < n, and x cannotbe written in this way for any m < n. Now it suffices to show that for all n, A⊕Bdoes not have a chain of order type κ in which all entries have length n. Supposethat n is minimum such that this is not true; we want to get a contradiction.Say 〈x(α) : α < κ〉 is a strictly increasing sequence of elements of A ⊕ B, eachx(α) being of length n, say x(α) =

∑i<n aαi · bαi with ∀i < n[0 �= aαi ∈ A and

0 �= bαi ∈ B], and bαi · bαj = 0 for distinct i, j < n.

If n = 1, then α < β < κ implies that aα0 · bα0 < aβ0 · bβ0 , hence aα0 ≤ aβ0 and

bα0 ≤ bβ0 . Since A does not have a chain of order type κ, it follows that there is aγ < κ such that aα0 = aγ0 for all α > γ. Similarly there is a δ < κ such that bα0 = bδ0for all α > δ. Then x(α) = x(ε) for all α, ε ≥ γ, δ, contradiction.

Thus n > 1. Now for α < β < κ and i < n we have aαi · bαi ≤∑

i<n aβi · b

βi ≤∑

i<n bβi , and hence:

(1) If α < β < κ and i < n, then bαi ≤∑

i<n bβi .

Next we claim

(2) If α < β < κ, i, j < n, and bαi · bβj �= 0, then aαi ≤ aβj .

In fact, α < β < κ implies that aαi · bαi ·∏

k<n(−aβk +−bβk) = 0; multiplying by bβj

gives aαi · bαi · bβj · −a

βj = 0, and the conclusion of (2) follows.

Now we claim

(3) There is a g ∈ κn such that aαg(α) ≤ aβg(β) whenever α < β < κ.

In fact, if α < β < κ, then the set {g ∈ κn : aαg(α) ≤ aβg(β)} is closed in the spaceκn (with the discrete topology on n and the product topology on κn). Thus bycompactness of κn it suffices to show that if α1 < · · · < αm < κ then there is ag ∈ κn such that aα1

g(α1)≤ · · · ≤ aαm

g(αm). This is clear by induction from (1) and

(2).

Choose g as in (3). Then there is an α < κ such that aαg(α) = aβg(β) for all

β > α. Now for each β < κ let hβ be a permutation of n such that hβ(0) = g(α+β).Define for any β < κ and i < n

cβi = aα+βhβ(i)

and dβi = bα+βhβ(i)

;

y(β) =∑i<n

cβi · dβi .

[email protected]

Then for any β < κ we have

y(β) =∑i<n

cβi · dβi =

∑i<n

aα+βhβ(i)

· bα+βhβ(i)

=∑i<n

aα+βi · bα+β

i = x(α+ β).

Moreover,cβ0 = aα+β

hβ(0)= aα+β

g(α+β) = aα+γg(α+γ) = aα+γ

hγ(0)= cγ0

for any β < γ < κ.

Now for each β < κ the element y(β) · −c00 has length less than n, andy(β) · −c00 ≤ y(γ) · −c00 for β < γ. Hence by the choice of n there is a β < κsuch that y(β) · −c00 = y(γ) · −c00 for all γ > β. Consequently, y(γ) · c00 < y(δ) · c00whenever β ≤ γ < δ < κ. Now for each γ < κ and i < n we define

eγi = cβ+γi · c00 and z(γ) =

∑i<n

eγi · dβ+γi .

Note that eγ0 = c00 for all γ < κ. Now 〈z(γ) : γ < κ〉 is strictly increasing, andeγi ≤ e00 for all γ < κ and i < n. Each z(γ) has length at most n, and so bythe choice of n there are κ of them which do have length n. This gives a strictlyincreasing sequence 〈γξ : ξ < κ〉 of ordinals less than κ such that each z(γξ) haslength n.

Now we claim

(4) If ξ < η < κ and 0 < i < n, then dβ+γξ

0 · dβ+γη

i = 0.

In fact, otherwise we have bα+β+γξ

hβ+γξ(0) · b

α+β+γη

hβ+γη (i)�= 0, and hence by (2), a

α+β+γξ

hβ+γξ(0) ≤

aα+β+γη

hβ+γη (i). It follows that c

β+γξ

0 ≤ cβ+γη

i , so c00 = eγξ

0 ≤ eγη

i ≤ c00. It follows that

eγη

i = eγη

0 . But then z(γ−η) has length less than n, since eγη

0 ·dβ+γη

0 and eγη

i ·dβ+γη

i

can be combined. This contradiction proves (4).

(5) If ξ < η < κ, then dβ+γξ

0 ≤ dβ+γη

0 .

We have dβ+γξ

0 = bα+β+γξ

hβ(0)and d

β+γη

0 = bα+β+γη

hβ(0), so (5) follows from (1) and (4).

Now by (5) there is a θ < κ such that dβ+γξ

0 = dβ+γθ

0 for all ξ > θ. Thus forξ ≥ θ we have

z(γξ) = c00 · dβ+γθ

0 +∑

0<i<n

eγξ

i · dβ+γξ

i ,

and dβ+γθ

0 ·∑

0<i<n eγξ

i · dβ+γξ

i = 0, so⟨ ∑0<i<n

eγξ

i · dβ+γξ

i : θ ≤ ξ < κ

⟩

is strictly increasing with each entry of length less than n, contradiction. �

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Proposition 4.10. There is a BA A such that c(A) = Depth(A) = Depth(A⊕A) <c(A⊕A).

Proof. By Lemma 3.2 and Corollary 3.11, letB be a BA such that c(B) < c(B⊕B).Say c(B) = κ. Let A = Intalg(κ)×B. Since B⊕B can be isomorphically embeddedin A⊕A, it is clear that A satisfies the conditions of the proposition. �

We now briefly discuss depth and amalgamated free products. For the first resultwe need the following well-known set theoretical fact.

Lemma. There is a system 〈aξ : ξ < ω1〉 of infinite subsets of ω such that for allξ, η < ω1, if ξ < η then aξ\aη is finite and aη\aξ is infinite.

Proof. We define 〈aξ : ξ < ω1〉 by recursion. Let a0 be the set of all even naturalnumbers. Now suppose that α < ω1 and aξ has been defined for all ξ < α so thatthe following conditions hold:

(1) If ξ < η < α, then aξ\aη is finite and aη\aξ is infinite.

(2) If F is a finite subset of α, then ω\⋃

x∈F aξ is infinite.

Let 〈bi : i < ω〉 enumerate {aξ : ξ < α}, possibly with repetitions. Now for eachi < ω we define ji, ki recursively to be any two distinct members of the set

ω\(⋃

s<i

bs ∪ {js, ks : s < i}).

Then let aα = ω\{ji : i ∈ ω}. Now if s s} ⊆ ω\

(⋃i<s bi ∪ aα

).

Hence the conditions (1) and (2) hold for α + 1. Inductively, they also hold forlimit α. �

The following theorem is a special case of a theorem in McKenzie, Monk [82].

Theorem 4.11. Let A be the BA of finite and cofinite subsets of ω. Then thereexist B,C ≥ A both satisfying ccc such that Depth(B ⊕A C) = ω1.

Proof. Let M be the collection of all even integers, N the set of all odd integers.Then we take two sequences 〈aα : α < ω1〉 and 〈bα : α < ω1〉 such that

(1) Each aα is an infinite subset of M , and for α < β < ω1 we have aα\aβ finiteand aβ\aα infinite.

(2) Similarly for the bα’s, subsets of N .

These sequences exist by the lemma. Now let B = C = A×P(ω). For each a ∈ Alet g(a) = (a, a). Then g is an isomorphism of A into B = C. So it is enough toprove that Depth(B⊕g[A]C) = ω1. For each α < ω1 let cα = (0, aα)×(0, bα) [using

[email protected]

× rather than · to indicate which one is in B and which one in C]. We claim that〈cα : α < ω1〉 is as desired. Let α < β < ω1. Then

cα · −cβ = [((0, aα) · (1,−aβ))× (0, bα)] + [(0, aα)× ((0, bα) · (1,−bβ))]= [(0, aα\aβ)× (0, bα)] + [(0, aα)× (0, bα\bβ)].

Now ddef= aα\aβ is a finite subset of M , so d ∈ A. And (0, aα\aβ) ≤ (d, d), while

(0, bα) · (d, d) = (0, 0). Using a similar argument for the second summand, thisshows that cα · −cβ = 0.

Now suppose that α < β and cβ · −cα = 0. So (0, aβ\aα)× (0, bβ) = 0, hencethere is a d ∈ A such that (0, aβ\aα) ≤ (d, d) and (0, bβ) · (d, d) = (0, 0). Nowaβ\aα is infinite, so d is cofinite. Hence bβ ∩ d �= 0, contradiction. �

The following theorem solves Problem 2 of McKenzie, Monk [82]:

Theorem 4.12. Let A be the BA of finite and cofinite subsets of ω, and let κ bean uncountable cardinal. Then there exist B,C ≥ A such that |B| = |C| = κ,Depth(B) = Depth(C) = ω, and Depth(B ⊕A C) ≥ ω1.

Proof. First we choose B and C as in the proof of Theorem 4.11. In particular,B ⊕A C has a chain of the form 〈bα × cα : α < ω1〉, while B and C have size2ω. Clearly we may assume that |B| = |C| = ω1. Let B′ = B × Finco(κ) andC′ = C × Finco(κ). Thus |B′| = |C′| = κ. Set

A′ = {(a, 0) : a ∈ [ω]<ω} ∪ {(a, 1) : ω\a ∈ [ω]<ω}.

Clearly A′ is isomorphic to A. To prove the theorem it suffices to show thatB′ ⊕A′ C′ has depth ω1. Let b′α = (bα, 0), c

′α = (cα, 0), and dα = b′α × c′α for all

α < ω1. We claim that 〈dα : α < ω1〉 is a chain in B′ ⊕A′ C′, as desired. To provethis, suppose that α < β. Choose u, v ∈ A such that bα · −bβ ≤ u, cα ∩ u = 0,bα · v = 0, and cα · −cβ ≤ v. Now

dα · −dβ = (b′α · −b′β)× c′α + b′α × (c′α · −c′β).

Note that (b′α · −b′β) × c′α = (bα · −bβ, 0) × (cα, 0). Then for some ε ∈ {0, 1}we have (u, ε) ∈ A′, (bα · −bβ, 0) ≤ (u, ε), and (u, ε) · (cα, 0) = (0, 0). Therefore(b′α · −b′β) × c′α = 0. Similarly for the other summand, so dα · −dβ = 0. Supposethat also dβ ·−dα = 0. This easily gives (bβ ·−bα, 0)× (cβ , 0) = (0, 0). Hence thereis a (w, ε) ∈ A′ such that (bβ · −bα, 0) ≤ (w, ε) and (cβ , 0) · (w, ε) = (0, 0). Hencebβ · −bα ≤ w and cβ · w = 0, contradiction. �

The following result of Shelah [02] (Remark 1.2, 4) solves Problem 10 in Monk [96]:

There is a countable BA A such that for every strong limit cardinal μ of cofinalityℵ0 there exist B,C ≥ A such that Depth(B),Depth(C) ≤ μ while Depth(B ⊕A

C) ≥ μ+.

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Some results in Shelah [02] partially solve Problem 11 of Monk [96]. The easiestto state is the following part of Observation 1.8:

If λ is weakly compact, A ≤ B,C, |A| < λ, and Depth′(B ⊕A C) = λ+, thenDepth′(B) = λ+ or Depth′(C) = λ+.

We give a proof of this.

Theorem 4.13. If κ is weakly compact, A ≤ B,C, |A| < κ, and B ⊕A C has achain with order type κ, then B or C has such a chain.

Proof. Say 〈dα : α < κ〉 is strictly increasing in Ddef= B ⊕A C. Now for all α < κ

the set {−dβ : β < α} ∪ {dα} has fip, and so is contained in an ultrafilter Eα ofD. Since |P(A)| < κ, there exist a subset Γ of κ of size κ and an ultrafilter E∗

on A such that Eα ∩ A = E∗ for all α ∈ Γ. Let I∗ = {a ∈ A : −a ∈ E∗}; this is amaximal ideal of A. Let J = 〈I∗〉IdB , K = 〈I∗〉IdC , and K = 〈I∗〉IdD .

(1) 〈[dα]L : α ∈ Γ〉 is strictly increasing.

In fact, clearly [dα]L ≤ [dβ ]L if α < β, both in Γ. Now suppose that α, β ∈ Γ,α < β, and [dβ ]L ≤ [dα]L. It follows that there is an a ∈ I∗ such that dβ ·−dα ≤ a.Hence dβ ·−a ·−dα = 0. So there is an f ∈ A such that dβ ·−a ≤ f and −dα ≤ −f .Case 1. f ∈ I∗. Then dβ ≤ a + f ∈ I∗. Hence −a · −f ∈ E∗ = Eβ ∩ A. Alsodβ ∈ Eβ , so dβ · −a · −f �= 0, contradiction.

Case 2. −f ∈ I∗. Then f ∈ E∗ = Eβ ∩ A. Also −dα ∈ Eβ , so −dα · f �= 0,contradiction.

Thus (1) holds.

(2) D/L ∼= B/J ⊕ C/K.

In fact, it suffices to show that if b ∈ B\J and c ∈ C\K then b · c /∈ L. Supposeto the contrary that b · c ∈ L. Say b · c ≤ e ∈ I∗. Then b · −e · c = 0, so there is anf ∈ A such that b · −e ≤ f and c ≤ −f .Case 1. f ∈ I∗. Then b ≤ e+ f ∈ I∗, so b ∈ J , contradiction.

Case 2. −f ∈ I∗. Then c ∈ K, contradiction.

Thus (2) holds.

Now by (1), (2), and Theorem 4.9, B/J or C/K has a chain of order typeκ. Say by symmetry that 〈[eα]J : α < κ〉 is strictly increasing in B/J . Thus forα < β we have eα · −eβ ∈ J while eβ · −aα /∈ J . Say eα · −eβ ≤ aαβ ∈ I∗. Thuseα · −aαβ ≤ eβ · −aαβ while eβ · −eα · −aαβ �= 0. Let f({α, β}) = aαβ . Since κ isweakly compact, there exist a Γ ∈ [κ]κ and an a ∈ A such that aαβ = a for allα, β ∈ Γ such that α < β. Thus 〈eα · −a : α ∈ Γ〉 is strictly increasing. �

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The following form of Problem 11 of Monk [96] remains open.

Problem 28. Find necessary and sufficient conditions on a BA A for there to existextensions B,C of A such that Depth′(B ⊕A C) > max(Depth′(B),Depth′(C)).

Concerning unions, we note that Depth is an ordinary sup function with respectto the function P , where P (A) = {X ⊆ A : X is a well-ordered chain in A}, andso Theorem 3.16 applies.

For ultraproducts the situation is similar to that for cellularity. It is easy tosee that if F is a countably complete ultrafilter on an infinite set I and Ai is a BAwith depth ω for each i ∈ I, then

∏i∈I Ai/F has depth ω. The following problem

has not been considered:

Problem 29. Determine the possibilities for Depth(∏

i∈I Ai/F ) in terms of〈Depth(Ai) : i ∈ I〉 with F countably complete.

If F is a countably incomplete ultrafilter on I and each algebra is infinite, then∏i∈I Ai/F has depth > ω. This is easiest to see by recalling that

∏i∈I Ai/F is

ω1-saturated, and noting

(∗) If an infinite BA A is κ-saturated, then A has a chain of order type κ.

To prove (∗), we construct a ∈ κA by recursion. Suppose that aβ has been definedfor all β < α, so that if β is a successor ordinal γ+1, then A � −aγ is infinite. If βis a successor ordinal, it is clear how to proceed in order to still have the indicatedcondition. If β is limit, consider the set

{cxα < v0 : α < β} ∪ { “there are at least n”v1(v0 < v1) : n ∈ ω}.

This set is finitely satisfiable in A, and so an element satisfying all of these formulasgives the desired element aβ.

Now we consider regular ultrafilters. The first result follows easily from atheorem of W. Hodges, that if F is a regular ultrafilter on I then in I〈ω,>〉/Fthere is a chain of order type |I|+. We give a direct BA proof of the BA result:

Theorem 4.14. Let F be a |I|-regular ultrafilter on I, and suppose that Ai is aninfinite BA for every i ∈ I. Then in

∏i∈I Ai/F there is a chain of order type |I|+.

Proof. For brevity set κ = |I|. By the definition of regularity choose E ⊆ F suchthat |E| = κ and for all i ∈ I the set {e ∈ E : i ∈ e} is finite. Let G be a one-one function from E onto κ. For each i ∈ I choose a strictly increasing sequence〈xij : j < ω〉 in Ai, and let Xi = {xij : j < ω}. Then it suffices to show:

(∗) If gα ∈∏

i∈I Xi for all α < κ, then there is an f ∈∏

i∈I Xi such thatgα/F < f/F < 1 for all α < κ.

To define f , let i ∈ I. Let e(1), . . . , e(m) be all of the elements u of E suchthat i ∈ u. Then let f(i) be any element of Xi greater than all of the elementsgG(e(1))(i), . . . , gG(e(m))(i). This defines f . Now if α < κ and i ∈ G−1(α), we havegα(i) < f(i) < 1, as desired. �

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Theorem 4.15. Let I be an infinite set, and suppose that Ai is an infinite BAfor every i ∈ I. Then there is a proper filter G on I such that G contains allcofinite sets, and

∏i∈I Ai/F has a chain of order type 2|I| for every ultrafilter F

including G.

Proof. Again let κ = |I|. Let S ⊆ κω satisfy the following condition:

(1) |S| = 2κ, and for every finite sequence i0, . . . , ik−1 of natural numbers andevery sequence f0, . . . , fk−1 of distinct members of S of length k, there is an α < κsuch that ft(α) = it for all t < k.

For the existence of such a set, see Comfort, Negrepontis [74], pp. 75–77. Let〈fα : α < 2κ〉 enumerate S without repetitions. For α < β < 2κ, let Jαβ = {γ <κ : fα(γ) < fβ(γ)}. From (1) it is clear that the intersection of any finite numberof the sets Jαβ is infinite. Hence

{Jαβ : α < β < 2κ} ∪ {Γ ⊆ κ : |κ\Γ| < ω}

generates a proper filter G containing all cofinite sets. Clearly G is as desired. �

Now we give some results of Douglas Peterson; see Peterson [97]. Recall the notionof essential supremum, ess.sup., from just before Lemma 3.17.

Theorem 4.16. Suppose that 〈Ai : i ∈ I〉 is a system of infinite BAs, with I infinite,and F is an ultrafilter on I. Then Depth

(∏i∈I Ai/F

)≥ ess.supFi∈I(Depth(Ai)).

Proof. For any linearly ordered set L let Depth(L) be the supremum of the size ofwell-ordered subsets of L. Let λ = ess.supFi∈I(Depth(Ai)). If λ is a successor car-dinal, then {i ∈ I : Depth(Ai) = λ} ∈ F , and hence clearly Depth

(∏i∈I Ai/F

)≥

Depth(Iλ/F ) ≥ λ, as desired. If λ is a limit ordinal, then by similar reason-ing, Depth

(∏i∈I Ai/F

)≥ κ for every successor cardinal κ < λ, and so also

Depth(∏

i∈I Ai/F)≥ λ. �

Theorem 4.17. Suppose that 〈Ai : i ∈ I〉 is a system of infinite BAs, with Iinfinite, and that F is a regular ultrafilter on I. Let λ = ess.supFi∈I(Depth(Ai)),and assume that cf(λ) ≤ |I| < λ. Then Depth

(∏i∈I Ai/F

)≥ λ+.

Proof. Case 1. {i ∈ I : Depth(Ai) = λ} ∈ F . We may assume that Depth(Ai) = λfor all i ∈ I. By Lemma 3.18 we get a system 〈κi : i ∈ I〉 of infinite cardinalssuch that κi < λ for all i ∈ I, and ess.sup F

i∈I(κi) = λ. Let δi = κ+i for all

i ∈ I. Then Depth(∏

i∈I Ai/F)≥ Depth

(∏i∈I δi/F

), so it suffices to show that

Depth(∏

i∈I δi/F)≥ λ+. Suppose that {fα/F : α < λ} is a set of elements of∏

i∈I δi/F ; we shall find an element f ∈∏

i∈I δi such that f/F > fα/F for allα < λ, and this will clearly finish the proof. Let i ∈ I. Then {fα(i) : α < κi} isnot cofinal in δi, so we can let f(i) be an element of δi greater than each fα(i),α < κi. Then for any α < λ we have {i ∈ I : f(i) > fα(i)} ⊇ {i ∈ I : κi > α} ∈ F ,so f/F > fα/F , as desired.

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Case 2. {i ∈ I : Depth(Ai) < λ} ∈ F . Then we can assume that Depth(Ai) < λfor all i ∈ I. Then by Lemma 3.18 there is a system 〈κi : i ∈ I〉 of infinite cardinalssuch that κi < Depth(Ai) for all i ∈ I, and ess.sup F

i∈I(κi) = λ. Let δi = κ+i . Note

that Ai has a well-ordered subset of size δi. Hence the procedure of Case 1 can beused. �

Theorem 4.18 (GCH). Suppose that 〈Ai : i ∈ I〉 is a system of infinite BAs,with I infinite, and F is a regular ultrafilter on I. Then Depth

(∏i∈I Ai/F

)≥∣∣∏

i∈I Depth(Ai)/F∣∣.

Proof. Let λ = ess.sup Fi∈IDepth(Ai). Then we consider three cases:

Case 1. λ ≤ |I|. Then λ|I| = 2|I| = |I|+, and this case follows from Theorem 4.15.

Case 2. cf(λ) ≤ |I| < λ. Then λ|I| = λ+, and the result follows from Theorem4.17.

Case 3. |I| < cf(λ). Then λ|I| = λ and we are through by Theorem 4.16. �

By a result of Donder [88], every uniform ultrafilter is regular in the core model.Thus by Theorem 4.18 it is consistent that

Depth

(∏i∈I

Ai/F

)≥∣∣∣∣∣∏i∈I

Depth(Ai)/F

∣∣∣∣∣always holds. The question arises to find an example with > here. This was donein Shelah [02] (publication 652), assuming V = L or actually something weaker.

The following is Problem 12 of Monk [96]; but see Shelah [05].

Problem 30. Is an example with Depth(∏

i∈I Ai/F)>∣∣∏

i∈I Depth(Ai)/F∣∣ pos-

sible in ZFC?

We turn to the other direction. First we have:

Theorem 4.19. Suppose that 〈Ai : i ∈ I〉 is a system of infinite BAs, I infinite, Fis a uniform ultrafilter on I, and κ = max(|I|, ess.supFi∈I(Depth(Ai))).

Then Depth(∏

i∈I Ai/F)≤ 2κ.

Proof. Suppose that Depth(∏

i∈I Ai/F)> 2κ. Choose b ∈ F so that

supi∈b

Depth(Ai) = ess.supFi∈I(Depth(Ai)).

Suppose that 〈[fα] : α < (2κ)+〉 is strictly increasing in∏

i∈I Ai/F . For α < βlet F ({α, β}) be the least i ∈ b such that fα(i) < fβ(i). By the Erdos–Radotheorem (2κ)+ → (κ+)2κ let M be a subset of (2κ)+ of size κ+ such that for somei ∈ I, F ({α, β}) = i for all α < β, both in M . Since Depth(Ai) ≤ κ, this is acontradiction. �

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Concerning Theorem 4.18, it is consistent to have inequality in the other direction;this is a result of Shelah [90] which also solves Problem 4 of Monk [90]. The proofis very similar to the proof of Theorem 1.5.8 in McKenzie, Monk [82] (also dueto Shelah). Note that the theorem says that it is consistent to have a BA A suchthat Depth(ωA/F ) < |ωDepth(A)/F |. We give a complete proof here, except foran essential use of a result of Laver. For forcing notation we follow Kunen [80].

Theorem 4.20. Suppose M is a countable transitive model of ZFC+CH, let κ beany regular uncountable cardinal of M , let P be the partial order for adding κSacks reals side-by-side (explained below), and let G be P -generic over M . Thenin M [G] cardinals are preserved, 2ω ≥ κ, and there is a nonprincipal ultrafilterF on ω such that Depth(ωA/F ) = ω1, where A is the BA of finite and cofinitesubsets of ω.

Note by Theorem 3.19 that |ωDepth(A)/F | ≥ 2ω, so this theorem gives the desiredresult.

Proof. A perfect tree is a nonempty subset T of <ω2 such that if t ∈ T and m issmaller than the domain of t then t � m ∈ T , and such that for any t ∈ T thereis some s ∈ T with t ⊆ s such that s0, s1 ∈ T . We write p ≤ q in place of p ⊆ qfor perfect trees p, q. A branching point of p is a point t ∈ p such that t0 ∈ p andt1 ∈ p. An nth branching point is a branching point t such that there are exactlyn branching points < t. Thus

(1) A 0th branching point is a branching point t ∈ p such that there is no branchingat all from the root ∅ up to t.

(2) For any s ∈ p with at most n branching points below s there is an nth branchingpoint t of p such that s ≤ t.

For perfect trees p and q, p ≤n q means that p ⊆ q and every nth branching pointof q is a branching point of p. Clearly then

(3) If p ≤n q, then F ⊆ p, where

F = {t ∈ q : t ⊆ sε for some nth branching point s of q and some ε ∈ 2}.

(4) If p ≤n q, then p ≤i q for every i ≤ n.

We also note:

(5) If p ≤ q and n ∈ ω, then there is an nth branching point t of q such that t ∈ p.

For, let s be an nth branching point of p. Then it is an mth branching point of qfor some m ≥ n. Let t ≤ s be an nth branching point of q. Thus t ∈ p, as desired.

(6) If p ≤n q, then every nth branching point of q is an nth branching point of p.

Thus p ≤n q means that p ⊆ q, and any points of q thrown away to get p havemore than n branching points strictly below them.

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A fusion sequence is a sequence such that

p0 ≥0 p1 ≥1 p2 ≥2 · · · ≥n−1 pn ≥n · · ·

Lemma 4.21 (Fusion lemma). If 〈pn : n ∈ ω〉 is a fusion sequence, then pdef=⋂

n∈ω pn is a perfect tree, and p ≤n pn for all n ∈ ω.

Proof. Let n ∈ ω, and let s be an nth branching point of pn. If n ≤ m, thenpn ≥n pm, and so s is a branching point of pm, so that s, s0, s1 ∈ pm. Hences, s0, s1 ∈ p, and s is a branching point of p.

Thus we just need to see that p is a perfect tree. If t ∈ p and m < dmn(t),then obviously t � m ∈ p.

Now suppose that s ∈ p; we want to find t ≥ s such that t0, t1 ∈ p. Letm = dmn(s). Choose an mth branching point t of pm with s ≤ t. By the firstparagraph of this proof we have t, t0, t1 ∈ p, as desired. �

If p is a perfect tree and t ∈ p, we define

p � t = {u ∈ p : u and t are comparable}.

So p � t is a perfect tree; it does not branch up to the height of t, and from t andabove it is the tree p.

Now let p be a perfect tree, s an nth branching point of p, and t one of theimmediate successors of s in p. Suppose that q ≤ p � t. Then

rdef= q ∪ {u ∈ p : u and t are incomparable}

is a perfect tree called the amalgamation of q into p at t. It is obtained from p byreplacing the part of p above t by the part of q above t. Clearly r ≤n p. Also notethat (r � t) = q.

Let Q be the collection of all perfect trees. Q is called Sacks forcing. Itsgreatest element is <ω2, the full binary tree of height ω. The partial order P thatwe are concerned with is the σ-product of κ copies of Q; it consists of all p ∈ κQsuch that p(α) = 1 for all but countably many α ∈ κ, where 1 is the full binarytree of height ω. The support of an element p ∈ P is the set of all α ∈ κ such thatp(α) �= 1; it is denoted by supp(p). For p, q ∈ P we define p ≤ q iff p(α) ⊆ q(α)for all α < κ.

Lemma 4.22. Assuming CH, P has the ω2-chain condition.

Proof. Let 〈pα : α < ω2〉 be a sequence of members of P ; we want to find distinctα, β < ω2 such that pα and pβ are compatible. By CH, we may assume that〈supp(pα) : α < ω2〉 forms a Δ-system, say with kernel M . Now by CH thereare at most ω1 perfect trees, so there are at most ω1 functions from M into thecollection of all perfect trees. Hence we may assume that pα � M = pβ � M for allα, β < ω2. Then any distinct α, β < ω2 work. �

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The following lemma is essentially Theorem 5.12 of Jech [86], but we give a proofof it. The notation introduced in the proof will be used later.

Lemma 4.23. Let M be a countable transitive model of ZFC, and in M let P be theσ-product of κ-many Sacks forcings, where κ is any infinite cardinal. Suppose thatB ∈ M , p ∈ P , X is a P -name, and p � X : ω → B. Then there is a countableA ∈M and a p∞ ≤ p, p∞ ∈ P , such that p∞ � X : ω → A.

Proof. We assume given a well-ordering of all objects that play a role in thisproof. This is so we can make the construction very definite, implicitly choosingthe “first” object when we make an arbitrary choice. We need an auxiliary functiong : ω → ω × ω; it is the standard bijection. Let g(0) = (0, 0). If g(n) has beendefined, say g(n) = (i, j), let

g(n+ 1) =

{(i+ 1, j − 1), if j �= 0;

(0, i+ 1), otherwise.

Then g maps onto ω × ω, and if g(n) = (i, j), then i ≤ n.

We construct a sequence p = p0 ≥ p1 ≥ p2 ≥ · · · of members of P , and finitesets Ai, Si, Fi, Gij for i, j ∈ ω. Let p0 = p, A0 = ∅, S0 = supp(p), 〈G0j : j < ω〉the first system of finite sets such that

⋃j∈ω G0j = S0, and F0 = G00.

Now suppose that pi, Ai, Si, Fi, Gij have been defined for all i < n and allj ∈ ω. For each i ∈ Fn−1 let Ein be the set of all successors of all nth branchingpoints of the tree pn−1(i). Let σ0n, . . . , σlnn be all of the functions σ with domainFn−1 such that σ(i) ∈ Ein for all i ∈ Fn−1. We construct qln+1,n ≤ · · · ≤ q1n ≤ q0nin P and An = {a0n, . . . , alnn} as follows. Let q0n = pn−1. Assume that qkn hasbeen defined so that qkn ≤ q(k−1)n ≤ · · · ≤ q0n, and qkn(i) ≤n q(k−1)n(i) ≤n

· · · ≤n q0n(i) if i ∈ Fn−1. Now σkn(i) is a successor of an nth branching point ofqkn(i) if i ∈ Fn−1. Let

q′kn(i) ={qkn(i) � σkn(i), if i ∈ Fn−1,

qkn(i), otherwise.

So q′kn ≤ qkn. Hence there is an rkn ≤ q′kn and an akn ∈ B such that rkn � X(n) =akn. Let

q(k+1)n(i) =

{amalgamation of rkn(i) into qkn(i) if i ∈ Fn−1,

rkn(i), otherwise.

Thus q(k+1)n(i) ≤n qkn(i) if i ∈ Fn−1. Let pn = qln+1,n. Thus pn(i) ≤n pn−1(i)for all i ∈ Fn−1.

Let Sn = supp(pn), and let 〈Gnj : j ∈ ω〉 be the first system of finite setswith union Sn. Note that if i ≤ n and g(i) = (k, l), then k ≤ i ≤ n, so that Gg(i)

has been defined. Let Fn =⋃

i≤n Gg(i). This finishes the construction.

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Let S =⋃

n∈ω Sn. Hence for all i ∈ S, p∞(i)def=⋂

n∈ω pn(i) is a perfect tree,by the fusion lemma, since for i ∈ Fn we have

p0(i) ≥ · · · ≥ pn−1(i) ≥n pn(i) ≥n+1 pn+1(i) ≥ · · · .

Let p∞(i) = 1 for i /∈ S. Define A =⋃

n∈ω An. Now we prove that p∞ � X :ω → A, which will finish the proof. And to do this it suffices to show that, for anyn ∈ ω,

p∞ � X(n) = a0n ∨ · · · ∨ X(n) = alnn.

In turn, to do this it suffices to take an arbitrary q ≤ p∞ and find q ≤ q and ksuch that q � X(n) = akn. Choose q ≤ q and b ∈ B such that q � X(n) = b.Consider Fn−1 and σ1n . . . σlnn as above. For each i ∈ Fn−1 let τ(i) ∈ Ein ∩ q(i);it exists since q(i) ≤ q(i) ≤ p∞(i) ≤n pn−1(i). Say τ = σkn. Then if rkn is asabove, we have q(i) � σkn(i) ≤ q(k+1)n(i) � σkn(i) = rkn(i) for i ∈ Fn−1. Thus ifq(i) = q(i) � σkn(i) for i ∈ Fn−1 and q(i) = q(i) otherwise, then q ≤ rkn, q. Infact, clearly q ≤ q, and q(i) ≤ rkn(i) for i ∈ Fn−1. For i /∈ Fn−1,

q(i) = q(i) ≤ q(i) ≤ p∞(i) ≤ pn(i) ≤ q(k+1)n(i) = rkn(i),

as desired. q � X(n) = akn, as desired. �

We now begin the proof of Theorem 4.20 itself. Let G be P -generic over M . ByLemma 4.22, cardinals ≥ ω2 are preserved in M [G]. Applying Lemma 4.23 withB = ω1 we see that ω1 is also preserved. So cardinals are preserved in M [G].

Next, for α < κ and n ∈ ω let

Mαn = {p ∈ P : pα has no branching up the nth level}.

Clearly each such set Mαn is dense in P . Letting gα =⋂

p∈G pα, it follows thatgα ∈ ω2. For α �= β, let

Nαβ = {p ∈ P : ∃n∃s ∈ n2[s ∈ pα�pβ ]}.

We claim that each set Nαβ is dense. In fact, let p ∈ P be arbitrary, and chooses so that s0, s1 ∈ pα. If s0 /∈ pβ or s1 /∈ pβ, then p ∈ Nαβ , as desired. Supposethat s0, s1 ∈ pβ . Choose q ≤ p such that s1 /∈ qα but s1 ∈ qβ . Then q ∈ Nαβ , asdesired.

It follows that κ ≤ 2ω in M [G].

For each p ∈ P and each subset Γ of κ, let p � Γ be the function which agrees withp on Γ and is the 1 of P otherwise. By a result of Laver [84], let F ′ be a Ramseyultrafilter in M which generates a Ramsey ultrafilter F in M [G]. By Theorem4.14, we only need to show that ωA/F has no chain of type ω2, in M [G]. Notethat A is still the BA of finite and cofinite subsets of ω in M [G], by absoluteness.

We argue by contradiction. Thus there is a sequence fdef= 〈fα : α < ω2〉 ∈M [G] of

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members of ωA such that 〈fα/F : α < ω2〉 is strictly increasing. Let f be a namefor f . Then there is a p ∈ P such that

p �∀α < ω2(fα ∈ ωA)

∧ ∀α, β < ω2∃a ∈ F ′[α < β → a ⊆ {n ∈ ω : fα(n) < fβ(n)}].

Thus for each α < ω2 we have p � fα : ω → A, so we can apply the proof ofLemma 4.23 to p. We thus obtain for each α certain constructed objects in V ; withan obvious correspondence with that proof, they are, for all n, j ∈ ω,

pαn, Gαnj ,

Fαn , Eα

in for all i ∈ Fαn−1,

Aαn , lαn ,

and, for all i = 1, . . . , lαn ,σαin, qαin,

(qαin)′, rαin,

and aαin;

finally, we have pα∞. Now we claim

(1) ∀α < ω2∀u ≤ pα∞∀n ∈ ω∀j ∈ ω[[u � j ∈ fα(n) iff u � supp(pα∞) � j ∈ fα(n)]and [(u � j /∈ fα(n) iff u � supp(pα∞) � j /∈ fα(n)]].

Suppose that α < ω2, u ≤ pα∞, n ∈ ω, j ∈ ω, u � j ∈ fα(n), and u � supp(pα∞) �j ∈ fα(n); we want to get a contradiction. Choose v ≤ u � supp(pα∞) such thatv � j /∈ fα(n). For each i ∈ Fα

n−1 let τ(i) ∈ Eαin ∩ v(i). Such a τ(i) exists

since v(i) ≤ u(i) ≤ pα∞(i) ≤ pαn−1(i); then (5) gives an nth branching point wof pαn−1(i) such that w ∈ v(i), and a successor of w in v(i) is as desired. Say

τ = σαkn. Thus rαkn � fα(n) = aαkn, and rαkn ≤ (qαkn)

′. Also, if i ∈ Fαn−1, then

v(i) � σαkn(i) ≤ qαk+1,n � σα

kn(i) = rαkn(i). Let v(i) = v(i) � σαkn(i) for i ∈ Fα

n−1

and v(i) = v(i) otherwise. Then v ≤ rαkn, v, so j /∈ aαk . On the other hand,u(i) � σα

kn ≤ rαkn(i) for i ∈ Fαn−1. Let u(i) = u(i) � σα

kn(i) for i ∈ Fαn−1 and

u(i) = u(i) otherwise. Then u ≤ rαkn, u. So j ∈ aαk , contradiction. The other partof (1) is similar.

Now we may assume that 〈supp(pα∞) : α < ω2〉 forms a Δ-system, say withkernel Δ. Note that for each α < ω2, p

α∞ � Δ : Δ → P(<ω2); the set of all

such functions has, by CH in V , ω1 elements. Hence we may assume that for allα, β < ω2 we have pα∞ � Δ = pβ∞ � Δ. Next, for each α < ω2, the set supp(pα∞)\Δhas a certain countable order type. There are ω1 countable order types, so we mayassume that all such order types are the same. Thus for any α, β < ω2 there is aunique order isomorphism παβ from supp(pα∞)\Δ onto supp(pβ∞)\Δ. We extendπαβ to a permutation of κ, still denoted by παβ , by letting it be πβα (= π−1

αβ ) on

supp(pβ∞)\Δ and the identity elsewhere. Thus παβ = πβα. And this permutationπαβ extends to other objects; for example, if p ∈ P , then παβ(p) is the member q

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of P such that q(i) = p(παβ(i)) for all i ∈ κ. Note here that if p has support Γ,then παβ(p) has support παβ [Γ]. Now consider the objects

πα0(pαn), πα0[G

αnj ],

πα0[Fαn ], 〈Eα

(πα0(i))n: i ∈ πα0[F

αn−1]〉,

Aαn , lαn ,

and, for all i = 1, . . . , lαn ,

σαin ◦ π0α, πα0(q

αin),

πα0((qαin)

′), πα0(rαin),

and aαin;

and, finally, πα0(pα∞). By CH, there are only ω1 of these things, so we may assume

that they are the same for all α ∈ ω2\{0}. Now take any two distinct α, β ∈ ω2\{0}.Thus, for example,

(παβ(pα∞))(i) = (π0β(πα0(p

α∞)))(i)

= (πα0(pα∞))(π0β(i))

= (πβ0(pβ∞))(π0β(i))

= pβ∞(i).

Hence παβ(pα∞) = pβ∞. Another useful fact now is that if i ∈ Fα

n−1 then E(παβ(i))n =

Eαin for all i ∈ Fα

n−1. In fact, πα0(i) ∈ πα0[Fαn−1] = πβ0[F

βn−1] and Eβ

(παβ(i))n=

Eβ(πβ0(πα0(i)))n

= Eα(πα0(πα0(i)))n

= Eαin.

Next,

(2) ∀u ≤ pα∞∀n ∈ ω∀j ∈ ω(u � j ∈ fα(n) iff παβ(u) � j ∈ fβ(n)).

For, suppose that u � j ∈ fα(n) but παβ(u) � j ∈ fβ(n). By (1) we may assume

that supp(u) ⊆ supp(pα∞). Choose v ≤ παβ(u) such that v � j /∈ fβ(n). Now if

i ∈ F βn−1, then

v(i) ≤ (παβ(u))(i) ≤ (παβ(pα∞))(i) = pβ∞(i),

so there is a τ(i) ∈ Eβin ∩ v(i). Say τ = σβ

kn. Thus rβkn � fβ(n) = aβkn. As above,

let v(i) = v(i) � σβkn(i) for i ∈ F β

n−1 and v(i) = v(i) otherwise. Then v ≤ rβkn, v, so

j /∈ aβkn. Now πβαv ≤ u. Furthermore, if i ∈ Fαn−1 then παβi ∈ F β

n−1, and so

σαkn(i) = (πβα(σ

βkn))(i) = σβ

kn(παβ(i)) ∈ Eβ(παβ(i))n

∩v(παβ(i)) = Eαin∩(πβα(v))(i).

Now let w(i) = (πβαv)(i) � σαkn(i) for i ∈ Fα

n−1 and w(i) = (πβαv)(i) otherwise.Then w ≤ rαkn and w ≤ πβαv ≤ u, so j ∈ ak, contradiction. This proves (2).

Now let s be the member of P which agrees with pα∞ and pβ∞ on their supportsand is 1 otherwise. Clearly παβ(s) = s. We may assume that α < β. Then

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(3) s � ∃X ∈ F ′∀i ∈ X∀j ∈ ω(j ∈ fα(i)→ j ∈ fβ(i)).

We claim that

(4) s � ∃X ∈ F ′∀i ∈ X∀j ∈ ω(j ∈ fβ(i)→ j ∈ fα(i)).

This is a clear contradiction. So, it suffices to prove (4). By (3), there is a u ≤ sand an X ∈ F ′ such that

(5) u � ∀i ∈ X∀j ∈ ω(j ∈ fα(i)→ j ∈ fβ(i)).

It suffices now to show

παβ(u) � ∀i ∈ X∀j ∈ ω(j ∈ fβ(i)→ j ∈ fα(i)).

So, let v ≤ παβ(u), i ∈ X , j ∈ ω, and assume that v � j ∈ fβ(i). Since v ≤s ≤ pβ∞, from (2) we get παβ(v) � j ∈ fα(i). And παβ(v) ≤ u, so by (5) we get

παβ(v) � j ∈ fβ(i). But παβ(v) ≤ s ≤ pα∞, so by (2) again, v � j ∈ fα(i), asdesired. �

Shelah has a more recent construction providing examples with

Depth(∏

i∈IAi/D

)<∏

i∈IDepth(Ai)/D,

and this construction applies to some other functions too. To formulate this result,we need a definition.

Suppose that O is an operation on sequences of BAs, and inv is a cardinalinvariant on BAs such that |inv(B)| ≤ |B| for every infinite BA B. Then we saythat the property �O holds provided that if μ is a cardinal and Bi is a BA foreach i < μ+, then

supi<μ+

inv(Bi) ≤ inv(Oi<μ+Bi

)≤ μ+ sup

i<μ+

invBi.

Roslanowski, Shelah [98] proved the following result:

Suppose that inv is a cardinal invariant on BAs satisfying �⊕ or �Πw ; supposethat inv(B) ≤ |B| for any BA B. Suppose that for each infinite cardinal χ there isa BA B such that χ < inv(B) and there is no inaccessible cardinal in the interval(χ, |B|]. Assume further that

' 〈λi : i < κ〉 is a sequence of weakly inaccessible cardinals λi > κ+, D is anℵ1-complete ultrafilter on κ, and

∏i<κ(λi, <)/D is μ+-like.

Then there exist BAs Bi for i < κ such that

inv(Bi) = λi and inv

(∏i<κ

Bi/D

)≤ μ.

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As a corollary of this and Magidor, Shelah [98] one has:

Suppose that inv is a cardinal invariant on BAs such that the following threeconditions hold:

(i) inv(B) ≤ |B| for all infinite BAs B.

(ii) supi<μ+ inv(Bi) ≤ inv(∏w

i<μ+ Bi

)≤ μ + supi<μ+ inv(Bi) for every system

〈Bi : i < μ+〉 of BAs.Then it is consistent to have a system 〈Bi : i < κ〉 of BAs such that

inv(∏

i<κ Bi/D)<∏

i<κ inv(Bi)/D.

This corollary applies not only to depth, but also to length, independence, π-character, and tightness.

For subdirect products the situation is similar to that for cellularity, with essen-tially the same proof: there is a BA with depth ω which is a subdirect product ofBAs having high depth.

Depth of moderate products can easily be described using the arguments forproducts:

Theorem 4.24. Depth(∏B

i∈I Ai) = max{Depth(B), supi∈I Depth(Ai)}.

Proof. We use the notation introduced in Chapter 1 for moderate products. Clearly≥ holds. Now let κ = max{Depth(B), supi∈I Depth(Ai)}, and, to get a contradic-tion, suppose that 〈h(bα, Fα, a

α) : α < κ+〉 is a strictly increasing sequence. With-out loss of generality each h(bα, Fα, a

α) is normal. Then bα ⊆ bβ for α < β < κ+.

Hence there is a Γ ∈ [κ+]κ+

such that bα = bβ for all α, β ∈ Γ. Then aα < aβ forα < β, both in Γ, and this easily gives a contradiction. �

Depth for one-point gluing behaves like arbitrary products if all algebras havemore than two elements and I is infinite or some Ai is infinite. In fact, assumethe notation from Chapter 1, and let B be the one-point gluing. We claim thatDepth(B) = |I|+supi∈I Depth(Ai). ≤ holds since B is a subalgebra of the product.Now suppose that i ∈ I and 〈aα : α < κ〉 is strictly increasing in Ai, κ an infinitecardinal. If some aα ∈ Fi, we may assume that all aα ∈ Fi, and then we can define

bα(j) =

{aα if j = i,

1 otherwise,

and we obtain a strictly increasing sequence of length κ in B. If all aα are not in Fi

we can work similarly, with 0 instead of 1. This shows that Depth(Ai) ≤ Depth(B).To show that |I| ≤ Depth(B) when I is infinite, let 〈fα : α < κ〉 be a one-oneenumeration of I. For each α < κ choose aα ∈ A+

f(α)\Ff(α). Now for each β < κ

define

bβ(α) =

{aα if α < β,

0 otherwise.

Then 〈bβ : β < κ〉 is a strictly increasing sequence of elements of B.

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Note that a one-point gluing of a product of two-element algebras still hasjust two elements.

For the Alexsandroff duplicate we have Depth(Dup(A)) = Depth(A). In fact,if 〈aα : α < κ〉 is strictly increasing in A, then 〈(aα,S(aα)) : α < κ〉 is strictlyincreasing in Dup(A). Now suppose that 〈(aα, Xα) : α < κ〉 is strictly increasingin Dup(A), but A does not have a strictly increasing sequence of length κ. Notethat κ > ω. Define α ≡ β iff α, β < κ and aα = aβ . Then there are fewerthan κ equivalence classes under ≡, so some equivalence class M is uncountable.Let 〈αξ : ξ < δ〉 be a strictly increasing enumeration of M . Now if ξ < δ, thenXαξ

\S(aα0) is finite, so there is an η < ω1 such thatXαξ\S(aα0) = Xαη\S(aα0) for

all ξ ∈ [η, ω1). Similarly there is a ρ ∈ [η, ω1) such that S(aα0)\Xαξ= S(aα0 )\Xαρ

for all ξ ∈ [ρ, ω1). Now Xαρ ⊂ Xαρ+1 . Take any b ∈ Xαρ+1\Xαρ . Thus

b ∈ S(aα0 )→ b ∈ (S(aα0 )\Xαρ)\(S(aα0 )\Xαρ+1);

b /∈ S(aα0 )→ b ∈ (Xαρ+1\S(aα0))\(Xαρ\S(aα0));

contradiction.

Concerning the exponential, from Lemma 1.22(i),(vi) it is clear that if 〈aα :α < κ〉 is strictly increasing in A, then 〈V (S(aα)) : α < κ〉 is strictly increasingin Exp(A). Hence Depth(A) ≤ Depth(Exp(A)). We do not know whether there isa BA A with < here.

Next we discuss derived functions with respect to depth. The first result is thatDepthH+ is the same as tightness. To prove this, we need an equivalent form oftightness due to Arhangelskiı and Shapirovskiı. It involves the notion of a freesequence in a topological space. Let X be a topological space. A free sequence inX is a sequence 〈xξ : ξ < α〉 (α an ordinal) of elements of X such that for all

ξ < α we have {xη : η < ξ} ∩ {xη : ξ ≤ η < α} = ∅. For an arbitrary topologicalspace X and a point x ∈ X , the tightness t(x) of x in X is, by definition, the leastcardinal κ such that if Y ⊆ X and x ∈ Y , then there is a subset Z ⊆ Y such that|Z| ≤ κ and x ∈ Z. And the tightness t(X) of X itself is supx∈X t(x). Clearly thismeans that t(A) = t(Ult(A)) for any BA A. The equivalent form of tightness dueto Arhangelskiı (based on proofs of Shapirovskiı) is given in the following theorem.

Theorem 4.25. Let X be a compact Hausdorff space, and let κ be a regular cardinal.

(i) If there is a free sequence of length κ, then there is a y ∈ X such that t(y) ≥ κ.

(ii) If there exist y ∈ X and Y ⊆ X such that y ∈ Y and ∀Z ∈ [Y ]<κ[y /∈ Z]],then there is a free sequence of length κ.

(iii) t(X) = sup{λ : there is a free sequence of length λ}.

Proof. First we note that (i) gives ≥ in (iii). For ≤ in (iii), suppose that μ < t(X).Then the hypothesis of (ii) holds for κ = μ+, and so (ii) yields a free sequence oflength μ+. This gives ≤ in (iii). So it suffices to prove (i) and (ii).

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(i): Suppose that 〈xξ : ξ < κ〉 is a free sequence, where κ is regular; we shallfind a point y ∈ X such that t(y) ≥ κ. First note:

(1) There is a y ∈ X such that |U ∩ {xξ : ξ < κ}| = κ for each neighborhoodU of y.

In fact, otherwise for every y ∈ X let U(y) be an open neighborhood of y suchthat |U(y) ∩ {xξ : ξ < κ}| < κ. Thus {U(y) : y ∈ X} is an open cover of X . LetU(y0), . . . , U(yn−1) be a finite subcover. Then

{xξ : ξ < κ} =⋃i<n

(U(yi) ∩ {xξ : ξ < κ}),

and the right side has cardinality < κ, contradiction. So (1) holds.

Take y as in (1). Assume that t(y) < κ. Now y ∈ {xξ : ξ < κ}. Hence bythe definition of tightness, choose a subset Γ of κ of power < κ such that y ∈{xξ : ξ ∈ Γ}. Let η = sup(Γ) + 1. Hence y ∈ {xξ : ξ < η}, so by freeness y /∈{xξ : η ≤ ξ}. So there is a neighborhood U of y such that U ∩ {xξ : η ≤ ξ} = 0.This contradicts (1). Thus (i) holds.

(ii): Suppose that κ is regular and ∃y ∈ X∃Y ⊆ X [y ∈ Y and ∀Z ∈ [Y ]<κ[y /∈Z]]. We will construct a free sequence of length κ. Set

Y ′ = {x : there is a Z ∈ [Y ]<κ such that x ∈ Z}.

Thus Y ⊆ Y ′, so y ∈ Y ′. Note

(2) If Z ∈ [Y ′]<κ, then y /∈ Z.

In fact, suppose that Z ∈ [Y ′]<κ. For each x ∈ Z choose Wx ∈ [Y ]<κ such thatx ∈Wx. Let V =

⋃x∈Z Wx. Then V ∈ [Y ]<κ and Z ⊆ V . So y /∈ Z.

By the same proof we have

(3) If Z ∈ [Y ′]<κ and z ∈ Z, then z ∈ Y ′.

We now construct xξ, Fξ, Uξ for ξ < κ such that xξ ∈ Y ′, y ∈ Fξ ⊆ Uξ withUξ open and Fξ a closed neighborhood of y, by recursion. Suppose these have

been constructed for all η < ξ, where ξ < κ. Since y /∈ {xη : η < ξ}, let Uξ bean open neighborhood of y such that Uξ ∩ {xη : η < ξ} = 0. Let Fξ be a closedneighborhood of y such that Fξ ⊆ Uξ. Then we claim

(4) Y ′ �⊆⋃

η≤ξ(X\Fη) ∪ {xη : η < ξ}.

For, suppose not; then we show that y ∈ {xη : η < ξ} (contradiction). For, let Ube an open neighborhood of y and let F ′ be a closed neighborhood of y whichis included in U . Let W be the closure of the set {Fη : η ≤ ξ} ∪ {F ′} under

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finite intersections. Since y ∈ Y ′, for all H ∈ W choose zH ∈ Y ′ ∩ H . ThenH ′ ∩ {zH : H ∈W} �= 0 for all H ′ ∈ W . Choose

t ∈⋂

H′∈W

H ′ ∩ {zH : H ∈ W}.

By (3), t ∈ Y ′. Now t ∈ Fη for all η ≤ ξ, so by the “suppose not” for (4),

t ∈ {xη : η < ξ}. Since t ∈ F ′ ⊆ U , it follows that U ∩{xη : η < ξ} �= 0, as desired.

So (4) holds; choose xξ in the left side of (4) but not in the right side. Thiscompletes the construction.

Suppose ξ < κ+ and s ∈ {xη : η < ξ} ∩ {xη : ξ ≤ η < κ}. Then s /∈ Uξ, sos /∈ Fξ. Thus s ∈ X\Fξ, which is open, so there is an η with ξ ≤ η < κ such thatxη ∈ X\Fξ, contradiction. �

By Theorem 4.25, if t(X) is regular and is attained in the free sequence sense thenit is attained in the defined sense, i.e., there is a point y with tightness t(X).

Theorem 4.26. For any infinite BA A we have DepthH+(A) = t(A).

Proof. For ≥, let 〈Fξ : ξ < κ〉 be a free sequence; we produce a quotient A/I of Ahaving a strictly increasing sequence of order type κ. For brevity let Y = {Fξ : ξ <κ}. For every ξ < κ there is an element aξ of A such that {Fη : η < ξ} ⊆ S(aξ)and S(aξ) ∩ {Fη : ξ ≤ η < κ} = 0.

Consider the following ideal on A: I = {x ∈ A : Y ⊆ S(−x)}. Suppose ξ <η < κ. Then S(aξ ·−aη)∩Y = 0: if Fν ∈ S(aξ ·−aη), then Fν ∈ S(aξ), hence ν < ξ,and −aη ∈ Fν , hence η ≤ ν, so η < ξ, contradiction. This shows that [aξ] ≤ [aη]for ξ < η < κ. Still suppose that ξ < η < κ. Then Fξ ∈ S(aη)\S(aξ) = S(aη ·−aξ).Thus Y �⊆ S(−aη + aξ), so aη · −aξ /∈ I, which means that [aη] < [aξ], as desired.

For ≤, let I be an ideal in A, and let 〈[aξ] : ξ < κ〉 be a strictly increasingsequence in A/I. For each ξ < κ, the set {x : −x ∈ I}∪ {aξ+1,−aξ} has the finiteintersection property, since aξ+1 · −aξ /∈ I. Let Fξ be an ultrafilter including thisset. Then, we claim, 〈Fξ : ξ < κ〉 is a free sequence. To prove this it suffices toshow that for any ξ < κ we have

(1) {Fη : η < ξ} ⊆ S(aξ) and S(aξ) ∩ {Fη : ξ ≤ η < κ} = 0.

If η < ξ < κ, then aη+1 ·−aξ ∈ I, and hence −aη+1+aξ ∈ Fη; but also aη+1 ∈ Fη,so aξ ∈ Fη and so Fη ∈ S(aξ), proving the first part of (1). For the second part,suppose that ξ ≤ η < κ and Fη ∈ S(aξ). Now aξ · −aη ∈ I, so −aξ + aη ∈ Fη; butalso −aη ∈ Fη, so −aξ ∈ Fη, contradiction. �

Corollary 4.27. DepthH+ and t (for free sequences) have the same attainmentproperties, i.e., for any BA A and any infinite cardinal κ, A has a homomorphicimage with a chain of order type κ iff Ult(A) has a free sequence of type κ. �

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Note that, as in the relation between spread and cellularity, DepthH+ involves twosups, while t for free sequences involves only one; we return to this below.

Since Depth(A) ≤ c(A), it is clear that DepthH−(A) = ω. It is also easy tosee that DepthS+(A) = Depth(A) and DepthS−(A) = ω. Depthh+ is a little moreinteresting:

Theorem 4.28. Depthh+(A) = s(A) for any infinite BA A.

Proof. For ≥, suppose that Y is a discrete subspace of Ult(A); clearly Y , since itis discrete, has an increasing sequence of closed-open sets of order type |Y |. For≤, suppose that Y is a subspace of Ult(A) and 〈Uκ : κ < κ〉 is a strictly increasingsystem of clopen subsets of Y . For each κ < κ choose yκ ∈ Uκ+1\Uκ. Clearly{yκ : κ < κ} is a discrete subspace of Ult(A). �

The proof shows that Depthh+(A) and s(A) have the same attainment properties.

Since Depthh−(A) ≤ DepthH−(A), we have Depthh−(A) = ω for any infiniteBA A. Obviously dDepthS+(A) = Depth(A) for any BA A.

The status of the derived function dDepthS− is not clear. Note that for Athe interval algebra on a cardinal κ we have dDepthS−(A) = ω: this follows uponconsidering the subalgebra of A generated by {{α} : α a non-limit ordinal < κ}.Also, Koppelberg and Shelah have independently observed that if A is atomlessand λ-saturated (in the model-theoretic sense), then dDepthS−(A) ≥ λ. To showthis, suppose that B is a dense subalgebra of A. By induction choose elementsaα ∈ A and bα ∈ B for α < λ so that α < β implies that aα > bα > aβ > 0; theaα’s can be chosen by λ-saturation, and the bα’s by denseness. So the sequence〈bα : α < λ〉 shows that the depth of B is at least λ.

Problem 31. Is it true that for all infinite cardinals κ ≤ λ there is a BA A of sizeλ with dDepthS−(A) = κ?

We now discuss small depth towspect and tow. These are defined as follows. Atower in a BA A is a sequence 〈aα : α < κ〉 of elements of A\{1} which is strictlyincreasing and has sum 1. We define

tow(A) = min{κ : A has a tower of length κ};towspect(A) = {κ : κ is regular and A has a tower of length κ}.

For convenience we define tow(A) = ∞ if A does not have a tower. Note thattowspect is a set of regular cardinals. tow and towspect are discussed in Monk[01a], Monk [02], Monk [07], and in several articles concerning the special caseA = P(ω)/fin. We indicate some of the general results about BAs.

In some cases we consider towers in a reversed sense: a strictly decreasingsequence of nonzero elements with product 0; we call these ≥-towers. Obviouslytowers in one sense exist iff they exist in the other sense.

Note that tow(A) <∞ for all atomless BAs, but not for all BAs. For example,the finite-cofinite algebra on an uncountable set does not have any towers. There

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are atomic BAs with towers, though, for example an interval algebra on a cardinal.The following simple characterization of BAs that have towers is sometimes useful.

Proposition 4.29. For any infinite BA A the following conditions are equivalent:

(i) A has a tower.

(ii) Every maximal ideal I on A has one of the following properties:

(a) There is a strictly increasing sequence of elements of I with sum 1.

(b) There is a strictly decreasing sequence of elements in I\{0} with prod-uct 0.

Proof. (i)⇒(ii): suppose that 〈aα : α < κ〉 is strictly increasing with sum 1, κ aninfinite cardinal. Let I be any maximal ideal in A. If all aα ∈ I, then (a) holds.If aα /∈ I for some α, then 〈−aβ : α ≤ β < κ〉 is a strictly decreasing sequence ofelements of I\{0} with product 0.

(ii)⇒(i): obvious. �Corollary 4.30. Each of the following BAs fails to have a tower:

(i) Finco(κ), with κ an uncountable cardinal;

(ii)∏w

i∈I Ai, with I uncountable, and such that each Ai fails to have a tower.

Proof. (i): Consider the maximal ideal of all finite subsets of κ.

(ii): Consider the maximal ideal consisting of all a ∈∏w

i∈I Ai such that{i ∈ I : ai �= 0} is finite. �Corollary 4.31.

(i) Every atomless BA has a tower.

(ii) If a(A) = ω, in particular if c(A) = ω or |A| = ω, then A has a tower.

(iii) If A has a tower, then so do A×B and A⊕B. �

A weak tower is a strictly increasing sequence 〈aα : α < κ〉 of elements of Adifferent from 1, with

∑α<κ aα = 1, where κ is a cardinal, not necessarily regular.

For brevity let

towwspect(A) = {κ : κ is the length of a weak tower in A}.

Clearly towspect(A) ⊆ towwspect(A).

The following general set theoretic fact is useful in discussing this function.

Lemma 4.32. If κ and λ are infinite cardinals, with λ < κ and cf(κ) = cf(λ), thenthere is a strictly increasing function f : λ→ κ with rng(f) cofinal in κ.

Proof. Let 〈μξ : ξ < cf(κ)〉 be a strictly increasing continuous sequence of ordinalswith supremum λ, with μ0 = 0, and let 〈νξ : ξ < cf(κ)〉 be a strictly increasingcontinuous sequence of ordinals with supremum κ, with λ ≤ |νξ+1\νξ| for eachξ < cf(κ). Then f can be taken as the union of strictly increasing functions fromμξ+1\μξ into νξ+1\νξ for ξ < cf(κ). �

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Corollary 4.33. If A has a weak tower of order type κ, and λ is a cardinal suchthat λ < κ and cf(λ) = cf(κ), then A has a weak tower of order type λ.

Proof. Let 〈aξ : ξ < κ〉 be a weak tower in A. Let f be as in Proposition 4.32.Then 〈af(ξ) : ξ < λ〉 is also a weak tower in A. �

Problem 32. Let C be a nonempty set of infinite cardinals such that the followingcondition holds:

(∗) If κ ∈ C, λ < κ, and cf(λ) = cf(κ), then λ ∈ C.

Is there an atomless BA A such that towwspect(A) = C?

For this problem, see Theorem 4.44.

A very weak tower is a strictly increasing sequence 〈aα : α < β〉 of elementsof A different from 1, with

∑α<β aα = 1, where β is a limit ordinal, not necessarily

a cardinal.

Problem 33. Characterize those sets K of limit ordinals for which there is anatomless BA A such that K is the collection of all order types of very weak towersof A.

Now we consider our special subalgebras and tow.

Proposition 4.34. If A ≤reg B, then towspect(A) ⊆ towspect(B) and tow(B) ≤tow(A). The conclusion also holds if A ≤free B, A ≤proj B, A ≤rc B, A ≤π B. �

This proposition suggests the following question.

Problem 34. Given nonempty sets K ⊆ L of cardinals, are there BAs A ≤reg Bsuch that towspect(A) = K and towspect(B) = L? There is a similar question forA ≤free B, A ≤proj B, A ≤rc B, A ≤π B.

See also Proposition 4.45.

Proposition 4.35. If A ≤σ B and tow(A) > ω, then towspect(A) ⊆ towspect(B)and tow(B) ≤ tow(A).

Proof. The proof is much like that of Proposition 3.35. Thus let T be a towerin A, but suppose that it is not one in B. Then there is a b ∈ B\{1} such thata ≤ b for all a ∈ T . By hypothesis, A � b is a countably generated ideal of A; sayit is generated by the countable set Y . Let 〈aα : α < κ〉 enumerate T in strictlyincreasing order; thus κ is a regular uncountable cardinal by hypothesis. For eachα < κ let bα ∈ Y be such that aα ≤ bα. Since κ is regular uncountable and Y iscountable, there is a c ∈ Y such that bα = c for κ many α < κ. Hence c is aboveeach member of T , contradiction. �

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Problem 35. Given nonempty sets K ⊆ L of cardinals with ω /∈ K, are there BAsA ≤σ B such that towspect(A) = K and towspect(B) = L?

If tow(B) > ω, then for any countably infinite subalgebra of B we have A ≤σ Band tow(A) < tow(B).

As in the case of a, there are BAs A,B with A ≤u B and tow(A) < tow(B).

There are BAs A,B with A ≤mg B and tow(A) < tow(B). Namely, let B =Intalg(ω1) and let A be a countably infinite subalgebra of B appearing in a chainshowing that A is minimally generated; see Theorem 2.52.

It is possible, in the other direction, to have BAs A,B with A ≤mg B andtow(A) > tow(B). Namely let L be an η1 set, and let M = L + ω. Set B =Intalg(M) and A = Intalg(L). Then tow(B) = ω, since 〈[−∞, i) : i ∈ ω〉 is atower in B. Let C = {[−∞, u) : u ∈ L}. Then C generates A, and the functionf which is the identity on C and maps 1A to 1B extends to an isomorphism ofA into B by Remark 15.2 of the handbook. By the proof of Proposition 2.52 wehave A ≤mg B. Finally, we claim that tow(A) > ω. For, suppose that 〈ti : i < ω〉is a tower in A, with t0 = 0. Let E be the set of all intervals [a, b) occurring inthe standard representations of members of {ti : i < ω}. Note that E is infinite.If a �= −∞ for all [a, b) ∈ E, then by the η1 property there exist c < d with d < afor all [a, b) ∈ E. Hence [c, d) is disjoint from each member of D, contradiction.So [−∞, b) ∈ E for some b. Then by the same argument, there is some element[b, c) ∈ E. Continuing, we get elements [−∞, b0), [b0, b1), . . . of E. Then by theη1 property there exist [c, d) with each bi < c < d and d < e for any element[e, f) ∈ E such that bi < e for all i ∈ ω. Hence again [c, d) is disjoint from eachmember of D, contradiction. This proves the claim.

Proposition 4.36. If A ≤m B, then towspect(A) ⊆ towspect(B) and tow(B) ≤tow(A).

Proof. See the proof of Proposition 3.34. �

Again we have a problem in connection with this proposition:

Problem 36. Given nonempty sets K ⊆ L of cardinals, are there BAs A ≤m Bsuch that towspect(A) = K and towspect(B) = L?

There are BAs A,B with A ≤free B, tow(B) = ω, and tow(A) arbitrarily large.Similarly for ≤proj, ≤rc, ≤σ, ≤reg, and ≤u.

One can have A ≤π B, hence also A ≤reg B, with tow(A) arbitrary and tow(B) =ω. Simply take B = A.

There are BAs A,B with A ≤s B and tow(B) < tow(A). In fact, let A be ω1-saturated. Clearly tow(A) ≥ ω1. Let 〈aξ : ξ < ω1〉 be strictly increasing in A. LetI = {b ∈ A : an · b = 0 for all n ∈ ω}. Then I is an ideal in A. Let J = {0}. Thenlet B = A(x) with A � x = I and A � −x = J ; see Proposition 2.28. We claim that

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〈an ·−x+x : n ∈ ω〉 is a tower in B. If m < n, then clearly am ·−x+x ≤ an ·−x+x.Suppose that (an ·−x+x)·−(am·−x+x) = 0. Thus an ·−am ·−x = 0, so an ·am ∈ I.Hence an · −am = an · an · −am = 0, contradiction. Thus 〈an · −x+ x : n ∈ ω〉 isstrictly increasing. Suppose that (an · −x+ x) · (b · x + c · −x) = 0 for all n ∈ ω;we want to show that b · x+ c · −x = 0. We have

0 = (an · −x+ x) · (b · x+ c · −x) = b · x+ an · c · −x,

so b · x = 0 and an · c · −x = 0 for all n ∈ ω. Hence b ≤ −x and so b = 0. Alsoan · c ∈ I for all n ∈ ω; so an · c = an · an · c = 0 for all n ∈ ω, so that c ∈ I. Hencec ≤ x, so c · −x = 0.

Problem 37. Does A ≤s B imply that towspect(A) ⊆ towspect(B) or tow(B) ≤t(A)?

Problem 38. Are there BAs A,B such that A ≤m B and tow(B) < tow(A)?

Proposition 4.37. Let A and B be infinite BAs. Then

(i) towspect(A×B) = towspect(A) ∪ towspect(B).

(ii) tow(A×B) = min(tow(A), tow(B)).

(Note that since we allow tow to have ∞ as a value, (ii) includes the statementthat if neither tow(A) nor tow(B) have towers, then also A × B does not have atower.)

Proof. (i): Suppose that 〈aα : α < κ〉 is a tower in A. We may assume that a0 �= 0.Then

〈(0, 1)〉�〈(aα, 1) : α < κ〉

is a tower in A × B. Similarly, a tower in B yields a tower in A × B. Thus ⊇ in(i) holds.

Now suppose that 〈(aα, bα) : α < κ〉 is a tower in A × B, κ regular. Defineα ≡ β iff aα = aβ . Thus ≡ is an equivalence relation on κ. If there are κ equivalenceclasses, this gives a tower in A of order type κ. If there are fewer than κ equivalenceclasses, then some equivalence class has κ elements, and this gives a tower in B oforder type κ.

(ii): Note that the argument for (i) shows that A×B has a tower iff A or Bhas a tower; so (ii) follows from (i). �

Proposition 4.38. Suppose that 〈Ai : i ∈ I〉 is a system of infinite BAs, with Iinfinite.

(i) If |I| = ω, then towspect(∏w

i∈I Ai) = {ω} ∪⋃

i∈I towspect(Ai).

(ii) If |I| > ω, then towspect(∏w

i∈I Ai) =⋃

i∈I towspect(Ai).

(iii) If |I| = ω, then tow(∏w

i∈I Ai) = ω.

(iv) If |I| > ω, then tow(∏w

i∈I Ai) = mini∈I tow(Ai).

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Proof. Clearly (iii) and (iv) follow from (i) and (ii). From Proposition 4.37 itfollows that towspect(Ai) ⊆ towspect(

∏wi∈I Ai) for every i ∈ I. Now suppose that

〈xα : α < κ〉 is a tower in∏w

i∈I Ai, κ regular. We want to show that either|I| = ω = κ, or κ ∈

⋃i∈I towspect(Ai).

Case 1. There is an α < κ such that {i ∈ I : xα(i) �= 1} is finite. Then κ ∈⋃i∈I towspect(Ai) by Proposition 4.37.

Case 2. Mαdef= {i ∈ I : xα(i) �= 0} is finite for all α < κ. If κ < |I|, then there is an

i ∈ I such that xα(i) = 0 for all α < κ, contradicting∑

α<κ xα = 1. So |I| ≤ κ. Wemay assume that κ is uncountable and regular. Clearly Mα ⊆ Mβ if α < β < κ.Hence there is an α < κ such that Mα = Mβ for all β ∈ [α, κ). This again givesan i ∈ I such that xα(i) = 0 for all α < κ, contradicting

∑α<κ xα = 1. �

Kevin Selker has generalized Proposition 4.38 to moderate products:

Let 〈Ai : i ∈ I〉 be a system of BAs, and B a BA, subject to the conditions for a

moderate product. Then tow(∏B

i∈I Ai) = min(tow(B),mini∈I tow(Ai)).

Proposition 4.39. If I is infinite, then tow(∏

i∈I Ai) = ω for infinite BAs Ai. �

Concerning towers and free products we have the following result from Monk [07].

Theorem 4.40. towspect(A⊕B) = towspect(A) ∪ towspect(B) for any infinite BAsA,B. Hence tow(A⊕B) = min{tow(A), tow(B)}.

Proof. Clearly every tower in A is a tower in A ⊕ B; and similarly for B. So ⊇holds. Now suppose that 〈cα : α < κ〉 is a tower in A ⊕ B. Thus κ is a regularcardinal, and we want to show that κ ∈ towspect(A)∪ towspect(B). For each α < κwrite

(1) cα =∑i<nα

(aαi · bαi ),

where ∀i < nα[0 �= ai ∈ A and 0 �= bi ∈ B], and for all distinct i, j < nα, bαi ·bαj = 0.

We call nα the length of cα.

First we take the case κ > ω, where we adapt arguments of McKenzie, Monk[82]. Here we first take the case in which nα = n does not depend on α, and weproceed by induction on n. The statement that we prove by induction is as follows:

(2) For every positive integer n and every pair (w, x) ∈ A×B, if 〈cα : α < κ〉 is atower in (A⊕B) � (w · x) and each cα is given by (1), with nα = n for all α < κ,aαi ≤ w and bαi ≤ x for all i < n and α < κ, then A or B has a tower of ordertype κ.

For n = 1 we have cα = aα0 · bα0 for each α < κ. Clearly α < β implies that

aα0 ≤ aβ0 , bα0 ≤ bβ0 , and aα0 < aβ0 or bα0 < bβ0 . Define α ≡ β iff aα0 = aβ0 . This is an

equivalence relation on κ. If there are κ many equivalence classes, then there is a

strictly increasing subsequence 〈aα(ξ)0 : ξ < κ〉 of order type κ. It is easily checked

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that 〈aα(ξ)0 + −w : ξ < κ〉 is a tower in A. If there are fewer than κ equivalenceclasses, then one of them has size κ, and we get a strictly increasing sequence

〈bα(ξ)0 +−x : ξ < κ〉 which is a tower in B. This takes care of the case n = 1.

Now we assume inductively that n > 1. Obviously we have

(3) If α < β and i < n, then aαi · bαi ·∏

j<n(−aβj +−bβj ) = 0.

(4) If α < β and i < n, then bαi ≤∑

j<n bβj .

In fact, from (3) we have bαi ·∏

j<n(−bβj ) = 0, giving (4).

(5) If α < β, i, k < n, and bαi · bβk �= 0, then aαi ≤ aβk .

For, multiplying (3) by bβk we get aαi · bαi · bβk · −a

βk = 0, giving (5).

(6) If M is a finite subset of κ, then there is a γ ∈ Mn such that aαγ(α) ≤ aβγ(β)whenever α, β ∈M and α < β.

This follows by an easy induction from (4) and (5).

(7) There is a γ ∈ κn such that ∀α, β ∈ κ[α < β ⇒ aαγ(α) ≤ aβγ(β)].

This follows from (6) and the compactness of the space κn, since for any α < β

the set {γ ∈ κn : aαγ(α) ≤ aβγ(β)} is clearly closed.

Define α ≡ β iff aαγ(α) = aβγ(β). If there are κ equivalence classes, we get a

tower in A of order type κ, as in the argument for n = 1. So, assume otherwise,let M be an equivalence class with κ elements, and let u = aαγ(α) for any α ∈ M .Then cα = cα · −u+ cα · u. By Lemma 4.2 we now have two cases.

Case 1. There is an N ∈ [M ]κ such that 〈cα · −u : α ∈ N〉 is strictly increasing.Each cα · −u clearly has length less than n; so there is a P ∈ [N ]κ such that eachcα · −u has the same length less than n for each α ∈ P . Let 〈α(ξ) : ξ < κ〉 bethe strictly increasing enumeration of P . We claim that 〈cα(ξ) · −u : ξ < κ〉 is atower in (A ⊕ B) � (w · −u · x). To show this, it suffices to take d ∈ A � (w · −u),e ∈ B � x, assume that cα(ξ) · −u · d · e = 0 for every ξ < κ and show that d = 0 ore = 0. Since 〈cα(ξ) : ξ < κ〉 is a tower in (A ⊕ B) � (w · x) we have −u · d · e = 0.Hence either e = 0 or −u · d = 0. In the latter case d = 0 since d ≤ −u. Thisproves the claim. Now the induction hypothesis gives the desired result.

Case 2. There is an N ∈ [M ]κ such that 〈cα · u : α ∈ N〉 is strictly increasing.If {α ∈ N : cα · u can be written as in (1) with a smaller n} has size κ, thenwe get our result as in Case 1. So, suppose that this set has size less than κ. LetP = {α ∈ N : cα · u cannot be written as in (1) with a smaller n}; so |P | = κ.

(8) For all α, β ∈ P , if α < β then bαγ(α) ≤ bβγ(β).

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For, by (4) it suffices to show that bαγ(α) · bβi = 0 for all i �= γ(β). Suppose that this

product is nonzero for some i �= γ(β). Then by (5), aαγ(α) ≤ aβi , i.e., u ≤ aβi . Hence

cβ · u = u ·∑j<n

(aβj · bβj )

= u · aβγ(β) · bβγ(β) + u · aβi · b

βi + u ·

∑j �=i,γ(β)

(aβj · bβj )

= u · (bβγ(β) + bβi ) + u ·∑

j �=i,γ(β)

(aβj · bβj ),

which contradicts β ∈ P . Thus (8) holds.

Again, define α ≡ β iff α, β ∈ P and bαγ(α) = bβγ(β). This is an equivalence

relation on P . If there are κ equivalence classes, this yields a tower in B of ordertype κ. So, suppose that there are fewer than κ equivalence classes, and let Q bea class with κ elements. Say bαγ(α) = v for all α ∈ Q. Thus for any α ∈ Q we have

cα = u ·v+∑

i�=γ(α)(aαi ·bαi ). It follows that the sequence 〈

∑i�=γ(α)(a

αi ·bαi ) : α ∈ Q〉

is strictly increasing. Let 〈α(ξ) : ξ < κ〉 be the strictly increasing enumeration ofQ.

We claim that 〈∑

i�=γ(α(ξ))(aα(ξ)i · bα(ξ)i ) : ξ < κ〉 is a tower in (A⊕B) � (w ·x ·−v).

To prove this it suffices to take any d ∈ A � w and e ∈ B � (x · −v), assume that

d · e ·∑

i�=γ(α(ξ))(aα(ξ)i · bα(ξ)i ) = 0 for all ξ < κ, and show that d = 0 or e = 0.

But e · v = 0, so d · e · cα(ξ) = 0 for all ξ < κ, so d = 0 or e = 0. Now the desiredconclusion follows by the inductive hypothesis.

This finishes the induction on n. Now we consider the general case, wherenα depends upon α. But since κ is regular, there is an m ∈ ω such that {α < κ :nα = m} has κ elements, and so our inductive result applies.

So this completely finishes the case κ uncountable and regular.

Now we consider the case κ = ω. It is convenient to consider the dual oftowers for this proof. So, suppose that 〈ci : i < ω〉 is a strictly decreasing sequenceof elements of A⊕B with meet 0; we want to get a contradiction. For each i < ωwrite

ci =∑k<mi

ai,k · bi,k

with 0 �= ai,k ∈ A, 0 �= bi,k ∈ B, and bi,k · bi,l = 0 if k �= l; we do not assumethat mi is minimal. Now we define by recursion on i an element m′

i ∈ ω, a′ik ∈ Afor k < m′

i, and b′ik ∈ B for k < m′i so that ci =

∑k<m′

ia′ik · b′ik. Let m′

0 = m0,

a′0k = a0k, and b′0k = b0k for k < m0. Suppose that the definition has been madefor i. Note that if k < mi+1 then ai+1,k ·bi+1,k ·

∏l<m′

i(−a′il+−b′il) = 0, and hence

bi+1,k ≤∑

l<m′ib′il. It follows that

ci+1 =∑

k<mi+1

ai+1,k ·∑{bi+1,k · b′il : l < m′

i, bi+1,k · b′il �= 0}.

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Rearranging this sum, we get

(9) ci+1 =∑

k<m′i+1

a′i+1,k · b′i+1,k,

where b′i+1,k · b′i+1,l = 0 if k �= l, and for each k < m′i+1 there is an l < m′

i suchthat b′i+1,k ≤ b′il.

Next we claim:

(10) ∃k < m0∀i > 0∃l < mi(b′i,l ≤ b′0,k).

In fact, otherwise for every k < m0 there is an i(k) > 0 such that for everyl < mi(k) we have b

′i(k),l �≤ b′0,k; by (9) we have b′i(k),l ≤ b′0s for some s < m0, hence

s �= k and so b′i(k),l · b′0k = 0. So ci(k) · b′0,k = 0. If j > i(k) for all k < m0, we then

get cj · c0 = 0, contradiction. So (10) holds. We fix such a k.

(11) For every i > 0 there is an l ∈∏

0<j≤i mj such that b′0,k ≥ b′1,l(1) ≥ · · · ≥b′i,l(i).

In fact, let l(0) = k. By (10), choose l(i) < mi such that b′i,l(i) ≤ b′0,k. Then for

0 < j < i, by (9) choose l(j) < mj such that b′i,l(i) ≤ b′j,l(j). Now suppose that

0 ≤ j < n ≤ i. Then b′i,l(i) ≤ b′j,l(j) · b′n,l(n), so by (9) we get b′n,l(n) ≤ b′j,l(j). So(11) holds.

By (11) and Konig’s tree lemma we get l ∈∏

0<j<ω mj such that b′0,k ≥b′1,l(1) ≥ · · · ≥ b′i,l(i) ≥ · · · . Now assume that B has no tower of length ω. It follows

that the meet of all b′i,l(i)’s is not zero; say that d �= 0 and d ≤ b′i,l(i) for everypositive integer i. Now if 0 0 wehave ci · d = al(i) · d. Hence 0 =

∏0<i<ω(ci · d) =

∏0<i<ω(al(i) · d), and it follows

that 〈al(i) : 0 < i < ω〉 is a decreasing sequence with meet 0, and hence somesubsequence is a (dual) tower in A. �

Proposition 4.41. If 〈Ai : i ∈ I〉 is a system of BAs each with at least 4 elements,with I infinite, then tow(⊕i∈IAi) = ω.

Proof. Choose j ∈ ωI one-one, and for each k ∈ ω choose ck ∈ Ajk with 0 < ck < 1.Let b′i =

∑k≤i ck for each i ∈ ω. Clearly 〈b′i : i ∈ ω〉 is strictly increasing. We

claim that∑

i∈ω b′i = 1. For, suppose that x ∈ ⊕i∈IAi with b′i ≤ x < 1 for all

i ∈ ω. We can write x =∑

k∈M dk with M a finite subset of I and dk ∈ A+k for

all k ∈ M . Choose l ∈ ω such that jl /∈ M . Let f be a homomorphism of ⊕i∈IAi

into 2 such that f(dk) = 0 for all k ∈ M and f(cjl) = 1. Then f(b′l) = 1 whilef(x) = 0, contradiction. �

Problem 39. Describe the behaviour of tow under ultraproducts.

Now we work towards a characterization of towspect, taken from Monk [07].

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Lemma 4.42. Let A be any infinite BA, and let S be the Stone isomorphism. Thenno tower in S[A] remains a tower in P(Ult(A)).

Proof. Suppose that 〈aα : α < κ〉 is a tower in A, with κ a regular cardinal. LetX =

⋃α<κ S(aα). By compactness, X �= Ult(A). �

Lemma 4.43. Let A be an atomless BA, and let κ be a regular cardinal. ThenA ≤ B for some atomless BA B such that towspect(B) = {κ}.

Proof. We define a sequence 〈Cα : α < κ〉 of BAs as follows. Let C0 = A. If Cα hasbeen defined, let Cα ⊆ Cα+1, where Cα+1 is obtained from Cα by applying Lemma4.42 and then extending to an atomless BA. For α < κ limit, let Cα =

⋃β<α Cβ .

Finally, let B =⋃

α<κ Cα. We claim that B is as desired.

Every atomless BA has a tower, so it suffices to show that B has no tower ofregular length different from κ. Suppose that 〈bα : α < λ〉 is a tower in B, whereλ �= κ is a regular cardinal. For each α < λ there is a ξα < κ such that bα ∈ Cξα .

First suppose that λ < κ. Choose β < κ such that ξα < β for all α < λ.Then 〈bα : α < λ〉 is a tower in Cβ . By construction and Lemma 4.42 it is not atower in Cβ+1 and hence also not in B, contradiction.

Second suppose that κ < λ. There is an β < κ such that ξα = β for λ manyα < λ. Then 〈bα : α < λ〉 has a cofinal subsequence in Cβ ; this gives a tower inCβ , and a contradiction follows as above. �

According to the following theorem, there are no restrictions on the possibilitiesfor towspect.

Theorem 4.44. If M is a nonempty set of regular cardinals, then there is a BA Asuch that towspect(A) = M .

Proof. Let f be a function mapping some uncountable set I onto M . For eachi ∈ I let Ai be a BA such that towspect(Ai) = {f(i)}, using Lemma 4.43. Then∏w

i∈I Ai is as desired, by Proposition 4.38(ii). �

Theorem 4.45. Let K and L be nonempty sets of regular cardinals. Then there areatomless BAs A,B such that A ≤ B, towspect(A) = K, and towspect(B) = L.

Proof. Choose A such that towspect(A) = K, by Theorem 4.44. Let κ be the leastmember of L. let C be such that A ≤ C and towspect(C) = {κ}, by Lemma 4.43.

Choose D such that towspect(D) = L, by Theorem 4.44. Now Bdef= C × D is as

desired. �

Proposition 4.46. For any atomless BA A, if ω ≤ κ < tow(A), then 2κ ≤ c(A).

Proof. We define elements at for each t ∈⋃

α≤κα2 by recursion on α. for α = 0

and t ∈ 02 we have t = ∅, and we set a0 = 1. If aτ has been defined for a certaint ∈ α2 with α < κ, we split at into two nonzero disjoint elements at�〈0〉 andat�〈1〉. If α is a limit ordinal (including the possibility that α = κ, and at has

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been defined for all t ∈⋃

β<αβ2, and if s ∈ α2, let as be a nonzero element ≤ as�β

for all β < α; such an element exists since κ < tow(A).

This finishes the construction. Clearly the elements 〈at : t ∈ κ2〉 are nonzeroand pairwise disjoint. �

We now consider some other “small” cardinal functions related to tow: a (alreadydiscussed in Chapter 3),

spl, h. and p. The last three are defined as follows. A subset X of a BA A issplitting iff for all a ∈ A there is an x ∈ X such that a · x �= 0 �= a · −x.

spl(A) = min{|X | : X is a splitting subset of A};h(A) = min{κ : A is not (κ,∞)-distributive};

p(A) = min{|X | :

∑X = 1 and

∑Y �= 1 for every finite subset Y of X

};

pspect(A) ={|X | :

∑X = 1 and

∑Y �= 1 for every finite subset Y of X

}.

These are well-known functions in the case A = P(ω)/fin, where spl(A) is denotedby s. In the case of general BAs they are discussed in Monk [01a].

Proposition 4.47.

(i) aspect(A) ⊆ pspect(A).

(ii) towspect(A) ⊆ pspect(A).

(iii) p(A) ≤ a(A), tow(A).

Proof. Clearly every infinite partition of unity is a p-set for A, and similarly fortowers. �Proposition 4.48. If p(A) = ω, then tow(A) = ω = a(A).

Proof. Let Y = {ai : i ∈ ω} satisfy the condition defining p(A). Then clearly〈∑

j≤i aj : i ∈ ω〉 has a subsequence which is a tower; so tow(A) = ω. Furthermore,let bi = ai ·

∏j ω, then a(A⊕B) > ω.

Proof. If a(A⊕B) = ω, then tow(A⊕B) = ω, by Corollary 4.49. Hence tow(A) = ωor tow(B) = ω, by Theorem 4.40. So a(A) = ω or a(B) = ω by Corollary 4.49again. �

This corollary is relevant to Problem 8.

Now we consider the relationship between a and tow when they are uncountable.

Theorem 4.51. If ω < κ and M is a nonempty collection of regular cardinals eachgreater than κ, then there is a BA A such that a(A) = κ and towspect(A) = M .

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Proof. By Theorem 4.44 there is a BA B such that towspect(B) = M . Note thatthe proofs of Lemma 4.43 and Theorem 4.44 also give a(B) greater than eachmember of M . Now the weak product of κ many copies of B gives the algebra Adesired, by Proposition 3.38(ii) and Proposition 4.38(ii). �

Note that one cannot hope to have aspect(A) = {κ} here, since a tower of sizegreater than κ clearly gives rise to a partition of size greater than κ.

It is more difficult to get examples with t less than a. See McKenzie, Monk[04] for the following result:

If κ and λ are regular cardinals with ω < κ < λ, then there is a BA A such thattow(A) = spl(A) = κ and a(A) = λ.

Problem 40. Describe in cardinal number terms the pairs M,N of cardinals suchthat there is a BA A with towspect(A) = M and aspect(A) = N .

For our special kinds of subalgebras, the facts and problems concerning p are verysimilar to those for a and tow.

Proposition 4.52. If A ≤reg B, then pspect(A) ⊆ pspect(B) and hence p(B) ≤ p(A).The conclusion also holds if A ≤free B, A ≤proj B, A ≤rc B, A ≤π B. �

Problem 41. Given nonempty sets K ⊆ L of cardinals, are there BAs A ≤reg Bsuch that pspect(A) = K and pspect(B) = L? There is a similar question forA ≤free B, A ≤proj B, A ≤rc B, A ≤π B.

Proposition 4.53. If A ≤σ B and p(A) > ω, then pspect(A) ⊆ pspect(B), andp(B) ≤ p(A).

Proof. See the proof of Proposition 4.35. �

Problem 42. Given nonempty sets K ⊆ L of cardinals with ω /∈ K, are there BAsA ≤σ B such that pspect(A) = K and pspect(B) = L?

One can have A ≤σ B with p(A) = ω < p(B). For example, let B have a largepseudo-intersection number, and let A be a countable subalgebra.

There are BAs A,B with A ≤u B and p(A) < p(B); see the argument for a.

There are BAs A,B with A ≤mg B and p(A) < p(B).

Proposition 4.54. If A ≤m B, then pspect(A) ⊆ pspect(B), and hence p(B) ≤p(A). �

Problem 43. Given nonempty sets K ⊆ L of cardinals with ω /∈ K, are there BAsA ≤mm B such that pspect(A) = K and pspect(B) = L?

There are BAs A,B with A ≤free B, p(B) = ω, and p(A) arbitrarily large. Similarlyfor ≤proj, ≤rc, ≤σ, ≤reg, and ≤u.

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One can have A ≤π B, hence also A ≤reg B, with p(A) arbitrary and p(B) = ω.Simply take B = A.

One can have A ≤s B with p(B) < p(A). See the corresponding example for tow.

Problem 44. Does A ≤s B imply that pspect(A) ⊆ pspect(B)?

There are BAs A,B such that A ≤mg B and p(B) < p(A). See the constructionfollowing Problem 35.

Problem 45. Are there BAs A,B such that A ≤m B and p(B) < p(A)?

Proposition 4.55. p(A×B) = min(p(A), p(B)).

Proof. If X ⊆ A,∑

X = 1, and∑

F < 1 for every finite subset F of A, then∑({(x, 0) : x ∈ X} ∪ {(0, 1)}) = 1 and

∑({(x, 0) : x ∈ F} ∪ {(0, 1)}) < (1, 1) for

every finite subset F of X . Similarly for B, so ≤ holds.

Now suppose that Y ⊆ A × B,∑

Y = (1, 1), and∑

F < (1, 1) for everyfinite subset F of Y . Let U = {a ∈ A : ∃b ∈ B[(a, b) ∈ Y ]} and V = {b ∈ B : ∃a ∈A[(a, b) ∈ Y ]}. Then

∑U = 1 and

∑V = 1. If

∑F = 1 and

∑G = 1 for some

finite subsets F of U and G of V , this clearly yields a finite subset H of Y suchthat

∑H = (1, 1), contradiction. �

The following two propositions have proofs just like the proofs of Propositions 3.38and 3.39.

Proposition 4.56. Suppose that 〈Ai : i ∈ I〉 is a system of infinite BAs, with Iinfinite. Then

(i) pspect

(w∏i∈I

Ai

)= {|I|} ∪

⋃i∈I

pspect(Ai)


∃J ⊆ I∃λ ∈∏j∈J

pspect(Aj)

(J �= ∅ and κ = sup

j∈Jλj

)}.

(ii) p

(w∏i∈I

Ai

)= min(|I|,min

i∈Ip(Ai)). �

Proposition 4.57. Let 〈Ai : i ∈ I〉 be a system of infinite BAs, with I infinite.

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Then:

(i) pspect

(∏i∈I

Ai

)= [ω, |I|] ∪

⋃i∈I

pspect(Ai)


∃J ⊆ I∃λ ∈∏j∈J

pspect(Aj)

[J �= ∅ and κ = sup

j∈Jλj

]}.

(ii) a

(∏i∈I

Ai

)= ω. �

Kevin Selker has generalized Proposition 4.56 to moderate products. In particular,he showed:

For a moderate product, p(∏B

i∈I Ai) = min(p(B),mini∈I p(Ai)).

Proposition 4.58. If p(A), p(B) > ω, then p(A⊕B) > ω, for infinite BAs A,B.

Proof. Suppose that p(A ⊕ B) = ω. Then a(A ⊕ B) = ω by Corollary 4.48. Soa(A) = ω or a(B) = ω by Corollary 4.50. Then p(A) = ω or p(B) = ω by Corollary4.49. �

Problem 46. Is it true that for all infinite BAs A,B we have p(A ⊕ B) =min(p(A), p(B))?

Proposition 4.59. p(⊕i∈IAi) = ω for any infinite system 〈Ai : i ∈ I〉 of infiniteBAs.

Proof. By Propositions 4.41 and 4.49. �

Concerning the following theorem, see Proposition 4.47(iii).

Theorem 4.60. Let κ and λ be regular cardinals, with ω1 < κ ≤ λ. Then there isa linear order L such that for A = Intalg(L) we have p(A) = κ, tow(A) = λ, anda(A) ≥ λ.

Proof. By Hausdorff’s theorem (Hausdorff [1908]) let L have the following prop-erties:

(1) Every element of L has character (λ, λ∗).

(2) The gaps of L have the following characters:

(a) (ω1, κ∗) and (κ, ω∗

1).

(b) (ν, λ∗) and (λ, ν∗) for all ν ∈ [ω, λ).

(3) L has coinitiality λ∗ and cofinality λ.

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Let A = Intalg(L). Later in this section we characterize those μ such that aninterval algebra has a tower of length μ. Applying this to A, we get tow(A) = λ.In Chapter 14 we prove that p(A) = κ. It remains to show that a(A) ≥ λ. LetX be any infinite partition of A. Without changing |X | we may assume that eachelement of X has the form [a, b) with a, b ∈ L ∪ {−∞,∞}.

If [a, b) ∈ X with b < ∞, and ∀[c, d) ∈ X [b �= c], then there is a sequence〈[xξ, yξ) : ξ < λ of elements of X , strictly decreasing under < (between disjointsubsets of L), with inf b. Thus we may assume that

(1) ∀[a, b) ∈ X∃c ∈ L ∪ {∞}[[b, c) ∈ X ].

If ∀[a, b) ∈ X [a �= −∞], then there is a sequence 〈[xξ , yξ) : ξ < λ of elements of X ,strictly decreasing under <, with inf −∞. Hence we may assume that [−∞, a) ∈ Xfor some a ∈ L. Then by (1) we get an increasing sequence 〈[bi, ci) : i ∈ ω〉 ofmembers of X such that a = b0 and ci = bi+1 for all i ∈ ω. Then {x ∈ L : x ≤ bifor some i ∈ ω} determines a gap in L with lower character ω. Its upper characteris λ∗, and hence there is a sequence 〈[xξ , yξ) : ξ < λ of elements of X , strictlydecreasing under <, coinitial in this gap. �

However, the following two questions arise.

Problem 47. Given cardinals κ, λ, μ with ω < κ ≤ λ, μ, is there a BA A such thatp(A) = κ, tow(A) = λ, and a(A) = μ?

Problem 48. Is

p(A) = min{|X | : X is a maximal ramification set in A}?

Recall here the notion of a ramification set from Chapter 2.

We also mention the following well-known problem.

Problem 49. Is it consistent that p(P(ω)/fin) < tow(P(ω)/fin)?

For some results on this problem, see Shelah [09]. The following simple lemma isfound in this paper specialized to P(ω)/fin. A strong p-set for A is a subset X ofA+ closed under · such that

∏X = 0 and |X | = p(A).

Proposition 4.61. Suppose that p(A) < tow(A), and X is a strong p-set for A.Then there exist a regular κ < p(A) and a system 〈aα : α < κ〉 of elements of Asuch that the following conditions hold:

(i) If α < β < κ, then aβ ≤ aα.

(ii) aα · b �= 0 for all α < κ and b ∈ X.

(iii) if c is a nonzero lower bound for {aα : α < κ}, then there is a b ∈ X suchthat b · c = 0.

Proof. Write X = {bα : α < p(A)}. We define 〈aα : α < β〉 by recursion, whereβ ≤ p(A) is to be determined. Let a0 = 1. Suppose that aα has been defined so

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that aα · c �= 0 for all c ∈ X . Let aα+1 = aα · bα. Clearly aα+1 · c �= 0 for all c ∈ X .Now suppose that aα has been defined for all α < γ, with γ a limit ordinal. Ifthere is a nonzero x such that x ≤ aα for all α < γ and x · c �= 0 for all c ∈ X , welet aγ be such an x. Otherwise the construction stops, with β = γ.

If the construction continues all the way to p(A), then clearly {aα : α < p(A)}is a downwards tower, contradicting p(A) < tow(A). Thus it ends at some limitordinal β < p(A). Clearly the conclusion of the proposition holds with κ = cf(α)and a subsequence of 〈aα : α < β〉. �

Turning to h, we recall some facts about distributivity; see the Handbook. h(A) isdefined for any non-atomic BA A. An equivalent condition for (κ,∞)-distributivityis that every set of at most κ partitions of A has a common refinement.

Proposition 4.62. tow(A) ≤ h(A) for every atomless BA A.

Proof. Let κ = h(A). Let 〈Pα : α < κ〉 be a system of partitions of A which do nothave a common refinement. We define by recursion another system 〈Qα : α < κ〉of refinements of A. Let Q0 = P0. If Qα has been defined for all α < β, whereβ < κ, let Qβ be a common refinement of all the Qα’s for α < β together withPβ . Clearly a common refinement of all the Qα’s with α < κ would be a commonrefinement of all the Pα’s; so there is no such refinement of the Qα’s.

Let R be a maximal family of pairwise disjoint nonzero elements of A suchthat ∀x ∈ R∀α < κ∃y ∈ Qα[x ≤ y]. Then R cannot be a partition, so there issome z ∈ A+ such that x ·z = 0 for all x ∈ R. Now we define a decreasing sequencew0, w1, . . . of elements of A. Choose w0 ∈ Q0 such that z · w0 �= 0 Suppose thatwα ∈ Qα has been defined so that wα · z �= 0. Choose wα+1 ∈ Qα+1 such thatwα · z · wα+1 �= 0. Then wα+1 ≤ wα since Qα+1 refines Qα. Now suppose that βis a limit ordinal less than κ and wα has been defined for all α < β. If there is anonzero y ≤ wα · z for all α < β, then there is a v ∈ Pβ such that v · y �= 0, andwe let wβ be such a v. Note that wβ · z �= 0, and wβ ≤ wα for all α < β since Qβ

refines Qα. If there is no such y, the construction stops, and we have a ≥-towerwhich is a cofinal subsequence of 〈wα · z : α < β of length β < κ.

Suppose that the construction does not stop for any β < κ. Then∏

α<κ wα ·z = 0, since otherwise this product would contradict the maximality of R. Againthis yields a ≥-tower of length at most κ. �

Proposition 4.63. h(A) ≤ spl(A) for any atomless BA A.

Proof. Suppose that spl(A) < h(A), and let S be a splitting set in A of size spl(A).We may assume that 0, 1 /∈ S. For each s ∈ S let Ps = {s,−s}. So Ps is a partition.Let Q be a common refinement of all the Ps’s. Take any q ∈ Q, and choose s ∈ Ssuch that q · s �= 0 �= q · −s. But Q refines Ps, contradiction. �

The above facts describe all of the relationships between the functions p, a, tow,h, and spl; see the diagrams at the end of the book. To see that no other relationshold, and that the differences can be large, we need some examples. Recall here

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the result of McKenzie, Monk [04] mentioned earlier, showing that spl(A) < a(A)is possible, with an arbitrarily large difference. For a(A) < tow(A), see Proposition4.51. Examples with tow(A) < h(A) and h(B) < spl(B) will be given shortly.

Proposition 4.64. h(A) is regular, for any atomless BA A.

Proof. Suppose that κdef= h(A) is singular. Let P be a collection of κ many

partitions with no common refinement. Write P =⋃

α<cf(κ) Qα, with each Qα

of size less than κ. Then each Qα has a common refinement Qα, and {Qα : α <cf(κ)} has a common refinement P . Clearly P is a common refinement of P ,contradiction. �Proposition 4.65. Suppose that κ is regular. Assume one of the following:

(i) A is a κ-saturated atomless BA.

(ii) With κ = ℵα, there is an ηα-set L such that A = Intalg(L).

Then h(A) ≥ κ.

Proof. It suffices to show that any system of partitions 〈Pξ : ξ < λ〉 has a refine-ment, where λ is an infinite cardinal less than κ.

In case (ii), the set Xdef= {[u, v) : u, v ∈ L, u < v} is dense in A, and so there

is a partition with elements from X . Taking a refinement of X with Pξ for eachξ < λ, we may assume that each Pξ refines X .

We construct partitions 〈Qξ : ξ ≤ λ〉 such that for each ξ ≤ λ, Qξ refinesall Qη for η < ξ, and for each ξ < λ, Qξ refines Pξ. Let Q0 = P0. If Qξ has beendefined, with ξ < λ, let Qξ+1 be a refinement of Qξ and Pξ+1. Suppose that Qξ hasbeen defined for all ξ < η satisfying these conditions, where η is a limit ordinal≤ λ.Let R be maximal disjoint subject to the condition ∀ξ < η∀x ∈ R∃y ∈ Qξ[x ≤ y].We claim that

∑R = 1. For, take any z �= 0.

Now we split the proof into the cases corresponding to the assumptions (i)or (ii).

(i): We construct 〈aξ : ξ ≤ η〉. Choose a0 ∈ Q0 so that a0 · z �= 0. If aξ hasbeen defined, ξ < η, so that aξ ·z �= 0, choose aξ+1 ∈ Qξ+1 such that aξ+1 ·aξ ·z �= 0.Since Qξ+1 refines Qξ, it follows that aξ+1 ≤ aξ. Now suppose that ξ ≤ η is limitand aσ has been defined for all σ < ξ so that σ < τ < ξ implies that aτ ≤ aσ, andso that aσ · z �= 0 for all σ < ξ. Consider the following set of formulas in one freevariable v:

(∗) {v �= 0} ∪ {v ≤ z} ∪ {v ≤ aσ : σ < ξ}.

A finite subset of this set is contained in a finite subset of the form

(∗∗) {v �= 0} ∪ {v ≤ z} ∪ {v ≤ aσ : σ ∈ F}

for some finite nonempty subset F of ξ. Let σ be the largest member of F . Thenby assumption, aσ · z satisfies (∗∗).

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So by κ-saturatedness, we can take a b ∈ A such that a �= 0, a ≤ z, anda ≤ aξ for all σ < ξ. Choose aξ ∈ Qξ so that aξ · b �= 0. Then aξ · z �= 0. Thisfinishes the construction of 〈aξ : ξ ≤ η〉. We have aη ∈ R and aη · z �= 0. So∑

R = 1, as claimed. If η < λ we let Qη be a refinement of R and Pη. If η = λ,

then Qηdef= R is the desired refinement of each Pξ for ξ < λ.

(ii): We may assume that z = [s, t) for some s, t ∈ L with s < t. We wantto find [x, y) ∈ R such that [s, t) ∩ [x, y) �= ∅. To do this we define two sequences〈vξ : ξ < η〉 and 〈wξ : ξ < η〉 of elements of L such that 〈vξ : ξ < η〉 is increasing,〈wξ : ξ < η〉 is decreasing, vξ < wγ for all ξ, γ < η, [vξ, wξ) ∈ Qξ, and max(s, vξ) <min(t, wξ) for all ξ < η. Choose v0 < w0 so that [v0, w0) ∈ Q0 and [v0, w0) ∩[s, t) �= ∅. Thus max(s, v0) < min(t, w0). Now suppose that vξ and wξ have beenconstructed for all ξ < γ, where γ < η, so that 〈vξ : ξ < γ〉 is increasing, 〈wξ : ξ <γ〉 is decreasing, and ∀ξ < γ[max(s, vξ) < min(t, wξ)]. Then by the ηα propertythere are elements x, y of L such that max(s, vξ) < x < y < min(t, wδ) for allξ, δ < γ. Choose [vγ , wγ) ∈ Qγ such that [vγ , wγ) ∩ [x, y) �= ∅. Now if ξ < γ, then[vξ, wξ)∩[vγ , wγ) �= ∅ so, since Qγ refines Qξ, we must have [vγ , wγ) ⊆ [vξ, wξ), andso vξ ≤ vγ and wγ ≤ wξ. Now choose u ∈ [vγ , wγ)∩ [x, y). Then s < x < u < y < tand vγ ≤ u < wγ , so max(s, vγ) ≤ u < min(t, wγ). This finishes the constructionof 〈vξ : ξ < η〉 and 〈wξ : ξ < η〉.

By the ηα property choose p, r ∈ L such that max(s, vξ) < p < r < min(t, wξ)for all ξ < η. Thus [p, r) ⊆ [vξ, wξ) for all ξ < η. By the maximality of R it followsthat there is a q ∈ R such that q ∩ [p, r) �= ∅. If f is in this set, then f ∈ q ∩ [s, t),as desired.

If η < λ we let Qη be a refinement of R and Pη. If η = λ, then Qηdef= R is

the desired refinement of each Pξ for ξ < λ. �

Corollary 4.66. (GCH)For each regular cardinal there is a BAA such that h(A)=κ.�

Problem 50. Can one prove in ZFC that for each regular cardinal κ there is a BAA such that h(A) = κ?

Proposition 4.67. Suppose that κ and λ are cardinals, κ regular, with κ ≤ λ. Thenthere is an interval algebra A such that tow(A) ≤ κ and h(A) ≥ λ.

Proof. Let M be a dense linear order in which every element has character (κ, κ∗),while the gaps have character (μ, κ∗) and (κ, μ∗) for ω ≤ μ < κ, μ regular; and Mhas coinitiality κ∗ and cofinality κ.

Write λ = ℵε, and let L be an ηε-set. Finally, letN = M×L, lexicographicallyordered, and set A = Intalg(N). We claim that tow(A) = κ and h(A) ≥ λ.

Let 〈mβ : β < κ〉 be strictly increasing and cofinal in M , and let u be anymember of L. Then 〈[−∞, (mβ , u)) : β < λ〉 is a tower in A of order type κ. Sotow(A) ≤ κ.

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To prove that h(A) ≥ λ, take a system 〈Pα : α < μ〉 of partitions, with μ < λ;we want to find a refinement of them. Now the set

Rdef= {[(m, s), (m, t)) : m ∈M, s, t ∈ L, s < t}

is dense in A, and hence there is a partition of A containing only elements of thisset. Taking a refinement of each Pα with such a partition, we may assume thateach Pα consists of elements of R.

Now we define partitions 〈Qα : α ≤ μ〉 by recursion. Let Q0 = P0. If Qα hasbeen defined, let Qα+1 be a refinement of Qα and Pα. Now suppose that β is a limitordinal ≤ μ and 〈Qα : α < β〉 has been defined so that Qγ is a refinement of Qα

if α < γ < β. Let S be maximal disjoint such that ∀v ∈ S∀α < β∃q ∈ Qα[v ≤ q].We claim that

∑S = 1.

To prove this it suffices to take any m ∈ M and s, t ∈ L with s < tand find q ∈ S such that q ∩ [(m, s), (m, t)) �= ∅. To do this we define two se-quences 〈vα : α < β〉 and 〈wα : α < β〉 of elements of L such that 〈vα :α < β〉 is increasing, 〈wα : α < β〉 is decreasing, vα < wγ for all α, γ < β,[(m, vα), (m,wα)) ∈ Qα, and max(s, vα) < min(t, wα) for all α < β. Choosev0 < w0 so that [(m, v0), (m,w0)) ∈ Q0 and [(m, v0), (m,w0))∩ [(m, s), (n, t)) �= ∅.Thus max(s, v0) < min(t, w0). Now suppose that vα and wα have been constructedfor all α < γ, where γ < β so that 〈vα : α < γ〉 is increasing, 〈wα : α < γ〉 isdecreasing, and ∀α < γ[max(s, vα) < min(t, wα)]. Then by the ηε property thereare elements x, y of L such that max(s, vα) < x < y < min(t, wδ) for all α, δ < γ.Choose [(m, vγ), (m,wγ)) ∈ Qγ such that [(m, vγ), (m,wγ)) ∩ [(m,x), (m, y)) �= ∅.Now if α < γ, then [(m, vα), (m,wα)) ∩ [(m, vγ), (m,wγ)) �= ∅ so, since Qγ re-fines Qα, we must have [(m, vγ), (m,wγ)) ⊆ [(m, vα), (m,wα)), and so vα ≤ vγand wγ ≤ wα. Now choose (m, z) ∈ [(m, vγ), (m,wγ)) ∩ [(m,x), (m, y)). Thens < x < z < y < t and vγ ≤ z < wγ , so max(s, vγ) ≤ z < min(t, wγ). This finishesthe construction of 〈vα : α < β〉 and 〈wα : α < β〉.

By the ηε property choose p, r ∈ L such that max(s, vα) < p < r < min(t, wα)for all α < β. Thus [(m, p), (m, r)) ⊆ [(m, vα), (m,wα)) for all α < β. By themaximality of S it follows that there is a q ∈ S such that q ∩ [(m, p), (m, r)) �= ∅.If f is in this set, then f ∈ q ∩ [(m, s),m, t)), as desired.

Hence we can let Qβ = S, finishing the recursive definition; Qμ is as desired.�

We postpone giving an example with h(A) < spl(A) until after discussing algebraicoperations concerning these two functions.

Proposition 4.68. For any atomless BAs A and B we have

(i) h(A ×B) = min(h(A), h(B)).

(ii) spl(A×B) = max(spl(A), spl(B)).

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Proof. For (i), suppose that 〈Pα : α < h(A)〉 is a system of partitions of A withno common refinement. For each α < h(A) let

Qα = {(x, 0) : x ∈ Pα} ∪ {(0, 1)}.

Thus Qα is a partition of A×B. Clearly a refinement of all the Qα’s would yielda refinement of the Pα’s. This shows that h(A×B) ≤ h(A). similarly h(A×B) ≤h(B).

Now suppose that 〈Rα : α < κ〉 is a system of partitions of A × B, whereκ < min(h(A), h(B)). We will find a common refinement of them, proving theequality. For each α < κ let

Sα = {a ∈ A+ : ∃b ∈ B[(a, b) ∈ Rα]};Tα = {b ∈ B+ : ∃a ∈ A[(a, b) ∈ Rα]}.

Clearly each Sα is a partition of A, and each Tα is a partition of B. Let U be acommon refinement of the Sα’s, and V a common refinement of the Tα’s. Let

W = {(x, 0) : x ∈ U} ∪ {(0, y) : y ∈ V }.

Then W is a partition of A×B which refines each Rα.

(ii): If X splits A and Y splits B, then X × Y splits A × B. Hence spl(A ×B) ≤ max(spl(A), spl(B)). Now suppose that Z ⊆ A× B splits A× B and |Z| <max(spl(A), spl(B)). Say by symmetry that max(spl(A), spl(B)) = spl(A). Then{a ∈ A : ∃b[(a, b) ∈ Z]} has size less than spl(A), so there is an a ∈ A+ such thatfor all (u, v) ∈ Z, a · u = 0 or a · −u = 0. Then (a, 0) ∈ (A × B)+ and for all(u, v) ∈ Z, (a, 0) · (u, v) = (0, 0) or (a, 0) · −(u, v) = (0, 0, contradiction. �

Proposition 4.69. If 〈Ai : i ∈ I〉 is a system of atomless BAs, with I infinite, then

(i) h(∏w

i∈I Ai

)= mini∈I h(Ai).

(ii) spl(∏w

i∈I Ai

)= max(|I|, supi∈I spl(Ai)).


i∈I Ai. For each i ∈ I and a ∈ Ai define upi(a) ∈ Bby setting, for any j ∈ I,

upi(a)j =

{a if i = j,

0 otherwise.

(i): By Proposition 4.68, ≤ holds. Now suppose that κ < mini∈I h(Ai) and 〈Pα :α < κ〉 is a system of partitions of B. For each i ∈ I and α < κ let Qi

α = {xi :x ∈ Pα}\{0}. Then Qi

α is a partition of Ai, and so there is a refinement Ri of allQi

α’s. Let

S = {upi(x) : i ∈ I, x ∈ Ri}.

Then S is a partition of B which refines each Pα. Thus (i) holds.

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(ii): For each i ∈ I let Xi be a splitting set for Ai of size spl(Ai). Then

{upi(x) : i ∈ I, x ∈ Xi}

is a splitting set for B, and its size is max(|I|, supi∈I spl(Ai)).

Now suppose that Z splits B and |Z| < max(|I|, supi∈I spl(Ai)). If |Z| < |I|,then there is an i ∈ I such that zi = 0 or zi = 1 for all z ∈ Z, contradiction.So |I| ≤ |Z|. Choose i ∈ I such that |Z| < spl(Ai). Then {zi : z ∈ Z} splits Ai,contradiction. �

The proof of Proposition 4.69 also works for moderate products.

Proposition 4.70. If 〈Ai : i ∈ I〉 is a system of atomless BAs, with I infinite, then

(i) h(∏

i∈I Ai

)= mini∈I h(Ai).

(ii) spl(∏w

i∈I Ai

)= supi∈I spl(Ai).

Proof. Again let B =∏

i∈I Ai for brevity. (i): as in the proof of Proposition 4.69.(ii): For each i ∈ I let Xi be a splitting set for Ai with |Xi| = spl(Ai). Letκ = supi∈I spl(Ai), and for each i ∈ I let 〈xi

α : α < κ〉 be an enumeration of Xi,repetitions allowed. For each α < κ define yα ∈ B by setting yαi = xi

α for anyi ∈ I. Clearly {yα : α < κ} splits B. So ≤ holds.

Now suppose that Z splits B and |Z| < κ. Choose i ∈ I such that |Z| <spl(Ai). Then {zi : z ∈ Z} splits Ai, contradiction. �

Concerning free products we have the following problem.

Problem 51. What is the relationship, if any, between h(A⊕B) and h(A), h(B)?

Clearly if X splits A, then also X splits A⊕B. Thus we have

Proposition 4.71. spl(A⊕B) ≤ min(spl(A), spl(B)) for atomless A,B. �

Problem 52. Is spl(A⊕B) = min(spl(A), spl(B)) for atomless A,B?

Problem 53. What is the relationship between h of algebras and their ultraproduct?

Problem 54. What is the relationship between spl of algebras and their ultraprod-uct?

Proposition 4.72. If κ is regular and 2κ ≤ λ, then there is an atomless BA A suchthat κ ≤ h(A) ≤ 2κ and spl(A) = λ.

Proof. Let B be κ-saturated of size at most 2κ. and let Cα = B for all α < λ. Welet A =

∏wα<λ Cα. Then κ ≤ h(A) ≤ 2κ and spl(A) = λ by Proposition 4.69. �

Problem 55. Suppose that κ is regular and κ ≤ λ. Is there an atomless BA A suchthat h(A) = κ and spl(A) = λ?

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We return to the discussion of other derived functions for depth. Clearly [ω, t(A)) ⊆DepthHs(A) ⊆ [ω, t(A)], by an argument very similar to that used for the func-tions c and s. Like for cellularity, there is a problem whether t(A) ∈ DepthHs(A).This is trivially true if t(A) is a successor cardinal or a limit cardinal of cofinal-ity ω by Corollary 4.27 and Theorem 12.2 below. For each singular cardinal κwith cf(κ) > ω there is a BA A such that |A| = Depth(A) = t(A) and Ult(A)has no free sequence of length κ, hence by Corollary 4.27 A has no homomor-phic image B such that Depth(B) = t(A) and Depth(B) is attained. Namely, let〈μα : α < cf(κ)〉 be a strictly increasing sequence of infinite cardinals with sup κ,let A =

∏wα<cfκ Intalg(μα), and use Theorem 12.1.

The following result of Roslanowski, Shelah [00] is relevant here (see Conclu-sion 7.6 in this paper):

It is consistent that there is a Boolean algebra A of size λ such that there is anultrafilter of A of tightness λ, there is no free sequence of length λ in A, andt(A) = λ /∈ DepthHs(A).

Of course it would be of interest to construct such an example in ZFC; this is aversion of Problem 13 in Monk [96].

Problem 56. Can one construct in ZFC a BA A such that t(A) /∈ DepthHs(A)?

Clearly DepthSs = [ω,DepthA] for any infinite BA A.

Next comes the relation DepthSr. It is easy to see that parts (i)–(ii) and (iv)–(v)of Theorem 3.51 hold with cellularity replaced by depth. This is expressed in partof the following proposition.

Proposition 4.73. For any infinite BA A the following conditions hold:

(i) If (κ, λ) ∈ DepthSr(A), then κ ≤ λ ≤ |A| and κ ≤ Depth(A).

(ii) For each κ ∈ [ω,Depth(A)] we have (κ, κ) ∈ DepthSr(A).

(iii) If (κ, λ) ∈ DepthSr(A) and κ ≤ μ ≤ λ, then (κ, μ) ∈ DepthSr(A).

(iv) (DepthA, |A|) ∈ DepthSr(A).

(v) If ω ≤ λ ≤ |A| then (κ, λ) ∈ DepthSr(A) for some κ.

(vi) If (κ+, λ) ∈ DepthSr(A), then (ω, κ+) ∈ DepthSr(A).

(vii) If (κ+, λ) ∈ DepthSr(A) and ω ≤ μ < κ+, then (μ, κ+) ∈ DepthSr(A).

Proof. (i)–(v) are easy. For (vi), assume that (κ+, λ) ∈ DepthSr(A). Then A hasa subalgebra B isomorphic to Finco(κ+), and Depth(B) = ω.

For (vii), assume the hypotheses. Let B be a subalgebra of A such that|B| = λ and Depth(B) = κ+. Let 〈bα : α < κ+〉 be a strictly increasing sequencein B. Then B ∼= (B � bμ)× (B � −bμ). Since bα+1 · −bα ≤ −bμ for all α ∈ [μ, κ+),it follows that B � −bμ has a subalgebra isomorphic to Finco(κ+). Hence B has asubalgebra of size κ+ with depth μ. �

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Concerning (iii) of Theorem 3.51, the following result of Roslanowski, Shelah [01](Conclusion 18) is relevant:

It is consistent that there is a cardinal κ such that there is a BA A of size (2κ)+

with Depth(A) = κ while (ω, (2κ)+) /∈ DepthSr(A).

Again it would be of interest to construct such an example in ZFC. This is aversion of Problem 14 of Monk [96].

Problem 57. Can one show in ZFC that there is a cardinal κ such that there is aBA A of size (2κ)+ with Depth(A) = κ while (ω, (2κ)+) /∈ DepthSr(A)?

Problem 58. Characterize the relation DepthSr.

This is Problem 15 of Monk [96].

The following theorem and corollary are relevant to this problem:

Theorem 4.74. Let κ be an infinite cardinal, and let μ be minimum such thatωμ > κ. Thus μ ≤ κ. Let L be the linearly ordered set μQ ordered lexicographically,where Q is the set of all rationals in the interval [0, 1). Then

(i) L does not have any strictly increasing sequence of order type μ+ or anystrictly decreasing sequence of order type μ+.

Further, set D = {f ∈ μQ : there is an α < μ such that f(β) = 0 for all β > α}.It is clear that |D| ≤ κ and D is dense in L. Let M be a subset of L of size κ+

which includes D, and let A be the interval algebra on M . Then

(ii) Any subset N of A of size κ+ contains a linearly ordered subset of size κ+

which can be isomorphically embedded in M or M∗.

Proof. For (i), assume the contrary. Assume that 〈fα : α < μ+〉 is strictly increas-ing. (The strictly decreasing case is treated similarly.) Now a contradiction willfollow rather easily from the following statement:

(1) If γ ≤ μ, Γ ∈ [μ+]μ+

, and fα � γ < fβ � γ for all α, β ∈ Γ for which α < β,

then there exist δ < γ and Δ ∈ [Γ]μ+

such that fα � δ < fβ � δ for all α, β ∈ Δfor which α < β.

Assume the hypothesis of (1). For each α ∈ Γ let f ′α = fα � γ. Clearly Γ does not

have a largest element. For each α ∈ Γ let α′ ∈ Γ be minimum with α < α′, andlet χ(α) < γ be minimum such that fα(χ(α)) �= fα′(χ(α)). Now

Γ =⋃δ<γ

{α ∈ Γ : χ(α) = δ},

so there exist δ < γ and Θ ∈ [Γ]μ+

such that χ(α) = δ for all α ∈ Θ. Define α ≡ βiff α, β ∈ Θ and fα � δ = fβ � δ. Note that fα(δ) < fβ(δ) if α ≡ β and α < β.

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Hence each equivalence class under ≡ is countable. Let Δ ⊆ Θ have exactly oneelement from each equivalence class. Clearly the conclusion of (1) now holds.

From (1) we get a strictly decreasing sequence μ > δ(0) > δ(1) > · · · ofordinals, contradiction. So (i) holds.

For (ii), let N ∈ [A]κ+

. For each x ∈ A write

x = [a(1, x), b(1, x)) ∪ · · · ∪ [a(mx, x), b(mx, x)),

where a(1, x), b(1, x), . . . , a(mx, x), b(mx, x) are inM∪{+∞} and a(1, x) < b(1, x)< · · · < a(mx, x) < b(mx, x). Note that if x ∈ A and a(1, x) > 0, then a(1,−x) = 0.If |{x ∈ N : a(1, x) = 0}| < κ+, let N1 = {x : −x ∈ N, a(1,−x) > 0}; otherwiselet N1 = {x ∈ N : a(1, x) = 0}. So |N1| = κ+ and ∀x ∈ N1[a(1, x) = 0].

Next,

N1 =⋃n∈ω

{x ∈ N1 : mx = n}.

It follows that there is an n ∈ ω such that N2def= {x ∈ N1 : mx = n} has size κ+.

Now for each x ∈ N2 we choose c(1, x), . . . , c(n, x), d(1, x), . . . , d(n, x) ∈ D so thata(i, x) < c(i, x) < b(i, x) < d(i, x) < a(i + 1, x) for all i = 1, . . . , n (omitting theterm a(i+ 1, x) for i = n, and also omitting d(n, x) if b(n, x) =∞). Now

N2 =⋃

e,f∈E

{x ∈ N2 : ∀i = 1, . . . , n[c(i, x) = e(i) and d(i, x) = f(i)]},

where E is the set of all functions mapping {1, . . . , n} into D. Since |D| ≤ κ, itfollows that there exist e, f ∈ E so that

N3def= {x ∈ N2 : ∀i = 1, . . . , n[c(i, x) = e(i) and d(i, x) = f(i)]}

has size κ+.

Now we construct subsets P0, . . . , Ps of N3 by recursion, where s is to bedetermined. Let P0 = N3. Suppose that Pk has been defined, so that Pk ⊆ N3 and|Pk| = κ+. If k < n, define x ≡k y iff x, y ∈ Pk and a(k+1, x) = a(k+1, y). This isan equivalence relation on Pk. If there are κ

+ equivalence classes under ≡k, we letPk+1 be a subset of Pk containing exactly one element from each ≡k class, and welet s = k + 1. If there are less than κ+ equivalence classes, then some equivalenceclass Pk+1 has κ

+ elements. If n ≤ k < 2n, we work with b instead of a, in the samefashion. Namely, define x ≡k y iff x, y ∈ Pk and b(k − n+ 1, x) = b(k − n+ 1, y).This is an equivalence relation on Pk. If there are κ+ equivalence classes under≡k, we let Pk+1 be a subset of Pk containing exactly one element from each ≡k

class, and we let s = k+1. If there are less than κ+ equivalence classes, then someequivalence class Pk+1 has κ+ elements. This construction must stop for somes ≤ 2n. Otherwise ∀k = 1, . . . , n∀x, y ∈ P2n[a(k, x) = a(k, y) and b(k, x) = b(k, y)],contradiction.

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If 1 ≤ s ≤ n, then the elements a(s, x) for x ∈ Ps are all distinct, while ifn < s ≤ 2n, then the elements b(s− n, x) for x ∈ Ps are all distinct.

Case 1. 1 ≤ s ≤ n. We define a homomorphism f of 〈Ps〉A into the BA of allsubsets of L ∩ [ds−1, cs) by setting f(x) = x ∩ [ds−1, cs) for all x ∈ 〈Ps〉A. Nowif x, y ∈ Ps and x < y, then f(x) = [a(s, x), cs) ⊆ f(y) = [a(s, y), cs), hencea(s, x) > a(s, y). Thus Ps is isomorphic to a subset of M∗.

Case 2. n < s ≤ 2n. Similarly: Ps is isomorphic to a subset of M . �

Corollary 4.75. (GCH) For every infinite cardinal λ there is an interval algebra Aof size λ++ such that every subalgebra of A of size λ++ has depth λ+.

Proof. We can apply Theorem 4.74 with κ = λ+; then μ = λ+. If B is a subalgebraof A of size λ++ and X is a chain in B of order type λ++, then by (ii), X has alinearly ordered subset of size λ++ which can be isomorphically embedded in Mor M∗, contradicting D dense in M . By the Erdos–Rado theorem B has a chainof order type λ+. �

Problem 59. Can one prove in ZFC that for every infinite cardinal λ there is aninterval algebra A such that |A| = λ+ and every subalgebra of A of size λ+ hasdepth λ?

Now we consider systematically the possibilities for DepthSr for algebras of sizeat most ω2, following the lexicographic order used in discussing cSr, and omittingmost easy cases. Note in particular that (ω, ω) is always a member of DepthSr.

(S1) DepthSr(A) = {(ω, ω), (ω, ω1)} for A = Fr(ω1).

(S2) There does not exist a BA A such that DepthSr(A) = {(ω, ω), (ω1, ω1)}, byProposition 4.73(vi).

(S3) DepthSr(A) = {(ω, ω), (ω, ω1), (ω, ω2)} for A = Fr(ω2).

(S4) DepthSr(A) = {(ω, ω), (ω, ω1), (ω1, ω1)} for A = Intalg(ω1).

(S5) DepthSr(A) = {(ω, ω), (ω, ω1), (ω1, ω1), (ω1, ω2)} for the algebra A of Corol-lary 4.75, assuming GCH.


DepthSr(A) = {(ω, ω), (ω, ω1), (ω1, ω1), (ω1, ω2)}?

(S6) DepthSr(A) = {(ω, ω), (ω, ω1), (ω1, ω1), (ω2, ω2)} is ruled out by Theorem4.73(vi).

(S7) DepthSr(A) = {(ω, ω), (ω, ω1), (ω, ω2), (ω1, ω1), (ω1, ω2)} for

Fr(ω2)× Intalg (ω1).

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(S8) DepthSr(A) = {(ω, ω), (ω, ω1), (ω, ω2), (ω1, ω1), (ω2, ω2)} is ruled out by The-orem 4.73(vii).

(S9) DepthSr(A) = {(ω, ω), (ω, ω1), (ω, ω2), (ω1, ω1), (ω1, ω2), (ω2, ω2)} for A =Fr(ω2)× Intalg(ω2).

We turn to DepthHr.

Theorem 4.76.

(i) If (κ, λ) ∈ DepthHrA, then κ ≤ λ ≤ |A| and κ ≤ t(A).

(ii) If κ ∈ [ω, t(A)) then there is a λ ≤ 2κ such that (κ, λ) ∈ DepthHrA.

(iii) (Depth(A), |A|) ∈ DepthHr(A).

(iv) If (κ′, λ′) ∈ DepthHr(A) where κ′ is a successor cardinal or a limit cardinalof cofinality ω, then there is a κ′′ ≥ κ′ such that (κ′′, |A|) ∈ DepthHr(A).

Proof. (i)–(iii) are clear. For (iv), we follow the proof of Theorem 3.51(v). Forbrevity let λ = |A|. There is nothing to prove if κ′ = λ, so suppose that κ′ < λ.Let f be a homomorphism from A onto a BA B such that Depth(B) = κ′. ByTheorem 4.8 Depth(B) is attained. Let 〈bξ : ξ < κ′〉 be a strictly increasingsequence in B. For each ξ < κ′ choose aξ ∈ A such that f(aξ) = bξ. We nowconsider two cases.

Case 1. |A � aξ| < λ for all ξ < κ′. Let J be the ideal in A generated by {aξ ·−aη :ξ < η < κ′}. Then

(∗) |J | < λ.

In fact, a ∈ J if and only if there is a finite set Γ of ordered pairs (ξ, η) withξ < η < κ′ such that

(∗∗) a ≤∑

(ξ,η)∈Γ

(aξ · −aη).

Take any such set Γ. For any a satisfying (∗∗) let g(a) = 〈a · aξ · −aη : (ξ, η) ∈ Γ〉.Clearly g is a one-one function from the set of a satisfying (∗∗) into

∏(ξ,η)∈Γ(A �

aξ). Hence by our case condition, there are fewer than λ such a’s. Since the numberof sets Γ is κ′ < λ, (∗) follows.

From (∗) we see that |A/J | = λ.

Now we claim that if α < β < κ′, then aβ · −aα /∈ J . [Hence 〈aα/J : α < κ′〉is a strictly increasing sequence in A/J , as desired.] For, suppose that α < β < κ′

and aβ ·−aα ∈ J . Then there is a finite set Γ of ordered pairs (ξ, η) with ξ < η < κ′

such that

(∗∗∗) aβ · −aα ≤∑

(ξ,η)∈Γ

(aξ · −aη).

Applying f , we get bβ · −bα = 0, contradiction. So the claim holds, and this caseis finished.

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Case 2. There is a ξ0 < κ′ such that |A � aξ0 | = λ. Then if we take the homomor-phism

g : A ∼= (A � aξ0)× (A � −aξ0)→ (A � aξ0)× (B � −bξ0)

where the second homomorphism is determined by the identity and f � (A � −aξ0),we get a homomorphism from A onto the algebra C

def= (A � aξ0) × (B � −bξ0),

which has size λ. The depth of C is at least κ′, since 〈bα · −bξ0 : ξ0 ≤ α < κ′〉 isclearly strictly increasing in B � −bξ0 . �

Also, the following examples are relevant:

(1) If A is the finite-cofinite algebra on κ, then DepthHrA = {(ω, λ) : ω ≤ λ ≤ κ}.

(2) If A is free on κ, then DepthHrA = {(λ, μ) : ω ≤ λ ≤ μ ≤ κ}.

(3) If L is a linear order, then every homomorphic image A of Intalg(L) of size(2κ)+ has a chain of order type κ+. This is true by the Erdos–Rado theorem, sinceA is isomorphic to an interval algebra. This fact gives a trivial solution of Problem16 of Monk [96].

We give full details on the next examples.

Theorem 4.77. Suppose that ω ≤ ρ ≤ κ. Let A = 〈[κ]≤ρ〉P(κ). Let

S = {(μ, ν) : ω ≤ μ ≤ ν ≤ 2ρ, νω = ν};T = {(μ, νρ) : ρ+ ≤ μ ≤ 2ρ, 2ρ < νρ, ν ≤ κ}.

Then DepthHr(A) = S ∪ T .

Proof. First suppose that (μ, ν) ∈ S. The mapping a !→ a ∩ ρ gives a homomor-phism of A onto P(ρ). Now by the theorem of Fichtenholz, Kantorovich and Haus-dorff, P(ρ) has an free subalgebra of size 2ρ; hence by Sikorski’s extension theoremthere is a homomorphism of P(ρ) onto an algebra B such that Fr(ν) ≤ B ≤ Fr(ν).Since νω = ν and free algebras have ccc, it follows that |B| = ν. Now by Sikorski’sextension theorem again, there is a homomorphism of B onto an algebra C suchthat Fr(ν)× Intalg(μ) ≤ C ≤ Fr(ν)× Intalg(μ). Thus |C| = ν and Depth(C) = μ,so (μ, ν) ∈ DepthHr(A).

Second, suppose that ρ+ ≤ μ ≤ 2ρ, 2ρ < νρ, and ν ≤ κ. The mappinga !→ a ∩ [ν]≤ρ is a homomorphism from A onto 〈[ν]≤ρ〉P(ν). Then the mappinga !→ (a∩ρ, a\ρ) is an isomorphism from 〈[ν]≤ρ〉P(ν) onto P(ρ)×〈[ν\ρ]≤ρ〉P(ν\ρ).Now ρ < ν, as otherwise νρ ≤ ρρ = 2ρ, contradiction. It follows that |ν| = |ν\ρ|,and so 〈[ν\ρ]≤ρ〉P(ν\ρ) ∼= 〈[ν]≤ρ〉P(ν). As above there is a homomorphism fromP(ρ) onto an algebra with depth μ. Putting this together with the identity on〈[ν\ρ]≤ρ〉P(ν\ρ), we get a homomorphism from P(ρ) × 〈[ν\ρ]≤ρ〉P(ν\ρ) onto analgebra B with depth max(μ, ρ+) = μ and size νρ. This shows that (μ, νρ) ∈DepthHr(A).

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Now suppose conversely that (μ, ν) ∈ cHr(A). Since A is σ-complete, by atheorem of Koppelberg we have νω = ν. So if ν ≤ 2ρ, then (μ, ν) ∈ S.

Suppose that 2ρ < ν. By Juhasz, Shelah [98], νρ = ν. Let σ be minimumsuch that σρ = ν. If κ < σ, then κρ ≤ σρ = ν ≤ κρ, so κρ = ν, contradiction.Hence σ ≤ κ. Thus we have ν = σρ with σ ≤ κ. So now it suffices to show thatρ+ ≤ μ ≤ 2ρ. Let I be an ideal on A such that |A/I| = ν and Depth(A/I) = μ. LetB = A/I. Define by recursion aξ ∈ [κ]≤ρ such that ξ < η implies that [aξ]I < [aη]I ,continuing as long as possible. Say that this produces a sequence 〈aξ : ξ < α〉. We

claim that ρ+ ≤ α. For, suppose that α < ρ+. Then bdef=⋃

ξ<α aξ has size ≤ ρ.Now B ∼= (B � [b]I) × (B � −[b]I), and |B � [b]I | ≤ 2ρ < ν. It follows that thereis a c ∈ [κ]≤ρ such that 0 < [c]I < −[b]I . Let aα = b ∪ (c\b). So [aξ]I ≤ [aα]I forall ξ < α. Moreover, aα\aξ ⊇ c\b and c\b /∈ I, so we could have extended oursequence further, contradiction. Hence ρ+ ≤ α. This proves that ρ+ ≤ μ.

Now we show that μ ≤ 2ρ. In fact, suppose that 2ρ < μ. Let 〈[aξ]I : ξ < μ〉be strictly increasing.

Case 1. {ξ < (2ρ)+ : |aξ| ≤ ρ} has size (2ρ)+. We may assume that ∀ξ <

(2ρ)+[|aξ| ≤ ρ]. By the general Δ-system lemma, let M ∈ [(2ρ)+](2ρ)+

be suchthat 〈aξ : ξ ∈M〉 is a Δ-system, say with kernel c. Take ξ < η < θ in M . Then

0 �= [aη\aξ] = [aη\c] = [aη\aθ] = 0,

contradiction.

Case 2. {ξ < (2ρ)+ : |aξ| ≤ ρ} has size less than (2ρ)+. This case is treatedsimilarly to Case 1. �

Both sets S and T are needed here. Thus for ρ = ω and κ = (2ω)+ we have(ω, 2ω) ∈ S\T and (ω1, (2

ω)+) ∈ T \S.

The theorem simplifies as follows assuming that 2ρ = ρ+.

Corollary 4.78. Suppose that ω ≤ ρ ≤ κ and 2ρ = ρ+. Let A = 〈[κ]≤ρ〉P(κ). Let

S = {(μ, ν) : ω ≤ μ ≤ ν ≤ ρ+, ω < cf(ν)};T = {(ρ+, νρ) : ρ+ < νρ, ν ≤ κ}.

Then DepthHr(A) = S ∪ T .

Corollary 4.79. (CH) DepthHr([ω2]≤ω) = {(ω, ω1), (ω1, ω1), (ω1, ω2)}.

Corollary 4.80. Suppose that ω ≤ ρ ≤ κ. Then 〈[κ]≤ρ〉P(κ) has an independentsubset of size 2ρ, but none of size (2ρ)+.

Proof. Clearly 〈[κ]≤ρ〉P(κ) has an independent subset of size 2ρ. Suppose thatX is an independent subset of size (2ρ)+. Then by Sikorski’s extension theorem〈[κ]≤ρ〉P(κ) has a homomorphic image with depth (2ρ)+, contradicting Theorem4.77. �

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Theorem 4.81. Suppose that λ is an uncountable regular limit cardinal and λ ≤ κ.Let A = 〈[κ]<λ〉P(κ). Define

S = {(μ, ν) : ω ≤ μ ≤ ν < 2<λ, νω = ν};T = {(μ, ν<λ) : λ ≤ μ ≤ 2<λ ≤ ν ≤ κ};U = {(μ, 2<λ) : ω ≤ μ ≤ 2<λ}.

Then:

(i) S ∪ T ⊆ DepthHr(A);

(ii) If there is a ρ < λ such that 2ρ = 2<λ, then DepthHr(A) = S ∪ T ∪ U .

(iii) If 2ρ < 2<λ for all ρ < λ, then DepthHr(A) = S ∪ T .

Proof. To prove that S ⊆ DepthHr(A), suppose that ω ≤ μ ≤ ν < 2<λ and νω = ν.Choose ρ < λ such that ν < 2ρ. The mapping a !→ a ∩ ρ gives a homomorphismof A onto P(ρ), and by Theorem 4.77 with κ = ρ we get (μ, ν) ∈ DepthHr(A).

To prove that T ⊆ DepthHr(A), suppose that λ ≤ μ ≤ 2<λ ≤ ν ≤ κ. Nowfirst suppose also that μ < 2<λ. Choose ρ < λ such that μ < 2ρ. By Theorem4.77 there is a homomorphic image B of P(ρ) with depth μ. Then the desiredhomomorphic image of A is obtained from the following obvious composition ofhomomorphisms:

A ∼= P(ρ)× 〈[κ\ρ]<λ〉→ B ×A

→ B × 〈[ν]<λ〉.If μ = 2<λ a more complicated construction is needed. Let 〈Mα : α < λ〉 be adisjoint sequence of infinite members of [κ]<λ such that 〈|Mα| : α < λ〉 is strictlyincreasing, with supremum λ. For each α < λ let fα be a homomorphism ofP(Mα) onto an algebra Bα whose size and depth are both equal to 2|Mα|; thisis possible by Theorem 4.77. Let N =

⋃α<λ Mα, and let g : A → C

def= 〈[N ]<λ〉

be the homomorphism (onto) given by g(a) = a ∩ N for all a ∈ A. Next, defineh : C →

∏α<λ P(Mα) by setting (h(a))α = a ∩Mα for all a ∈ C and α < λ.

Clearly h is a homomorphism. Let D = rng(h).

(1) |D| ≤ 2<λ.

For, note that if a ∈ [N ]<λ, then a ⊆⋃

β<α Mβ for some α < λ; then (h(a))β = ∅for all β ∈ (α, λ). Hence

|D| ≤∑α<λ

∏β<α

2|Mβ | =∑α<λ

2∑

β<α |Mβ | ≤ 2<λ.

So (1) holds.

Now define k :∏

α<λ P(Mα) →∏

α<λ Bα by setting (k(x))α = fα(xα) forevery x ∈

∏α<λ P(Mα). So k is a homomorphism. Let E = rng(k ◦ h ◦ g). So by

(1) we have |E| ≤ 2<λ.

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(2) Depth(E) = 2<λ.

In fact, we construct a strictly increasing sequence of members of E, as follows.For α < λ, let 〈aαξ : ξ < 2|Mα|〉 be a sequence of subsets of Mα such that〈fα(aαξ) : ξ < 2|Mα|〉 is a strictly increasing sequence of elements of Bα. Then forα < λ and ξ < 2|Mα| let

cαξ =⋃β<α

Mβ ∪ aαξ.

Clearly if (α, ξ) < (β, η) lexicographically, then k(h(g(cαξ))) < k(h(g(cβη))). Thus(2) holds.

Next, for each a ∈ A let l(a) = a ∩ ν. So l is a homomorphism from A onto〈[ν]<λ〉P(ν). Finally, for each a ∈ A let s(a) = (k(h(g(a))), l(a)). Clearly s is a

homomorphism from A onto a BA with depth μ and size ν<λ. Thus (i) holds.

For (ii), suppose that ρ < λ such that 2ρ = 2<λ. For ⊇, by (i) we just needto show that U ⊆ DepthHr(A). The mapping a !→ a ∩ ρ is a homomorphism fromA onto P(ρ), and so U ⊆ DepthHr(A) by Theorem 4.77 with κ = ρ.

For ⊆, suppose that (μ, ν) ∈ DepthHr(A). Let I be an ideal of A such that|A/I| = ν and Depth(A/I) = μ. To reach a contradiction, suppose that (μ, ν) /∈S ∪ T ∪ U . Since (μ, ν) /∈ U , we have 2<λ < ν. Hence by Juhasz, Shelah [98] wehave ν<λ = ν. Let σ be minimum such that σ<λ = ν. As above it follows thatσ ≤ κ. So now since (μ, ν) /∈ T we get μ < λ or 2<λ < μ.

Suppose that μ < λ. Define by recursion aξ ∈ [κ]<λ such that ξ < η impliesthat [aξ]I < [aη]I , continuing as long as possible. Say that this produces a sequence

〈aξ : ξ < α〉. Thus by our supposition, α < λ. Then bdef=⋃

ξ<α aξ has size less

than λ. Now B ∼= (B � [b]I)× (B � −[b]I), and |B � [b]I | ≤ 2<λ < ν. It follows thatthere is a c ∈ [κ]≤ρ such that 0 < [c]I < −[b]I . Let aα = b∪ (c\b). So [aξ]I ≤ [aα]Ifor all ξ < α. Moreover, aα\aξ ⊇ c\b and c\b /∈ I, so we could have extended oursequence further, contradiction.

The proof that μ ≤ 2<λ is very similar to the argument in the proof ofTheorem 4.77.

For (iii), ⊇ is given by (i). Now suppose that (μ, ν) ∈ DepthHr(A)\(S ∪ T ).Then 2<λ ≤ ν ≤ κ and μ < λ or 2<λ < μ. The above arguments then give acontradiction. �

Corollary 4.82. If λ is an uncountable strongly inaccessible cardinal ≤ κ, then

DepthHr(〈[κ]<λ〉) = {(μ, ν) : ω ≤ μ ≤ ν < λ, νω = ν}∪{(λ, ν<λ) : λ ≤ ν ≤ κ}. �

Problem 61. Characterize the relation DepthHr.

This is Problem 17 of Monk [96].

For small cardinals there are many open questions, and we will not surveythe situation here.

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Now we turn to the depth and small depth in special classes of BAs. Depthis the same as cellularity for complete BAs, while every infinite complete BA hasa tower of length ω. For any complete BA A, towspect(A) contains all regularcardinals in the interval [ω, c(A)), and also has c(A) as a member if it is attained.

It is possible to have Depth(A) < c(A) for an interval algebra. For example,let τ be the order type of the real numbers, let L be an ordered set of type0 + (ω + ω∗) · τ , and let A be the interval algebra on L. It is easily seen thatDepth(A) = ω while c(A) = 2ω.

Concerning small depth for interval algebras we have the following result,which was given in Monk [01] in a somewhat weaker form.

Theorem 4.83. Suppose that L is an infinite linear order with first element 0, andκ is a regular cardinal. Then the following conditions are equivalent.

(i) Intalg(L) has a tower of order type κ.

(ii) One of the following holds:

(a) There is a c ∈ L and a strictly decreasing sequence 〈aα : α < κ〉 ofelements of (c,∞) coinitial with c.

(b) There is a c ∈ L ∪ {∞} and a strictly increasing sequence 〈aα : α < κ〉of elements of [0, c) cofinal in c.

(c) There exist a strictly increasing sequence 〈bα : α < κ〉 of elements of Land a strictly decreasing sequence 〈cα : α < κ〉 of elements of L suchthat bα < cβ for all α, β < κ, and there is no element d ∈ L such thatbα < d < cβ for all α, β < κ.

Proof. (ii)⇒(i): Assume (ii)(a). Then 〈[0, c) ∪ [aα,∞) : α < κ〉 is a tower inIntalg(L).

Assume (ii)(b). Then 〈[0, aα) ∪ [c,∞) : ξ < κ〉 is a tower in Intalg(L).

Assume (ii)(c). Then 〈[0, bα) ∪ [cα,∞) : α < κ〉 is a tower in Intalg(L).

(i)⇒(ii): Let 〈aα : α < κ〉 be a tower in Intalg(L). Write

aα = [bα0 , cα0 ) ∪ · · · ∪ [bαmα−1, c

σmα−1),

where 0 ≤ bα0 < cα0 < · · · < bαmα−1 < cαmα−1 ≤ ∞. Clearly bα0 ≥ bβ0 if α < β < κ. Ifeach bα0 �= 0, then clearly 0 is the g.l.b. of 〈bα0 : α < κ〉, and (ii)(a) holds with c = 0.

Hence we may assume that bα0 = 0 for all α < κ. Clearly cα0 ≤ cβ0 if α < β < κ. Ifsupα<κ c

α0 = ∞, then (ii)(b) holds for ∞ and some subsequence of 〈cα0 : α < κ〉.

So we may assume that supα<κ cα0 �=∞. We now consider several cases.

Case 1. There is an α < κ such that cβ0 = cα0 for all β ∈ [α, κ). Clearly then eachmα > 1 and there is a subsequence of 〈bα1 : α < κ〉 which gives (ii)(a).

Case 2. There is no α as in Case 1, but supα<κ cα0 exists. This gives (ii)(b).

Case 3. There is no α as in Case 1, but supα<κ cα0 does not exist. Clearly then

there is a subsequence of 〈bα1 : α < κ〉 which gives (ii)(c). �

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As an application we give an example of an infinite linear order whose intervalalgebra does not have any towers. The construction depends on some more-or-lessstandard notation, which we now introduce.

A gap in a linear order L is a pair (A,B) of nonempty subsets of L such thatL = A∪B, ∀a ∈ A∀b ∈ B(a < b), A has no largest element, and B has no smallestelement. The lower character of such a gap is the least cardinality of a subset ofA cofinal in A, and similarly for upper character; these are both regular cardinals.The character of (M,N) is the pair of these characters.

Theorem 4.84. There is an infinite linear order L such that Intalg(L) does nothave a tower.

Proof. We start out with a dense linear order M such that every point of M hascharacter (ω1, ω1), the gaps of M have characters (ω, ω1) and (ω1, ω), M has nofirst or last element, and M has cofinality and coinitiality ω1. The existence ofM follows from a theorem in Hausdorff [1908]. We replace each element of M byω∗+ω, put ω to the left of the result, and ω∗ to the right. For definiteness let a and bbe one-one functions such that rng(a)∩rng(b) = ∅ and (rng(a)∪rng(b))∩(M×Z) =∅. Let

L = rng(a) ∪ rng(b) ∪ (M × Z),

and for m,n ∈ ω, p, q ∈ Z, and c, d ∈M define

am < an iff m < n;

bm < bn iff n < m;

am < (c, p);

(c, p) < (c, q) iff p < q;

(c, p) < (d, q) iff c < d, when c �= d;

am < bn;

(c, p) < bm.

Clearly this gives a linear order, and the elements do not have infinite characters.Thus it suffices by the preceding theorem to show that the characters of the gapsof L are (ω, ω1) or (ω1, ω).

Suppose that (A,B) is a gap in L. Then there are these possibilities:

Case 1. A = {an : n ∈ ω}. Then the character of (A,B) is (ω, ω1).

Case 2. There is a c ∈ M such that A = {x ∈ L : x < (c,m) for every m ∈ Z}.Then the character of (A,B) is (ω1, ω).

Case 3. There is a c ∈ M such that A = {x ∈ L : x < (d,m) for every d > c andevery m ∈ Z}. Then the character of (A,B) is (ω, ω1).

Case 4. There is a gap (C,D) in M such that A = {(c,m) : c ∈ C,m ∈ Z}∪rng(a).Then the character of (A,B) is the same as the character of (C,D), and thus is(ω, ω1) or (ω1, ω).

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Case 5. A = {x ∈ L : x < bm for all m ∈ ω}. Then the character of (A,B) is(ω1, ω). �

Another application of Theorem 4.82 is to the tower spectrum.

Theorem 4.85. If K is a nonempty set of regular cardinals, then there is a denselinear order L such that towspect(Intalg(L)) = K.

Proof. Let μ be any successor cardinal greater than each member of K, and alsofix λ ∈ K. By the main theorem of Hausdorff [08], let L be a dense linear ordersuch that L has an element with character (κ, κ) for each κ ∈ K, L has coinitialityλ∗ and cofinality λ, and the set of gap characters of L is

{(ρ, μ) : ρ regular, ρ < μ} ∪ {(μ, ρ) : ρ regular, ρ < μ}.

Then L is as desired. �

Now we turn to tree algebras. By Proposition 16.20 in the Handbook, if T is aninfinite tree then Depth(Treealg(T )) is equal to max(sup{|C| : C is a chain inT }, ω).

We turn to towers in tree algebras.

Lemma 4.86. Suppose that T is a tree with a single root r. Suppose that κ is aregular cardinal, and T has a chain of order type κ with finitely many immediatesuccessors.

Then Treealg(T ) has a tower of order type κ.

Proof. Let 〈tα : α < κ〉 be a chain in T with the finite set M of immediatesuccessors. For each α < κ let

aα = [(T ↑ r)\(T ↑ tα)] ∪⋃s∈M

(T ↑ s).

Clearly 〈aα : α < κ〉 is strictly increasing. To show that it is a tower, it sufficesto take an element b of the form b = (T ↑ s)\

⋃u∈N (T ↑ u) with N a finite set of

elements of (T ↑ s)\{s} and find α < κ such that b ∩ aα �= ∅. If tα �≤ s for someα < κ, then s ∈ b ∩ [(T ↑ r)\(T ↑ tα)] ⊆ b ∩ aα, as desired. If tα ≤ s for all α < κ,then there is an u ∈M such that u ≤ s, and a0 ∩ b �= ∅. �

Theorem 4.87. Let T be an infinite tree with a single root r. Then the followingconditions are equivalent:

(i) Treealg(T ) has a tower of order type ω.

(ii) One of the following conditions holds:

(a) T has an element with exactly ω immediate successors.

(b) T has a chain of countable limit length with at most ω immediate suc-cessors.

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Proof. (ii)⇒(i): Assume (ii)(a); let t be an element with 〈si : i ∈ ω〉 an enumerationof all distinct immediate successors. For each i ∈ ω let

ai = [(T ↑ r)\(T ↑ t)] ∪⋃j≤i

(T ↑ sj).

Clearly 〈ai : i ∈ ω〉 is a strictly increasing sequence of elements of Treealg(T ).To show that it is a tower, it suffices to take an element b of Treealg(T ) of theform b = (T ↑ u)\

⋃v∈M (T ↑ v), where M is a collection of pairwise incomparable

elements of (T ↑ u)\{u} and show that b ∩ ai �= ∅ for some i ∈ ω. If t �≤ u, theni = 0 works. If t = u, then i = 0 still works, since (T ↑ si) ⊆ b for any si whichis not below any member of M . If t < u, then there is an i such that si ≤ u, andhence b ≤ ai, as desired.

Now assume (ii)(b); let 〈ti : i ∈ ω〉 be strictly increasing with a set M ofimmediate successors, |M | ≤ ω. For M finite, Lemma 4.85 gives the desired result.For M infinite, say with M = {si : i ∈ ω}, define for each i ∈ ω

ai = [(T ↑ r)\(T ↑ ti)] ∪⋃j≤i

(T ↑ si).

The argument for Lemma 4.85 just needs a slight change to work for this case.

Now suppose that both conditions (ii)(a) and (ii)(b) fail, but Treealg(T ) hasa tower of order type ω; we want to get a contradiction. Say that 〈ai : i ∈ ω〉 is atower in Treealg(T ). For each i < ω write ai in full normal form:

ai =⋃

t∈Mi

eit;

eit = (T ↑ t)\⋃

s∈Nit

(T ↑ s);

whereMi is a finite subset of T ,Nit is a finite set of pairwise incomparable elementsof (T ↑ t)\{t}, eit ∩ eiu = ∅ for t �= u, and t /∈ Niu for t �= u.

(1) If i < ω, t ∈Mi, and s ∈ Nit, then s /∈ ai.

In fact, suppose that i < ω, t ∈Mi, s ∈ Nit, and s ∈ ai. Choose u ∈Mi such thats ∈ eiu. Since s /∈ eit, it follows that u �= t. Moreover, s �= u by normality. Nowu < s and t < s, giving two cases:

Case 1. t < u. Then u ∈ eit (since u < s) and u ∈ eiu, contradicting disjointness.

Case 2. u < t. Then t ∈ eiu (since t < s ∈ eiu) and t ∈ eit, contradictingdisjointness.

Thus (1) holds.

(2) T =⋃

i<ω ai.

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In fact, suppose that t ∈ T \⋃

i<ω ai. If t has a finite setM of immediate successors,then {t} = (T ↑ t)\

⋃s∈M (T ↑ s), so {t}∩ai �= ∅ for some i < ω, contradiction. So

t has infinitely many immediate successors. Since (ii)(a) fails, it has uncountablymany.

Let Q be the set of immediate successors of t below some member of⋃

i<ω Mi.So Q is countable. Let u be an immediate successor of t not in Q. Choose i < ωsuch that ai ∩ (T ↑ u) �= ∅. Say s ∈ ai ∩ (T ↑ u). Choose v ∈ Mi such thatv ≤ s. Then u ≤ s and v ≤ s, so u and v are comparable. If u ≤ v, then u ∈ Q,contradiction. Hence v < u, and so v ≤ t. So t ∈ [v, s] and v, s ∈ ai, hence t ∈ ai,contradiction. Thus (2) holds.

By (2) we may assume that r ∈ ai for all i < ω. Hence r ∈Mi for all i < ω.

(3) If i < j and s ∈ Njr, then there is a unique t ∈ Nir such that t ≤ s.

In fact, under the hypotheses of (3), if s /∈⋃

t∈Nir(T ↑ t), then s ∈ eir ⊆ ai ⊆ aj .

Since s /∈ ejr , it follows that s ∈ ejt for some t ∈ Mj\{r}. So t ≤ s. Since t ∈ ejt,it follows that t /∈ ejr. Hence t = s, contradicting normality of aj. So it followsthat there is a t ∈ Nir with t ≤ s. This t is unique, since the elements of Nir arepairwise incomparable. Thus (3) holds.

(4) If i < j, then eir ⊆ ejr.

For, suppose that s /∈ ejr. Then there is a t ∈ Njr such that t ≤ s. By (3) we getu ∈ Nir such that u ≤ t, and hence s /∈ eir.

Now let T ′ = {(s, i) : i < ω and s ∈ Nir}. We define (s, i) < (t, j) iffi < j and s ≤ t. This makes T ′ into a tree of height ω, by (3). For each i < ω,the set N ′

i = {(s, i) : s ∈ Nir} is the set of all elements of level i, and it isfinite. Hence by Konig’s tree lemma, there is a branch (s0, 0) < (s1, 1) < · · ·through T ′. By (4), there is a subsequence (si(0), i(0)) < (si(1), i(1)) such thati(0) < i(1) < · · · and si(0) < si(1) < · · · . Hence by the assumption that (ii)(b)fails, the sequence 〈si(k) : k < ω〉 has uncountably many immediate successors.Let Q be the collection of all immediate successors t of 〈si(k) : k < ω〉 such thatt ≤ u for some u ∈

⋃γ<ω Mγ . So Q is countable. Let t be an immediate successor

of 〈si(k) : k < ω〉 which is not in Q. Choose k < ω such that (T ↑ t) ∩ ai(k) �= ∅.Choose u ∈ (T ↑ t) ∩ ai(k). Since u ∈ ai(k), choose v ∈ Mi(k) such that u ∈ ei(k)v.Thus v, t ≤ u, so v and t are comparable. If t ≤ v, the choice of t is contradicted.So v < t. Choose h < ω such that v < si(h). We may assume that k < h. Nowsi(h) ∈ ai(k) since v, u ∈ ei(k)v and v < si(h) < t ≤ u. But ai(k) ⊆ ai(h), sosi(h) ∈ ai(h), contradicting (1). �

Theorem 4.88. Suppose that T is an infinite tree with a single root r, and let κ bean uncountable regular cardinal. Then the following conditions are equivalent.

(i) Treealg(T ) has a tower of order type κ.

(ii) There is a strictly increasing sequence 〈tα : α < κ〉 of elements of T whichhas only finitely many immediate successors.

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Proof. (ii)⇒(i) holds by Lemma 4.85. Now suppose that (i) holds but (ii) fails.Let 〈aα : α < κ〉 be a tower in Intalg(T ). For each α < κ write aα in full normalform:

aα =⋃

t∈Mα

eαt;

eαt = (T ↑ t)\⋃

s∈Nαt

(T ↑ s);

where Mα is a finite subset of T , Nαt is a finite set of pairwise incomparableelements of (T ↑ t)\{t}, eαt ∩ eαu = ∅ for t �= u, and t /∈ Nαu for t �= u.

(1) T =⋃

α<κ aα.

For, suppose that w ∈ T \⋃

α<κ aα. Then as in the previous proof, w must haveinfinitely many immediate successors. For each α < κ let P (α) consist of allimmediate successors of w which are below some t ∈Mα. Suppose that α < β < κ.Let s ∈ P (α); say s ≤ t ∈ Mα. Then t ∈ aα ⊆ aβ, so there is a u ∈ Mβ suchthat t ∈ (T ↑ u). Thus u ≤ t. So s, u ≤ t and hence s and u are comparable. Ifu < s, then u ≤ w < t and u, t ∈ aβ, so w ∈ aβ, contradiction. It follows thats ≤ u. This proves that P (α) ⊆ P (β). Now κ uncountable implies that there isan α < κ such that P (α) = P (β) for all β ∈ [α, κ). Choose β < κ with somet ∈ aβ ∩ (T ↑ w)\

⋃s∈P (α)(T ↑ s). Then w < t since w /∈ aβ , so there is an

immediate successor s of w such that s ≤ t. Choose u ∈Mβ such that u ≤ t. Sinces, u ≤ t, they are comparable. If u < s, then u ≤ w and hence w ∈ [u, t] ⊆ aβ ,contradiction. So s ≤ u, hence s ∈ P (β) and s ≤ t, contradiction. This proves (1).

By (1) we may assume that r ∈ aα for all α < κ.

(2) If α < β and s ∈ Nβr, then there is a unique t ∈ Nαr such that t ≤ s.

This is clear.

(3) If α < β, then eαr ⊆ eβr.

For, suppose that s /∈ eβr. Then there is a t ∈ Nβr such that t ≤ s. By (2) we getu ∈ Nαr such that u ≤ t, and hence s /∈ eαr.

Now let T ′ = {(s, α) : α < κ and s ∈ Nαr}. We define (s, α) < (t, β) iff α < βand s ≤ t. This makes T ′ into a tree of height κ. Now for each finite subset F ofκ let

MF = {f ∈ Pα<κNαr : ∀α, β ∈ F [α < β → f(α) < f(β)]}.

Clearly MF is a closed subset of Pα<κNαr. By (2), each MF is nonempty. Henceby the Tychonoff product theorem we obtain a function f ∈ Pα<κNαr such thatf(α) < f(β) whenever α < β.

(4) There do not exist α < κ and s ∈ T such that f(β) = (s, β) for all β ≥ α.

This is true by (1). It follows that there are strictly increasing sequences 〈αξ :ξ < κ〉 of ordinals and 〈sξ : ξ < κ〉 of elements of T such that sξ ∈ Nαξr for all

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ξ < κ. Let β = supξ<κ αξ. So β is a limit ordinal of cofinality κ. Moreover, by thehypothesis that (ii) fails, it follows that 〈sξ : ξ < κ〉 has infinitely many immediatesuccessors. Let S be the set of all immediate successors of 〈sξ : ξ < κ〉.

(5) If w ∈ S, then there is a ξ < κ such that w ∈ aαξ, and for any such ξ we have

w ∈Mαξ.

In fact, by (1) T =⋃

ξ<κ aαξ, so there is a ξ < κ such that w ∈ aαξ

. Take anysuch ξ. Then there is a t(w) ∈Mαξ

such that t(w) ≤ w. If t(w) < w, choose η < κsuch that αξ < αη and t(w) < sη. Now sη ∈ [t(w), w] ⊆ aαξ

⊆ aαη , contradiction.It follows that t(w) = w, and so w ∈Mαξ

.

Now let 〈w(n) : n ∈ ω〉 be a one-one enumeration of some elements of S. Foreach n ∈ ω choose by (5) a ξ(n) < κ such that ∀η ∈ [ξ(n), κ)[w(n) ∈ Mαη ]. Letη = supn∈ω ξ(n). Then Mαη is infinite, contradiction. �

Theorem 4.89. Let T be an infinite tree. Then the following conditions are equiv-alent:

(i) Treealg(T ) has a tower.

(ii) One of the following conditions holds:

(a) T has an element with exactly ω immediate successors.

(b) T has a chain of countable limit length with at most ω immediate suc-cessors.

(c) For some uncountable regular cardinal κ, T has an chain of order typeκ with only finitely many immediate successors. �

Theorem 4.90. If K is a nonempty set of regular cardinals, then there is an atom-less tree algebra A such that towspect(A) = K.

Proof. Let λ be the smallest member of K. Let Tω be the tree <ωω1. For eachuncountable regular cardinal, let Tκ be the tree determined by the following con-ditions. T has a unique root. T has height κ. Each element of T , and each initialchain of T of limit ordinal type less than κ has exactly ω1 immediate successors.Now the tree desired in the theorem is formed by adjoining a new root beneaththe following trees: ω1 copies of Tλ and, for each κ ∈ K\{λ}, a copy of Tκ. By theabove theorems, T is as desired.

Although these theorems are fairly definitive, they use unusual trees – trees thathave many elements directly above certain infinite chains. Let us call a tree T limit-normal iff every initial chain of T of limit ordinal length has at most one immediatesuccessor. Concerning towers in such trees we have the following results.

Theorem 4.91. Suppose that T is a limit-normal tree with a single root andTreealg(T ) is atomless. Then the following conditions are equivalent:

(i) Treealg(T ) has a tower.

(ii) Treealg(T ) has a tower of length ω.

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(iii) T has an element with exactly ω immediate successors or T has infiniteheight. �

Theorem 4.92. Suppose that T is a limit-normal tree with a single root, withTreealg(T ) atomless. Suppose that κ is an uncountable regular cardinal. Then thefollowing conditions are equivalent:

(i) Treealg(T ) has a tower of order type κ.

(ii) T has height at least κ. �

Theorem 4.93. Suppose that T is a limit-normal tree with a single root, withTreealg(T ) atomless. Then towspect(Treealg(T )) has the form [ω, λ)reg for someλ > ω. �

Theorems 4.85 and 4.86 generalize to pseudo-trees; see Monk [09].

Concerning superatomic BAs we mention a result of Dow, Monk [97]: if λ isa Ramsey cardinal, then every superatomic BA with depth at least λ+ has depthat least λ. This result has been strengthened by Shelah, Spinas [99].

We finish this chapter by giving two theorems concerning depth and relatedfunctions in the algebra P(ω)/fin.

Theorem 4.94. Under MA, p(P(ω)/fin) = 2ω.

Proof. It clearly suffices to prove the following statement:

(∗) If X ⊆ [ω]ω, |X | < 2ω, and ω\⋃F is infinite for every finite F ⊆ X , then

there is an infinite set d ⊆ ω such that x∩d is finite for all x ∈ X and ω\(d∪⋃

F )is infinite for every finite F ⊆ X .

This is immediate from Kunen [80], II.2.15. �

Corollary 4.95. Under MA, each of the following functions, applied to P(ω)/fin,is equal to 2ω: c, a, Depth, tow, p, h, spl. �

Actually the proof of Theorem 4.93 shows that if MA(κ), then p(P(ω)/fin) > κ.Hence we get the following additional corollary, by MA(ω):

Corollary 4.96. (In ZFC) Each of the following functions, applied to P(ω)/fin, isgreater than ω: c, a, Depth, tow, p, h, spl. �

For both corollaries, recall that c(P(ω)/fin) = 2ω in ZFC.

The second result is of a folklore nature.

Theorem 4.97. There is a model of ZFC in which 2ω > ω1 while P(ω)/fin hasdepth ω1. In fact, we can take M [G], where M satisfies CH and G adds Cohenreals.

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Proof. We follow Kunen [80] for the forcing framework. Assume that (in M) CHholds, and κ is a regular uncountable cardinal. Let P = {p : p is a finite functionwith domain contained in κ and range contained in 2}. And let G be M -genericover P. Let ϕ be the formula

〈Tα : α < ω2〉 is a sequence of subsets of ω,

and ∀α, β ∈ ω2[α < β ⇒ Tα\Tβ is finite and Tβ\Tα is infinite],

and suppose that M [G] |= ϕ; we want to get a contradiction. Choose p ∈ G sothat p � ϕ. From now on we work in M .

Temporarily fix α < ω2 and n ∈ ω. Now p � n ∈ Tα ∨n /∈ Tα, so ∀q ≤ p∃r ≤q(r � n ∈ Tα ∨ r � n /∈ Tα). Hence there is a maximal pairwise incompatible setAαn ⊆ {q : q ≤ p} such that ∀q ∈ Aαn(q � n ∈ Tα ∨ q � n /∈ Tα).

Now for any α ∈ ω2 let

Cα = dmn(p) ∪⋃n∈ω

⋃q∈Aαn

dmn(q).

Thus Cα is countable. By CH there is an X ∈ [ω2]ω2 such that 〈Cα : α ∈ X〉 is a

Δ-system, say with kernel C. We may also assume that there is a γ < ω1 such thatCα\C has order type γ for all α ∈ X . For all α, β ∈ X , let jαβ be the permutationof Cα ∪Cβ such that jαβ is the identity on C, and is the unique order preservingmap from Cα\C onto Cβ\C and from Cβ\C onto Cα\C. Thus jαα is the identityon Cα, and jαβ = jβα. Extend each jαβ to a permutation j′αβ of κ by letting j′αβbe the identity outside of Cα ∪Cβ . Note that j′αβ ◦ j′αβ is the identity mapping onκ. Obviously also

(1) (j′βγ ◦ j′αβ) � Cα = j′αγ � Cα for any α, β, γ ∈ X .

Now j′αβ naturally induces an automorphism j′′αβ of P : for any r ∈P , the domainof j′′αβ(r) is j

′αβ [dmn(r)] and for any γ ∈ dmn(r), (j′′αβ(r))(j

′αβ (γ)) = r(γ). Clearly

(2) j′′αβ = j′′βα.

(3) j′′αβ ◦ j′′αβ is the identity on P.

We will use the notation (j′′αβ)∗ introduced in Kunen [80] VII.7.12; the propertyVII.7.13 is important. Now for each α ∈ X define Pα = {q ∈ P : dmn(q) ⊆ Cα

and q ≤ p}. Note that Pα is countable.

(4) j′′αβ [Pα] = Pβ for any α, β ∈ X .

In fact, suppose that q ∈ Pα. Then dmn(j′′αβ(q)) = j′αβ [dmn(q)] ⊆ j′αβ [Cα] = Cβ .Next, dmn(p) ⊆ C ⊆ Cβ . If γ ∈ dmn(p), then γ ∈ dmn(q) ∩ C, so γ = j′αβ(γ) ∈dmn(j′′αβ(q)) and (j′′αβ(q))(γ) = j′′αβ(q))(j

′αβ(γ)) = q(γ) = p(γ). This shows that

j′′αβ(q) ∈ Pβ .

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Hence j′′αβ [Pα] ⊆ Pβ. By symmetry, j′′αβ [Pβ ] ⊆ Pα. So

Pβ = j′′αβ [j′′αβ [Pβ ]] ⊆ j′′αβ [Pα].

So (4) holds.

(5) (j′′βγ � Pβ) ◦ (j′′αβ � Pα) = (j′′αγ � Pα).

To prove this, take any q ∈ Pα. Then

dmn(j′′βγ(j′′αβ(q))) = j′βγ [dmn(j′′αβ(q))] = j′βγ [j

′αβ [dmn(q)]] = j′αγ [dmn(q)],

using (1). Now take any δ ∈ dmn(q). Then

(j′′βγ(j′′αβ(q)))(j

′αγ (δ)) = (j′′βγ(j

′′αβ(q)))(j

′βγ(j

′αβ(δ)))

= (j′′αβ(q))(j′αβ(δ)) = q(δ) = j′′αγ(j

′αγ(δ)),

proving (5).

We define hα : Pα → ω3 by

(hα(q))(n) =

⎧⎨⎩0, if q � n /∈ Tα,

1, if q � n ∈ Tα,

2, otherwise.

Define α ≡ β iff α, β ∈ X and hβ ◦ (j′′αβ � Pα) = hα. It is straightforward to checkthat ≡ is an equivalence relation on X . Then:

(6) There are at most ω1 equivalence classes under ≡. �

In fact, suppose that Γ ∈ [X ]ω2 consists of pairwise inequivalent ordinals. Fixα ∈ Γ. Then for any distinct β, γ ∈ Γ we have hβ ◦ (j′′αβ � Pα) �= hγ ◦ (j′′αγ � Pα),since otherwise

hβ◦(j′′γβ � Pγ) = hβ◦(j′′αβ � Pα)◦(j′′γα � Pγ) = hγ◦(j′′αγ � Pα)◦(j′′γα � Pγ) = hγ ,

contradiction. But this gives ℵ2 members of Pα(ω3), which by CH has cardinalityℵ1, contradiction. Thus (6) holds.

By (6), let X ′ be an equivalence class with ℵ2 elements. Fix α, β ∈ X ′ withα < β. Now p � (Tβ\Tαis infinite), so, since dmn(p) ⊆ C,

(7) p � ((j′′αβ)∗(T ))β\(j′′αβ)∗(T ))α is infinite.

Next we claim:

(8) p � ∀n ∈ ω(n /∈ Tα → n /∈ ((j′′αβ)∗(T ))β).

To prove this, suppose that q ≤ p and q � n /∈ Tα; we want to show that q �n /∈ ((j′′αβ)∗(T ))β. Suppose that this is not true. Then there is an r ≤ q such that

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r � n ∈ ((j′′αβ)∗(T ))β . Since r ≤ q, we also have r � n /∈ Tα, so there is a t ∈ Aαn

such that r and t are compatible. Thus t ∈Pα and (hα(t))(n) = 0. Let s = j′′αβ(t).Since α ≡ β, we have hβ(s) = hβ(j

′′αβ(t)) = hα(t). Hence (hβ(s))(n) = 0, so

s � n /∈ Tβ . Hence t � n /∈ ((j′′αβ)∗(T ))β . Since r and t are compatible andr � n ∈ (j′′αβ)∗(T ))β, this is a contradiction. So, (8) holds.

Similarly:

(9) p � ∀n ∈ ω(n ∈ Tβ → n ∈ ((j′′αβ)∗(T ))α).

Next, since p � (Tα\Tβ is finite), choose m ∈ ω and q ≤ p so that

(10) q � ∀n ≥ m(n ∈ Tα → n ∈ Tβ).

By (7), q � ∃n ≥ m(n ∈ ((j′′αβ)∗(T ))β ∧ n /∈ ((j′′αβ)∗(T ))α), so choose n ≥ m andr ≤ q so that r � n ∈ ((j′′αβ)∗(T ))β ∧ n /∈ ((j′′αβ)∗(T ))α). By (9), r � n ∈ Tα, so by(10), r � n ∈ Tβ. Then by (9), r � n ∈ (j′′αβ)∗(T ))α, contradiction. �

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5 Topological Density

We begin with some equivalents of this notion. A set X of non-zero elementsof a BA A is said to be centered provided that it satisfies the finite intersectionproperty. And A is called κ-centered if A\{0} is the union of κ centered sets.

Theorem 5.1. For any infinite BA A, d(A) is equal to each of the following cardi-nals:

min{κ : A is isomorphic to a subalgebra of P(κ)};min{κ : A is κ-centered};min{κ : there is a system of κ centered subsets of A whose union

is dense in A}.min{κ : A\{0} is a union of κ proper filters};min{κ : A\{0} is a union of κ ultrafilters}.

Proof. Call the six cardinals mentioned κ0, . . . , κ5 respectively, starting with d(A)itself. κ0 ≤ κ1: Let g be an isomorphism of A into P(κ). For each α < κ letFα = {a ∈ A : α ∈ g(a)}. Then, as is easily checked, Fα is an ultrafilter on A. LetY = {Fα : α < κ}. We claim that Y is dense in Ult(A). For, let U be a non-emptyopen set in Ult(A). We may assume that U = S(a) for some a ∈ A. Thus a �= 0,so choose α ∈ g(a). Then a ∈ Fα, and so Fα ∈ Y ∩ U , as desired.

κ1 ≤ κ2: Suppose that A\{0} =⋃

α<κ Xα, where eachXα is centered. Extendeach Xα to an ultrafilter Fα. For each a ∈ A let f(a) = {α < κ : a ∈ Fα}. Clearlyf is an isomorphism of A into P(κ), as desired.

κ2 ≤ κ3: Suppose that 〈Xα : α < κ〉 is a system of centered sets whose unionis dense in A. For each α < κ let Yα be the set of all elements above some memberof Xα. Clearly Yα is centered, and

⋃α<κ Yα = A\{0}.

Obviously κ3 ≤ κ4 ≤ κ5.

κ5 ≤ κ0: Let X be a dense subset of Ult(A). Then obviously A\{0} =⋃F∈X F , as desired. �

Corollary 5.2. c(A) ≤ d(A) and |A| ≤ 2d(A) for any infinite BA A.

Proof. The first inequality is clear from Theorem 5.1(i). For the second inequality,let D be a dense subset of Ult(A) of size d(A). Define f(a) = S(a) ∩ D for any

, ,OI 10.1007/978-3- - - _ ,


014


Basel 26

219

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220 Chapter 5. Topological Density

a ∈ A, where S is the Stone isomorphism. If a �= b, then S(a�b) �= ∅, henceD ∩ S(a�b) �= ∅ and so D ∩ S(a) �= D ∩ S(b), showing that f is one-one. �

We begin the discussion of algebraic operations for d. If A is a subalgebra of B,then d(A) ≤ d(B), and the difference can be arbitrarily large. If A ≤π B, thend(A) = d(B). In fact, let f be an isomorphism from A into P(κ). By Sikorski’sextension theorem, f extends to a homomorphism from B into P(κ), and thishomomorphism is one-one since A ≤π B. If A ≤s B, then d(A) = d(B). In fact, ifB\{0} is the union of a set F of ultrafilters, then A\{0} =

⋃F∈F (F ∩A. Clearly

the difference can be large when A ≤mg B or A ≤free B, and so the same appliesto ≤proj, ≤rc, ≤σ, ≤reg, and ≤u.

If A is a homomorphic image of B, then d can change either direction in goingfrom B to A. Thus if B is a large free BA and A is a countable homomorphic imageof B, then d goes down. On the other hand, if B = P(ω), and A = P(ω)/fin,then d(B) = ω while d(A) = 2ω, since in A there is a disjoint set of size 2ω.

Next, d(A × B) = max(d(A), d(B)) for infinite BAs A,B. To see this, notethat ≥ is clear, since A and B are isomorphic to subalgebras of A × B. For theother inequality, suppose that f (resp. g) is an isomorphism of A (resp. B) intoP(κ) (resp. P(λ)). Let

X = {(0, α) : α < κ} ∪ {(1, α) : α < λ}.

We define h mapping A×B into P(X) by setting

h(a, b) = {(0, α) : α ∈ f(a)} ∪ {(1, α) : α ∈ g(b)}

for all (a, b) ∈ A× B. It is easily verified that h is an isomorphism of A× B intoP(X), and this proves ≤. A similar idea works for products and weak productsin general:

Theorem 5.3. If 〈Ai : i ∈ I〉 is a system of non-trivial BAs and∏w

i∈I Ai ≤ B ≤∏i∈I Ai, then d(B) =

∑i∈I d(Ai).

Proof. Clearly |I| ≤ d(B). Each Ai can be isomorphically embedded in∏w

i∈I Ai,hence in B, so d(Ai) ≤ d(B). Thus ≥ holds.

Now for each i ∈ I let fi be an isomorphism from Ai into P(d(Ai)). We defineg : B → P({(i, α) : i ∈ I and α ∈ d(Ai)} by setting g(b) = {(i, α) : α ∈ fi(ai)}for each b ∈ B. Clearly g is an isomorphic embedding, and this gives ≤. �

Clearly d(A ⊕ B) = max(d(A), d(B)): if f is an isomorphism of A into P(κ)and g is an isomorphism of B into P(λ), then the following function clearlyextends to an isomorphism of A ⊕ B into P(κ × λ): for a ∈ A and b ∈ B,h(a) = f(a)×λ and h(b) = κ× g(b). For free products of several algebras there isa much more general topological result. To prove it, we need the following lemma.

Lemma 5.4. Let κ be an infinite cardinal. Then the product space(κ2)

κ has density≤ κ (where κ has the discrete topology).

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Chapter 5. Topological Density 221

Proof. Let D = {f ∈ (κ2)κ: there is a finite subset M of κ such that for all

x, y ∈ κ2, if x � M = y � M , then f(x) = f(y)}. We show that |D| ≤ κ. First,

D =⋃

M∈[κ]<ω

{f ∈ (κ2)κ : for all x, y ∈ κ2( if x � M = y � M, then f(x) = f(y))}.

So, it suffices to take any finite M ⊆ κ and show that Ndef= {f ∈ (κ2)

κ : for allx, y ∈ κ2, if x � M = y � M then f(x) = f(y)} has power at most κ. For any

f ∈ N , let f ′ ∈ (M2)κ be defined as follows: for any x ∈ M2, choose any y ∈ κ2

such that x ⊆ y and let f ′(x) = f(y). Clearly the assignment f !→ f ′ is one-one.So |N | ≤ κ, as desired.

To show that D is dense in(κ2)

κ, let U be a nonempty open set in(κ2)

κ. Wemay assume that U has a very special form, namely that there is a finite subsetF of κ2 and a function g mapping F into κ such that

U = {f ∈ (κ2)κ : g ⊆ f}.

Now let G be a finite subset of κ such that f � G �= h � G for distinct f, h ∈ F .

Define k ∈ (κ2)κ in the following way: for any x ∈ κ2, set k(x) = g(f) if x � G =

f � G for some f ∈ F , otherwise let k(x) be 0. Clearly k ∈ D ∩U , as desired. �

Note that if f is a continuous function from X onto Y , then d(Y ) ≤ d(X).

Theorem 5.5. Let 〈Xi : i ∈ I〉 be a system of topological spaces each having atleast two disjoint non-empty open sets. Then d(

∏i∈I Xi) = max(λ, supi∈Id(Xi)),

where λ is the least cardinal such that |I| ≤ 2λ.

Proof. Clearly d(Xi) ≤ d(∏

i∈I Xi) for each i ∈ I. Suppose that D is dense in∏i∈I Xi but |D| < λ. Then 2|D| < |I|. Let U0

i and U1i disjoint non-empty open

sets in Xi for all i ∈ I. For each i ∈ I let

Vi =

{x ∈

∏j∈I

Xj : xi ∈ U0i

}.

Then our supposition implies that there are distinct i, j ∈ I such that Vi ∩ D =Vj ∩D. Let W = {x : xi ∈ U0

i and xj ∈ U1j }. Choose x ∈ W ∩D. Then x ∈ Vi but

x /∈ Vj , contradiction.

Up to this point we have proved the inequality ≥. Now for each i ∈ I, letDi be dense in Xi with |Di| = d(Xi). Set κ = max(λ, supi∈I |Di|). Then for eachi ∈ I there is a function fi mapping κ onto Di. Since |I| ≤ 2λ, we then get a

continuous function from(κ2)

κ onto∏

i∈I Di. Namely, let g be a one-one function

from I into κ2. For each x ∈ (κ2)κ and each i ∈ I let (h(x))i = fi(xg(i)). Then

h is the desired continuous function. To see that h is continuous, let U be basicopen in

∏i∈I Di. Then there is a finite F ⊆ I such that pri[U ] is open in Di for

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all i ∈ F and pri[U ] = Di for all i ∈ I\F . Let L = {l ∈ g[F ]κ : ∀i ∈ F (fi(lg(i)) ∈pri[U ])}. For each l ∈ L the set Wl

def= {k ∈ (κ2)

κ : l ⊆ k} is open in(κ2)

κ, andh−1[U ] =

⋃l∈L Wl. So h is continuous. Clearly h maps onto

∏i∈I Di.

Now Lemma 5.4 yields the desired result. �

The second part of the following corollary was observed by Sabine Koppelberg.

Corollary 5.6. Let A be a free BA on κ free generators. Then d(A) is the smallestcardinal λ such that κ ≤ 2λ. More generally, if B is an infinite subalgebra of A,then d(B) is the least cardinal μ such that |B| ≤ 2μ.

Proof. For each infinite cardinal ν let log2ν be the least cardinal μ such thatν ≤ 2μ. For A itself the corollary is true directly by Theorem 5.5. Now let B be aninfinite subalgebra of A. Note that B is a subalgebra of a subalgebra of A generatedby |B| free generators, and so d(B) ≤ log2|B|. If |B| = ω, the desired conclusionis obvious. If ω < |B| and |B| is regular, the conclusion follows from Theorem9.16 of the BA handbook. Finally, suppose that |B| is a singular cardinal. Thenfor each regular ν < |B| we have log2ν ≤ d(B) by Theorem 9.16. Since clearlylog2|B| = supν<|B| log2ν, this case now follows too. �

Theorem 5.5 characterizes the topological density of free products of BAs. Wedo not have a characterization of the topological density of amalgamated freeproducts of BAs. Note by the example P(ω)⊕Finco(ω) P(ω) that the topologicaldensity can be larger than that of the constituent algebras; see the discussion ofcellularity of amalgamated free products.

Problem 62. Describe the topological density of amalgamated free products of BAs.

Next we treat the topological density of the union of a well-ordered chain.

Proposition 5.7. Let 〈Bα : α < κ〉 be a strictly increasing sequence of BAs withunion A. Then:

(i) supα<κ d(Bα) ≤ d(A) ≤∑

α<κ d(Bα) ≤ κ·supα<κ d(Bα) ≤ (2supα<κ d(Bα))+.

(ii) κ ≤ |A| ≤ 2d(A).

Proof. If A\{0} =⋃K with K a set of ultrafilters, and if β < κ, then Bβ\{0} =⋃

F∈K(Bα ∩ F ). This shows that d(Bα) ≤ d(A). Now for each α < κ let Xα be aset of ultrafilters on A such that |Xα| = d(Bα) and {F ∩Bα : F ∈ Xα} is dense inUlt(Bα). Now

⋃α<κXα is dense in Ult(A). In fact, given a ∈ A+, choose α < κ

such that a ∈ Bα. Then there is an F ∈ Xα such that F ∩ Bα ∈ SBα(a), hencea ∈ F and F ∈ SA(a). So

d(A) ≤∣∣∣∣∣⋃α<κ

Xα

∣∣∣∣∣ ≤∑α<κ

d(Bα) ≤ κ · supα<κ

d(Bα).

Since|Bβ | ≤ 2d(Bβ) ≤ 2supα<κ d(Bα)

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for each β < κ, we must have κ ≤ (2supα<κ d(Bα))+, since otherwise

(2supα<κ d(Bα))+ ≤ |Bβ | ≤ 2supα<κ d(Bα)

with β = (2supα<κ d(Bα))+, contradiction. Since 〈Bα : α < κ〉 is strictly increasing,clearly κ ≤ |A|. |A| ≤ 2d(A) by Corollary 5.6. �

Proposition 5.7 is clarified somewhat by the following example. Let λ be an un-countable regular cardinal, and let B = Finco(λ). For each α < λ let Aα = 〈{{ξ} :ξ < α}〉B. Then 〈Aα : α < λ〉 is a strictly increasing system of subalgebras withunion B. We have d(Aα) < λ for each α < λ, while d(B) = λ.

Corollary 5.8. Let 〈Bα : α < κ〉 be a strictly increasing sequence of BAs withunion A. Suppose that supα<κ d(Bα) < d(A). Then d(A) ≤ κ. �A connection between cellularity and topological density is given in Shelah [80];we can use it to get another result about unions. Shelah’s Theorem depends ona combinatorial result which we prove first. For this combinatorial result, seeComfort, Negrepontis [74], Theorem 3.16.

Theorem 5.9. Suppose that λ = λ<κ. Then there is a system 〈fα : α < 2λ〉 ofmembers of λλ such that

∀M ∈[2λ]<κ ∀g ∈ Mλ∃β < λ∀α ∈M [fα(β) = g(α)].

Proof. Let

F ={(F,G, s) : F ∈ [λ]<κ, G ∈ [P(F )]<κ, and s ∈ Gλ

}.

Now if F ∈ [λ]<κ, say |F | = μ, then∣∣[P(F )]<κ∣∣ ≤ (2μ)<κ| ≤ (λμ)<κ ≤ λ<κ = λ,

and if G ∈ [P(F )]<κ then |Gλ| ≤ λ<κ = λ. It follows that |F | = λ. Let h bea bijection from λ onto F , and let k be a bijection from 2λ onto P(λ). Now foreach α < 2λ we define fα ∈ λλ by setting, for each β < λ, with h(β) = (F,G, s),

fα(β) =

{s(k(α) ∩ F ) if k(α) ∩ F ∈ G,

0 otherwise.

Now to prove that 〈fα : α < 2λ〉 is as desired, suppose that M ∈ [2λ]<κ andg ∈ Mλ. For distinct members α, β of M choose γ(α, β) ∈ k(α)�k(β). Then let

F = {γ(α, β) : α, β ∈M,α �= β} and G = {k(α) ∩ F : α ∈M}.

Moreover, define s : G → λ by setting s(k(α) ∩ F ) = g(α) for any α ∈ M .This is possible since k(α) ∩ F ) �= k(β) ∩ F ) for distinct α, β ∈ M . Finally, letβ = h−1(F,G, s). Then for any α ∈M we have

fα(β) = s(k(α) ∩ F ) = g(α). �

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Corollary 5.10. Suppose that λ<κ = λ. Then there is a system 〈fα : α < λ〉 of func-tions mapping λ+ into [λ]<κ such that ∀M ∈ [λ+]<κ∀g ∈ M ([λ]<κ)∃β < λ[g ⊆ fβ].

Proof. Let k : λ → [λ]<κ be a bijection, and let 〈f ′α : α < 2λ〉 be obtained from

Theorem 5.9. For each β < λ define fβ : λ+ → [λ]<κ by setting fβ(α) = k(f ′α(β))

for any α < λ+. Now suppose that M ∈ [λ+]<κ and g : M → [λ]<κ. Then k−1◦g ∈Mλ, so there is a β < λ such that for all α ∈ M we have f ′

α(β) = (k−1 ◦ g)(α).Hence for all α ∈M , g(α) = k(f ′

α(β)) = fβ(α. Thus g ⊆ fβ. �

Theorem 5.11. Suppose that κ is uncountable and regular, and λ<κ = λ. Let A bea BA of size λ+ satisfying the κ-cc. Then d(A) ≤ λ.

In more detail, A\{0} is the union of a family F of proper filters with |F | ≤λ, and every proper filter generated by < κ elements is contained in some memberof F .

Proof. Note that the final condition of the theorem implies that A\{0} is the unionof F , so we really just need to establish that condition.

By Corollary 10.5 of the BA handbook we have |A| = λ+, so it clearly sufficesto assume that A is complete. For any subset X of A and any a ∈ A let

high(a,X) = inf{b ∈ 〈X〉cmA : a ≤ b}.

Write A = {bi : i < λ+} without repetitions. We now define 〈Bi : i < λ+〉. LetB0 = 2. For i limit, let Bi =

⋃j<i Bj . Now suppose that Bi has been defined,

with |Bi| ≤ λ. Choose Mi ⊆ A\Bi with bi ∈ Bi ∪Mi and |Mi| = λ. Let Bi+1 =〈Bi ∪Mi〉cm. Then |Bi| = λ by the Handbook Corollary 10.5. Write Bi+1\Bi ={aiα : α < λ} without repetitions. Clearly

(1)⋃

i<λ+ Bi = A, and if δ < λ+ and cf(δ) ≥ κ then Bδ is complete.

Let 〈fα : α < λ〉 be as in Corollary 5.10.

Now we define ultrafilters Diζ on Bi for i ≤ λ+ and ζ < λ, by recursion on

i. Let D0ζ = {1} for all ζ < λ. For i limit and any ζ < λ let Di

ζ =⋃

j<i Djζ . Now

suppose that Diζ has been defined. If Di

ζ ∪ {aiα : α ∈ fζ(i)} has fip, extend it

to an ultrafilter Di+1ζ on Bi+1. Otherwise let Di+1

ζ be any extension of Diζ to an

ultrafilter on Bi+1. This completes the definition of the Diζ .

For each ζ < λ let Eζ = Dλ+

ζ . For each i < λ+ define

Si =

⎧⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎩

{i} if i = 0, i is a successor, or i is a limit ordinal

with cf(i) ≥ κ,

{α(i, j) : j < cf(i)} if i is limit, cf(i) < κ, and 〈α(i, j) : j < cf(i)〉is a strictly increasing sequence of successor

ordinals with supremum i

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Also, for each a ∈ A let

i(a) =

{the unique i such that a ∈ Bi+1\Bi if a �= 0, 1,

0 if a = 0 or a = 1.

Now suppose that C ∈ [A]<κ has fip; we want to show that C ⊆ Eζ for someζ < λ. Let C = {cβ : β < γ} for some γ < κ. Now we define 〈Fn : n ∈ ω〉 byrecursion. Let F0 = C. Then let

Fn+1 = 〈Fn ∪ {high(a,Bi) : a ∈ Fn, i ∈ Si(a)}〉A.

Now each set Si has size less than κ, so by induction, also each Fn has sizeless than κ. Let G =

⋃n∈ω Fn. So also |G| < κ. Let H be an ultrafilter on G

such that C ⊆ H . Let M = {i(a) : a ∈ G} and define g with domain M bysetting g(i) = {α < λ : aiα ∈ H}. Since |H | < κ, we have |g(i)| < κ. Thusg : M → [λ]<κ. Moreover, |M | < κ. Hence by the choice of 〈fα : α < λ〉 above(using Corollary 5.10), choose ξ < λ such that g ⊆ fξ. Now the rest of theproof consists in proving by induction on i ≤ λ+ that H ∩ Bi ⊆ Di

ξ. (Hence

C ⊆ H = H ∩ A =⋃

i<λ+(H ∩Bi) ⊆ Dλ+

ξ = Eξ.)

Trivially H ∩B0 ⊆ D0ξ . The inductive step to a limit ordinal i is obvious. So

now assume thatH∩Bk ⊆ Dkξ for all k ≤ i; we want to prove thatH∩Bi+1 ⊆ Di+1

ξ .

If Bi+1 ∩G = Bi, then H ∩Bi+1 = H ∩Bi ⊆ Diξ ⊆ Di+1

ξ , as desired. So, supposethat G ∩ Bi+1\Bi �= ∅. It follows that i ∈ M , and hence g(i) = fξ(i). Also, ifα ∈ fξ(i), then aiα ∈ H .

(2) Diξ ∪ {aiα : α ∈ fξ(i)} has fip.

To prove (2), suppose that w is a finite subset of fξ(i) and let b =∏

α∈w aiα; wewant to show that b · c �= 0 for every member c of Di

ξ. Clearly b ∈ H . If b ∈ Bi,

then b ∈ Bi ∩H ⊆ Diξ, so obviously b · c �= 0 for every member c of Di

ξ. Now each

aiα is in Bi+1, so if b /∈ Bi then b = aiβ for some β. Since b ∈ H it follows that

β ∈ fξ(i). Suppose that c ∈ Diξ and b · c = 0.

Suppose that i is a successor, or i is limit and cf(i) ≥ κ. Then i ∈ Si(b),and hence high(b, Bi) ∈ G. Moreover, Bi is complete. Now from b · c = 0 we getb ≤ −c, hence high(b, Bi) ≤ −c, so high(b, Bi) · c = 0. But b ≤ high(b, Bi) ∈ Bi,so high(b, Bi) ∈ H ∩Bi ⊆ Di

ξ, contradiction.

Now suppose that i is limit and cf(i) < κ. Then there is a j < cf(i) such thatc ∈ Bα(i,j). Since α(i, j) is a successor ordinal, it follows that Bα(i,j) is complete.Hence high(b, Bα(i,j)) ∈ Bα(i,j). As above we get high(b, Bα(i,j)) · c = 0. Hence

high(b, Bα(i,j)) ∈ H ∩Bα(i,j) ⊆ Dα(i,j)ξ , contradiction.

This finishes the proof of (2).

Now suppose that u ∈ H ∩ Bi+1. If u ∈ Bi, then u ∈ Diξ ⊆ Di+1

ξ , as desired.

Suppose that u /∈ Bi. So u ∈ Bi+1\Bi, and hence u = aiα for some α < λ. Sinceu ∈ H , it follows that α ∈ g(i) = fξ(i). Hence u = aiα ∈ Di+1

ξ , as desired. �

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Corollary 5.12. Let 〈Bα : α < κ〉 be a strictly increasing sequence of BAs whoseunion is A.

Suppose that d(Bα) ≤ μ for all α < κ, νμ = ν, κ = ν+, and 2μ = μ+. ThenA satisfies the μ+-cc, |A| = ν+, and d(A) ≤ ν.

In particular, if κ = ω2, d(Bα) = ω for all α < ω2 and CH holds, then A hasccc, |A| = ω2, and d(A) = ω1.

Proof. Let α < κ. Since 2μ = μ+ and d(Bα) ≤ μ we have |Bα| ≤ μ+. Also, sinceνμ = ν we have μ+ ≤ ν. So μ+ < ν+ = κ. Thus if X ⊆ A, |X | = μ+, and X ispairwise disjoint, then X ⊆ Bα for some α < κ, hence μ+ ≤ c(Bα) ≤ d(Bα) ≤ μ,contradiction. So A satisfies the μ+-cc. Also, κ = ν+ ≤ |A| ≤ μ+ · ν+ = ν+, so

|A| = ν+. Now ν<μ+

= νμ = ν, so by Theorem 5.11, d(A) ≤ ν.

The second part of the corollary follows from the first part, with κ = ω2,μ = ω, ν = ω1. �

We turn to ultraproducts.

Proposition 5.13. If 〈Ai : i ∈ I〉 is a system of infinite BAs, and F an ultrafilteron I. Then d(

∏i∈I Ai/F ) ≤ |

∏i∈I d(Ai)/F |.

Proof. Let fi be an isomorphism of Ai into P(d(Ai)) for each i ∈ I. It is easyto see that there is a function g mapping

∏i∈I Ai/F into P(

∏i∈I d(Ai)/F ) such

that for any w ∈∏

i∈I Ai,

g(w/F ) =

{y/F : y ∈

∏i∈I

d(Ai) and {i ∈ I : yi ∈ fi(wi)} ∈ F

}.

To see that g is one-one, suppose that w,w′ ∈∏

i∈I Ai with w/F �= w′/F . So

Mdef= {i ∈ I : wi �= w′

i} ∈ F . For each i ∈ M choose yi ∈ fi(wi)�fi(w′i). Define

yi = 0 for all i ∈ I\M . Note that M ⊆ {i ∈ I : yi ∈ fi(wi)}∪ {i ∈ I : yi ∈ fi(w′i)},

so {i ∈ I : yi ∈ fi(wi)} ∈ F or {i ∈ I : yi ∈ fi(w′i)} ∈ F .

Case 1. Ndef= {i ∈ I : yi ∈ fi(wi)} ∈ F . Thus y/F ∈ g(w/F ). Suppose that

y/F ∈ g(w′/F ). Then there exists y′ ∈∏

i∈I d(Ai) such that y/F = y′/F and

Pdef= {i ∈ I : y′i ∈ fi(w

′i)} ∈ F . Also, Q

def= {i ∈ I : yi = y′i} ∈ F . Thus

N ∩P ∩Q ∈ F , and so N ∩P ∩Q �= ∅. Take any i ∈ N ∩P ∩Q. Then yi ∈ fi(wi),y′i ∈ fi(w

′i), and yi = y′i, contradiction. Thus y/F /∈ g(z′), so g(z) �= g(z′).

Case 2. {i ∈ I : yi ∈ fi(x′i)} ∈ F . This is treated similarly. �

Roslanowski, Shelah [98] constructed a system 〈Bi : i ∈ I〉 of BAs in ZFC suchthat d

(∏i∈I Bi/F

)<∣∣∏

i∈I d(Bi)/F∣∣; this is a positive solution of Problem 9 of

Monk [90].

We give this construction here. It depends on a finite version of topologicaldensity which is interesting in itself. For any positive integer n, a subset X of a BAA satisfies the n-intersection property iff the product of any n or fewer members

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of X is always nonzero. We define dn(A) to be the least cardinal κ such thatA\{0} is the union of κ sets each satisfying the n-intersection property. Note thatdn(A) ≤ d(A).

The following proposition generalizes Proposition 5.13.

Proposition 5.14. Suppose that κ is an infinite cardinal, 〈Bi : i < κ〉 is a systemof infinite BAs, D is an ultrafilter on κ, f : κ → ω, and ∀i ∈ ω[{j < κ : f(j) ≥i} ∈ D]. Then

d

(∏i<κ

Bi/D

)≤∣∣∣∣∣∏i<κ

df(i)(B)/D

∣∣∣∣∣ .Proof. For each i ∈ ω. let 〈Mij : j < df(i)(Bi)〉 be a system of sets satisfying

the f(i)-intersection property such that B+i =

⋃j<df(i)

Mij . Now for each z ∈∏i<κ df(i)(B)/D let

Nz =

{y ∈

∏i<κ

Bi/D : ∃x ∈ y∃u ∈ z∀i < κ[xi ∈Miu(i)]

}.

(1) ∀z ∈∏

i<κ df(i)(B)/D[Nz satisfies the fip].

To prove (1), let z ∈∏

i<κ df(i)(B)/D and F ∈ [Nz]<ω. Let |F | = m, and define

M = {i < κ : f(i) ≥ m}; so M ∈ D. For each y ∈ F choose xy ∈ y anduy ∈ z such that ∀i < κ[xy

i ∈ Miuy(i)]. Now if y, y′ ∈ F , then uy, uy′ ∈ z, and so

{i < κ : uy(i) = uy′(i)} ∈ D. Thus the set

Pdef= M ∩

⋂y,y′∈F

{i < κ : uy(i) = uy′(i)} ∈ D

is inD. Take any i ∈ P . Then for any y ∈ F , xyi ∈Miuy(i), and hence

∏y∈F xy

i �= 0.It follows that

∏F �= 0. So (1) holds.

(2)⋃{

Nz : z ∈∏

i<κ df(i)(B)/D}= A+, where A =

∏i<κ Bi/D.

To prove (2), take any y ∈ A+. Choose x ∈ y. Then Mdef= {i < κ : xi �= 0} ∈ D.

For any i ∈ M choose u(i) < df(i) so that xi ∈ Miu(i), and let u(i) = 0 fori ∈ κ\M . Let z = u/D. Then y ∈ Nz, as desired in (2). �

Proposition 5.15. Suppose that κ is an infinite cardinal, 〈λi : i < κ〉 is a systemof cardinals, 2κ <

∏i<κ λi, and 2 < n < ω.

Then there is a BA B such that

dn−1(B) ≤∑α<κ

∏i<α

λi and dn(B) = |B| =∏i<κ

λi.

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Proof. Let A be freely generated by 〈xη : η ∈∏

i<κ λi〉, and let I be the ideal inA generated by{

xη0 · . . . · xηn−1 : ∃α < κ

[∀l < n

[ηl ∈

∏i<κ

λi

], ∀l,m < n[ηl � α = ηm � α],

〈ηl(α) : l < n〉 is one-one]}

.

Let B = A/I. The basic property of B is as follows:

(1) For any l < m < ω and any distinct η0, . . . , ηm ∈∏

i<κ λi the followingconditions are equivalent:

(a) [xη0 ] · . . . · [xηl] · −[xηl+1

] · . . . · −[xηm ] = 0.

(b) There exist α < κ and ν0, . . . , νn−1 ∈∏

i<κ λi such that ∀s, t < n[νs � α =νt � α], 〈νs(α) : s < n〉 is one-one, and {νs : s < n} ⊆ {η0, . . . , ηl}.

In fact, obviously (b) implies (a), and an easy argument using freeness gives theother implication.

Now define

X ={[xη0 ] · . . . · [xηl

] · −[xηl+1] · . . . · −[xηm ] : 0 ≤ l < m < ω,

∀k ≤ m[ηk ∈

∏i<κ λi

], and 〈η0, . . . , ηm〉 is one-one

}\{0}.

Clearly X is dense in B. Now if 0 ≤ l < m < ω, α < κ, ∀k ≤ m[ϕk ∈∏

i<α λi],and 〈ϕk : k ≤ m〉 is one-one, let

Dlmα〈ϕk:k≤m〉 =

{[xη0 ] · . . . · [xηl

] · −[xηl+1] · . . . · −[xηm ] :

∀k ≤ m[ϕk ⊆ ηk ∈

∏i<κ λi

], and 〈η0, . . . , ηm〉 is one-one

}\{0}.

Note that there are at most∑

α<κ

∏i<α λi of the sets Dlmα

〈ϕk:k≤m〉. Hence for thefirst conclusion of the proposition it suffices to show that the union of all such setsis X , and each such set satisfies the (n− 1)-intersection property.

To show that the union of all such sets is X , let 0 ≤ l < m < ω and let〈ηk : k ≤ m〉 be a sequence of distinct members of

∏i<κ λi such that

[xη0 ] · . . . · [xηl] · −[xηl+1

] · . . . · −[xηm ] �= 0.

Then clearly there is an α < κ such that the functions ηk � α are distinct fork ≤ m. Let ϕk = ηk � α for all k ≤ m. Then

[xη0 ] · . . . · [xηl] · −[xηl+1

] · . . . · −[xηm ] ∈ Dlmα〈ϕk:k≤m〉,

as desired.

Now we take any set Dlmα〈ϕk:k≤m〉 and show that it satisfies the (n − 1)-

intersection property. So, suppose we are given 〈ηtk : k ≤ m, t < n − 1〉 such

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that ∀t < n − 1∀k ≤ m[ϕk ⊆ ηtk ∈∏

i<κ λi], ∀t < n− 1[〈ηtk : k ≤ m〉 is one-one],and

∀t < n− 1[[xηt

0] · . . . · [xηt

l] · −[xηt

l+1] · . . . · −[xηt

m] �= 0

].

We want to show that∏t<n−1

[xηt0] · . . . · [xηt

l] · −[xηt

l+1] · . . . · −[xηt

m] �= 0.

Suppose that this product is 0. Then by (1) choose β < κ and ν0, . . . , νn−1 ∈∏i<κ λi such that ∀s, t < n[νs � β = νt � β], 〈νs(β) : s < n〉 is one-one, and

{νs : s < n} ⊆ {ηsk : k ≤ l, s < n − 1}. Then there is an s < n − 1 and distinctu, v < n and distinct a, b ≤ l such that νu = ηsa and νv = ηsb .

Case 1. α ≤ β. Now ϕa ⊆ ηsa = νu and ϕb ⊆ ηsb = νv and νu � α = νv � α, soϕa = ϕb, contradiction.

Case 2. β < α. For each s < n choose t(s) < n − 1 and k(s) ≤ l such that

νs = ηt(s)k(s). Since ϕk(s) ⊆ η

t(s)k(s) and β < α, it follows that νs � (β+1) ⊆ ϕk(s). Now

for s �= u we have νs � (β + 1) �= νu � (β + 1), so it follows that k(s) �= k(u). Thisbeing true for all s < n, it follows from (1) that Dlmα

〈ϕk:k≤m〉 = ∅, contradiction.It remains only to show that dn(B) =

∏i<κ λi. Suppose to the contrary

that B+ =⋃

j<θ Cj , where θ <∏

i<κ λi and each Ci satisfies the n-intersection

property. Fix j < θ, and let Dj ={η ∈

∏i<κ λi : [xη] ∈ Cj

}. We claim that |Dj | ≤

2κ. Suppose not. Define f : [Dj ]2 → κ by setting

f({σ, τ}) = minα<κ

[σ(α) �= τ(α)]

for any two distinct σ, τ ∈ Dj . By the Erdos–Rado theorem (2κ)+ → (κ+)2κ we geta subset E of D of size κ+ and an α < κ such that σ � α = τ � α and σ(α) �= τ(α)for all distinct σ, τ ∈ E. But then by (1), any set of n or more elements [xσ ],σ ∈ E, has zero product, contradiction. Thus |Cj | ≤ 2κ, so

∏i<κ λi =

⋃j<θ Dj

has size at most θ · 2κ <∏

i<κ λi, contradiction. �

Corollary 5.16. If λ is a strong limit singular cardinal and 2 < n < ω, then thereis a BA B such that dn(B) = d(B) = |B| = 2λ and dn−1(B) ≤ λ.

Proof. Let 〈μi : i < cf(λ)〉 be a strictly increasing sequence of infinite cardinalswith supremum λ. Then

2λ = 2∑

i<cf(λ) μi =∏

i<cf(λ)

2μi =∏

i<cf(λ)

μi ≤∏

i<cf(λ)

λ = λcf(λ) ≤ λλ = 2λ;

Hence∏

i<cf(λ) μi = 2λ. Hence Proposition 5.15 applies to give the desired result.�

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Theorem 5.17. Let λ be a strong limit singular cardinal, and κ < cf(λ). Supposethat D is an ultrafilter on κ which is not countably complete. Then there is asystem 〈Bi : i < κ〉 of BAs such that

d

(∏i<κ

Bi/D

)≤ λ < 2λ =

∣∣∣∣∣∏i<κ

d(Bi)/D

∣∣∣∣∣ .Proof. Let 〈Mi : i ∈ ω〉 be a sequence of members of D such that

⋂i∈ω Mi /∈ D,

with M0 �= κ. Let Ni =⋂

j≤i Mj for every i ∈ ω. Then 〈Ni : i ∈ ω〉 is a decreasingsequence of sets but it is not eventually constant. So there is a function g : ω → ωsuch that g(0) = 0 and Ng(i) ⊃ Ng(i+1) for all i ∈ ω. Define P0 = (κ\N0)∪

⋂i∈ω Ni

and Pi+1 = Ng(i)\Ng(i+1) for all i ∈ ω. Thus 〈Pi : i ∈ ω〉 is a partition of κ withall members not in D. For each α < κ let f(α) = i + 2, where i is minimum suchthat α ∈ Pi. Then

(1) For all i ∈ ω we have {α ∈ κ : f(α) ≥ i} ∈ D.

Now by Corollary 5.16, for each i < κ let Bi be a BA such that df(i)(Bi) ≤ λ

while |Bi| = d(Bi) = df(i)+1(Bi) = 2λ. Then by Proposition 5.14

d(∏

i<κBi

)≤∣∣∣∏

i<κdf(i)(Bi)/D

∣∣∣ ≤ λκ = λ.

On the other hand, since d(Bi) = 2λ for all i < κ, it is clear that∣∣∣∏i<κ

d(Bi)/D∣∣∣ = 2λ. �

If B is a subdirect product of 〈Ai : i ∈ I〉, then d(B) ≤∑

i∈I d(Ai), by theargument used for products. Unlike full products and weak direct products, how-ever, it is not necessary that |I| ≤ d(B), or that d(Ai) ≤ d(B) for all i ∈ I. For

example, the subalgebra Bdef= {f ∈ ω1A : f(α) = f(β) for all α, β < ω1} of

ω1A, where A is countable, is isomorphic to A and ω1 �≤ d(B). The subalgebra

Bdef= {([a], a) : a ∈ P(ω)} of P(ω)/fin×P(ω) is a subdirect product isomorpic

to P(ω), and d(P(ω)/fin) = 2ω while d(B) = ω.

Moderate products are taken care of by Theorem 5.3.

For the one-point gluing B of the pair (〈Ai : i ∈ I〉, 〈Fi : i ∈ I〉) we haved(B) =

∑i∈I d(Ai). Here we use the notation of Chapter 1, and we assume that

each Ai is infinite. In fact, ≤ holds since B is a subalgebra of∏

i∈I Ai. To showthat |I| ≤ d(B), for each i ∈ I let aii be a nonzero element of Ai not in Fi, andlet aij = 0 for j �= i. This gives a system 〈ai : i ∈ I〉 of nonzero pairwise disjointelements of B, and so |I| ≤ d(B). Finally, we show that Ai is isomorphic to asubalgebra of B for every i ∈ I: for each a ∈ Ai let

(fi(a))(j) =

{a if j = i,

a/Fi otherwise.

Clearly d(Dup(A)) = |Ult(A)|.

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A special case of a result in Mizokami [79] is that d(Exp(A)) = d(A) for anyinfinite BA A. In fact, d(Exp(A)) ≤ d(A) by Proposition 2.5 and the above remarkson products and free products. On the other hand, let X ⊆ Ult(A) be dense, with|X | = d(A). We claim that [X ]<ω is dense in Exp(A). For, take a basis elementV (S(a1), . . . ,S(am)) of Exp(A). For each i = 1, . . . ,m choose Fi ∈ X ∩ S(ai).Then {Fi : 1 ≤ i ≤ m} ∈ V (S(a1), . . . ,S(am)).

We turn to derived functions for topological density. We shall show that dH+ = hd,but to do this we need two results about tightness and spread which are corollariesof Theorems 4.26 and 3.30.

Theorem 5.18 (Shapirovskiı). t(A) ≤ s(A) for any infinite BA A.

Proof. By Theorem 4.25 it suffices to note that if 〈Fξ : ξ < κ〉 is a free sequence,with κ regular, then 〈Fξ : ξ < κ〉 is one-one and {Fξ : ξ < κ} is discrete. Letξ < κ. There exist clopen sets S(a),S(b) such that {Fη : η < ξ} ∩ S(a) = 0, {Fη :ξ ≤ η < κ} ⊆ S(a), {Fη : η < ξ + 1} ⊆ S(b), and {Fη : ξ + 1 ≤ η < κ} ∩ S(b) = 0.Clearly then S(a · b) ∩ {Fη : η < κ} = {Fξ}, as desired. �

Theorem 5.19. s(A) ≤ dH+(A) for any infinite BA A.

Proof. By Theorem 3.30 and Corollary 5.2, s(A) = cH+(A) ≤ dH+(A). �

Theorem 5.20. dH+(A) = dh+(A) = hd(A) for any infinite BA A.

Proof. dh+(A) = hd(A) by definition, and dH+(A) ≤ dh+(A) since homomorphicimages correspond to closed sets in Ult(A). So it remains to show that hd(A) ≤dH+(A). Let κ = dH+(A), and suppose that κ < hd(A). Choose Y ⊆ Ult(A) suchthat κ < d(Y ). Let Z be a dense subset of Y of size ≤ κ. For each z ∈ Z we havez ∈ Y , and so z ∈Wz for some Wz ∈ [Y ]≤κ by Theorems 5.18 and 5.19. We claimnow that

⋃z∈Z Wz is dense in Y ; since

∣∣⋃z∈Z Wz

∣∣ ≤ κ, this will be a contradiction.

Let U be an open set in Ult(A) such that U ∩ Y �= 0. Then U ∩ Y �= 0, so choosez ∈ U ∩ Z. Since z ∈Wz , we get U ∩Wz �= 0, as desired. �

Notice that attainment in the dH+ sense obviously implies attainment in the hdsense.

We have dH−(A) = dh−(A) = ω for infinite A, since by Sikorski’s extension theo-rem there is a homomorphism of A onto an infinite subalgebra of P(ω). ClearlydS+(A) = d(A), dS−(A) = ω, and ddS+(A) = d(A) for any infinite BA A. Further-more, ddS−(A) = d(A): if B is a dense subalgebra of A and f is an isomorphismof B into P(κ), then f can be extended to an isomorphism of A into P(κ), asdesired.

Concerning the spectrum function dHs we mention the following problem,Problem 18 in Monk [96].

Problem 63. Is it true that [ω, hd(A)) ⊆ dHs(A) for every infinite BA A?

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Problem 64. Completely describe dHs.

This is problem 19 in Monk [96],

Concerning dSs we have the following theorem and example, due to S. Koppelberg,solving Problem 11 in Monk [90].

Theorem 5.21. (GCH) dSs(A) = [ω, d(A)] for every infinite BA A.

Proof. Suppose that ω ≤ κ < d(A); we want to find a subalgebra B of A such thatd(B) = κ. If κ is a limit cardinal, then any subalgebra of A of size κ will do, byGCH. So we may assume that κ is a successor cardinal, and hence is regular. If Ahas a disjoint subset of size κ, then the subalgebra generated by such a subset isisomorphic to Finco(κ), which has topological density κ. So we may assume thatA satisfies the κ-cc. Now μ<κ < κ+ for every μ < κ+ by GCH. Hence by Theorem10.1 of the BA Handbook, Part I, A has a free subalgebra B of size κ+. By GCH,d(B) = κ, as desired. �

The equality in Theorem 5.21 cannot be proved in ZFC. Namely, if for example2ω = 2ω1 = ω2 and 2ω2 = ω4, then for A the free BA on ω4 free generators wehave dSsA = {ω, ω2} by Corollary 5.6.

We now consider dSr. First we have the following vague problem.

Problem 65. Characterize in cardinal number terms the sets dSr.

Here are some general remarks about dSr:

(1) (ω, ω) is in every dSr.

(2) [ω, |A|] ⊆ rng(dSr(A))

Proposition 5.22.

(i) If (κ, κ) /∈ dSr(A), then c′(A) ≤ κ.

(ii) If in addition (2κ)+ ≤ |A|, then for every infinite λ ≤ (2κ)+ we have (μ, λ) ∈dSr(A), where μ is the least cardinal such that λ ≤ 2μ.

Proof. (i): If c′(A) > κ, then A has a disjoint subset of size κ, and hence asubalgebra isomorphic to Finco(κ). This subalgebra has topological density κ.

(ii): If in addition (2κ)+ ≤ |A|, then by Corollary 10.9 of the Handbook, Ahas an independent subset of size (2κ)+. Then the conclusion follows by Corollary5.6. �

There are many open questions concerning dSr(A) for |A| ≤ ω2, so we do notsurvey them. For example, the following question is open.

Problem 66. Is there a BA A such that dSr(A) = {(ω, ω), (ω1, ω1), (ω1, ω2)}?

For dHr we have as usual the following vague problem.

Problem 67. Characterize in cardinal number terms the sets dHr.

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Here are some general remarks concerning dHr.

(1) For every infinite BA A there is a κ ∈ [ω, |A|] such that (ω, κ) ∈ dHr(A).

In fact, A has a subalgebra isomorphic to Finco(ω), and so gives a homomorphicimage B of A such that d(B) = ω. By this argument we have the following moregeneral result:

(2) [ω, c′(A)) ⊆ dmn(dHr(A)) for every infinite BA A.

(3) Assuming CH, if A is ccc and |A| = ω2, then d(A) = ω1.

This holds by Theorem 5.11.

(4) Since subalgebras of interval algebras are retractive, if A is a subalgebra of aninterval algebra and B is a homomorphic image of A, then d(B) ≤ d(A).

(5) Every infinite homomorphic image of Finco(κ) (κ an infinite cardinal) is iso-morphic to Finco(λ) for some infinite cardinal λ.

Again there are many problems concerning dHr(A) for |A| ≤ ω2, so we do notsurvey them. For example,

Problem 68. Is it consistent to have a BA A such that dHr(A) = {(ω, ω), (ω1, ω2)}?

The difference between c(A) and d(A) can be arbitrarily large, for example in freeBAs. In fact, Rabus and Shelah [99] have shown that for each uncountable cardinalκ there is a ccc BA A such that d(A) = κ. Hence for ω ≤ λ ≤ κ there is a BA Bsuch that c(B) = λ and d(B) = κ. Namely, the case κ = ω is easy, and for κ > ωone can take B = Finco(λ) ×A, with A as above.

Turning to topological density for special classes of BAs, note first that if A isatomic, then d(A) is the number of atoms of A. Hence if A is the finite-cofinitealgebra on κ, then dSr(A) = {(λ, λ) : ω ≤ λ ≤ κ} = dHr(A).

For any infinite BA A we have d(A) = d(A). In fact, let κ = d(A). ByTheorem 5.1, let f be an isomorphism from A into P(κ). By Sikorski’s extensiontheorem, there is an extension g of f to a homomorphism from A into P(κ). If

x ∈ A+, choose a ∈ A+ such that a ≤ x. Then 0 �= f(a) = g(a) ≤ g(x). So g is

one-one, as desired.

For interval algebras, we have one interesting inequality not true for BAs ingeneral. It is actually true for linearly ordered spaces in general, and we give thatgeneral form, due to Kurepa [35]. This result has evidently been rediscovered bymany people independently; see, e.g., Juhasz [75].

Theorem 5.23. If L is an infinite linearly ordered space, then d(L) ≤ (c(L))+.

Proof. Assume the contrary. Set κ = (c(L))+. Let ≺ be a well-ordering of L. Nowwe set

N = {p ∈ L : p is the ≺ -least element of some neighborhood of p}.

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Clearly N is dense in L. Hence |N | > κ. Now for each p ∈ N let Ip be the unionof all open intervals having p as their ≺-first element. Then, we claim,

(1) If p, p′ ∈ N and p ≺ p′, then Ip ∩ Ip′ = 0 or Ip′ ⊂ Ip.

In fact, suppose p ≺ p′ and Ip ∩ Ip′ �= 0. This means that there exist an openinterval U with ≺-first element p and an open interval U ′ with ≺-first element p′

such that U ∩ U ′ �= 0; hence U ∪ U ′ is an open interval with both p and p′ asmembers, and with p as ≺-first member. So, if V is any open interval with ≺-firstelement p′, then V ∪ U ∪ U ′ is an open interval with ≺-first element p, and henceV ⊆ Ip. This shows that Ip′ ⊆ Ip. Since p ∈ Ip\Ip′ , (1) then follows.

Next, set

N0 = {p ∈ N : Ip is not contained in any other Ip′}.

Now Ip ∩ Ip′ = 0 for all distinct p, p′ ∈ N0, so |N0| < κ. We continue inductivelyfor all ξ < κ:

Hξ = N\⋃η<ξ

Nη;

Nξ = {p ∈ Hξ : Ip is not contained in any other Ip′ for p′ ∈ Hξ}.

Note that |Nξ| < κ, and hence always Hξ �= 0. Hence |⋃

ξ<κ Nξ| ≤ κ, so there isa p ∈ N\

⋃ξ<κ Nξ. Thus p ∈ Hξ for all ξ < κ. But then for each ξ < κ there is a

p(ξ) ∈ Nξ such that Ip ⊂ Ip(ξ). In fact, there is a q ∈ Hξ such that Ip ⊂ Iq. Takingthe smallest such q under ≺, we get the desired p(ξ). Hence for all ξ, η < κ we haveIp(ξ) ⊂ Ip(η) or Ip(η) ⊂ Ip(ξ). By the partition relation κ→ (κ, ω)2 we may assumethat p(ξ) ≺ p(η) whenever ξ < η < κ, and hence the sequence 〈Ip(ξ) : ξ < κ〉 isstrictly decreasing. For each ξ < κ choose xξ ∈ Ip(ξ)\Ip(ξ+1). Let

K l = {xξ : xξ < p(ξ + 1)}, Kr = {xξ : xξ > p(ξ + 1)}.

Now

(2) if ξ < η and xη, xξ ∈ K l, then xξ < xη.

In fact, suppose that xη ≤ xξ. Now xξ < p(ξ + 1) ∈ Ip(ξ+1), xξ /∈ Ip(ξ+1), andIp(ξ+1) is convex. Hence xξ is less than all members of Ip(ξ+1). Since ξ < η, we haveξ + 1 ≤ η, hence p(ξ + 1) ( p(η), and so Ip(η) ⊆ Ip(ξ+1). It follows that xξ < p(η).Thus xη ≤ xξ < p(η). Since xη and p(η) are in Ip(η) and Ip(η) is convex, it followsthat xξ ∈ Ip(η). Now ξ + 1 ≤ η, so Ip(η) ⊆ Ip(ξ+1). So xξ ∈ Ip(ξ+1), contradiction.Thus (2) holds

Similarly, if ξ < η and xη, xξ ∈ Kr, then xξ > xη. Now one of K1,K2 hasκ elements, so we obtain a strictly increasing or strictly decreasing sequence oflength κ, and hence κ disjoint open intervals, contradiction. �

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Some care must be taken in translating from topology to Boolean algebras forinterval algebras. For example, the linear order 2×R with the anti-lexicographicorder has density ω as a linearly ordered space, but Intalg(2×R) has a family ofsize 2ω of pairwise disjoint elements, and so has topological density 2ω. However,the following fact should be noted.

Proposition 5.24. If L is a dense linear order, then d(L) = d(Intalg(L).

Proof. First suppose that X ⊆ L is dense of size d(L). Let Y = {[a, b) : a, b ∈ Xand a < b. We claim that Y is dense in Intalg(L). For, suppose that [c, d) ∈Intalg(L). Choose e ∈ X ∩ (c, d) and f ∈ X ∩ (e, d). Then [e, f) ∈ Y and [e, f) ⊆[c, d). This proves the claim. So d(Intalg(L)) ≤ d(L).

Second suppose that Y is dense in Intalg(L). We may assume that the mem-bers of Y have the form [a, b). Let X = {b : [a, b) ∈ Y for some a}. We claim thatX is dense in L. Suppose that c < d. Choose [a, b) ∈ Y such that [a, b) ⊆ [c, d).Then b ∈ (c, d) ∩X , as desired. �

The interval algebra of a dense Suslin line gives an example of an interval algebraA in which c(A) < d(A); on the other hand, Martin’s axiom plus ¬CH impliesthat for an interval algebra, c(A) = ω ⇒ d(A) = ω (see Kunen [80], TheoremII4.2).

Since interval algebras are retractive, it follows that if B is a homomorphicimage of an interval algebra A, then d(B) ≤ d(A). Hence d(A) = hd(A) for everyinterval algebra A.

For a minimally generated BA A we also have d(A) ≤ (c(A))+. In fact, byCorollary 2.60 A is co-complete with an interval algebra B. Hence

d(A) = d(A) = d(B) = d(B) ≤ (c(B))+ = (c(A))+.

An example of a complete BA A for which c(A) < d(A) can be obtained by takingA to be the completion of a large free BA.

Theorem 5.25. If T is a tree, then d(Treealg(T )) = |T |.

Proof. For brevity let κ = |T | and A = Treealg(T ). Suppose that d(A) < κ. Eachlevel of T gives rise to a disjoint set in A, and so has size at most d(A), sincec(A) ≤ d(A). Also, each chain in T has size at most d(A), since a chain gives riseto a disjoint subset of A of its size. Hence κ is regular, and T has κ levels. Let Fbe a collection of ultrafilters on A such that |F | = d(A) and A\{0} =

⋃F . For

each F ∈ F let s(F ) = {t ∈ T : T ↑ t ∈ F}. Then s(F ) is an initial chain of Tand so has length ≤ d(A). Let α be a level greater than the lengths of all s(F ) forF ∈ F . Take any t ∈ T of height α. Then T ↑ t /∈

⋃F , contradiction. �

We do not have a clear description of d(A) for A a pseudo-tree algebra:

Problem 69. Describe d(A) for A a pseudo-tree algebra.

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6 π-Weight

If A is a subalgebra of B, then π(A) can vary either way from π(B); for clearlyone can have π(A) < π(B), while if we take B = P(ω) and A the subalgebra ofB generated by an independent subset of size 2ω, then we have π(B) = ω andπ(A) = 2ω. Similarly, if A is a homomorphic image of B: it is easy to get suchA and B with π(A) < π(B), and if we take B = P(ω) and A = B/fin, thenπ(B) = ω while π(A) = 2ω since A has a disjoint subset of size 2ω.

Now we discuss special subalgebras.

Proposition 6.1. If A ≤π B, then π(A) = π(B).

Proof. Assume that A ≤π B. If X is dense in A, then it is also dense in B; soπ(B) ≤ π(A). Now suppose that Y is dense in B. For each y ∈ Y + choose ay ∈ A+

such that ay ≤ y. Then {ay : y ∈ Y } is dense in A; so π(A) ≤ π(B). �Proposition 6.2. If A ≤m B, then π(A) = π(B).

Proof. Write B = A(x). If A � x and A � −x are both nonprincipal, then A isdense in B by Proposition 2.44, and so π(A) = π(B) by Proposition 6.1. If oneof these ideals is principal, then by Proposition 2.45 we can write B = A(y) forsome atom y of B. Then A is isomorphic to B � −y, and it is easy to see thatπ(A) = π(B). �Proposition 6.3. If A ≤reg B, A ≤free B, A ≤proj B, or A ≤rc B, then π(A) ≤π(B).

Proof. By general results it suffices to take the case A ≤reg B. Let X be dense inB, with |X | = π(B). For each nonzero x ∈ X there is a lower bound 0 �= ax ∈ A

of the set {a ∈ A : x ≤ a}. In fact, otherwise we have∏A{a ∈ A : x ≤ a} = 0,

hence∏B{a ∈ A : x ≤ a} = 0; but 0 �= x ≤

∏B{a ∈ A : x ≤ a}, contradiction.Clearly {ax : x ∈ X} is dense in A. �

Clearly one can have π(A) much smaller than π(B) if A ≤free B; similarly forA ≤reg B, A ≤proj B, A ≤rc B, A ≤u B, or A ≤σ B.

An example with A ≤s B and π(A) < π(B) is as follows. Let A = P(ω),I = [ω]<ω, J = {∅}, and let B = A(x) be the simple extension of A associated withI, J . (See Proposition 2.28.) Thus A � x = I and A � −x = J . Clearly π(A) = ω.

, ,OI 10.1007/978-3- - - _ ,


014


Basel 27

237

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238 Chapter 6. π-Weight

Let 〈yα : α < 2ω〉 be a system of infinite pairwise almost disjoint subsets of ω.Then 〈yα · −x : α < 2ω〉 is a system of 2ω nonzero pairwise disjoint elements of B.

Since every superatomic BA is minimally generated, one can have A ≤mg Bwith π(A) < π(B); it suffices to take B superatomic with uncountably many atomsand A a countable subalgebra which is the ωth algebra in some representing chainfor B over 2.

On the other hand, one can have A ≤mg B with π(A) > π(B). Let X be anuncountable independent set in P(ω), and let A = 〈X〉. Thus π(A) > ω. We definea sequence 〈Cα : α < ω · ω〉 of subalgebras of P(ω). Let C0 = A. Suppose thatCα has been defined, where α = ω ·m+n with m,n ∈ ω. Let Cα(x) be the simpleextension of Cα such that Cα � x = {c ∈ Cα : n /∈ c} and Cα � −x = {0}. Thisextension exists since {c ∈ Cα : n /∈ c} is a maximal ideal in Cα, using Proposition2.28; and Cα ≤m Cα(x) by Proposition 2.32. Now by Sikorski’s extension criterion,there is a homomorphism f of Cα(x) into P(ω) such that f(c) = c for all c ∈ Cα,and f(x) =

⋃{c ∈ Cα : n /∈ c}. Let Cα+1 = f [Cα(x)]. Then Cα ≤m Cα+1 by

Proposition 2.42. For α a limit ordinal ≤ ω · ω, let Cα =⋃

β<αCβ .

Then A (mg Cω·ω. For m,n ∈ ω let

dmn =⋃{c ∈ Cω·m+n : n /∈ c}.

Thus by construction dmn ∈ Cω·m+n+1 ⊆ Cω·ω. Clearly dmn �= ω, so −dmn �= 0.We claim that {−dmn : m,n ∈ ω} is dense in Cω·ω. For, suppose that c ∈ C+

ω·ω.Say c ∈ Cω·m+n with m,n ∈ ω. Since c �= 0 we have −c �= ω. Say p ∈ ω\c. Nowc ∈ Cω·(m+1)+p, so −c ⊆ dm+1,p. Hence −dm+1,p ⊆ c, as desired.

These facts concerning special subalgebras leave the following questions open.

Problem 70. Are there BAs A,B such that A ≤σ B and π(A) > π(B)?

Problem 71. Are there BAs A,B such that A ≤s B and π(A) > π(B)?

Problem 72. Are there BAs A,B such that A ≤u B and π(A) > π(B)?

Turning to products, we have π(∏

i∈I Ai) = max(|I|, supi∈Iπ(Ai)) for any system〈Ai : i ∈ I〉 of infinite BAs. For, ≥ is clear; now suppose Di is a dense subset ofAi for each i ∈ I. Let

E =

{f ∈

∏i∈I

(Di ∪ {0}) : f(i) �= 0 for only finitely many i ∈ I

}.

Clearly E is dense in∏

i∈I Ai, and |E| = max(|I|, supi∈Iπ(Ai)), as desired. Theequation π(

∏wi∈I Ai) = max(|I|, supi∈Iπ(Ai)) is proved by the same argument.

Next, we discuss free products.

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Chapter 6. π-Weight 239

An easy argument shows that π(⊕i∈IAi) = max(|I|, supi∈Iπ(Ai)) for anysystem 〈Ai : i ∈ I〉 of Boolean algebras. In fact, if Di is dense in Ai for each i ∈ I,then

Edef= {d0 · . . . · dn−1 : ∃ distinct i0, . . . , in−1 ∈ I such that ∀j < n(dj ∈ Dij )}

is clearly dense in ⊕i∈IAi, and it has the indicated cardinality. On the other hand,supposeX is dense in ⊕i∈IAi. We may assume that each element ofX is a productof members of

⋃i∈I Ai, with distinct factors coming from distinct Ai’s. For each

i ∈ I let Yi = {x ∈ X : x ≤ a for some a ∈ Ai}. For each x ∈ Yi, let x+i be

obtained from x by replacing each factor of x which is not in Ai by 1. Clearly then{x+

i : x ∈ Yi} ⊆ Ai and this set is dense in Ai, so π(Ai) ≤ |{x+i : x ∈ Yi}| ≤ |Yi| ≤

|X |. It is also clear that |I| ≤ |X |; so |X | ≥ max(|I|, supi∈Iπ(Ai)), as desired.

We do not have a characterization of the density of amalgamated free prod-ucts of BAs; see the corresponding discussions for cellularity and topological den-sity.

Problem 73. Characterize the density of amalgamated free products of BAs.

We turn to the discussion of unions. The following theorem takes care of somepossibilities.

Theorem 6.4. Suppose that 〈Aα : α < κ〉 is a strictly increasing sequence of BAs,with union B, where κ is regular. Let λ = supα<κπ(Aα). Then

(i) π(B) ≤∑

α<κ π(Aα) ≤ max(κ, λ).

Now assume that, in addition, Aα =⋃

β<αAβ for all limit α < κ. Then

(ii) κ ≤ 2λ,

(iii) π(B) ≤ λ+.

Proof. For (i), if Xα is dense in Aα for each α < κ, then⋃

α<κ Xα is dense in B.Now we make the additional assumption indicated, and prove (ii). Assume thatκ ≥ (2λ)+. Let μ = (2λ)+. Since clearly |Aα| ≤ 2λ for all α < κ, we have κ = μ.Let S = {α < μ : cf(α) = λ+}. Thus S is stationary in μ. For each α ∈ S, Aα

has a dense subset Dα of size ≤ λ. Since cf(α) = λ+, it follows that there is anf(α) < α such that Dα ⊆ Af(α). Now f is regressive on S, so f is constant ona stationary subset S′ of S. Let β be the constant value of f on S′. Then Dβ isdense in B. But |B| ≥ (2λ)+, contradiction. So, (ii) holds. For (iii), if π(B) > λ+,then by (i) we have κ > λ+, and we can use an argument similar to that for (ii),taking S = {α < κ : cf(α) = λ+}. �

Note that the upper bound λ+ mentioned in Theorem 6.4 can be attained: use afree algebra.

Turning to ultraproducts, it is clear that π(∏

i∈I Ai/F ) ≤ |∏

i∈I π(Ai)/F |. InKoppelberg, Shelah [95] there is a forcing construction in which < holds; this

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answers Problem 12 of Monk [90]. Several results about ultraproducts and π holdmore generally for the sup-min functions defined in the introduction. These resultsare due to Douglas Peterson [97].

Theorem 6.5. Let k be a sup-min function, 〈Ai : i ∈ I〉 a sequence of infinite BAswith I infinite, and F a regular ultrafilter on I. Suppose that k(Ai) ≥ ω for alli ∈ I, λ = ess.sup F

i∈Ik(Ai), and cf(λ) ≤ |I| < λ. Then k(∏

i∈I Ai/F)≥ λ+.

Proof. For brevity let B =∏

i∈I Ai/F . Let b = {i ∈ I : k(Ai) > ω}. Now b ∈ F ,as otherwise ess.supF

i∈Ik(Ai) = ω.

(1) There is a system 〈κi : i ∈ I〉 such that {i ∈ I : ω < κi < k(Ai)} ∈ F andess.supFi∈Iκi = λ.

To prove (1) we apply Lemma 3.18 with κ replaced by λ to obtain a sequence〈κi : i ∈ I〉 such that ∀i ∈ I[κi < λ] and ess.supFi∈Iκi = λ. Choose d ∈ F so thatsupi∈d κi = ess.supFi∈Iκi. Then

supi∈b∩d

κi ≤ supi∈d

κi

and ∀i ∈ b ∩ d[ω < κi < k(Ai)], as desired in (1).

Now fix i ∈ I. By (1) and (1) in the definition of sup-min functions, we canfind Gi ⊆ Ai such that (Ai, Gi) |= ψ and min{|P | : (Ai, Gi, P ) |= ϕ} > κi. LetH =

∏i∈I Gi/F . For each i ∈ I there is a Pi ⊆ Ai such that (Ai, Gi, Pi) |=

ϕ. Hence (B,H,∏

i∈I Pi/F ) |= ϕ. Also, (B,H) |= ψ. We claim that min{|P | :(B,H, P ) |= ϕ} ≥ λ+; this will prove the theorem, by (1) in the definition ofsup-min functions. To prove the claim, suppose that P = {fα/F : α < λ} ⊆ Band (B,H, P ) |= ∀x ∈ P(x �= 0 ∧ ϕ′(F, x)); we will show

(∗) (B,H, P ) |= ¬∀x ∈ F ∃y ∈ Pϕ′′(F, x, y).

We may assume that fα(i) �= 0 for all α < λ and i ∈ I. Now for any α < λ we have(B,H) |= ϕ′(F, fα/F ), and hence {i ∈ I : (Ai, Gi) |= ϕ′(F, fα(i))} ∈ F . Hence wecan assume that (Ai, Gi) |= ϕ′(F, fα(i)) for all α < λ and i ∈ I. (If (Ai, Gi) �|=ϕ′(F, fα(i)), replace fα(i) by a nonzero element ai such that (Ai, Gi) |= ϕ′(F, ai);ai exists by (3) of the definition of sup-min function.) Now fix i ∈ I again. Then

(Ai, Gi, {fα(i) : α < κi}) |= ∀x ∈ P(x �= 0 ∧ ϕ′(F, x)),

so, since (Ai, Gi, {fα(i) : α < κi}) �|= ϕ, we can choose ai ∈ Gi such that (Ai, Gi) |=¬ϕ′′(F, ai, fαi) for all α < κi. Now for any α < λ we have {i ∈ I : α < κi} ∈ F ,and hence (B,H) |= ¬ϕ′′(F, a0/F, fα/F ], and this proves (∗). �Theorem 6.6. Suppose that k is a sup-min function, 〈Ai : i ∈ I〉 is a system ofinfinite BAs, with I infinite, and F is an ultrafilter on I. Then k

(∏i∈I Ai/F

)≥

ess.sup Fi∈Ik(Ai).

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Proof. Let λ = ess.sup Fi∈Ik(Ai) and B =

∏i∈I Ai/F . Take any κ < λ. Then the

set Kdef= {i ∈ I : k(Ai) > κ} is in F . For each i ∈ K choose Gi ⊆ Ai such that

(Ai, Gi) |= ψ and min{|P | : (Ai, Gi, P ) |= ϕ} > κ. For i ∈ I\K let Gi = Ai.Set H =

∏i∈I Gi/F . Then (B,H) |= ψ. We claim that min{|P | : (B,H, P ) |=

ϕ} > κ; this will prove the theorem. Suppose that P = {fα/F : α < κ} ⊆ B and(B,H, P ) |= ∀x ∈ P(x �= 0 ∧ ϕ′(F, x)). As in the proof of Theorem 6.5 we canassume that (Ai, Gi) |= ϕ′(F, fα(i)) and fα(i) �= 0 for all α < λ and i ∈ I. Fixi ∈ K. Then

(Ai, Gi, {fα(i) : α < κ}) |= ∀x ∈ P (x �= 0 ∧ ϕ′(F, x)),

so, since (Ai, Gi, {fα(i) : α < κ}) �|= ϕ, we can choose ai ∈ Gi such that (Ai, Gi) |=¬ϕ′′(F, ai, fα(i)) for all α < κ. Then (B,H) |= ¬ϕ′′(F, a/F, fα/F ], as desired. �

Theorem 6.7. Suppose that k is a sup-min function such that for every infiniteBA A there is a G ⊆ A such that (A,G) |= ψ and min{|P | : (A,G, P ) |= ϕ} ≥ ω.Assume that 〈Ai : i ∈ I〉 is a system of infinite BAs, with I infinite, and F is aregular ultrafilter on I. Then k

(∏i∈I Ai/F

)≥ |I|+.

Proof. Let B =∏

i∈I Ai/F . For each i ∈ I choose Gi ⊆ Fi such that (Ai, Gi) |= ψand min{|P | : (Ai, Gi, P ) |= ϕ} ≥ ω. Let H =

∏i∈I Gi/F . Thus (B,H) |= ψ.

We want to show that min{|P | : (B,H, P ) |= ϕ} ≥ |I|+; this will prove thetheorem. To this end, suppose that P = {fα/F : α < |I|} ⊆ B and (B,H, P ) |=∀x ∈ P (x �= 0 ∧ ϕ′(F, x)). As in the proof of Theorem 6.5 we may assume that(Ai, Gi) |= ϕ′(F, fα(i)) and fα(i) �= 0 for all α < |I| and i ∈ I. Let {aα : α < |I|}be a regular family for F . Now fix i ∈ I. Then the set Ki

def= {α < |I| : i ∈ aα} is

finite. Since min{|Q| : (Ai, Gi, Q) |= ϕ} ≥ ω and Pidef= {fα(i) : α ∈ Ki} is finite, it

follows that (Ai, Gi, Pi) |= ¬ϕ. But (Ai, Gi, Pi) |= ∀x ∈ P (x �= 0 ∧ ϕ′F, x)), so wecan choose ai ∈ Gi such that (Ai, Gi) |= ¬ϕ′′(F, ai, fα(i)] for all α ∈ Ki. Therefore{i ∈ I : (Ai, Gi) |= ¬ϕ′′(F, ai, fα(i))} ⊇ aα ∈ F , so (B,H) |= ¬ϕ′′(F, a/F, fα/F ),as desired. �

Theorem 6.8. Suppose that k is a sup-min function such that the formula ψ inthe definition is ∀xPx and the formula ϕ′ is x = x. Assume that 〈Ai : i ∈ I〉is a system of infinite BAs, with I infinite, and F is an ultrafilter on I. Thenk(∏

i∈I Ai/F)≤∣∣∏


Proof. For each i ∈ I choose Pi ⊆ Ai such that |Pi| = k(Ai) and (Ai, Ai, Pi) |= ϕ.Then, we claim,

(∏i∈I Ai/F,

∏i∈I Ai/F,

∏i∈I Pi/F

)|= ϕ, which will prove the

theorem. So, suppose that x ∈∏

i∈I Ai. For each i ∈ I choose yi ∈ Pi such that(Ai, Ai) |= ϕ′′(F, x(i), yi). Then(∏

i∈I

Ai/F,∏i∈I

Ai/F

)|= ϕ′′[F, x/F, y/F ] �

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Theorem 6.9 (GCH). Suppose that k is a sup-min function such that the formulaψ in the definition is ∀xPx and the formula ϕ′ is x = x. We also suppose thatfor every infinite BA A there is a G ⊆ A such that (A,G) |= ψ and min{|P | :(A,G, P ) |= ϕ} ≥ ω. If 〈Ai : i ∈ I〉 is a system of infinite BAs, with I infinite,and F is a regular ultrafilter on I, then k

(∏i∈I Ai/F

)=∣∣∏


Proof. Let λ = ess.sup Fi∈Ik(Ai). We consider several cases.

Case 1. λ ≤ |I|. Then, using Keisler, Prikry [74] and Theorems 6.7 and 6.8 wehave

2|I| = |I|+ ≤ k

(∏i∈I

Ai/F

)≤∣∣∣∣∣∏i∈I

k(Ai)/F

∣∣∣∣∣ ≤ λ|I| = 2|I|.

Case 2. cf λ ≤ |I| < λ. Then by Theorem 6.5 we get λ+ ≤ k(∏

i∈I Ai/F)≤∣∣∏

i∈I k(Ai)/F∣∣ ≤ λ|I| = λ+.

Case 3. |I| < cf λ. Then by Theorem 6.6 we have

λ ≤ k

(∏i∈I

Ai/F

)≤∣∣∣∣∣∏i∈I

k(Ai)/F

∣∣∣∣∣ ≤ λ|I| = λ. �

By a result of Donder [88], V = L implies that every uniform ultrafilter isregular, and hence that the equality in Theorem 6.9 always holds; thus this answersProblem 12 of Monk [90] in a different way from the solution of Koppelberg andShelah mentioned at the outset.

The formula for the algebraic density of products and weak products given aboveno longer works for subdirect products in general. For example, let f : P(ω) →(P(ω)/fin) × P(ω) be defined by f(x) = ([x], x). Then rng(f) is a subdirectproduct of P(ω)/fin and P(ω), and π(P(ω)/fin) = 2ω while π(P(ω)) = ω.

That formula does hold for moderate products, by the same argument givenfor products.

Concerning one-point gluing we have the following result.

Proposition 6.10. If B is the one-point gluing of the pair (〈Ai : i ∈ I〉, 〈Fi : i ∈ I〉),then π(B) = π(

∏i∈I Ai).

Proof. For each i ∈ I let Xi ⊆ A+i be dense in Ai and of size π(Ai). Clearly

u ∈ Xi for each atom u of Ai. If Fi is the principal ultrafilter generated by {u},let X ′

i = {x ∈ Xi : x · u = 0}; otherwise let X ′i = Xi.

We now consider two cases.

Case 1. Every Fi is principal, say generated by {ui}, with ui an atom in Ai. Foreach i ∈ I and each x ∈ X ′

i, define zxi ∈ B by

zxi(j) =

{x if i = j,

0 otherwise.

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Clearly {u} ∪ {zxi : i ∈ I, x ∈ Xi, x · ui = 0} is dense in B and it is of the desiredsize.

Case 2. Some Fi is nonprincipal.

(1) If Fi is nonprincipal, then for each x ∈ Xi there is a nonzero yx ≤ x such thatyx /∈ Fi.

In fact, assume the hypothesis. If x /∈ Fi, we can take yx = x. Otherwise, withx ∈ Fi there is a z < x with z ∈ Fi, and we can take yx = x · −z.

Now for each i ∈ I and x ∈ X ′i, define zxi ∈ B by

zxi(j) =

⎧⎨⎩0 if j �= i,

yx if j = i and Fi is nonprincipal

x if j = i and Fi is principal.

We claim that {zxi : i ∈ I, x ∈ X ′i} is dense in B. For, suppose that b ∈ B+. Then

there are two possibilities:

Subcase 2.1. ∀i ∈ I[bi ∈ Fi]. Take i ∈ I such that Fi is nonprincipal. Now bi ∈ Fi,and so bi �= 0. Choose x ∈ Xi such that x ≤ bi. Since Fi is nonprincipal, we haveXi = X ′

i. Then zxi ≤ b, as desired.

Subcase 2.2. ∀i ∈ I[bi /∈ Fi]. Choose i ∈ I such that bi �= 0, and choose x ∈ Xi

such that x ≤ bi. Then x ∈ X ′i and zxi ≤ b, as desired.

Thus we have now shown that B has a dense subset of size max{|I|, supi∈I π(Ai)}.Now let X be dense in B. To show that |I| ≤ |X |, suppose that i ∈ I. Choosea ∈ Ai\Fi, and let x be the member of B which is equal to a at the ith placeand is 0 elsewhere. There is some nonzero element of X below x. This shows that|I| ≤ |X |. To show that π(Ai) ≤ |X |, it suffices to show that {xi : x ∈ X} is densein Ai. Suppose that a ∈ A+

i . There is an element y of B such that yi = a, and thedesired conclusion follows. �

Clearly π(Dup(A)) = |Ult(A)| for any infinite BA A.

Proposition 6.11. π(Exp(A)) = π(A) for any infinite BA A.

Proof. First we show that π(Exp(A)) ≤ π(A). Let X be dense in A of size π(A).Define

Y = {V (S(x0), . . . ,S(xm−1)) : m ∈ ω, x ∈ mX} .We claim that Y is dense in Exp(A), giving the desired inequality. For, by Theorem1.20 it suffices to take a nonzero element of Exp(A) of the form V (S(b0), . . . ,S(bm−1)) and find a nonzero element of Y below it. For each i < m let xi bea nonzero element of X below bi. Then V (S(x0), . . . ,S(xm−1)) ⊆ V (S(b0), . . . ,S(bm−1)), as desired.

To show that π(A) ≤ π(Exp(A)), let Y be dense in Exp(A), of sizeπ(Exp(A)). By Theorem 1.20 we may assume that for each element y ∈ Y wecan write

y = V (S(ay0), . . . ,S(aymy−1)).

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Let X = {ayi : y ∈ Y, i < my}. We claim that X is dense in A; this will provethe above inequality. Suppose that b is a nonzero element of A. Choose y ∈ Ysuch that y ⊆ V (S(b)). We claim that ayi ≤ b for some i < my. Suppose that thisis not true. For each i < my let Ui be an ultrafilter such that ayi · −b ∈ Ui. LetF = {Ui : i < m}. Then F ∈ y\V (S(b)), contradiction. So the claim holds, givingthe desired inequality. �

derived functionsTurning to the functions derived from π, we first work towardproving that πH+ = πh+ = hd.

We call a sequence 〈xξ : ξ < κ〉 of elements of a space X left-separatedprovided that κ is a cardinal, and for every ξ < κ there is an open set U in Xsuch that U ∩ {xη : η < κ} = {xη : ξ ≤ η}. The definition of hd given in theintroduction obviously extends as follows to arbitrary spaces X :

hd(X) = sup{d(S) : S a subspace of X}.

We now need the following important fact relating left separation to hd:

Theorem 6.12. For any infinite Hausdorff space X, hd(X) is the supremum of allcardinals κ such that there is a left-separated sequence in X of type κ.

Proof. If 〈xξ : ξ < κ〉 is a left-separated sequence in X and κ is infinite andregular, then clearly the density of {xξ : ξ < κ} is κ. Hence the inequality ≥holds. Now suppose that Y is a subspace of X , and set d(Y ) = κ. We construct aleft-separated sequence 〈xξ : ξ < κ〉 as follows: having constructed xη ∈ Y for allη < ξ, where ξ < κ, it follows that {xη : η < ξ} is not dense in Y , and so we can

choose xξ ∈ Y \{xη : η < ξ}. This proves the other inequality. �

Note that the proof of Theorem 6.12 shows that if hd(X) is attained, then it isalso attained in the left-separated sense, and conversely if hd(X) is regular.

There is an algebraic version of left-separated sequences. We call a sequence〈aξ : ξ < α〉 of elements of A left-separated if for all ξ < α and all finite F ⊆ αsuch that ξ < β for all β ∈ F we have aξ ·

∏η∈F −aη �= 0.

Lemma 6.13. Let κ be an infinite cardinal. Then a BA A has a left-separatedsequence of length κ iff Ult(A) has a left-separated sequence of length κ.

Proof. ⇒: Let 〈aξ : ξ < κ〉 be a left-separated sequence of elements of A. Foreach ξ < κ, let Fξ be an ultrafilter on A containing the set {aξ} ∪ {−aη : ξ < η}.For any ξ < κ we have {Fη : η < κ} ∩

⋃ξ≤η S(aη) = {Fη : ξ ≤ η}. In fact, if

Fη ∈⋃

ξ≤ρ S(aρ), choose ρ with ξ ≤ ρ and Fη ∈ S(aρ). Then aρ ∈ Fη, so η �< ρ,so ξ ≤ ρ ≤ η. If ξ ≤ η, then aη ∈ Fη, hence Fη ∈ S(aη) and so Fη ∈

⋃ξ≤ρ S(aρ).

Thus 〈Fξ : ξ < κ〉 is a left-separated sequence in Ult(A).

⇐: Let 〈Fξ : ξ < κ〉 be a left-separated sequence in Ult(A). For each ξ < κchoose an element aξ such that Fξ ∈ S(aξ) and S(aξ) ∩ {Fη : η < κ} ⊆ {Fη : ξ ≤η < κ}. To check that 〈aξ : ξ < κ〉 is a left-separated sequence, suppose on the

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contrary that aξ ≤∑

η∈F xη, where ξ < η for all η ∈ F , F a finite subset of κ.Since aξ ∈ Fξ, it follows that aη ∈ Fξ for some η ∈ F . This contradicts the choiceof aη. �

The essential step in proving that πH+ = hd is as follows; we follow the proof ofvan Douwen [89], 10.1.

Lemma 6.14. Let A be an infinite BA. Then there exists a left-separated sequence〈xξ : ξ < π(A)〉 such that {xξ : ξ < π(A)} is dense in A.

Proof. The major part of the proof consists in proving

(1) There is a sequence 〈aξ : ξ < π(A)〉 of non-zero members of A such that {aξ :ξ < π(A)} is dense in A and for each η < π(A), |{ξ < η : aξ ·aη �= 0}| < π(A � aη).

To prove this, call an element b ∈ A π-homogeneous provided that π(A � c) = π(A �b) for every non-zero c ≤ b. Clearly the collection of all π-homogeneous elementsof A is dense in A. Let A be a maximal disjoint collection of π-homogeneouselements of A. Let κ = |A |; then κ ≤ c(A) ≤ π(A). For each b ∈ A let Mb be adense subset of A � b of cardinality π(A � b). Then

⋃{Mb : b ∈ A } is dense in A.

Now let 〈Nb : b ∈ A 〉 be a partition of π(A) into disjoint subsets of power π(A).For each b ∈ A let fb be a one-one function from Mb onto a subset of Nb of ordertype π(A � b). Now for each ξ < π(A), let

aξ =

{0, if ξ /∈

⋃b∈A rng(fb);

f−1b (ξ), if ξ ∈ rng(fb), b ∈ A .

Note that if aξ �= 0, then aξ ∈ Mb for a unique b. If aξ · aη �= 0, then there isa unique b such that aξ, aη ∈ Mb. Now suppose that η < π(A) and aη �= 0. Sayη ∈ rng(fb). Then

|{ξ < η : aξ · aη �= 0}| ≤ |{ξ ∈ rng(fb) : ξ < η}| < π(A � b).

Thus we have (1), except that some of the aη’s are zero. If we renumerate thenon-zero aη’s in increasing order of their indices, we really get (1).

Now we construct a sequence 〈bα : α < π(A)〉 of non-zero elements of A sothat the following conditions hold: (2α) bα ≤ aα for all α < π(A),

and

(3α) for all ξ < α and every finite F ⊆ (ξ, α] we have bξ ·∏

η∈F −bη �= 0 for allα < π(A).

Assume that β < π(A), and bα has been constructed for all α < β so that (2α)and (3α) hold. Then by (1) and the assumption that (2α) holds for all α < β we

see that the set Γdef= {α < β : bα · aβ �= 0} has power < π(A � aβ). Hence there is

a non-zero bβ in A such that bβ ≤ aβ and for all ϕ ∈ Γ and all finite G ⊆ Γ, if

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bϕ ·∏

γ∈G−bγ �= 0, then bϕ ·∏

γ∈G−bγ �⊆ bβ. Thus (2β) and (3β) hold, and theconstruction is complete.

It is clear from (2α) and (3α) that 〈bα : α < π(A)〉 is the desired densesequence. �

The following theorem is due to Shapirovskiı.

Theorem 6.15. πH+(A) = πh+(A) = hd(A) for any infinite BA A.

Proof. It is obvious that πH+(A) ≤ πh+(A). Now if O is a π-base for Y ⊆ Ult(A)with |O| = π(Y ), without loss of generality O has the form {S(a) ∩ Y : a ∈ A }for some A ⊆ A. Let f(x) = S(x) ∩ Y for any x ∈ A. Then f is a homomorphismonto some algebra B of subsets of Y , and O is dense in B. This shows thatπh+(A) ≤ πH+(A). It is also trivial that hd(A) ≤ πh+(A), since if S ⊆ Ult(A)then d(S) ≤ π(S). It remains just to show that πH+(A) ≤ hd(A). Suppose thatf is a homomorphism of A onto B, where B is infinite. Apply 6.14 to B to get asystem 〈bξ : ξ < π(B)〉 of elements of B such that for any ξ < π(B) and any finitesubset G of (ξ, π(B) we have bξ ·

∏η∈G−bη �= 0. For each ξ < π(B) choose aξ so

that f(aξ) = bξ. Clearly 〈aξ : ξ < π(B)〉 is a left-separated sequence of elementsof A, which by 6.12 and 6.13 is as desired. �

The proof of Theorem 6.15 shows that πH+ and πh+ have the same attainmentproperties; also, if πH+ is attained, then hd is attained in the left-separated sense.Also, if hd is attained in the defined sense then it is attained in the πh+ sense.

Clearly πH−(A) = πh−(A) = ω for any infinite BA A. The cardinal function πS+ isof some interest, since it does not coincide with any of our standard ones. ObviouslyπA ≤ πS+(A) for any infinite BA A. Moreover, πS+(A) ≤ πH+(A); this followsfrom the following fact: for every subalgebra B of A there is a homomorphic imageC of A such that π(B) = π(C). To see this, by the Sikorski extension theoremextend the identity function from B into B to a homomorphism from A onto asubalgebra C of B. Since B ⊆ C ⊆ B, it is clear that π(B) = π(C). Thus we haveshown that π(A) ≤ πS+(A) ≤ πH+(A) for any infinite BA A. It is possible to haveπ(A) < πS+(A): let A = P(κ); then π(A) = κ, while πS+(A) = 2κ, since A hasa free subalgebra B of power 2κ, and clearly π(B) = 2κ. It is more difficult tocome up with an example of an algebra where the other inequality is proper (thisexample is due to Monk):

Example 6.16. There is an infinite BA A such that πS+(A) < πH+(A).

To see this, let B be the interval algebra on the real numbers, and let A = B⊕B.Now, we claim, πS+(A) = ω, while πH+(A) = 2ω. To prove that πH+(A) = 2ω, by5.19, 5.20, and 6.15 it suffices to show that s(A) = 2ω. For each real number r letcr = br · b′r, where br = [−∞, r) (as a member of the first factor of B ⊕ B) andb′r = [r,∞) (as a member of the second factor ofB⊕B). Note that we have adjoined−∞ as a member of R in order to fulfill the requirement for interval algebras thatthe ordered set in question always has a first element. To show that 〈cr : r ∈ R〉 is

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an ideal independent system of elements, suppose that cr ≤ cs1 + · · · + csm withr /∈ {s1, . . . , sm}. Let Γ = {i : si < r} and Δ = {i : r < si}. Then

cr · −cs1 · . . . · −csm ≥ br · b′r ·∏i∈Γ

−bsi ·∏i∈Δ

−b′si �= 0,

contradiction. To prove that πS+(A) = ω, we proceed as follows. Let C be anysubalgebra of A. We want to show that π(C) = ω. Now for each element c of Cwe choose a representation of c of the form∑

i<m(c)

x0ic · x1ic,

where x0ic and x1ic are half-open intervals in B; x0ic is in the first factor of B⊕Band x1ic in the second. Let T = {(m, r, s) : m ∈ ω\{0} and r, s ∈ mQ}. An element(m, r, s) of T is a frame for c ∈ C provided that m(c) = m, and ri ∈ x0ic, si ∈ x1ic

for all i ∈ m. For each (m, r, s) ∈ T let Dmrs be the set of all c ∈ C with frame(m, r, s). Since C is the union of all sets Dmrs, it suffices to take an arbitrary(m, r, s) ∈ T and find a countable subset of C dense in Dmrs.

For each c ∈ Dmrs and each i < m write

x0ic = [aic, bic) and x1ic = [dic, eic).

Thus aic ≤ ri < bic and dic ≤ si < eic for all c ∈ Dmrs and i < m. For each i < mlet N0i be a countable subset of {aic : c ∈ Dmrs} cofinal in that set. Similarlychoose N1i coinitial for the bic’s, N2i cofinal for the dic’s, and N3i coinitial for theeic’s. Let M be the set of all products∏

i<m,j<4

uij

with u ∈ m×4⋃

i<m(N0i∪N1i∪N2i∪N3i). Clearly all such products are nonzero.Weclaim that M is dense in Dmrs. For, let c ∈ Dmrs. For each i < m choose ui0 ∈ N0i

such that aic ≤ ui0; similarly for uij , j = 1, 2, 3. Then clearly∏

i<m,j<4 uij ≤ c,as desired. �Shelah [96] showed that it is consistent to have a BA A with πS+(A) not attained;this answers Problem 13 in Monk [90]. On the other hand, under GCH the cardinalπS+(A) is always attained. In fact, if |A| is a limit cardinal, then π(A) = |A| andso of course πS+(A) = |A| is attained. If |A| = κ+, then π(A) is either κ or κ+,and it follows that πS+(A) is attained. This answers Problem 20 in Monk [96].

Clearly πS−A = ω for any infinite BA A. Furthermore, dπS+(A) = dπS−(A) =π(A) for any infinite BA A.

We now consider weak density and the cardinal function r. Several generalizationsof r have been intensively studied in the literature, and we survey this area. We

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present various results from Dow, Steprans, Watson [96], Balcar, Simon [92], Pe-terson [98], Bozeman [91], and Brown [05]. A subset X of A+ is weakly dense ifffor any a ∈ A there is an x ∈ X such that x ≤ a or x ≤ −a. The weak density orreaping number of A is the smallest size r(A) of a weakly dense subset of A.

We generalize this notion as follows. A weak partition of a BA A is a system〈xi : i ∈ I〉 of pairwise disjoint elements of A, with sum 1. We call it “weak”because it is not assumed that each xi is nonzero. Of course there is at least onenonzero xi. For any integer m ≥ 2, we call a subset X of A+ m-dense iff for everyweak partition 〈ai : i < m〉 of A, ∃x ∈ X∃i < m[x ≤ ai]. Let

rm(A) = min{|X | : X is m-dense in A}.

The following proposition should be noted, although we will not use it.

Proposition 6.17. Let A be an infinite BA, m an integer, and X ⊆ A. Then thefollowing are equivalent:

(i) X is m-dense.

(ii) For every partition 〈ai : i < m〉 of A there exist an x ∈ X and an i < msuch that x ≤ ai.

Proof. Obviously (i)⇒(ii). Now suppose that (ii) holds, and 〈ai : i < m〉 is a weakpartition. Let G = {i < m : ai = 0}. We may assume that G is nonempty. Fixj < m such that A � aj is infinite. Let 〈bi : i < |G| + 1〉 be a system of nonzeropairwise disjoint elements of A � aj with sum aj , and let f be a bijection from Gonto |G|. Define

a′k =

⎧⎨⎩ak if k /∈ G ∪ {j},bf(k) if k ∈ G,

b|G| if k = j.

Clearly 〈a′k : k < m〉 is a partition, and an application of (ii) yields x ∈ X andi < m such that x ≤ ai. �

Proposition 6.18.

(i) If a is an atom of A, then {a} is weakly dense, and hence r(A) = 1.

(ii) If A is atomless, then r(A) is infinite.

(iii) r2(A) ≤ r3(A) ≤ · · · for any BA A.

Proof. (i) is obvious. For (ii), suppose that A is atomless and X ⊆ A is finite andweakly dense. Let a0, . . . , an−1 list all of the atoms of 〈X〉, and for each i < nchoose bi with 0 < bi < ai. Let c =

∑i<n bi. Choose a nonzero x ∈ X such that

x ≤ c or x ≤ −c. Say x · ai �= 0. Then since ai is an atom of 〈X〉, we have ai ≤ x.This contradicts x ≤ c or x ≤ −c.

(iii) is clear. �

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A finer analysis of weak density can be made as follows. Suppose that m and n areintegers, with 2 ≤ n ≤ m. A subsetX of A+ is (m,n)-dense provided that for everyweak partition 〈ai : i < m〉 there is an x ∈ X such that |{i < m : x · ai �= 0}| < n.Then we define

rmn(A) = min{|X | : X is (m,n)-dense}.

We now work towards a result of Dow, Steprans, Watson [96] to the effect thatrn(A) ≤ r2(A)

+ for every integer n ≥ 2.

Some simple relationships between these notions of weak density are as fol-lows.

Proposition 6.19. Suppose that 2 ≤ n ≤ m. Then for any BA A,

(i) If also n+ 1 ≤ m, then rm,n+1(A) ≤ rmn(A).

(ii) rmn(A) ≤ rm+1,n(A).

(iii) rm+1,n+1(A) ≤ rmn(A).

(iv) If ∃k[m ≤ i ≤ mk and (n− 1)k < j], then rij(A) ≤ rmn(A).

(v) r2(A) ≤ rmn(A).

(vi) rn(A) = rn,2(A).

(vii) If m/(n− 1) ≤ 2, then rmn(A) = r2(A).

Proof. (i) is clear, since every (m,n)-dense subset is also (m,n + 1)-dense. (ii)holds since we can adjoin 0 to any weak partition of length m to obtain one oflength m+ 1; so every (m+ 1, n)-dense set is also (m,n)-dense.

For (iii), let X be (m,n)-dense, with |X | = rmn(A). Take any weak partition〈as : s < m+ 1〉. Define for s < m

a′s ={as if s < m− 1,

am−1 + am if s = m− 1.

Choose x ∈ X such that {s < m : x · a′s �= 0}| < n. Then |{s < m + 1 : x · as �=0}| < n+ 1.

For (iv), suppose that X is (m,n)-dense, with |X | = rmn(A). Take any weakpartition 〈as : s < i〉. Since m ≤ i ≤ mk, let 〈ul : l < m〉 be a partition of i intononempty subsets, each of size at most k. For each l < m let bl =

⋃s∈ul

as. Thus〈bl : l < m〉 is a weak partition. Choose x ∈ X such that |{l < m : x · bl �= 0}| < n.Then

|{s < i : x · as �= 0}| =∑l<m

x·bl �=0

|{s ∈ ul : x · as �= 0}|

≤ (n− 1)k < j.

For (v), suppose that |X | < r2(A). Then

(∗) For each i ≥ 2 there is a (non-weak) partition 〈as : s < i〉 such that ∀x ∈X∀s < i[x · as �= 0].

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We prove (∗) by induction. Take a such that ∀x ∈ X [x · a �= 0 �= x · −a]. Thenthe partition 〈a,−a〉 proves (∗) for i = 2. Now assume the conclusion of (∗) for i.Now the set {x · a0 : x ∈ X} has size less than r(A), and so there is a b such that∀x ∈ X [x · a0 · b �= 0 �= x · a0 · −b]. Replacing a0 by 〈a0 · b, a0 · −b〉, we get (∗) fori+ 1.

Now take i = n. If X is (m,n)-dense, then |{x ∈ X : x · as �= 0}| < n,contradiction.

(vi) is obvious.

For (vii), note by (v) that r2(A) ≤ rmn(A). Now we apply (iv) withm,n, i, j, kreplaced by 2, 2,m, n, n− 1 to get rmn(A) ≤ r22(A) = r2(A). �

To proceed we need some finite combinatorics. First we prove a simple fact aboutelementary algebra.

Proposition 6.20. If n is a prime and Zn is the group of integers modulo n, thenfor any x, k ∈ Zn with k �= 0, the elements x, x + k, x + 2k, . . . , x + (n − 1)k areall distinct (addition and multiplication mod n).

Proof. Suppose that there are i, j with 0 ≤ i < j < n and x + ik = x + jk. Then(j − i)k = 0. This is mod n, and with ordinary multiplication it says that (j − i)kis divisible by n. Since j − i, k < n and n is prime, this is not possible. �

Now for positive integers m,n, i, j, k, q we write

(m

n

)�→(j

i

)1,1

k,q

to mean that there is a function h : m × n → k such that ∀a ∈ [n]i∀b ∈ [m]j [h �(a× b) takes on more than q values]. Note that this implies that q + 1 ≤ k.

Lemma 6.21. For every prime n and positive integer k < n,

(n

n

)�→(2

k

)1,1

n,k

.

Proof. Define h : n × n → n by h(i, j) = i + j (mod n). Suppose that a ∈ [n]k

and b ∈ [n]2. Say b = {s, t} with s < t, and a = {xi : i < k}. Now the integersxi + t are all distinct for i < k. (Again, addition mod n). Similarly the integersxi+ s are all distinct for i < k. Suppose that h � (a× b) takes on at most k values.Then each integer xi + t is equal to xj + s for some j. By Proposition 6.20, theintegers x0 + u(t− s) are all distinct for u < n. Hence there is a smallest u suchthat x0+u(t−s) /∈ a. Obviously u > 0. Say x0+(u−1)(t−s) = xj . Choose v suchthat xj + t = xv + s. Then xj + (t− s) = xv ∈ a. But xj + (t− s) = x0 + u(t− s),contradiction. �

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Lemma 6.22. Suppose that i, j, k,m, n, q are positive integers such that

(m

n

)�→(j

i

)1,1

k,q

.

Also assume that A is a BA with rk,q+1(A) < rn,i(A). Then rm,j(A) ≤ rk,q+1(A)+.

Proof. Let h be as in the definition of(mn

)�→(ji

)1,1k,q

. Suppose that A is a BA with

rk,q+1(A) < rn,i(A). Let X be a subset of A of size rk,q+1(A) which is (k, q + 1)-dense. Let κ = rk,q+1(A)

+. We now define a sequence 〈Bα : α < κ〉 of subalgebrasof A, each of size rk,q+1(A). Let B0 = 〈X〉. Suppose that Bα has been defined.Then |Bα| < rn,i(A), so there is a weak partition 〈b(α, l) : l < n〉 of A such that∀x ∈ Bα[|{l < n : x · b(α, l) �= 0}| ≥ i]. Let Bα+1 = 〈Bα ∪ {b(α, l) : l < n}〉. For βlimit < κ let Bβ =

⋃α<β Bβ .

Let Bκ =⋃

α<κ Bα. Assume now, to get a contradiction, that κ < rm,j(A).Let 〈b(κ, l) : l < m〉 be a weak partition such that ∀x ∈ Bκ[|{l < m : x · b(κ, l) �=0}| ≥ j].

For all α < κ and l < k let

c(α, l) =∑

h(ζ,ξ)=l

(b(α, ξ) · b(κ, ζ)).

Then for each α < κ, the sequence 〈c(α, l) : l < k〉 is a weak partition of A. Infact, suppose that l < s < k. If h(ζ, ξ) = l and h(ζ′, ξ′) = s, then (ζ, ξ) �= (ζ′, ξ′),and hence b(α, ξ) · b(κ, ζ) · b(α, ξ′) · b(κ, ζ′) = 0. Also,∑

l<k

c(α, l) =∑ξ<nζ<m

(b(α, ξ) · b(κ, ζ)) = 1.

Now X is (k, q + 1)-dense in A, so for each α < κ there is an aα ∈ X such that|{l < k : aα ·c(α, l) �= 0}| < q+1. Let 〈leα : e < q〉 be such that {l < k : aα ·c(α, l) �=0} ⊆ {leα : e < q}. Thus aα ≤

∑e<q c(α, l

eα). Now

κ =⋃{α < κ : 〈leα : e < q〉 = s, aα = b},

the union being taken over all pairs (s, b) such that s ∈ qk and b ∈ X .

Hence one of the sets in the union has κ elements. In particular, there areelements α1 < α2 < · · · < αi < κ and integers le for e < q such that

leα1= leα2

= · · · = leαi= le and aα1 = aα2 = · · · = aαi = a

for each e < q. Now 〈b(α1, l) : l < n〉 is a weak partition, so there is an t1 < n suchthat a · b(α1, t1) �= 0. Suppose now that s < i and b(α1, t1), . . . , b(αs, ts) have beendefined so that a · b(α1, t1) · . . . · b(αs, ts) �= 0. Since 〈b(αs+1, l) : l < n〉 is a weak

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partition, there is then a ts+1 < n such that a · b(α1, t1) · . . . · b(αs+1, ts+1) �= 0.This finishes the definition of t1, . . . , ti.

Now by the definition of 〈b(κ, l) : l < m〉, the set

udef= {l < m : a · b(α1, t1) · . . . · b(αi, ti) · b(κ, l) �= 0}

has at least j elements. Let u′ be a subset of u with exactly j elements. Now weclaim

(∗) For any s ∈ {1, . . . , i} and any y ∈ u′ we have h(ts, y) ∈ {le : e < q}.

To prove (∗), let s and y be given as indicated. Then

a · b(αs, ts) · b(κ, y) ≥ a · b(α1, t1) · . . . · b(αi, ti) · b(κ, y) �= 0.

Now a ≤∑

e<q c(αs, le), and it follows that h(ts, y) = le for some e < q. This

proves (∗).By (∗), h maps {ts : 1 ≤ s ≤ i} × u′ into {le : e < q}. This contradicts the

basic property of h. �

Theorem 6.23. rn(A) ≤ r2(A)+ for every integer n ≥ 2 and every infinite BA A.

Proof. Since r2(A) ≤ r3(A) ≤ · · · , it suffices to take the case in which n is prime.If rn(A) = r2(A) the conclusion is obvious, so assume that rn(A) > r2(A). Thus byProposition 6.19(vi) we have rn2(A) > r2(A). Choose k ≤ n maximum such thatrnk(A) > r2(A); see Proposition 6.19(i). Now n

n−1 ≤ 2, so by Proposition 6.19(vii)we have k < n. Hence by Proposition 6.20 we have

(n

n

)�→(2

k

)1,1

n,k

.

Now the maximality of k implies by Proposition 6.19(i) that rn,k+1(A) = r2(A) <rnk(A). Hence Proposition 6.22 applies to give rn2(A) ≤ rn,k+1(A)

+ = r2(A)+.�

An example with ru(A) < ru+1(A) was given by Balcar, Simon [92]. To describetheir example we need two lemmas, the second one being of general interest.

Lemma 6.24. Let κ be an infinite cardinal. Then there is a system 〈Pn : n ∈ ω〉 ofpartitions of κ with the following properties:

(i) ∀n ∈ ω[|Pn| = κ].

(ii) ∀n ∈ ω∀X ∈ Pn[|X | = κ].

(iii) ∀n ∈ ω∀f ∈∏

i≤n Pi

[∣∣∣⋂i≤n f(i)∣∣∣ = κ

].

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Proof. Let P0 be a partition of κ into κ sets, each of size κ; so (i)–(iii) hold.Now suppose that P0, . . . , Pn have been defined so that (i)–(iii) hold. For eachf ∈

∏i≤n Pn let 〈Xfα : α < κ〉 be a partition of

⋂i≤n f(i) into κ sets, each of size

κ. Let

Pn+1 =

⎧⎨⎩

⋃f∈∏

i≤n Pi

Xfα : α < κ

⎫⎬⎭ .

Clearly each member of Pn+1 has size κ, and Pn+1 itself has size κ. If β < κ,define f ∈

∏i≤n Pi by letting f(i) be the member X of Pi such that β ∈ X .

Then β ∈⋂

i≤n f(i), and hence β ∈ Xfα for some α < κ. Thus⋃Pn+1 = κ.

Clearly the members of Pn+1 are pairwise disjoint. Finally, if f ∈∏

i≤n+1 Pi, then⋂i≤n+1 f(i) = X(f�(n+1))α for some α < κ, so that

∣∣∣⋂i≤n+1 f(i)∣∣∣ = κ. �

Lemma 6.25. If κ is an infinite cardinal, then there is a set A of independentpartitions of κ, with each member of A of size κ, and with |A | = 2κ.

Proof. Let 〈Pn : n ∈ ω〉 be as in Lemma 6.24, and let A = {Pn : 0 < n ∈ ω}.By Lemma 13.9 of the Handbook, let X ⊆ P0A be finitely distinguished, of sizeωκ = 2κ. For f ∈ X and u ∈ P0 we have f(u) ∈ A, and hence we can writef(u) = {w(f, u, α) : α < κ}, without repetitions. Now for any f ∈ X and α < κlet R(f, α) =

⋃u∈P0

(u ∩ w(f, u, α)).

(1) If α < β < κ, then R(f, α) ∩R(f, β) = ∅.

In fact,

R(f, α) ∩R(f, β) =

( ⋃u∈P0

(u ∩w(f, u, α))

)∩( ⋃

v∈P0

(v ∩ w(f, v, β))

)

=⋃

u,v∈P0

(u ∩ v ∩ w(f, u, α) ∩w(f, v, β));

and u ∩ v ∩ w(f, u, α) ∩w(f, v, β) = ∅ if u �= v and also if u = v.

(2)⋃

α<κ R(f, α) = κ.

In fact, ⋃α<κ

R(f, α) =⋃α<κ

⋃u∈P0

(u ∩ w(f, u, α))

=⋃

u∈P0

⋃α<κ

(u ∩ w(f, u, α))

=⋃

u∈P0

u = κ.

Thus S(f)def= {R(f, α) : α < κ} is a partition. Clearly each member of S(f) has

size κ Let A = {S(f) : f ∈ X}. Thus |A | = 2κ.

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It remains only to show that A is independent. To this end, let F be a finitesubset ofX , and for each f ∈ F let αf < κ; we want to show that

⋂f∈F R(f, αf ) �=

∅. Since X is finitely distinguished, choose v ∈ P0 so that 〈f(v) : f ∈ F 〉 is one-one.Then ⋂

f∈F

R(f, αf ) =⋂f∈F

⋃u∈P0

(u ∩ w(f, u, αf ))

⊇ v ∩⋂f∈F

w(f, v, αf );

since f(v) �= g(v) for distinct f, g ∈ F , it follows that this last intersection isnonempty. �

Independent partitions in arbitrary BAs have been considered by Robin Chest-nut [12].

Now we give the example from Balcar, Simon [92].

Theorem 6.26. For every infinite cardinal κ and every integer u ≥ 2 there is a BAB such that ru(B) = κ and ru+1(B) = κ+.

Proof. Let F be freely generated by 〈a(α, 0) : α < κ+〉, and set a(α, 1) = −a(α, 0)for all α < κ+. Let C = κFw (the weak κ-power of F ). The desired algebra Bwill be a certain algebra with C ≤ B ≤ C. If b ∈ F and α < κ let δαb be defined by

δαb (β) =

{b if α = β,

0 otherwise,

for all β < κ.

Now by Lemma 6.25 we have:

(1) There is a family 〈Aαβ : α < κ+, β < κ〉 with the following two properties:

(a) ∀α < κ+[〈Aαβ : β < κ〉 is a partition of κ.

(b) For every finite J ⊆ κ+ and every f ∈ Jκ, the set⋂

α∈J Aαf(α) is infinite.

For each α < κ+ let Ψα be the set of all functions ψ having the following twoproperties:

(2) There is a finite K ⊆ α such that dmn(ψ) = K(u + 1).

(3) rng(ψ) ⊆ 2(u+ 1)\{(0, 0), (1, 1), . . . , (u, u)}.

Note that Ψ0 has exactly (u + 1)2 − (u + 1) members. Clearly |Ψα| ≤ κ for allα < κ+. Let χα be a mapping of κ onto Ψα.

Now we define by recursion on α < κ+ elements b(α, 0), b(α, 1), . . . , b(α, u)of C. Suppose that they have been defined for all β < α (with α = 0 possible) sothat the following conditions hold:

(4) ∀β < α[〈b(β, i) : i ≤ u〉 is a partition of C].

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(5) ∀β < α∀i ≤ u∀γ < κ[b(β, i) · δγ1 ∈ 〈{δγa(ξ,0) : ξ ≤ β}〉].

We define b(α, i) by defining its product with each δγ1 . Say γ ∈ Aαβ . Let K be thefinite set such that dmn(χα

β) =K(u + 1). We define

b(α, i) ·δγ1 =∑⎧⎨

⎩δγa(α,j) ·∏β∈K

b(β, h(β)) : j < 2, h ∈ K(u + 1), (χαβ (h))(j) = i

⎫⎬⎭ .

We check (4) and (5) for α. For distinct i, k ≤ u we consider j, l < 2 and h, r ∈K(u+ 1) such that (χα

β (h))(j) = i and (χαβ(r))(l) = k; we want to show that

(6) δγa(α,j) ·∏β∈K

b(β, h(β)) · δγa(α,l) ·∏β∈K

b(β, r(β)) = 0.

If j �= l this is clear. If j = l, then h �= r and again (6) follows.

Next, we want to show that∑

i≤u(b(α, i) · δγ1 ) = δγ1 . To this end, note that

(7) 1 =∏α∈K

∑i≤u

b(α, i) =∑

h∈K(u+1)

∏β∈K

b(β, h(β)).

Hence

∑i≤u

(b(α, i) · δγ1 ) =∑{

δγa(α,j) ·∏β∈K

b(β, h(β)) : j < 2,

h ∈ K(u+ 1), (χαβ(h))(j) = i, i ≤ u

}

=∑⎧⎨

⎩δγa(α,j) ·∏β∈K

b(β, h(β)) : j < 2, h ∈ K(u+ 1)

⎫⎬⎭

= δγ1 .

This proves (4). (5) holds using the inductive hypothesis. This finishes the defini-tion of 〈b(α, i) : α < κ+, i ≤ u〉.

Let B be the subalgebra of C generated by C ∪ {b(α, i) : α < κ+, i ≤ u}.We show that ru+1(B) ≥ κ+. Suppose to the contrary thatX is a (u+1)-dense

subset of B of size ≤ κ. Now {δγ1 ·x : γ < κ, x ∈ X}\{0} is still (u+1)-dense and ofsize ≤ κ. So we may assume that ∀x ∈ X∃γ < κ[x ≤ δγ1 ]. Temporarily fix x ∈ X .There is a unique γ < κ such that x ≤ δγ1 . Then there is an α(x) < κ+ such thatx ∈ 〈{δγa(ξ,0) : ξ < α(x)〉C�δγ1 . Now unfix x. Choose ε < κ+ with ε > supx∈X α(x).

We claim that the weak partition 〈b(ε, i) : i ≤ u〉 contradicts X being (u + 1)-dense. To show this, we take any x ∈ X and find distinct i0, i1 ≤ u such thatx ·b(ε, i0) �= 0 �= x ·b(ε, i1). To do this, first choose γ < κ such that x ≤ δγ1 . Choose

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β < κ such that γ ∈ Aεβ . Say dmn(χεβ) = K(u + 1) with K ∈ [ε]<ω. Now, as

observed above, ∑h∈K(u+1)

∏ξ∈K

b(ξ, h(ξ)) = 1,

so we can choose h ∈ K(u+1) such that x·∏

ξ∈K b(ξ, h(ξ)) �= 0. Let χεβ(h) = (i0, i1),

with i0, i1 ≤ u and i0 �= i1. Now for any ξ ∈ K we have δγ1 · b(ξ, h(ξ)) ∈ 〈δγa(η,0) :

η ≤ ξ〉, so δγ1 ·∏

ξ∈K b(ξ, h(ξ)) ∈ 〈δγa(ξ,0) : η < ξ〉. Hence by freeness,

δγa(ε,0) · x ·∏ξ∈K

b(ξ, h(ξ)) �= 0 �= δγa(ε,1) · x ·∏ξ∈K

b(ξ, h(ξ)).

Hence by definition, x · b(ε, i0) �= 0 �= x · b(ε, i1). This proves that ru+1(B) ≥ u+1.

To show that ru(B) ≤ κ it suffices to prove

(8) For every weak partition 〈pi : i < u〉 of B there exist α < κ and i < u suchthat δα1 ≤ pi.

In fact, it suffices to prove the following more complicated statement:

(9) For every weak partition 〈pi : i < u〉 of B and every finite nonempty K ⊆ κ+,if⟨∏

α∈K b(α, h(α)) : h ∈ K(u+ 1)⟩refines 〈pi : i < u〉, then there is a j < u such

that

∀J ∈ [(max(K), κ+)]<ω∀f ∈ Jκ

[⋂ε∈J

Aεf(ε) ∩ {γ < κ : δγ1 ≤ pj} is infinite.

]

To see that (9) implies (8), assume (9), and let 〈pi : i < u〉 be any weak partitionof B. By the definition of B, there is a finite subset K of κ+ such that pi ∈〈C ∪ {b(α, j) : α ∈ K, j ≤ u}〉 for all i < u. Let F = K(u+ 1). Now we claim

(10) 〈C ∪ {b(α, j) : α ∈ K, j ≤ u}〉 ={x ∈ C : there is an d ∈ FC such that

x =∑h∈F

(d(h) ·

∏α∈K

b(α, h(α))

)}.

In fact, obviously the right side of (10) is a subset of the left, and the right sideis a subalgebra (using (7)). So it suffices to show that each b(α, j) is in the rightside. Let F ′ = {h ∈ F : h(α) = j}. If h ∈ F\F ′, then b(α, j) · b(α, h(α)) = 0; henceby (7), b(α, j) =

∑h∈F ′

∏α∈K b(α, h(α)). So if we define for h ∈ F

d(h) =

{1 if h ∈ F ′,0 otherwise,

we get b(α, j) =∑

h∈F

(d(h) ·

∏α∈K b(α, h(α))

). Thus (10) holds.

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By (10), for each i < u there is a di ∈ FC such that

pi =∑h∈F

(di(h) ·

∏α∈K

b(α, h(α))

).

Now for any h ∈ F let c0(h) = d0(h) +∏

j<u−dj(h), and for 0 < i < u letci(h) = di(h) ·

∏j<i−dj(h).

(11) For any i < u and h ∈ F we have pi ·∏

α∈K b(α,h(α))=ci(h)·∏

α∈K b(α,h(α)).

To prove (11), note that if h, h′ ∈ F with h �= h′, then∏α∈K

b(α, h(α)) ·∏α∈K

b(α, h′(α)) = 0,

and hencepi ·

∏α∈K

b(α, h(α)) = di(h) ·∏α∈K

b(α, h(α))

for any i < u. Hence if i, j < u and i �= j, then

0 = pi · pj ·∏α∈K

b(α, h(α)) = di(h) ·∏α∈K

b(α, h(α)) · dj(h) ·∏α∈K

b(α, h(α)).

Hence pi ·∏

α∈K b(α, h(α)) = di(h) ·∏

α∈K b(α, h(α)) ≤ di(h) ·∏

j �=i−dj(h); hence(11) holds for 0 < i. Clearly p0 ·

∏α∈K b(α, h(α)) ·

∏j<u−dj(h) = 0, so (11) holds

for i = 0 too.

By (11) and (7) we have

pi =∑h∈F

(ci(h) ·

∏α∈K

b(α, h(α))

),

and for each h ∈ F , the system 〈ci(h) : i < u〉 is a weak partition of unity of C.

Now for each i < u let Hi = {h ∈ F : (ci(h))(α) �= 1 for only finitely manyα < κ}. Note that if h ∈ F\Hi, then {α < κ : (ci(h))(α) �= 1} is infinite, andhence {α < κ : (ci(h))α) �= 0} is finite, since ci(h) ∈ C. Hence there is at least onei such that h ∈ Hi, as otherwise

∑i<u ci(h) �= 1. There is at most one, since the

ci(h)’s are pairwise disjoint. For each i < u let p′i =∑

h∈Hi

∏α∈K b(α, h(α)).

(12) 〈p′i : i < u〉 is a weak partition of B.

In fact, if i < j < u, h ∈ Hi, and k ∈ Hj , then h �= k by the remark preceding(12), and so

∏α∈K b(α, h(α))·

∏α∈K b(α, k(α)) = 0; hence p′i ·p′j = 0. Furthermore,

using (7), ∑i<u

p′i =∑i<u

∑h∈Hi

∏α∈K

b(α, h(α))

=∑h∈F

∏α∈K

b(α, h(α)) = 1.

So (12) holds.

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Now by (7), 〈∏

α∈K b(α, h(α)) : h ∈ F 〉 is a weak partition of B. It clearlyrefines 〈p′i : i < u〉. Hence by (9) we can choose j < u such that

∀J ∈ [(max(K), κ+)]<ω∀f ∈ Jκ

[⋂ε∈J

Aεf(ε) ∩ {γ < κ : δγ1 ≤ p′j} is infinite.

]

In particular, {γ < κ : δγ1 ≤ p′j} is infinite. For each h ∈ Hj let Fh = {γ < κ :(cj(h))(γ) �= 1}; so Fh is finite. Choose γ with δγ1 ≤ p′j and γ /∈

⋃h∈Hj

Fh. Thus

(cj(h))(γ) = 1 for all h ∈ Hj . It follows that

1 = p′j(γ) =∑h∈Hj

(∏α∈K

b(α, h(α))

)(γ)

=∑h∈Hj

((cj(h) ·

∏α∈K

b(α, h(α))

)(γ)

≤ pj(γ).

Hence δγ1 ≤ pj. This is as desired in (8).

Thus it remains only to prove (9), which we do by induction on |K|. Firstsuppose that K = {α}. Now the hypothesis of (9) says in this case that 〈b(α, i) :i ≤ u〉 refines 〈pi : i < u〉. Hence there is an i < u such that b(α, j) + b(α, k) ≤ pifor two distinct j, k ≤ u. Define ψ : 0(u + 1) → 2(u + 1) by setting ψ(0) = (j, k).Thus ψ ∈ Ψα; say χα

β = ψ. Take any γ ∈ Aαβ . Then

b(α, j) · δγ1 =∑{δγa(α,s) : s < 2, (χα

β(0))(s) = j}

=∑{δγa(α,s) : s < 2, (ψ(0))(s) = j}

= δγa(α,0),

and so δγa(α,0) ≤ b(α, j). Similarly, δγa(α,1) ≤ b(α, k). It follows that δγ1 = δγa(α,0) +

δγa(α,1) ≤ pi. This is true for any γ ∈ Aαβ . Thus if we take any finite J ⊆ (α, κ+)

and f ∈ Jκ, we have that Aαβ ∩⋂

ε∈J Aεf(ε) is infinite and δγ1 ≤ pi for everyγ ∈ Aαβ ∩

⋂ε∈J Aεf(ε), as desired in (9).

For the inductive step, suppose that |K| = s+1 and 〈pi : i < u〉 satisfies thehypothesis of (9). Let K = {β0, β1, . . . , βs−1, α} with β0 < β1 < · · · < βs−1 < α,and let L = {β0, β1, . . . , βs−1}. We now define subsets Hj of L(u+1) for j < u byrecursion:

Hj =

{h ∈ L(u+ 1) : h /∈

⋃k<j

Hk

and∣∣∣{i ≤ u :

∏β∈L

(b(β, h(β)) · b(α, i)) ≤ pj

}∣∣∣ ≥ 2

}.

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(13)⋃

j<u Hj =L(u + 1).

In fact, by the hypothesis of (9) we have

∀h ∈ L(u+ 1)∀i ≤ u∃j < u

⎛⎝∏

β∈L

(b(β, h(β)) · b(α, i)) ≤ pj

⎞⎠ ;

it follows that

∀h ∈ L(u+ 1)∃j < u

⎡⎣∣∣∣∣∣∣⎧⎨⎩i ≤ u :

∏β∈L

(b(β, h(β)) · b(α, i)) ≤ pj

⎫⎬⎭∣∣∣∣∣∣ ≥ 2

⎤⎦ ,

and (13) follows.

Now for each j < u let qj =∑

h∈Hj

∏β∈L b(β, h(β)). Clearly 〈qj : j < u〉 is

a weak partition of B and 〈∏

δ∈L b(δ, h(δ)) : h ∈ L(u+1)〉 refines it. Hence by the

inductive hypothesis there is a j0 < u such that Sdef= {γ < κ+ : δγ1 ≤ qj0} meets

every member of

{⋂ε∈J

Aεf(ε) : J ∈ [(βs−1, κ+)]<ω, f ∈ Jκ

}

in an infinite set. Now we define ψ : L(u+ 1)→ 2(u+ 1)\{(0, 0), (1, 1), . . . , (u, u)}by setting, for each h ∈ L(u + 1),

ψ(h) =

⎧⎪⎪⎨⎪⎪⎩some pair (j, k) such that j �= k and

[b(α, j) + b(α, k)] ·∏

β∈L b(β, h(β)) ≤ pj0 if∏

β∈L b(β, h(β)) ≤ qj0 ,

(0, 1) otherwise.

This makes sense by the definition of qj0 . Now there is a ϕ < κ such that χαϕ = ψ.

Now let J be a finite subset of (α, κ+), and f ∈ Jκ. By the definition of S,the set S ∩ Aαϕ ∩

⋂δ∈J Aδf(δ) is infinite. Hence to finish the induction it suf-

fices to take any γ ∈ S ∩ Aαϕ ∩⋂

δ∈J Aδf(δ) and show that δγ1 ≤ pj0 . Since∑h∈L(u+1)

∏β∈L b(β, h(β)) = 1, it suffices to take any h ∈ L(u + 1) such that

δγ1 ·∏

β∈L b(β, h(β)) �= 0 and show that this element is ≤ pj0 . Now since δγ1 ≤ qj0 ,it follows that

∏β∈L b(β, h(β)) ∩ qj0 �= ∅, hence

∏β∈L b(β, h(β)) ≤ qj0 , and hence

(14) [b(α, (ψ(h))(0)) + b(α, (ψ(h))(1))] ·∏β∈L

b(β, h(β)) ≤ pj0 .

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By the definition of the b’s we have

δγa(α,0) ·∏β∈L

b(β, h(β)) ≤ b(α, ((χαβ )(h))(0)) ·

∏β∈L

b(β, h(β))

= b(α, (ψ(h))(0)) ·∏β∈L

b(β, h(β))

≤ pj0 .

Similarly, δγa(α,1) ·∏

β∈L b(β, h(β)) ≤ pj0 . Hence δγ1 ·∏

β∈L b(β, h(β)) ≤ pj0 , as

desired.

Thus we have proved that ru(B) ≤ κ and ru+1(B) ≥ κ+. By Proposition6.18(iii) and Theorem 6.23, r2(B) = · · · = ru(B) = κ < κ+ = ru+1(B) =ru+2(B) = · · · = r2(B)+. �

We now give another important theorem from Balcar, Simon [92]. We call a setX ⊆ A+ finitely weakly dense iff for every finite weak partition P of A there existan x ∈ X and a p ∈ P such that x ≤ p.

Proposition 6.27. Let A be an infinite BA and X ⊆ A. Then X is finitely weaklydense in A iff X is dense in some ultrafilter on A.

Proof. ⇐: Suppose that F is any ultrafilter on A and X is dense in F . We claimthat X is finitely weakly dense. For, suppose that P is a finite weak partition.Then there is a p ∈ P such that p ∈ F , and then there is an x ∈ X such thatx ≤ p, as desired.

⇒: Suppose that X is finitely weakly dense. We will find an ultrafilter Fsuch that X is dense in F , thereby finishing the proof of the proposition. For eachfinite weak partition P let wP =

∑{p ∈ P : ∃x ∈ X+[x ≤ p]}. We claim that

{wP : P a finite weak partition of A} has the fip. For, suppose that P0, . . . , Pn−1

are finite weak partitions, n > 0. Let

Q =

{∏i<n

xi : x ∈∏i<n

Pi

}\{∅}.

Thus Q is a finite weak partition which refines each Pi. Choose x ∈ X+ andq ∈ Q such that x ≤ q. Clearly x ≤

∏i<n wPi . This proves the claim. Let F be

an ultrafilter containing {wP : P a finite weak partition of A}. To show that Xis dense in F , take any a ∈ F . Then wa,−a ∈ F , and so also a · wa,−a ∈ F . Bythe definition of wa,−a this means that there is an x ∈ X such that x ≤ a, asdesired. �Theorem 6.28. For any infinite BA A we have

supn∈ω,n≥2

rn(A) = min{|X | : X ⊆ A+, and X is finitely weakly dense in A}

= πχinf(A).

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Proof. Suppose that X is finitely weakly dense in A of smallest size. Then for anyn ∈ ω with n ≥ 2, the set X is n-dense in A. Hence rn(A) ≤ |X |.

Now for each n ∈ ω with n ≥ 2, let Xn be n dense with |Xn| = rn(A).Let Y =

⋃n∈ω,n≥2 Xn. Clearly Y is finitely weakly dense in A. Together with

the preceding paragraph, this proves the first equality of the theorem. The secondequality is immediate from Proposition 6.27. �Corollary 6.29. For any infinite BA A there is an m ∈ ω such that πχinf(A) =rm(A).

Proof. Immediate from Theorems 6.23 and 6.28. �

It is natural to consider arbitrary weak partitions as well as finite ones, in general-izing denseness. We call X ⊆ A infinitely weakly dense iff for every weak partitionY in A there exist x ∈ X+ and y ∈ Y such that x ≤ y.

Proposition 6.30.

min{|X | : X is infinitely weakly dense in A} = min{π(A � a) : a ∈ A+}.

Proof. Suppose thatX is infinitely weakly dense in A, of smallest size, and supposethat |X | < min{π(A � a) : a ∈ A+}. Then for all a ∈ A+, X ∩ (A � a) is not densein A � a; so there is a nonzero b ≤ a such that there is no x ∈ X such that0 �= x ≤ b. Thus the set {b ∈ A : there is no x ∈ X such that 0 �= x ≤ b} isdense in A. Let P be a weak partition with elements in this set. Choose b ∈ P andx ∈ X+ such that x ≤ b; this contradicts the choice of P . Hence ≥ holds in ourproposition.

Conversely, let κ = min{π(A � a) : a ∈ A+}, choose a ∈ A+ such thatκ = π(A � a), and choose X ⊆ (A � a) dense in A � a and of size κ. Suppose thatY is a weak partition of A. Choose y ∈ Y such that y · a �= 0, and choose x ∈ X+

with x ≤ y · a. This shows that X is infinitely weakly dense in A. �

We now consider algebraic operations and the reaping numbers rm. As with den-sity, one can have A ≤ B with rm(A) < rm(B) or rm(B) < rm(A). If A is aninfinite homomorphic image of B, then rm(A) < rm(B) is obviously possible. Theother inequality is possible also: for B = P(ω) and A = B/fin we have rm(B) = 1while rm(A) ≥ ω1, as we now prove.

Proposition 6.31. Let A = P(ω)/fin. Then r(A) ≥ ω1.

Proof. Let I = fin. Suppose that X ⊆ A+ is weakly dense and countable. SayX = {[x(i)]I : i < ω}. Now we define ai, bi ∈ ω for i < ω by recursion. Letf : ω → ω × ω be a bijection. Then pick

ai ∈ x(1st(f(i)))\({aj : j < i} ∪ {bj : j < i});bi ∈ x(1st(f(i)))\({aj : j ≤ i} ∪ {bj : j < i}).

Let y = {ai : i < ω}. Choose j < ω such that [x(j)]I ≤ [y]I or [x(j)]I ≤ −[y]I .

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Case 1. [x(j)]I ≤ [y]I . So x(j)\y is finite. But for any k ∈ ω, if i = f−1(j, k), thenbi ∈ x(j)\y, so x(j)\y is infinite, contradiction.

Case 2. [x(j)]I ≤ −[y]I . so x(j) ∩ y is finite, and we obtain a contradiction as inCase 1, using ai’s. �

One might want algebras A,B with B atomless, A a homomorphic image of B,and r(B) < r(A). We can modify the proof of Proposition 6.31 to obtain suchalgebras:

Proposition 6.32. Let B = ωFr(ω), and let

I = {b ∈ B : {i ∈ ω : b(i) �= 0} is finite}.

Let A = B/I. Then ω = r(B) < r(A).

Proof. Clearly r(B) = ω. Now suppose that X ⊆ A+ is weakly dense and count-able. Say X = {[x(i)]I : i < ω}. Now we define ai, bi ∈ ω for i < ω by recursion.Let f : ω → ω × ω be a bijection. Now x(1st(f(i))) /∈ I, and so we can chooseai ∈ ω\({aj : j < i} ∪ {bj : j < i}) so that (x(1st(f(i))))(ai) �= 0. Then choosebi ∈ ω\({aj : j ≤ i} ∪ {bj : j < i}) so that (x(1st(f(i))))(bi) �= 0. Now we definey ∈ B by

y(j) =

{1 if j = ai for some i,

0 otherwise.

Choose j < ω such that [x(j)]I ≤ [y]I or [x(j)]I ≤ −[y]I .Case 1. [x(j)]I ≤ [y]I . Thus x(j) ·−y ∈ I. But for any k ∈ ω, if i = f−1(j, k), then(x(j) · −y)(bi) �= 0, so x(j) · −y /∈ I, contradiction.

Case 2. [x(j)]I ≤ −[y]I . Similarly, using ai’s. �Proposition 6.33. For any m ∈ ω\2 and any system 〈Ai : i ∈ I〉 of infinite BAswe have rm(

∏i∈I Ai) = mini∈I rm(Ai).

Proof. Let mini∈I rm(Ai) = rm(Aj), and let X be m-dense in Aj of size rm(Aj).Define xa ∈

∏i∈I Ai for a ∈ X by setting

xa(i) =

{a if i = j,

0 otherwise.

Suppose that 〈bk : k < m〉 is a weak partition of∏

i∈I Ai. Then 〈bk(j) : k < m〉is a weak partition of Aj . Choose a ∈ X+ and k < m such that a ≤ bk(j). Thenxa ≤ bk. Thus {xa : a ∈ X} is m-dense in

∏i∈I Ai, proving ≤.

Now suppose thatX ⊆∏

i∈I Ai, |X | < rm(Aj), and X is m-dense in∏

i∈I Ai.Then for each i ∈ I we have |{a(i) : a ∈ X}| < rm(Ai), and so {a(i) : a ∈ X} isnot m-dense. So there is a weak m-partition 〈bik : k < m〉 of Ai such that there donot exist a ∈ X and k < m such that 0 �= a(i) ≤ bik. Define ck = 〈bik : i ∈ I〉 foreach k < m. Clearly 〈ck : k < m〉 is a weak partition of

∏i∈I Ai Choose k < m and

a ∈ X+ such that a ≤ ck. Say a(i) �= 0. Then 0 �= a(i) ≤ bik, contradiction. �

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Proposition 6.34. If 〈Ai : i ∈ I〉 is a system of infinite BAs, with I infinite, then

πχinf

(∏i∈I

Ai

)≤ min

i∈Iπχinf(Ai) ≤

(πχinf

(∏i∈I

Ai

))+

.

Proof. For all n ∈ ω\2 we have rn(∏

i∈I Ai) = mini∈I rn(Ai) ≤ mini∈I πχinf(Ai),and so the first inequality follows.

Next, choose m ∈ ω\2 such that πχinf(∏

i∈I Ai) = rm(∏

i∈I Ai), and thenchoose i0 so that rm(

∏i∈I Ai) = rm(Ai0). Thus

(1) rm(Ai0) = πχinf(∏

i∈I Ai).

Then

mini∈I

πχinf(Ai) ≤ πχinf(Ai0 ) ≤ (rm(Ai0 ))+ =

(πχinf

(∏i∈I

Ai

))+. �

According to Proposition 6.34, mini∈I πχinf(Ai) is either equal to πχinf

(∏i∈I Ai

)or to

(πχinf

(∏i∈I Ai

))+. Both cases are possible. Thus if all Ai are equal to some

atomless algebra B, then πχinf

(∏i∈I Ai

)= mini∈I πχinf(Ai). This is true since

by Corollary 6.29 there is an m ≥ 2 such that πχinf(B) = rm(B), by Proposition6.33 rm(

∏i∈I Ai) = rm(B), and by Proposition 6.34 πχinf(

∏i∈I Ai) ≤ πχinf(B) =

rm(B) = rm(∏

i∈I Ai) ≤ πχinf(∏

i∈I Ai). On the other hand, for each n ∈ ω\2 letAn be a BA with rn(An) = ω while rn+1(An) = ω1. Then minn∈ω\2 πχinf(An) =(πχinf

(∏n∈ω\2 An

))+. In fact, for any n ∈ ω\2 we have πχinf(An) ≥ rn+1(An) =

ω1, so that minn∈ω\2 πχinf(An) = ω1. Also, there is an m ∈ ω\2 such that

πχinf

(∏n∈ω\2 An

)= rm

(∏n∈ω\2An

),

and

rm

(∏n∈ω\2 An

)= minn∈ω\2 rm(An) = rm(Am) = ω.

Proposition 6.35. If 〈Ai : i ∈ I〉 is a system of atomless BAs, with I infinite, thenπχinf(

∏wi∈I Ai) = ω.

Proof. Let J be a countably infinite subset of I, and for each j ∈ J let aj bethe element of

∏wi∈I Ai) which is 1 at the jth position and 0 otherwise, and let

X = {aj : j ∈ J}. Let F be the ultrafilter on∏w

i∈I Ai) consisting of all elementsx such that {k ∈ I : x(k) �= 1} is finite. Clearly X is dense in F . �

Kevin Selker has shown that Proposition 6.35 also holds for moderate products.

Peterson [98] has determined how reaping numbers act under free products; westate these results without proof. πχinf(A ⊕ B) = max(πχinf(A), πχinf(B)). Ifπχinf(A) = πχinf(B), then πχinf(A⊕B) = r2(A⊕B).

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In the same article, Peterson also investigates derived functions of the reapingnumbers.

We now consider another variation of the reaping number. A subset X of aBA A is homogeneously weakly dense iff for any two distinct a, b ∈ A with a · b = 0there is an x ∈ X such that (x · a �= 0 = x · b) or (x · a = 0 �= x · b). Note thatneither a nor b is required to be nonzero. We define

hwd(A) = min{|X | : X ⊆ A is homogeneously weakly dense in A}.

We also define rel(X, a) = {x ∈ X : x · a �= 0}, the relevant points of X for a.

Proposition 6.36. hwd(A) is infinite, for every infinite BA A.

Proof. Suppose that X is finite and homogeneously weakly dense in A. Let B =〈X〉. Then B is finite. Let a be an atom of B such that A � a is infinite. Let b andc be nonzero disjoint elements of A � a. Choose x ∈ X so that x · b �= 0 = x · c orx · c �= 0 = x · b. This contradicts a being an atom of B. �

Proposition 6.37. For any infinite BA A and any X ⊆ A, X is weakly dense in Aiff ∀a ∈ A[rel(X, a) �= rel(X,−a)].

Proof. First suppose that X is weakly dense in A. Take any a ∈ A. Choose x ∈ X+

such that x ≤ a or x ≤ −a. Then x ∈ rel(X, a)�rel(X,−a).Conversely assume the condition of the proposition, and suppose that a ∈ A.

Choose x ∈ rel(X, a)�rel(X,−a). Then x ∈ X+, and x ≤ a or x ≤ −a. �

Proposition 6.38. For any infinite BA A and any X ⊆ A the following conditionsare equivalent:

(i) X is homogeneously weakly dense in A.

(ii) X � a is weakly dense in A � a for every a ∈ A+.

(iii) For every a ∈ A+ and every b ≤ a there is an x ∈ X such that x · a �= 0, andx · a ≤ b or x · a ≤ −b.

(iv) rel(X, a) �= rel(X, b) for all disjoint a, b ∈ A with a+ b �= 0.

Proof. (i)⇒(ii): Assume that X is homogeneously weakly dense in A, and a ∈ A+.Take any b ≤ a. If b = a, we apply the definition of homogeneous weak density tothe pair 0, a and get x ∈ X such that x · 0 �= 0 = x · a or x · 0 = 0 �= x · a. Thusx · a �= 0. So x · a is a nonzero element of X � a which is below b, since b = a. Ifb �= a, choose x ∈ X such that x · b �= 0 = x · a · −b or x · b = 0 �= x · a · −b. Thenx · a �= 0, and x · a ≤ b or x · a ≤ a · −b = −A�ab.

(ii)⇒(iii): Assume (ii), and suppose that a ∈ A+ and b ≤ a. Choose x ∈ Xsuch that x · a �= 0, and x · a ≤ b or x · a ≤ −A�ab. thus x · a ≤ b or x · a ≤ −b.

(iii)⇒(iv): Assume (iii), and suppose that a · b = 0 with a + b �= 0. Choosex ∈ X such that x · (a+ b) �= 0 and x · (a+ b) ≤ a or x · (a+ b) ≤ −a. So (x · b = 0and x · a �= 0) or (x · a = 0 and x · b �= 0). Thus x ∈ rel(X, a)�rel(X, b).

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(iv)⇒(i): Assume (iv), and suppose that a �= b with a · b = 0. Thus a+ b �= 0.By (iv) we have rel(X, a) �= rel(X, b). By symmetry say that x∈rel(X,a)\rel(X,b).Thus x · a �= 0 = x · b. �

It would be natural to generalize hwd like ordinary reaping was generalized to thenumbers rn. By the following result, nothing new is obtained in the case of hwd.Also, the value of hwd is not changed if we require the elements to be nonzero (aresult of Jennifer Brown [05]).

Proposition 6.39. Suppose that n ∈ ω\2, A is an infinite BA, and X ⊆ A. Letϕ(n,X) and ψ(n,X) be the following two statements:

(1) ϕ(n,X) : For every system 〈ai : i < n〉 of pairwise disjoint elements withat least one nonzero, there exist i < n and x ∈ X such that x · ai �= 0 and∀j �= i[x · aj = 0].

(2) ψ(n,X) : For every system 〈ai : i < n〉 of nonzero pairwise disjoint elements,there exist i < n and x ∈ X such that x · ai �= 0 and ∀j �= i[x · aj = 0].

Then for each integer n ≥ 2, hwd(A) = min{|X | : ϕ(n,X)} = min{|X | : ψ(n,X)}.

Proof. Let X be homogeneously weakly dense. So X is infinite, by Proposition6.36. Let Y be the closure of X under finite products. We prove ϕ(n, Y ) by induc-tion on n. The case n = 2 is given. Now suppose that the condition holds for n,and 〈ai : i ≤ n〉 is a system of pairwise disjoint elements with at least one nonzero.let b =

∑i<n ai. Since X is homogeneously weakly dense, choose x ∈ X such that

one of the following conditions holds: (1) x · an �= 0 = x · b. (2) x · an = 0 �= x · b.Now (1) gives the desired conclusion. So suppose that (2) holds.

Then 〈x · ai : i < n〉 is a system of pairwise disjoint elements, with at leastone nonzero. So by the inductive hypothesis choose y ∈ X and i < n such thaty · x · ai �= 0 = y · x · aj for all j < n. Clearly this is as desired. This proves thathwd(A) ≥ min{|X | : ϕ(n,X)}.

Clearly ϕ(n,X) implies ψ(n,X). Hence min{|X | : ϕ(n,X)} ≥ min{|X | :ψ(n,X)}.

Now we prove that ψ(n,X) implies that Y is homogeneously weakly densefor some Y with |Y | = |X |, by induction on n.

First we note that X is infinite, by the proof of Proposition 6.36. Supposethat n = 2. Set Y = X ∪ {1}. Let a and b be different disjoint elements of A. Ifboth are nonzero, then the desired conclusion is clear. If one is 0, the element 1 ofY can be used.

Now suppose that ϕ(n−1, X) implies that Y is homogeneously weakly densefor some Y with |Y | = |X |, and assume ϕ(n,X). Suppose that a0, . . . , an−2 aren− 1 nonzero pairwise disjoint elements of A.

Case 1. There is an i < n − 1 such that A � ai has more than one element.Write ai = b + c with b and c nonzero and disjoint. Applying ϕ(n,X) to the

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sequence a0, . . . , ai−1, b, c, ai+1, . . . , an−2, we get an element which clearly worksfor a0, . . . , an−2 as well.

Case 2. Suppose that each ai is an atom for i < n− 1. Suppose

¬∃i < n− 1∃x ∈ X [x · ai �= 0 ∧ ∀j �= i[x · aj = 0]].

Thus ϕ(n− 1, X) fails to hold. Now we claim

(1) Either ϕ(n− 1, Y ) holds for some Y with |Y | = |X |, or there are n− 1 atomsb0, . . . , bn−2 such that {a0, . . . , an−2} ∩ {b0, . . . , bn−2} = ∅ and

¬∃i < n− 1∃x ∈ X [x · bi �= 0 ∧ ∀j �= i[x · bj = 0]].

In fact, suppose that such atoms do not exist. Let X ′ = X ∪ {a0, . . . , an−1}.Suppose now that b0, . . . , bn−2 are n− 1 nonzero pairwise disjoint elements of A.If one of them is not an atom, then a desired element is obtained as in Case 1.Suppose that all of them are atoms. If {a0, . . . , an−2} ∩ {b0, . . . , bn−2} = ∅, thenthe desired conclusion follows from the assumption concerning (1). Suppose thatbi = aj . Then bi ∈ X ′ and it is a desired element for b0, . . . , bn−2. Thus ϕ(n−1, X ′)holds, as desired.

Thus (1) holds.

Assume that ϕ(n−1, Y ) does not hold, for any Y such that |Y | = |X |. Chooseb0, . . . , bn−2 by (1). Consider the n nonzero pairwise disjoint elements c0 = a0+b0,c1 = a1 + b1, . . . , cn−3 = an−3 + bn−3, cn−2 = an−2, cn−1 = bn−2. Choose i < nand x ∈ X such that x · ci �= 0 while x · cj = 0 for all j �= i. If i = n − 2, thenx · an−2 �= 0 while x · aj = 0 for all j �= n − 2, contradicting our assumptionon the ak’s. A similar contradiction is reached if i = n − 1. So i < n − 2. Thusx · (ai + bi) �= 0, while x · cj = 0 for all j �= i. Hence x · aj = 0 for all j �= i,and x · bj = 0 for all j �= i. So our assumption on the ak’s, or on the bk’s, iscontradicted.

What this means is that the initial assumption of this case is false, and thiscompletes the proof. �

Corollary 6.40. πχinf(A) ≤ hwd(A) for every infinite BA A.

Proof. Let n ∈ ω\2. By Proposition 6.39, let X ⊆ A be of size hwd(A) which is n-dense. Thus rn(A) ≤ hwd(A). Hence Proposition 6.28 gives the desired result. �

Proposition 6.41. Depth(A) ≤ hwd(A) for every infinite BA A.

Proof. Suppose that X ⊆ A with |X | < Depth(A); we show that X is not homo-geneously weakly dense. Let 〈aα : α < |X |+〉 be a strictly increasing sequence ofelements of A. Let

Y = {x ∈ X : ∀α < |X |+∃β ∈ (α, |X |+)[x · aβ · −aα �= 0]},

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and let Z = X\Y . Then for each x ∈ Z there is an αx < |X |+ such that ∀β ∈(αx, |X |+)[x · aβ · −aαx = 0]. Let β0 = supx∈Z αx. Then

(1) For all x ∈ Z, if β0 ≤ β < γ < |X |+, then x · aγ · −aβ = 0.

In fact, x · aγ · −aβ ≤ x · aγ · −aαx = 0.

(2) ∀α < |X |+∃β ∈ (α, |X |+)∀x ∈ Y [x · aβ · −aα �= 0].

In fact, for any x ∈ Y choose γx ∈ (α, |X |+) such that x · aγx · −aα �= 0. Letβ = supx∈Y γx. Clearly (2) holds for this β.

Now apply (2) with α replaced by β0 to obtain β1 ∈ (β0, |X |+) such that∀x ∈ Y [x · aβ1 · −aβ0 �= 0]; then apply (2) with α replaced by β1 to obtainβ2 ∈ (β1, |X |+) such that ∀x ∈ Y [x · aβ2 · −aβ1 �= 0]. Let c = aβ1 · −aβ0 andd = aβ2 ·−aβ1 . Then c ·d = 0, and c �= 0 �= d. Then by (1), ∀x ∈ Z[c ·x = 0 = d ·x],and by the choice of β1 and β2, ∀x ∈ Y [c · x �= 0 �= d · x]. Hence X is nothomogeneously weakly dense. �Proposition 6.42. If κ is regular and there is an isomorphic embedding of P(κ)into A preserving all sums, then κ ≤ hwd(A).

Proof. Suppose that X ∈ [A]<κ; we want to show that X is not homogeneouslyweakly dense. Let f be the assumed isomorphic embedding. For each x ∈ X letDx = {α < κ : f({α}) · x �= 0}. Now define

Y = {x ∈ X : |Dx| < κ}; Z = X\Y ; J =⋃x∈Y

Dx.

Thus |J | < κ and ∀x ∈ Z[|Dx| = κ]. Write Z = {zγ : γ < δ}, where δ < κ. Nowwe can choose by induction, for each γ < δ,

α0γ ∈ Dzγ\(J ∪ {α1

β : β < γ}),α1γ ∈ Dzγ\(J ∪ {α0

β : β ≤ γ}).

Now let I0 = {α0γ : γ < δ} and I1 = {α1

γ : γ < δ}. Then I0∩I1 = ∅ = I0∩J = I1∩J .Let a0 = f(I0) and a1 = f(I1). Since I0 ∩ I1 = ∅, we have a0 · a1 = 0. If x ∈ Y ,then Dx ⊆ J , hence I0 ∩Dx = ∅, so I0 ⊆ {α < κ : f({α}) · x = 0}. Hence

x · a0 = x · f(I0)≤ x · f({α < κ : f({α}) · x = 0})

= x · f(⋃

{{α} : α < κ, f({α}) · x = 0})

= x ·∑{f({α}) : α < κ, f({α}) · x = 0}

=∑{x · f({α}) : α < κ, f({α}) · x = 0}

= 0.

Similarly, x · a1 = 0.

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On the other hand, if x ∈ Z, then x = zγ for some γ < δ, and so α0γ ∈ I0∩Dx.

Hence 0 �= f({α0γ}) · x ≤ f(I0) · x = a0 · x. So a0 · x �= 0, and similarly a1 · x �= 0.

This proves that X is not homogeneously weakly dense. �

Proposition 6.43. hwd(A×B) = max(hwd(A), hwd(B)) for any infinite BAs A,B.

Proof. Suppose that X ⊆ A× B with |X | < hwd(A). Then {1st(x) : x ∈ X} is asubset of A of size less than hwd(A), so there exist disjoint a0, a1 ∈ A with a0 �= a1such that ∀x ∈ X [1st(x) · a0 = 0 iff 1st(x) · a1 = 0]. Hence (a0, 0) and (a1, 0) aredistinct disjoint elements of A×B such that ∀x ∈ X [x · a0 = 0 iff x · a1 = 0]. Thisproves that hwd(A) ≤ hwd(A×B). Similarly for B, so ≥ in the proposition holds.

Now let X ⊆ A be homogeneously weakly dense in A, with |X | = hwd(A),and let Y ⊆ B be homogeneously weakly dense in B, with |Y | = hwd(B). LetZ = {(x, 0) : x ∈ X}∪{(0, y) : y ∈ Y }. For ≤ of the proposition it suffices to showthat Z is homogeneously weakly dense in A×B. So, suppose that (a0, b0), (a1, b1)are distinct disjoint elements of A×B.

Case 1. a0 �= 0. Then a0, a1 are distinct and disjoint, so we can choose x ∈ X suchthat x · a0 �= 0 = x · a1 or x · a0 = 0 �= x · a1. Hence (x, 0) · (a0, b0) �= (0, 0) =(x, 0) · (a1, b1) or (x, 0) · (a0, b0) = (0, 0) �= (x, 0) · (a1, b1), as desired.Case 2. a1 �= 0. Symmetric to Case 1.

Case 3. a0 = 0 = a1. Then b0 �= 0 or b1 �= 0, and the arguments of Cases 1, 2apply. �

Proposition 6.44. Suppose that I is infinite and 〈Ai : i ∈ I〉 is a system of infiniteBAs. Then hwd(

∏i∈I Ai) = max{|I|, supi∈I hwd(Ai)}.

Proof. hwd(∏

i∈IAi)≥ supi∈Ihwd(Ai) by Proposition 6.43. Also, hwd(∏

i∈IAi)≥|I| by Proposition 6.42. Thus ≥ holds.

For the other direction, for each i ∈ I let Xi ⊆ Ai be homogeneously weaklydense in Ai, with |Xi| = hwd(Ai). For each i ∈ I and x ∈ Xi define yxi ∈∏

j∈I Aj by

yxi (j) =

{x if i = j,

0 otherwise.

Then Ydef= {yxi : i ∈ I, x ∈ Xi} has size

∑i∈I |Xi| = max{|I|, supi∈I hwd(Ai)},

and it is homogeneously weakly dense. In fact, if u, v are distinct disjoint elementsof∏

i∈I Ai, say by symmetry that i ∈ I with ui �= 0. Then ui and vi are distinctdisjoint elements of Ai, so there is an x ∈ Xi such that ui · x �= 0 = vi · x orui · x = 0 �= vi · x. Then u · yxi �= 0 = v · yxi or u · yxi = 0 �= v · yxi . �

We state some further results from Peterson [98] without proof.

• If 〈Ai : i ∈ I〉 is a system of infinite BAs, with I infinite, then hwd(∏w

i∈I Ai) =max{|I|, supi∈I hwd(Ai)}.

• If A and B are infinite BAs, then hwd(A⊕B) = max{hwd(A), hwd(B)}.

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• If 〈Ai : i ∈ I〉 is a system of infinite BAs, with I infinite, then hwd(⊕i∈IAi) =max{|I|, supi∈I hwd(Ai)}.

We now consider density and the above related notions for complete BAs.

Proposition 6.45. Let A be a an infinite complete BA, and X a subset of A closedunder products. Then the following conditions are equivalent:

(i) X is dense in some ultrafilter on A.

(ii) X is weakly dense in A.

Proof. (i)⇒(ii) is obvious; it is also part of Proposition 6.27.

For (ii)→(i), suppose that X is not dense in any ultrafilter; we show that Xis not weakly dense. By Proposition 6.27, X is not finitely weakly dense. Hencethere is an m ∈ ω\2 and a weak partition 〈ai : i < m〉 such that no member ofX+ is below any ai. For each x ∈ X+ let Ix = {i < m : ai · x �= 0}. Thus each setIx has at least two elements, and Iy ≤ Ix if y ≤ x. Let

Y = {y ∈ X+ : ∀z ∈ X+[z ≤ y → [Iz = Iy ]]}.

Now Y is dense in X . For, suppose that x ∈ X+. Let y ∈ X+ be such that y ≤ x,and |Iy| is smallest among all |Iz| with z ∈ X+ and z ≤ x. Then y ∈ Y , as desired.

Let D be a maximal disjoint set contained in Y . Then

(∗)∑

D =∑

X .

In fact, D ⊆ X , so ≤ holds. Suppose that < holds. Then∑

x∈X (x · −∑

D) =∑X · −

∑D �= 0, so there is an x ∈ X such that x · −

∑D �= 0. Choose y ∈ Y +

such that y ≤ x. Then y · d = 0 for all d ∈ D, contradicting the maximality of D.Hence (∗) holds.

Now for each d ∈ D let ϕ(d) < m be smallest such that d · aϕ(d) �= 0. Letu =

∑d∈D(d · aϕ(d)). We claim that there does not exist an x ∈ X+ such that

x ≤ u or x ≤ −u, thereby showing that X is not weakly dense.

So, take any x ∈ X+. By (∗), choose d ∈ D such that x · d �= 0. Now d ∈ Y ,so Ix·d = Id. Since d ·aϕ(d) �= 0, it follows that x ·d ·aϕ(d) �= 0, and so x ·u �= 0. Butd is not below aϕ(d), so there is an i < m with i �= ϕ(d) such that d ·ai �= 0. Hencefrom Ix·d = Id we get x · d · ai �= 0. Now the disjointness of D and of 〈aj : j < m〉implies that d · ai · u = 0. Hence x · −u �= 0. �Theorem 6.46. If A is an infinite complete BA, then r(A) = πχinf(A).

Proof. Let X be weakly dense in A, with |X | = r(A). Let Y be the closure ofX under products. By Proposition 6.45, Y is dense in some ultrafilter on A, soby Proposition 6.27, Y is finitely weakly dense in A. Then by Theorem 6.28,πχinf(A) ≤ |Y |. Thus our theorem follows from Theorem 6.28. �

We now work towards a result of Bozeman [91] which relates hwd to π.

A subset X of a BA A is nowhere relatively dense in A iff for all a ∈ A+,X � a is not dense in A � a.

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Proposition 6.47. X is nowhere relatively dense in B iff for all a ∈ B+ there is anonzero d ≤ a such that rel(X, d) ⊆ rel(X, a · −d).

Proof. ⇒: Given a ∈ B+, we know that X � a is not dense in B � a. Hence thereis a nonzero d ≤ a such that for all x ∈ X , if x ·a �= 0 then x ·a �≤ d. Suppose thatx ∈ rel(X, d). Thus x · d �= 0. Hence x · a �= 0, so x · a �≤ d, hence x · a · −d �= 0,and x ∈ rel(X, a · −d).

⇐. Let a ∈ B+. Choose d as indicated. We claim that X � a is not dense inB � a, in fact, ∀x ∈ X(x · a �= 0⇒ x · a �≤ d). For, suppose that x ∈ X , x · a �= 0,while x · a ≤ d. Thus x ∈ rel(X, d)\rel(X, a · −d), contradiction. �Proposition 6.48. Suppose that B is complete and X is weakly dense in B. Thenthere is a c ∈ B+ such that X � c is homogeneously weakly dense in B � c.

Proof. Suppose, on the contrary, that ∀c ∈ B+(X � c is not homogeneously weakly

dense in B � c). Hence by Proposition 6.38, Ydef= {a ∈ B+ : X � a is not weakly

dense in B � a} is dense in B. Let A be a maximal disjoint set ⊆ Y . Thus∑

A = 1.For each a ∈ A we haveX � a not weakly dense in B � a. Thus by Proposition 6.37,for every a ∈ A there is a ba ≤ a such that rel(X � a, ba) = rel(X � a,−B�aba).Hence

(∗) rel(X, ba) = rel(X, a · −ba).

In fact, let x ∈ X . Then

x ∈ rel(x, ba) iff x · ba �= 0

iff x · a · ba �= 0

iff x · a ∈ rel(X � a, ba)iff x · a ∈ rel(X � a,−B�aba)

iff x · a · −B�aba �= 0

iff x · a · −ba �= 0

iff x ∈ rel(X, a · −ba).

It follows that for any x ∈ X ,

x ·∑a∈A

ba �= 0 iff ∃a ∈ A[x · ba �= 0]

iff ∃a ∈ A[x · a · −ba �= 0]

iff x ·∑a∈A

(a · −ba) �= 0.

But∑

a∈A ba +∑

a∈A(a · −ba) =∑

A = 1 and(∑

a∈A ba)·(∑

a∈A(a · −ba))= 0,

so that∑

a∈A ba = −∑

a∈A(a ·−ba). This contradicts X being weakly dense in B.�

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A BA B is hwd-homogeneous iff hwd(B � a) = hwd(B) for every a ∈ B+; cf. theHandbook, page 198.

Proposition 6.49.

(i) If a ∈ B, then hwd(B � a) ≤ hwd(B).

(ii) {a ∈ B : B � a is hwd-homogeneous} is dense in B.

(iii) Any complete BA is isomorphic to a product of complete hwd-homogeneousBAs.

Proof. By Proposition 6.43. �Proposition 6.50. If B is complete and hwd-homogeneous, then hwd(B) = r(B).

Proof. Let X be weakly dense in B with |X | = r(B). By Proposition 6.48 choosec ∈ B+ such that X � c is homogeneously weakly dense in B � c. Then hwd(B) =hwd(B � c) ≤ |X � c| ≤ |X | = r(B) ≤ hwd(B) using Proposition 6,38, and sohwd(B) = r(B). �Theorem 6.51. Let B be a complete hwd-homogeneous BA. Suppose that X ={xτ : τ < κ} is homogeneously weakly dense in B, with each xτ nonzero, and|X | = hwd(B) = κ. For all ρ < κ let Yρ = {xτ : τ ≤ ρ}. We call W ⊆ X boundedif W is included in some Yρ. Let E = {

∑W : W is a bounded subset of X}. Let

D be the subalgebra of B generated by E.

The conclusion is that there is an a ∈ B+ such that D � a is dense in B � a.

Proof. Suppose not. Thus D is nowhere relatively dense in B. Now, we claim,

(1) For all ρ < κ and all a ∈ B+ there exist c, A, f such that c is a nonzero element≤ a, A is a disjoint subset of B � (a · −c), f : A→ κ, and the following conditionshold:

(i) sup f [rel(A, p)] = κ for all p ∈ rel(X, c).

(ii) rel(Yρ, c) = rel(Yρ,∑

A).

(iii)∑

A = a · −c.

To prove (1), fix ρ < κ and a ∈ B+. By Proposition 6.47 choose a nonzero s ≤ asuch that rel(D, s) ⊆ rel(D, a · −s). Let U = {x ∈ X : x · s = 0}, and letu = a ·

∑U . Now s ·

∑U = 0, so s ≤ a · −u. If x ∈ U , then x · a ≤ a ·

∑U = u,

hence x · a · −u · −s = 0 and x /∈ rel(X, a · −u · −s). Thus

(2) rel(X,−u · a · −s) ⊆ rel(X, s).

Now by the homogeneous weak density of X , X � (a · −u) is weakly dense inB � (a · −u). Hence there is a q ∈ X such that q · a · −u �= 0 and q · a · −u ≤ s orq ·a ·−u ≤ a ·−u ·−s. But the latter is impossible, since otherwise q ∈ rel(X, s) by(2), hence q ·s �= 0; since s ≤ a·−u, this implies that a·−u·−s ≥ q ·a·−u ≥ q ·s �= 0,which is impossible. Thus

(3) q · a · −u ≤ s.

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Now we make some definitions. Let c = q · a · −u, T = {x ∈ X : x · c = 0},t = a ·

∑T . For all α < κ let Tα = Yα ∩T and tα = a ·

∑Tα. Since Tα is bounded,

we have tα ∈ (D � a). Note that U ⊆ T since c ≤ s, and hence u ≤ t, so

(4) a · −t ≤ a · −u.

Define

t′α =

{tα · −

∑τ<α tτ if tα · −

∑τ<α tτ �= 0,

0 otherwiseA = {t′α : α < κ}

f(t′α) ={α if t′α �= 0,

0 otherwise.

Thus we have defined some objects c, A, and f , and we now show that they satisfy(i) and (ii) of (1). Clearly 0 �= c ≤ a, A is a disjoint subset of B � (a · −c), andf : A→ κ.

We prove (1)(i) by contradiction. Suppose that p ∈ rel(X, c) and sup f [rel(A, p)] <κ. Thus p·c �= 0 and there is an α < κ such that for all β > α, p·tβ ·−

∑τ<β tτ = 0.

Then

(5) p · tβ ≤ tα for all β < κ.

We prove (5) by induction on β. Since clearly tβ ≤ tα when β ≤ α, we only needto consider β > α. Assume that p · tγ ≤ tα for all γ < β, where α < β < κ. Thenp · tβ ≤

∑τ<β(p · tτ ) ≤ tα, as desired. So (5) holds. Hence

(6) p · t = p · tα.

Now, using (3) and (4),

p · q · a = p · q · a ·∑

T + p · q · a · −∑

T

= p · q · t+ p · q · a · −t≤ p · q · tα + p · q · a · −u≤ p · q · tα + s.

Let w = p ·q ·a ·−tα. Since p, q ∈ D and tα ∈ (D � a), we have w ∈ (D � a). By thedisplayed calculation above we have w ≤ s. Now t ·c = 0, and 0 �= p ·c = p ·q ·a ·−u,so p · q · a �≤ t. Hence, using (6), p · q · a · −t = p · q · a · −tα �= 0, i.e., w �= 0. Sop · q · −tα ∈ rel(D, s) but p · q · −tα /∈ rel(D, a · −s), contradicting the choice of s.This proves (1)(i).

For (1)(ii), first suppose that τ ≤ ρ and xτ ∈ rel(Yρ, c). Then xτ ∈ rel(X, c),so by (1)(i) we get xτ ·

∑A �= 0, and hence xτ ∈ rel(Yρ,

∑A). Conversely, suppose

that π ≤ ρ and xπ ∈ rel(Yρ,∑

A). Then by the definition of A there is an α > ρsuch that xπ ·tα ·−

∑τ<α tτ �= 0. Now xπ /∈ Tπ, since otherwise xπ ·tα ≤ a ·

∑Tπ =

tπ, a contradiction since π < α. Since Yπ ∩ T = Tπ and xπ ∈ Yπ, it follows thatxπ /∈ T , hence xπ · c �= 0, i.e., xπ ∈ rel(Yρ, c). So (1)(ii) holds.

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Thus we have constructed c, A, and f so that (1)(i) and (1)(ii) hold. We willobtain objects satisfying all three conditions (1)(i)–(iii) by applying Zorn’s lemma.Let P consist of all triples (c, A, f) such that (1)(i) and (1)(ii) hold. Partiallyorder P by setting (c, A, f) ≤ (c′, A′, f ′) iff c ≤ c′ and ∀t ∈ A∃t′ ∈ A′[t ≤ t′ andf(t) = f ′(t′)]. Suppose that Q is a chain in P. Let c′ =

∑(c,A,f)∈Q c. For each

α < κ let hα =∑

(c,A,f)∈Q

∑f−1[{α}]. If α and β are distinct ordinals less than

κ, then hα · hβ = 0: for assume that (c, A, f) ∈ Q and (c1, A1, f1) ∈ Q; we checkthat

∑f−1[{α}] ·

∑f−11 [{β}] = 0. To do this, suppose that x ∈ f−1[{α}] and

y ∈ f−11 [{β}]; we check that x · y = 0. By symmetry say (c, A, f) ≤ (c1, A1, f1).

Choose z ∈ A1 so that x ≤ z and f(x) = f1(z). Since f(x) = α and f1(y) = β we

have z �= y, and hence z · y = 0. So x · y = 0, as desired. Hence A′ def= {hα : α < κ}

is disjoint. Define f ′ : A′ → κ by

f ′(hα) =

{α if h(α) �= 0,

0 if h(α) = 0.

We check (1)(i) for (c′, A′, f ′): Suppose that p ∈ rel(X, c′). Choose (c, A, f) ∈ Qso that p ∈ rel(X, c). Thus by (1)(i) for (c, A, f) we have sup f [rel(A, p)] = κ. Weclaim that sup f ′[rel(A′, p)] = κ. For, given α < κ, choose t ∈ rel(A, p) such thatα < f(t). Let β = f(t). Thus t ∈ f−1[{β}], and hence t ≤ hβ . So hβ ∈ rel(A′, p),and f ′(hβ) = β, as desired.

Next we check (1)(ii) for (c′, A′, f ′): Suppose that τ ≤ ρ and xτ ∈ rel(Yρ, c′).

Choose (c, A, f) ∈ Q so that xτ ∈ rel(Yρ, c). By (1)(ii) for (c, A, f) we have xτ ∈rel(Yρ,

∑A). So there is a t ∈ A such that xτ · t �= 0. Say f(t) = α. Then t ≤ hα,

so xτ ∈ rel(Yρ,∑

A′). The converse is similar, so we have proved (1)(ii).

Clearly (c, A, f) ≤ (c′, A′, f ′) for all (c, A, f) ∈ Q. So at this point we havechecked the hypotheses of Zorn’s lemma; the first part of the proof of (1) was toshow that P �= 0. By Zorn’s lemma, let (c, A, f) be a maximal element of P . Wewill now show that (1)(iii) holds, completing the proof of (1). In fact, if it does not

hold, then a1def= a ·−c ·−

∑A is nonzero. Now we apply the first part of the proof

of (1) to ρ and a1 to get a nonzero c1 ≤ a1, a disjoint subset A1 of B � (a1 · −c1),and a function f1 : A1 → κ such that ∀p ∈ rel(X, c1)[sup f1[rel(A1, p)] = κ] andrel(Yρ, c1) = rel(Yρ,

∑A1). Then (c, A, f) < (c+ c1, A∪A1, f ∪ f1), contradiction.


We now construct by recursion a sequence 〈ci : i < ω〉 of nonzero elements of Bsuch that the following conditions hold for all n < ω:

(7) cn ·∑

k<n ck = 0.

(8) rel(X,∑

k<n ck)⊆ rel(X, cn).

(9) There exist a system A of nonzero pairwise disjoint elements of B � −∑

k≤n ck

and a function f : A→ κ such that ∀x ∈ rel(X,∑

k≤n ck

)[sup f [rel(A, x)] = κ].

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Applying (1) with anything for ρ and 1 for a, we get a nonzero c0 so that (9) holds,while (7) and (8) vacuously hold. Now suppose that ck has been constructed forevery k ≤ n. In particular, by (9) we have A and f as indicated there. Now foreach nonzero a ∈ A we apply (1) to f(a) in place of ρ to get a nonzero sa ≤ a, asystem Ca of nonzero pairwise disjoint elements of B � (a · −sa), and a functionga : Ca → κ so that the following conditions hold:

(10) sup ga[rel(Ca, p)] = κ for all p ∈ rel(X, sa).

(11) rel(Yf(a), sa) = rel(Yf(a),∑

Ca).

(12)∑

Ca = a · −sa.

Note:

(13) If a ∈ A, τ ≤ f(a), and xτ · a �= 0, then xτ · sa �= 0.

This follows from (11) and (12).

Now let cn+1 =∑

a∈A sa. Then (7) clearly holds for n+ 1. To prove (8) forn+1, suppose that xτ ·

∑k≤n ck �= 0. By (9) for n, choose a ∈ A such that a·xτ �= 0

and τ ≤ f(a). By (13) we have xτ · sa �= 0. Hence xτ · cn+1 �= 0, as desired: (8)holds for n+ 1.

Now let D =⋃

a∈ACa, and h =⋃

a∈A ga. Thus D is a collection of nonzeropairwise disjoint elements of B, and h : B → κ. Clearly in fact D ⊆ B �−∑

k≤n+1 ck. Now suppose that xτ ∈ rel(X,∑

k≤n+1 ck

).

(14) There is an a ∈ A such that xτ · sa �= 0.

In fact, ∑k≤n+1

ck =∑a∈A

sa +∑k≤n

ck,

so from xτ ∈ rel(X,∑

k≤n+1 ck

)we see that we have two cases.

Case 1. There is an a ∈ A such that xτ · sa �= 0. So (14) holds.

Case 2. xτ ·∑

k≤n ck �= 0. Then by the choice of A and f , there is an a ∈ A suchthat a · xτ �= 0 and f(a) ≥ τ . Hence by (13), condition (14) again holds.

Now by (14), xτ ∈ rel(X, sa), and so by (10), sup ga[rel(Ca, xτ )] = κ. Since

ga[rel(Ca, xτ )] = h[rel(Ca, xτ )] ⊆ h[rel(D, xτ )],

it follows that suph[rel(D, xτ )] = κ. Hence (9) holds, and we have finished theconstruction of 〈ci : i < ω〉.

Note that (8) implies that rel(X, cn) ⊆ rel(X, cm) whenever n < m. It followsthat rel

(X,∑

n<ω c2n)= rel

(X,∑

n<ω c2n+1

). But this contradicts the assump-

tion that X is homogeneously weakly dense. �

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Proposition 6.52. If B is an infinite complete BA which is hwd- and π-homo-geneous, then π(B) ≤ min{2<r(B), sup{λκ : κ < c′(B), λ < r(B)}}.

Proof. Let D and a be as in Theorem 6.51. Thus π(B) = π(B � a) ≤ |D|. Then theconclusion of our proposition follows by making two computations of |D|. First,for each ρ < r(B) we take subsets of Yρ; this gives |D| ≤ 2<r(B), using Proposition6.50. Second, we note that for each bounded subset W of X we have

∑W =

∑Y ,

where Y is maximal disjoint such that ∀y ∈ Y ∃w ∈W [y ≤ w]. �

Proposition 6.53. (GCH) If B is an infinite complete BA which is hwd- andπ-homogeneous, then r(B) = hwd(B) = πχinf(B) = πχ(B) = π(B).

Proof. We use Propositions 6.50 and 6.52:

r(B) ≤ πχinf(B) ≤ πχ(B) ≤ π(B) ≤ r(B) = hwd(B). �

Proposition 6.54. (GCH) If B is an atomless complete BA, then πχ(B) = π(B).

Proof. Let B ∼=∏

i∈I Ci, where each Ci is hwd- and π-homogeneous and I isinfinite. Then, using the remark after Problem 74 and Proposition 6.53,

π(B) = max(|I|, supi∈I

π(Ci)) = max(|I|, supi∈I

πχ(Ci)) ≤ πχ(B) ≤ π(B);

we also used the easily verified fact that max(|I|, supi∈I πχ(Ci)) ≤ πχ(B). �

The following problem remains; this was problem 40 in Monk [96].

Problem 74. Can one prove in ZFC that πχ(B) = π(B) for any atomless completeBA?

This finishes our treatment of weak density and related notions.

Concerning the function πSs we mention the following result from Shelah [96]:

Theorem 6.55. Let B be an infinite BA, and suppose that θ is an infinite regularcardinal less than πS+(B). Then there is a subalgebra A of B such that |A| =π(A) = θ.

Proof. Passing to a subalgebra of B if necessary, we may assume that ω < θ ≤π(B). We define a sequence 〈Aα : α < θ〉 of subalgebras of B, each of power lessthan θ, as follows. Let A0 be any denumerable subalgebra of B. For α a limitordinal < θ, let Aα be the union of preceding algebras. If Aα has been defined,then, since it has fewer than π(B) elements, there is a nonzero element b ∈ B suchthat for all x ∈ A+

α we have x �≤ b. Let Aα be the subalgebra of B generated byAα ∪ {b}. Clearly

⋃α<θ Aα is as desired. �

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Thus πSs(A) contains all regular cardinals in the interval [ω, πS+(A)). But Shelahalso showed in that paper that it is consistent to have a BA A with some ofthe singular cardinals in that interval not in πSs(A); and some special singularcardinals in that interval are always in πSs(A). These results answer problem 15in Monk [90].

We turn to πSr. First we note some easy facts:

(1) In general, |A| ≤ 2π(A), since every member of A is the join of a subset Xwhen X is dense in A. Hence (κ, λ) ∈ πSr(A) implies that λ ≤ 2κ.

(2) If (κ, λ) ∈ πSr(A) and ω ≤ μ ≤ λ, then there is some ν such that ω ≤ ν ≤ μand (ν, μ) ∈ πSr(A).

(3) If (κ, λ) ∈ πSr(A), then (κ, κ) ∈ πSr(A). In fact, if X is dense in B with|X | = π(B), and if Y ⊆ 〈X〉 is dense in 〈X〉, then Y is also dense in B, hence|Y | = |X |.

(4) If (κ, λ) ∈ πSr(A) and ω ≤ μ ≤ κ with μ regular, then (μ, μ) ∈ πSr(A). This isclear from Proposition 6.63 and its proof.

(5) (ω, ω) ∈ πSr(A) for any infinite BA A.

As for other functions, we have the following vague question.

Problem 75. Characterize πSr in cardinal number terms.

Many problems exist concerning the πSr relation for small algebras, so we do notsurvey this area.

Concerning πHr, we first mention the following general facts. We begin with anextension of Theorem 6.55.

Theorem 6.56. Let θ be an infinite regular cardinal less than πS+(B). Then B hasa homomorphic image C with π(C) = θ and |C| ≤ θ≤c(B).

Proof. By Theorem 6.55, letD be a subalgebra of B such that |D| = π(D) = θ. BySikorski’s extension theorem there is a homomorphism from B onto an algebra Csuch that D ≤ C ≤ D. Thus π(C) ≤ θ and |C| ≤ θ≤c(B). Suppose that X is densein C, with |X | < θ. For each x ∈ X+ we have x ∈ D, and so there is a dx ∈ D+

such that dx ≤ x. If c ∈ D+, choose x ∈ X+ such that x ≤ c. Then dx ≤ c. Thus{dx : x ∈ x} is dense in D and has size less than θ, contradiction. �

(1) If (κ, λ) ∈ πHr(A), then λ ≤ 2κ.

(2) For any infinite BA A there is a κ such that (ω, κ) ∈ πHr(A).

(3) (See Koppelberg [77]) Assuming MA, if A is an infinite BA and |A| < 2ω thenA has a countable homomorphic image.

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We have the usual problem concerning homomorphic spectrum:

Problem 76. Characterize πHr in cardinal number terms.

Many problems arise in considering πHr(A) for small A, so we will not try to surveythem.

One can see that d ≤ π by the following argument. Let A be a BA, and let X ⊆ A+

be dense in A with |X | = π(A). For each x ∈ X let Fx be an ultrafilter on A suchthat x ∈ Fx. For each a ∈ A let f(a) = {Fx : a ∈ Fx}. It is clear that f is anisomorphism from A into P(Y ), where Y = {Fx : x ∈ X}. Thus d(A) ≤ π(A) byTheorem 5.1.

The difference between d(A) and π(A) is small, however, since d(A) ≤ π(A) ≤|A| ≤ 2d(A) for any infinite BA A.

About π for special classes of BAs, note that π(A) = d(A) for any intervalalgebra A; in fact, π(A) is also equal to hd(A). To see this, note that d(A) =hd(A) for A an interval algebra, since any interval algebra is retractive. Nowπ(A) ≤ πH+(A) = hd(A) by Theorem 6.15, and d(A) ≤ π(A), so d(A) = π(A).

Another interesting fact about π and interval algebras was observed by Dou-glas Peterson: π(Intalg(L)) = d(L) · |M |, where d(L) is the density of L as atopological space and M is the set of atoms of Intalg(L). To prove ≤, let X bedense in L with |X | = d(L); we show that {[x, y) : x, y ∈ X, x < y} ∪ M isdense in Intalg(L). Take any nonzero a ∈ Intalg(L). Wlog a has the form [u, v).If [u, v) is finite, then b ≤ [u, v) for some atom b. If [u, v) is infinite, then thereexist x, y ∈ X with u < x < y < v, and so [x, y) ⊆ [u, v). For ≥, suppose tothe contrary that R is dense in Intalg(L) and |R| < d(L) · |M |. Clearly M ⊆ R.Wlog each member of R has the form [a, b). Since |M | ≤ |R|, we have |R| < d(L).Let R′ = {a ∈ L : ∃b([a, b) ∈ R)}. Thus L\R′ is a non-empty open set. Sayw ∈ (u, v) ⊆ L\R′. Then [w, v) ∈ Intalg(L), so [a, b) ⊆ [w, v) for some [a, b) ∈ R;but then a ∈ R′, contradiction.

If A is a minimally generated algebra, then π(A) = d(A). In fact, A is co-complete with an interval algebra B, π(B) = π(B), and d(B) = d(B) for any BAB, so this follows from the interval algebra result.

For A atomic, clearly π(A) is the number of atoms of A. Also note thatπS+(A) = |A| for A complete, and πS+(A) = hd(A) for A retractive.

If A is the completion of the free BA on ω1 free generators, then d(A) < π(A):clearly π(A) = ω1. The identity mapping from the free BA on ω1 free generatorsinto P(ω) can be extended to a homomorphism f from A into P(ω), and f mustbe one-one; so d(A) = ω.

For tree algebras we recall from Theorem 5.25 that π(A) = |A|. (Using d(A) ≤π(A).)

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7 Length

Recall that Length(A) is the sup of cardinalities of subsets of A which are simplyordered by the Boolean ordering. For references see the beginning of Chapter 4.The analysis of Length is similar to that for Depth. We begin by discussing ofLength, for which we need two small lemmas about orderings:

Lemma 7.1. Let L be a linear ordering of regular cardinality λ which has no strictlyincreasing or strictly decreasing sequences of length λ. Then there exist a < b inL such that |[a, b)| = λ.

Proof. If L has both a first and a last element, this is trivial. Suppose that L hasa first element a, but no last element. Let 〈bξ : ξ < κ〉 be strictly increasing andcofinal in L. Then κ < λ by assumption, and L =

⋃ξ<κ[a, bξ), so there is a ξ < κ

such that |[a, bξ)| = λ; then a, bξ are as desired. Other possibilities – last but nofirst element, neither a first nor a last element – are treated similarly. �Lemma 7.2. Let L be a linear ordering with first element 0, and with cardinalityκ+, where κ is infinite. Then there exist a < b in L such that |[a, b)| ≥ κ and|L\[a, b)| ≥ κ.

Proof. Suppose not. Then clearly

(1) in L there is no strictly increasing or strictly decreasing sequence of length κ+.

We define by induction a sequence 〈[aξ, bξ) : ξ < κ+〉 of half-open intervals in Lsuch that [aη, bη) ⊂ [aξ, bξ) for ξ < η, |[aξ, bξ)| = κ+, and |L\[aξ, bξ)| < κ for all ξ <α. By Lemma 7.1, choose a0 < b0 such that |[a0, b0)| = κ+. Then |L\[a0, b0)| < κ byour assumption that the Lemma fails. If [aξ, bξ) has been defined, then [aξ+1, bξ+1)can be defined: choose c with aξ < c < bξ; then |[aξ, c)| = κ+ or |[c, bξ)| = κ+,giving [aξ+1, bξ+1) in an obvious way, again using our assumption. Suppose that[aξ, bξ) has been defined for all ξ < β, where β is a limit ordinal < κ+. Then∣∣∣∣∣∣L\

⋂ξ<β

[aξ, bξ)

∣∣∣∣∣∣ =∣∣∣∣∣∣⋃ξ<β

L\[aξ, bξ)

∣∣∣∣∣∣ < κ+,

so by (1) and Lemma 7.1 applied to⋂

ξ<β[aξ, bξ), the interval [aβ , bβ) can bedefined. This finishes the construction.

, ,OI 10.1007/978-3- - - _ ,


014


Basel 28

279

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280 Chapter 7. Length

Now aξ ≤ aη and bξ ≥ bη for ξ < η, so one of {aξ : ξ < κ+} and {bξ : ξ < κ+}contains a suborder of L of size κ+. This contradicts (1). �

Theorem 7.3. If cf(LengthA) = ω, then LengthA is attained.

Proof. The proof should be fairly clear, following the lines of the proof of 4.8. Somemodifications: a is an ∞-element provided that for each i ∈ ω, some ordering ofsize λi is embeddable in A � a. When constructing ai, Lemma 7.2 is used to obtainelements c, d such that b = c+d, c ·d = 0, and both A � c and A � d contain chainsof size λi; then the new (∗) is applied. �

Later we will see that Theorem 7.3 is best possible.

Now we turn to products. The analog of the basic Theorem 4.3 for depth does nothold for length. For example, if A is any denumerable BA, then ωA has length 2ω.This is because P(Q) can be embedded in ωA, and R can be embedded in P(Q):for each r ∈ R, let f(r) = {q ∈ Q : q < r}. To generalize this example, let us calla subset D of a linear order L weakly dense in L provided that if a, b ∈ L anda < b, then there is a d ∈ D such that a ≤ d ≤ b. Now for any infinite cardinalκ let Ded(κ) = sup{λ: there is an ordering of size λ with a weakly dense subsetof size κ}. The following theorem from Kurepa [57] shows the connection of thisnotion with length in P(κ):

Theorem 7.4. Let κ and λ be cardinals such that ω ≤ κ ≤ λ. Then the followingtwo conditions are equivalent:

(i) There is an ordering L of size λ with a weakly dense subset of size κ.

(ii) In P(κ) there is a chain of size λ.

Proof. (i)⇒(ii). We may assume that κ < λ. Let D be weakly dense in L, with|D| = κ. Thus |L\D| = λ. Let f be a one-one function from κ onto D. For eacha ∈ L\D let g(a) = {α < κ : f(α) < a}. Clearly a < b implies that g(a) ⊆ g(b).Suppose a < b with a, b ∈ L\D; choose x ∈ D so that a ≤ x ≤ b. Hence a < x < b,and so f−1(x) ∈ g(b)\g(a) and g(a) �= g(b), as desired.

(ii)⇒(i). Let L be a chain in P(κ) of size λ. Define α ≡ β iff ∀c ∈ L[α ∈ c iffβ ∈ c]. Let Γ have one member from each ≡-class. For each α ∈ Γ let xα =

⋃{a ∈

L : α /∈ a}. Thus L\xα =⋂{a ∈ L : α ∈ a}. Then

(1) For all c ∈ L and α ∈ Γ, if α /∈ c then c ⊆ xα.

(2) For all c ∈ L and α ∈ Γ, if α ∈ c then xα ⊆ c.

In fact, (1) is clear. Under the assumptions of (2), if γ ∈ xα then there is a d ∈ Lsuch that α /∈ d and γ ∈ d, hence c �⊆ d, so d ⊆ c and γ ∈ c.

For any α, β ∈ Γ we have xα ⊆ xβ or xβ ⊆ xα. In fact, suppose that α, β ∈ Γand α �= β. Then α �≡ β, so there is an a ∈ L such that not(α ∈ a iff β ∈ a). Sayα ∈ a and β /∈ a. We claim then that xα ⊆ xβ . For, suppose that γ ∈ xα. Choose

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Chapter 7. Length 281

b ∈ L such that α /∈ b and γ ∈ b. Since α ∈ a, we have a �⊆ b, so b ⊆ a. Henceγ ∈ a. Since β /∈ a, it follows that γ ∈ xβ . This proves the claim.

Now by (1) and (2), the set L′ = L∪{xα : α ∈ Γ} is a chain. Clearly |L′| = λ.Let D be a subset of L′ of size κ containing {xα : α ∈ Γ}. We claim that D isweakly dense in L′. To prove this it suffices to take a, b ∈ L such that a < b andfind α ∈ Γ such that a ⊆ xα ⊆ b. Take any α ∈ b\a. We may assume that α ∈ Γ.Then xα is as desired, by (1) and (2). �

Because of this theorem, about all that we can say about the length of productsis this:

max(Ded(|I|), supi∈I

Length(Ai)) ≤ Length

(∏i∈I

Ai

)≤∏i∈I

Length(Ai).

Shelah [90] has shown that Length(∏

i∈I Ai) cannot be calculated purely from |I|and 〈Length(Ai) : i ∈ I〉.

For weak products, we have the following analogs of 4.6 and 4.7:

Theorem 7.5. Let κ = supi∈ILength(Ai), and suppose that cf(κ) > ω. Then thefollowing conditions are equivalent:

(i)∏w

i∈I Ai has no chain of size κ.

(ii) For all i ∈ I, Ai has no chain of size κ.

Proof. For the non-trivial direction (ii)⇒(i), suppose that X is a chain in∏w

i∈I Ai

of size κ. Wlog assume that for each x ∈ X , the 1-support Mx of x is finite. Definex ≡ y iff Mx = My. Then it is easy to see that ≡ is a convex equivalence relationon X ; there is an order induced on X/ ≡, and clearly that order is isomorphic toan interval of the ordered set ω. It follows from cf(κ) > ω that some equivalenceclass has cardinality κ. Then Lemma 4.2 gives a contradiction. �

Corollary 7.6. Length(∏w

i∈I Ai

)= supi∈ILengthAi. �

By 7.5 we see that 7.3 is best possible: if κ is a limit cardinal with cf(κ) > ω, thenit is easy to construct a weak product B such that Length(B) = κ but the lengthof B is not attained.

Now we discuss free products. First we mention some results whose proofs weresketched in McKenzie, Monk [82]; the proofs are similar to the correspondingresults about depth.

• If cf(κ) > ω, A has no chain of size cf(κ), and B has no chain of size κ, thenA⊕B has no chain of size κ.

• If κ is uncountable and regular, and for every i ∈ I the BA Ai has no chainof size κ, then ⊕i∈IAi has no chain of size κ.

• Length (⊕i∈IAi) = supi∈I Length(Ai) when each Ai is infinite.

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One of the results for depth does not carry over for length. Namely, it is possibleto have cf(κ) < κ with BAs A,B such that A has a chain of size cf(κ), B haslength κ but no chain of size κ, with A⊕B also having no chain of size κ. Namely,let κ = �ω1 , A = P(ω), and B =

∏wα<ω1

Intalg(�α). By remarks above, A hasa chain of size 2ω ≥ ω1 = cf(�ω1), while by Theorem 7.5, B has length �ω1 butno chain of size �ω1 . Finally, to show that A ⊕ B also has no chain of size �ω1 ,suppose that X is a chain in A ⊕ B. For each x ∈ X write x =

∑i<mx

axi · bxi ,where axi · axj = 0 for i �= j, axi ∈ A, and bxi ∈ B. For each n ∈ ω let

Tn = {bxi : x ∈ X, i < mx, n ∈ axi }.

We claim that Tn is a chain in B. In fact, suppose that x ∈ X , i < mx, n ∈ axi ,y ∈ X , j < my, and n ∈ ayj . By symmetry say x ≤ y. Then

axi · bxi∏

k<my

(−ayk +−byk) = 0;

multiplying by ayj we get axi · ayj · bxi · −b

yj = 0. Since n ∈ axi ∩ ayj , it follows

that bxi · −byj = 0, i.e., bxi ≤ byj . This proves the claim. Hence |Tn| < �ω1 . Let

T =⋃

n∈ω Tn. So also |T | < �ω1 . For any x ∈ X we have bxi ∈ T for all i < mx. Itfollows that |X | < �ω1 , as desired.

The following problem, mentioned in McKenzie, Monk [82], is still open. See thatarticle for additional relevant information.

Problem 77. Let ω < cf(κ) < κ, and let L be a dense linear ordering of size cf(κ)with no dense subset of size less than cf(κ), and with no family of cf(κ) pairwisedisjoint open intervals. Let A be the interval algebra on L, and suppose that B isa BA with Length(B) = κ. Does A⊕ B have a chain of size κ?

We have not investigated the length of amalgamated free products.

Problem 78. Investigate the length of amalgamated free products.

Concerning unions, Length is an ordinary sup function, and so Theorem 3.16applies.

Now we turn to ultraproducts, giving some results of Douglas Peterson. Sincelength is an ultra-sup function, Theorems 3.20–3.22 apply. Thus by Theorem 3.22,for F regular we have Length

(∏i∈I Ai/F

)≥∣∣∏

i∈I Length(Ai)/F∣∣. By Donder’s

theorem it is consistent that ≥ always holds. It is possible in ZFC to have >here, according to the following result of Shelah [99] (Conclusion 15.13 (2)), whichanswers Problem 22 in Monk [96]:

If D is a uniform ultrafilter on κ, then for a class of cardinals λ such that λκ = λ,there is a system 〈Bi : i < κ〉 of Boolean algebras such that Length(Bi) ≤ λ foreach i < κ, hence

∏i<κ Length(Bi)/D) ≤ λ, while Length(

∏i<κ Bi/D) = λ+.

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Magidor and Shelah have shown that it is consistent to have an example in whichthe length of an ultraproduct is strictly less than the size of the ultraproduct oftheir lengths; see Chapter 4 for details.

The following version of Theorem 3.23 holds:

Theorem 7.7. Let 〈Ai : i ∈ I〉 be a system of infinite BAs, with I infinite. Let Fbe a uniform ultrafilter on I, and let κ = max(|I|, ess.supFi∈ILength(Ai)).

Then Length(∏

i∈I Ai/F)≤ 2κ.

Proof. Let λ = ess.supFi∈ILength(Ai). We may assume that Length(Ai) ≤ λ for alli ∈ I. In order to get a contradiction, suppose that 〈fα/F : α < (2κ)+〉 is a systemof distinct comparable elements. Thus [(2κ)+]2 =

⋃i∈I{{α, β} : fα(i) and fβ(i)

are distinct comparable elements}, so by the Erdos–Rado theorem (2κ)+ → (κ+)2κwe get a homogeneous set which gives a contradiction. �

Concerning equality in Theorem 7.7, we note that it holds if F is regular andess.sup |Ai| ≤ |I|, since then

2|I| = (ess.supLength(Ai))|I|

=

∣∣∣∣∣∏i∈I

Length(Ai)/F

∣∣∣∣∣ ≤ Length

(∏i∈I

Ai/F

)

≤∣∣∣∣∣∏i∈I

Ai/F

∣∣∣∣∣ ≤ 2|I|.

On the other hand, if |A| = LengthA = κ and κω = κ, then Length (ωA/F ) < 2κ

for any nonprincipal ultrafilter F on ω.

The situation for length and subdirect products is similar to that for cellularityand depth: there is a BA with length ω which is a subdirect product of algebraswith large length.

For moderate products we have the following result, with essentially the sameproof as for depth:

Length

(B∏i∈I

Ai

)= max{Length(B), sup

i∈ILength(Ai)}.

Clearly Length(A) = Length(Dup(A)) for any infinite BA A.

Turning to the exponential, we first note that Length(A) ≤ Length(Exp(A)).In fact, if a, b ∈ A and a ≤ b, then clearly V (S(a)) ⊆ V (S(b)). Suppose that a < b.Let U be an ultrafilter on A such that b · −a ∈ U . Then {U} ∈ V (S(b))\V (S(a)).Hence V (S(a)) ⊂ V (S(b)).

These elementary facts show that Length(A) ≤ Length(Exp(A)).

Problem 79. Is there a BA A such that Length(A) < Length(Exp(A))?

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We turn to derived functions for length. The function LengthH+(A) seems to benew. Note that t(A) = DepthH+(A) ≤ LengthH+(A), using 4.26. It is possible tohave t(A) < LengthH+(A); this is true when A is the interval algebra on R, sincet(A) = ω, while obviously LengthH+(A) = Length(A) = 2ω. To see that t(A) = ω,use Theorem 4.26. An elementary characterization of LengthH+ can be given usingthe following notion of free chain from Monk [11]. A free chain for a BA A is anordered pair (L, a) such that L is a linear order, a ∈ LA, and for any F,G ∈ [L]<ω,if F < G then

∏ξ∈F aξ ·

∏η∈G−aη �= 0. We say that (L, a) is a free chain over L.

We define a related topological notion. Let X be a topological space. A freechain for X is an ordered pair (L, x) such that L is a linear order, x ∈ LX , andfor any ξ ∈ L,

{xη : η < ξ} ∩ {xη : ξ ≤ η} = ∅.

As in the case of free sequences, a BA A has a free chain (L, a) iff Ult(A) has afree chain (L, x).

Proposition 7.8. For any infinite BA A we have

LengthH+(A) = sup{|L| : A has a free chain (L, a)}.

Proof. The proof is just a modification of the proof of Theorem 4.26. For ≥,suppose that (L, a) is a free chain in A; we will find an ideal I of A such thatA/I has a chain of size |L|. For each ξ ∈ L let Fξ be an ultrafilter on A suchthat {aη : η < ξ} ∪ {−aη : ξ ≤ η ∈ L} ⊆ Fξ. Let Y = {Fξ : ξ ∈ L}, and letI = {x ∈ A : Y ⊆ S(−x)}. Clearly I is an ideal in A. We claim that

(1) ∀ξ, η ∈ L[ξ < η → aη/I < aξ/I].

To prove this, suppose that ξ < η. To show that aη · −aξ ∈ I, take any ρ ∈ I.If η < ρ, then also ξ < ρ and so aξ ∈ Fρ, and it follows that −aη + aξ ∈ Fρ, sothat Fρ ∈ S(−aη + aξ). If ρ ≤ η, then −aη ∈ Fρ and again Fρ ∈ S(−aη + aξ). Soaη · −aξ ∈ I. Thus aη ≤ I < aξ/I]. Also, aξ ∈ Fη and −aη ∈ Fη, so it follows thataη/I �= aξ/I. Thus (1) holds.

Conversely, suppose that I is an ideal in A and 〈aα/I : α ∈ L〉 is a chainin A/I. Let α <L β iff aα/I < aβ/I. This makes L into a linear order. We mayassume that a0/I �= 0 and no aα/I is equal to 1. We claim then that 〈−aα : α ∈ L〉is a free chain. For, suppose that F,G ∈ [L]<ω with F < G. Then if both F andG are nonempty, we have⎛

⎝∏α∈F

−aα ·∏β∈G

aβ

⎞⎠ /I =

∏α∈F

(−(aα/I)) ·∏β∈G

(aβ/I) = −(aα/I) · (aβ/I),

where α is the largest element of F and β is the smallest element of G. So −(aα/I)·(aβ/I �= 0, and hence

∏α∈F −aα ·

∏β∈G aβ �= 0. The case when one of F,G is

empty is treated similarly. �

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Problem 80. Is LengthH+(A) = t(A) · Length(A) for every infinite BA A?

Shelah has constructed an algebra A such that ω < Length(A) < |A| while A hasno homomorphic image of power smaller that |A|, assuming ¬CH (email messageof December 1990). This answers Problem 17 in Monk [90]. Since an infinite BAalways has a homomorphic image of size≤ 2ω, the assumption ¬CH is needed here.We present this result here. It depends on the following notation. If 〈an : n ∈ ω〉is a system of elements of a BA A and Y ⊆ ω, then {[x, an]if n∈Y : n ∈ ω} denotesthe following set of formulas:

{an ≤ x : n ∈ Y } ∪ {an · x = 0 : n ∈ ω\Y }.

If L is a chain, then a Dedekind cut of L is a pair (M,N) such that L = M ∪Nand u < v for all u ∈ M and v ∈ N . If in addition L is a subset of a BA A, thenan element a ∈ A realizes the Dedekind cut (M,N) if u ≤ a ≤ v for all u ∈ Mand v ∈ N .

For any BA A, Length′(A) is the smallest infinite cardinal κ such that everychain in A has size less than κ.

Shelah’s result will follow easily from the following lemma:

Lemma 7.9. Let ℵ0 ≤ μ < λdef= 2ℵ0 , and let A be a subalgebra of P(ω) containing

all singletons {i}, with |A| = μ. Then there is a BA B of size 2ℵ0 , a subalgebra ofP(ω), satisfying

⊗0 A is a dense subalgebra of B.

⊗1 If 〈an : n < ω〉 is a system of pairwise disjoint elements of B+, then for2ℵ0 subsets Y of ω there is an element aY ∈ B realizing {[x, an]if n∈Y : n ∈ ω}.

⊗2 If 〈an : n < ω} is a chain of members of B, then the number of Dedekindcuts of it realized in B is less than 2ℵ0 .

Proof. First we obviously have:

(1) there is an enumeration 〈〈aζn : n < ω〉 : ζ < λ〉 of all of the ω-tuples of subsetsof ω, each one repeated λ times.

Next we claim:

(2) There is a function h : λ\{0} → λ such that for every nonzero ε < λ we haveh(ε) < ε, and for each ζ < λ the set

Sζdef= {ε ∈ λ\{0} : h(ε) = ζ}

has power λ.

To see this, first choose a system 〈Dα : α < λ〉 of pairwise disjoint sets whoseunion is λ, each of power λ, with D0 the set consisting of 0 and all limit ordinalsless than λ. Define

Eα = (Dα ∪ {α+ 1})\⋃β<α

Eβ .

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Then

(3) Dα\Eα ⊆ α+ 1;

(4) Eα\Dα ⊆ {α+ 1}.

For, (4) is obvious. For (3), suppose that ζ ∈ Dα\Eα. Then there is a β < α suchthat ζ ∈ Eβ . Now ζ /∈ Dβ , so ζ = β + 1 by (4). Thus (3) holds.

By (3), |Eα| = λ for all α < λ. Next,

(5)⋃

α<λ Eα = λ.

For, suppose that ζ /∈⋃

α<λ Eα. Since E0 = D0 ∪ {1}, ζ is a successor ordinalα+ 1. Then ζ ∈ Eα, contradiction.

For each ε ∈ λ\{0} let h(ε) be the α such that ε ∈ Eα. If h(ε) = 0, obviouslyh(ε) < ε. Suppose that h(ε) �= 0, so ε is a successor ordinal γ + 1. Clearly ε ∈⋃

β≤γ Eβ , so h(ε) < γ + 1 = ε. This proves (2).

Now we define a BA Bε by induction on ε < λ such that:

a) Bε is a subalgebra of P(ω) containing all singletons, of cardinality < μ+ +|ε|+.

b) Bε is increasing and continuous in ε.

c) B0 = A.

d) If ρ < ε and 〈aρn : n < ω〉 is a linearly ordered system of elements of Bρ (notwo equal), then every Dedekind cut of it which is not realized in Bρ is alsonot realized in Bε.

e) Let Evens be the set of all even natural numbers. If h(ε) = ζ and 〈aζn : n < ω〉is a system of pairwise disjoint members ofBε (some possibly zero) with unionω, then for some Yε ⊆ Evens we have

(i) Bε+1 is generated by Bε ∪ {xε}, where xε =⋃

n∈Yεaζn.

(ii) If ψ < ε and h(ψ) = ζ, then Yψ �= Yε.

If 〈aζn : n < ω〉 is not such a system, then Bε+1 = Bε.

The construction is determined for ε = 0 and for limit ε. At stage ε → ε + 1, itthe hypothesis of e) does not hold, let Bε+1 = Bε. Now assume that hypothesisholds. Let κε = (μ+ |ε|)+, and let 〈Y ε

i : i < κε〉 be a sequence of almost disjointinfinite subsets of Evens. Now for each i < κε let xεi =

⋃n∈Y ε

iaζn and let Bεi =

〈B ∪ {xεi}〉. Let P = {i < κε : ∀ψ < ε[h(ψ) = ζ → Yψ �= Y εi ]}. So |P | = κε.

Suppose that d) fails for ε+ 1 with Bεi for Bε+1, for each i ∈ P ; we want to geta contradiction. Then for each i ∈ P there is a ρi ≤ ε such that d) fails; say that〈aiρn : n < ω〉 is a linearly ordered system of elements of Bρi and (Mi, Ni) is aDedekind cut of it which is not realized in Bρi but is realized in Bεi. We mayassume that ρ does not depend on i. By d) for ε, (Mi, Ni) is not realized in Bεi.Say ci realizes (Mi, Ni) in Bεi. We can write ci = b0i + b1i · xεi + b2i · −xεi, withb0i, b1i, b2i pairwise disjoint elements of Bε. We may assume that b0i, b1i, b2i do notdepend on i.

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If (Mi, Ni) = (Mj , Nj) for two distinct i, j ∈ P , we get a contradiction, asfollows. Note that xεi · xεj ∈ Bε, since

xεi · xεj =

⎛⎝ ⋃

m∈Y εi

aζm

⎞⎠ ∩

⎛⎝ ⋃

m∈Y εj

aζn

⎞⎠ =

⋃m∈Y ε

i ∩Y εj

aζm

and Y εi ∩ Y ε

j is finite. Now let

d = b0 + b1 · xεi · xεj + b2 · (−xεi +−xεj).

So d ∈ Bε. We claim that it realizes (Mi, Ni), which is a contradiction. For, ife ∈Mi and f ∈ Ni, then

e ≤ ci = b0 + b1 · xεi + b2 · −xεi;

e ≤ cj = b0 + b1 · xεj + b2 · −xεj ;

e ≤ (b0 + b1 · xεi + b2 · −xεi) · (b0 + b1 · xεj + b2 · −xεj)

≤ b0 + b1 · xεi · xεj + b2 · −xεi + b2 · −xεj

= d;

andb0 ≤ ci ≤ f ;

b1 · xεi · xεj ≤ b1 · xεi ≤ ci ≤ f ;

b2 · −xεi ≤ ci ≤ f

b2 · −xεj ≤ cj ≤ f

d ≤ f.

Thus we have shown that distinct i, j ∈ P give different Dedekind cuts. Wlog thetruth values of the following statements do not depend on i:

(6) S1i

def= {n ∈ Y ε

i : aζn · b1 �= 0} is infinite;

(7) S2i

def= {n ∈ Y ε

i : aζn · b2 �= 0} is infinite.

If both of these are false (for all i), take any i, and consider (Mi, Ni). Then, withc ∈Mi and d ∈ Ni,

c · b1 ≤ b1 · xεi = b1 ·∑n∈S1

i

aζn ≤ d · b1

and b2 · xεi = b2 ·∑

n∈S2iaζn and hence

c · b2 ≤ b2 · −xεi = b2 · −∑n∈S2

i

aζn ≤ d · b2;

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so (Mi, Ni) is realized by

b0 + b1 ·∑n∈S1

i

aζn + b2 · −∑n∈S2

i

aζn

in Bε, contradiction.

Thus either (6) or (7) is true. Take distinct Dedekind cuts (Mi, Ni) and(Mj, Nj); say Mi ⊂Mj . Choose e ∈Mj\Mi.

Case 1. (6) is true. Then in Bεi we have b0 + b1 · xεi + b2 · −xεi ≤ e, so

(∗) b1 · xεi ≤ e · b1.

And in Bεj we have e ≤ b0 + b1 · xεj + b2 · −xεj , so

(�) e · b1 ≤ xεj · b1.

Now if n ∈ S1i , then 0 �= aζn · b1 ≤ xεi · b1 ≤ e · b1 (by (∗)) ≤ xεj · b1 (by (�)), so

n ∈ Y εj , hence n ∈ S1

j . Thus S1i ⊆ S1

j , contradicting S1i ∩ S1

j ⊆ Y εi ∩ Y ε

j finite.

Case 2. (7) is true. Then in Bεi we have b0 + b1 · xεi + b2 · −xεi ≤ e, so

(∗∗) −xεi · b2 ≤ e · b2.

And

(��) e · b2 ≤ −xεj · b2.

If n ∈ S2j , then aζn · b2 · −xεj = 0, so by (��), aζn · b2 · e = 0, hence by (∗∗) aζn ·

b2 · −xεi = 0, so n ∈ Y εi , hence n ∈ S2

i . So S2j ⊆ S2

i , again giving a contradiction.

Thus there is an i ∈ P such that d) holds for Bεi as Bε+1. Clearly then thefirst part of e) also holds.

Thus the construction can be carried through. Let B =⋃

ε<λ Bε. Clearly⊗0 holds, by c). Now suppose that 〈an : n < ω〉 is a system of pairwise disjointelements of B+. Choose ε < λ such that an ∈ Bε for all n ∈ ω, extend {an :n ∈ ω} to a maximal disjoint set X in Bε, and enumerate X as 〈bn : n < ω}so that {an : n < ω} = {b2n : n < ω}. By (1), there is a ζ < λ such that〈bn : n < ω〉 = 〈aζn : n < ω〉. Then by (2) and e) we get 2ℵ0 subsets Y of ω suchthat {[x, an]if n∈Y : n ∈ ω} is realized in B. Next, suppose that 〈an : n < ω〉 is achain of members of B+. By (1), say 〈an : n < ω〉 = 〈aζn : n < ω〉. Choose ε < λsuch that all an are in Bε. Then by d), B realizes at most |Bε| Dedekind cuts of〈an : n < ω〉. �

Theorem 7.10. If ℵ0 ≤ μ < 2ℵ0 then there is a BA B such that:

(i) B is a subalgebra of P(ω) containing all singletons, and hence π(B) = ℵ0;(ii) μ+ ≤ Length′(B) ≤ 2ℵ0 ;

(iii) every infinite homomorphic image of B has size 2ℵ0 .

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Proof. We apply Lemma 7.9 to a subalgebra A of P(ω) containing all singletons,of size μ and with length μ. We obtain a BA B as a result.

We check that every infinite homomorphic image of B has size 2ℵ0 . Let f bea homomorphism from B onto C with C infinite. Let 〈cn : n < ω〉 be a system ofnonzero disjoint elements of C. Then there is a system 〈bn : n < ω〉 of non-zerodisjoint elements of B such that f(bn) = cn for all n < ω. Let D be a collectionof 2ℵ0 subsets of ω such that for each Y ∈ D there is an element bY realizing{[x, bn]if n∈Y : n < ω}. Clearly {f(bY ) : Y ∈ D} is a subset of C of size 2ℵ0 , asdesired.

Now suppose that J is a chain in B of size 2ℵ0 ; we shall get a contradiction.For each i ∈ ω let Mi = {b ∈ J : i /∈ b}. Clearly there is a countable subset Ki ofMi cofinal in Mi. Let I =

⋃i∈ω Ki. So, I is countable. Each element of J realizes

a Dedekind cut of I, so we can contradict ⊗2 of the Lemma by showing that anytwo distinct u, v ∈ J realize distinct Dedekind cuts of I. Suppose that u ⊂ v but{w ∈ I : u ⊆ w} = {w ∈ I : v ⊆ w}. Choose i ∈ v\u. Choose w ∈ Ki with u ⊆ w.Then v ⊆ w and hence i ∈ w, contradiction. �

The function LengthH− is also new. Note that ω ≤ LengthH−(A) ≤ 2ω, by aneasy argument using the Sikorski extension theorem. It is obviously possible tohave ω = LengthH−(A). If 2ω = ω2, the algebra B of Theorem 7.10 is such thatLengthH−(B) < CardH−(B). This answers Problem 18 in Monk [90]. ObviouslyLengthS+(A) = Length(A) and LengthS−A = ω. The following is Problem 23 inMonk [96]:

Problem 81. Is always Lengthh−(A) = ω?

Here Lengthh−(A) is defined as follows:

Lengthh−(A)=inf{sup{|X | :X is a chain of clopen subsets of Y } :Y ⊆Ult(A)}.

Clearly Lengthh+(A) ≥ Depthh+(A) = s(A) by 4.28.

Also, Lengthh+A ≥ LengthH+A; but it is possible to have Lengthh+A >LengthH+A. This is true, for example, if A is the finite-cofinite algebra on anuncountable cardinal κ. For then Lengthh+A = Ded(κ), while LengthH+A = ω.That Lengthh+A = Ded(κ) is seen like this: Ult(A) has a discrete subspace S ofsize κ, and so Theorem 7.4 applies for the chains of subsets of S, since every subsetis clopen.

Clearly dLengthS+(A) = Length(A). By the discussion of dDepthS− in Chap-ter 4 we see that dLengthS−(A) ≥ λ if A is atomless and λ-saturated (in themodel-theoretic sense). Thus Problem 20 of Monk [90] is answered.

Concerning LengthHs, note the following two examples. For A the free algebraof size κ, LengthHs(A) = [ω, κ]. If A is an interval algebra on an ordered set L,then LengthHs(A) = CardHs(A). In particular, LengthHs(Intalg(R)) = {ω, 2ω}; seeChapter 9. These examples suggest the following problem.

Problem 82. What are the possibilities for LengthHs(A)?

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Clearly LengthSs(A) = [ω,Length(A)]. The functions Lengthspect and Lengthmm

are studied in Monk [07]. The main result there is that under GCH, if K is anonempty set of regular cardinals, then there is a BAA such that Lengthspect(A) =K. This leaves the following question open.

Problem 83. Can one prove in ZFC that if K is a nonempty set of infinite cardinalsthen there is a BA A such that Lengthspect(A) = K?

Proposition 7.11. tow(A) ≤ Lengthmm(A) for every atomless BA A.

Proof. If X is a maximal chain, then X\{1} has no last element, and a cofinalsubset of it is a tower. �

Before considering LengthSr and LengthHr, we note an important connection oflength with depth. Obviously DepthA ≤ LengthA for any infinite BA A. Anotherclear relationship is that Length(A) ≤ 2Depth(A): if L is an ordered subset of Aof power (2Depth(A))+, let ≺ be a well-ordering of L; then by the Erdos–Radopartition relation (2κ)+ → (κ+)2κ we get a well-ordered or inversely well-orderedsubset of L of power (DepthA)+, contradiction.

Now for LengthSr we have the following adaptation of Theorem 4.73:


(i) If (κ, λ) ∈ LengthSr(A), then κ ≤ λ ≤ |A| and κ ≤ Length(A).

(ii) For each κ ∈ [ω,Length(A)] we have (κ, κ) ∈ LengthSr(A).

(iii) If (κ, λ) ∈ LengthSr(A) and κ ≤ μ ≤ λ, then (κ, μ) ∈ LengthSr(A).

(iv) (LengthA, |A|) ∈ LengthSr(A).

(v) If ω ≤ λ ≤ |A| then (κ, λ) ∈ LengthSr(A) for some κ.

(vi) If ((2κ)+, λ) ∈ LengthSr(A), then (ω, κ+) ∈ LengthSr(A).

Proof. (i)–(v) are obvious. For (vi), if B is a subalgebra of A with Length(B) =(2κ)+, then Depth(B) ≥ κ+ by the above, and so (ω, κ+) ∈ LengthSr(A) by theargument for 4.73(vi). �

As usual we have a vague question about LengthSr:

Problem 84. Characterize in cardinal number terminology the sets LengthSr(A).

We also have a vague question about LengthHr:

Problem 85. Characterize in cardinal number terminology the sets LengthHr(A).

Note that Length(A) > π(A) for A = P(ω); and c(A) > Length(A) for A thefinite-cofinite algebra on κ. If A is a tree algebra, then Length(A) = Depth(A) byProposition 16.20 of the Handbook.

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8 Irredundance

Clearly Irr(A) ≤ |A|. If A is a subalgebra of B, then Irr(A) ≤ Irr(B), and Irr canchange to any extent from B to A (along with cardinality). The same is true forA a homomorphic image of B.

Concerning our special kinds of subalgebras, recall that π(A) ≤ Irr(A) forevery infinite BA, by a theorem of McKenzie (Proposition 4.23 of the Handbook).Hence when B can have much larger size than A ≤ B, the irredundance canincrease. This applies to ≤rc, ≤free, ≤σ, ≤proj, ≤u, ≤reg, and ≤mg. Also, clearlyκ = Irr(Finco(κ) < 2κ = Irr(P(κ)).

Whether one can get < for the other subalgebra notions is open.

Problem 86. Can one have Irr(A) < Irr(B) for A ≤s B orA ≤m B?

The following equivalent formulation of irredundance will be useful.

Theorem 8.1. Let A be an infinite BA and X ⊆ A. Then the following conditionsare equivalent:

(i) X is irredundant.

(ii) For every x ∈ X there are ultrafilters F,G on A such that F ∩ (X\{x}) =G ∩ (X\{x}), x ∈ F , and −x ∈ G.

(iii) For every x ∈ X there are homomorphisms f, g : A → 2 such that f �(X\{x}) = g � (X\{x}) while f(x) �= g(x).

Proof. (ii) and (iii) are obviously equivalent. Now assume (ii). Then F∩〈X\{x}〉 =G ∩ 〈X\{x}〉. It follows that x /∈ 〈X\{x}〉.

Now assume that (i) holds, and suppose that x ∈ X . Now the set

(∗) {a ∈ 〈X\{x}〉 : −a ≤ x or − a ≤ −x}

has fip. For, otherwise we get

a0 · . . . · am−1 · b0 · . . . · bn−1 = 0

with each ai, bi ∈ 〈X\{x}〉 and −ai ≤ x, −bi ≤ −x. Then we may assume thatm,n > 0, then that m = n = 1. Hence a0 ≤ −b0 ≤ −x ≤ a0, so x = a0 ∈ 〈X\{x}〉,contradiction.

, ,OI 10.1007/978-3- - - _ ,


014


Basel 29

291

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292 Chapter 8. Irredundance

Let H be an ultrafilter on 〈X\{x}〉 containing the set (∗). Then H ∪{x} hasf.i.p. In fact, if a ∈ H and a · x = 0, then a ≤ −x, hence −a ∈ H , contradiction.Hence let F be an ultrafilter on A containing H ∪ {x}. Similarly, let G be anultrafilter on A containing H ∪ {−x}. This proves (ii). �

Roslanowski, Shelah [00] proved that Irr(A × B) = max(Irr(A), Irr(B)) if A × Bis infinite. This solves Problem 24 of Monk [96]. We give their proof here. Thefollowing construction will be used.

Fix an infinite cardinal λ. Let 〈xα : α < λ〉 be free generators of Fr(λ). For anyF ⊆ λ2 let IF be the ideal in Fr(λ) generated by all elements∏

α∈u

xε(α)α ,

where u ∈ [λ]<ω, ε ∈ u2, and ∀f ∈ F [ε �⊆ f ]. Then B[F ] is Fr(λ)/IF .

Proposition 8.2. If F ⊆ λ2 and f ∈ F , then there is a homomorphism fhom ofB[F ] into 2 such that fhom([xα]F ) = f(α) for every α ∈ λ.

Proof. If not, by Sikorski’s extension criterion we get finite disjoint u, v ⊆ λ suchthat

∏α∈u[xα]F ·

∏α∈v −[xα] = 0 while

∏α∈u f(α) ·

∏α∈v −f(α) �= 0. This means

that∏

α∈u xα ·∏

α∈v −xα ∈ IF and u ⊆ {α < λ : f(α) = 1} and v ⊆ {α < λ :f(α) = 0}. So there exist a finite set E ⊆ {(w, ε) : w ∈ [λ]<ω , ε ∈ w2} such thatε �⊆ g for all (ε, w) ∈ E and all g ∈ F , and∏

α∈u

xα ·∏α∈v

−xα ≤∑

(w,ε)∈E

∏α∈w

xε(α)α ,

Extending the mapping xα !→ f(α) to a homomorphism gives a contradiction. �Proposition 8.3. Suppose that F ⊆ λ2, u ∈ [λ]<ω, and ε ∈ u2. Then the followingconditions are equivalent:

(i)∏

α∈u[xα]ε(α) = 0.

(ii) ∀f ∈ F [ε �⊆ f ].

Proof. (i)⇒(ii): If ε ⊆ f for some f ∈ F , then the proof of Proposition 8.2 showsthat (i) fails to hold.

(ii)⇒(i): true by definition. �Theorem 8.4. Irr(B0 ×B1) = max(Irr(B0), Irr(B1)) for infinite B0 × B1.

Proof. Suppose that 〈(y0α, y1α) : α < λ〉 is an irredundant system in B0×B1. Henceby Theorem 8.1, for each α < λ we get homomorphisms f0

α, f1α : B0×B1 → 2 with

the following properties:

(1) f0α(y

0α, y

1α) �= f1

α(y0α, y

1α).

(2) ∀β ∈ λ\{α}[f0α(y

0β, y

1β) = f1

α(y0β, y

1β)].

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Chapter 8. Irredundance 293

Nowλ ={α < λ : f0

α(1, 0) = f1α(1, 0) = 0}

∪ {α < λ : f0α(1, 0) = f1

α(1, 0) = 1}∪ {α < λ : f0

α(1, 0) = 0 and f1α(1, 0) = 1}

∪ {α < λ : f0α(1, 0) = 1 and f1

α(1, 0) = 0}.Thus at least one of these four sets has size λ, so we have four cases:

Case 1. {α < λ : f0α(1, 0) = f1

α(1, 0) = 1} has size λ. We may assume thatf0α(1, 0) = f1

α(1, 0) = 1 for all α < λ. Now we claim:

(3) There exist homomorphisms glα of B0 into 2 such that glα(y0β) = f l

α(y0β , y

1β) for

all β < λ.

To prove (3), by Sikorski’s extension criterion it suffices to assume that u and vare disjoint finite subsets of λ such that

∏β∈u y

0β ·∏

β∈v−y0β = 0 and show that∏β∈u f

lα(y

0β , y

1β) ·

∏β∈v −f l

α(y0β , y

1β) = 0. We have

∏β∈u

f lα(y

0β .y

1β) ·

∏β∈v

−f lα(y

0β , y

1β) = f l

α(1, 0) ·∏β∈u

f lα(y

0β , y

1β) ·

∏β∈v

−f lα(y

0β , y

1β)

= f lα

⎛⎝(1, 0) ·

⎛⎝∏

β∈u

(y0β , y1β) ·

∏β∈v

−(y0β, y1β)

⎞⎠⎞⎠

= f lα

⎛⎝∏

β∈u

y0β ·∏β∈v

−y0β, 0

⎞⎠

= f lα((0, 0)) = 0.

Thus (3) holds.

By (1)–(3) we have g0α(y0α) �= g1α(y

0α), while ∀β ∈ λ\{α}[g0α(y0β) = g1α(y

1β)].

Hence by Theorem 1, 〈y0β : β < λ〉 is irredundant.Case 2. {α < λ : f0

α(1, 0) = f1α(1, 0) = 0} has size λ. This is similar to Case 1,

working with (0, 1) rather than (1, 0).

Case 3. {α < λ : f0α(1, 0) = 1 and f1

α(1, 0) = 0} has size λ. Hence we may assumethat

(4) ∀α < λ[f0α(1, 0) = 1 and f1

α(1, 0) = 0].

Now for each l ∈ 2 and α < λ define glα : λ→ 2 by setting glα(β) = f lα(y

0β , y

1β).

(5) If α �= β then g0α(β) = g1α(β).

(6) g0α(α) �= g1α(α).

These are immediate from (1)–(2).

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For l ∈ 2 let Fl = {glα : α < λ}. By Proposition 8.2 we have:

(7) For each l ∈ 2, the function (glα)hom is a homomorphism from B[Fl] into 2

such that (glα)hom([xβ ]Fl

) = glα(β) for every β < λ.

(8) For each l ∈ 2 there is a homomorphism kl from 〈{ylα : α < λ}〉 into B[Fl]such that kl(y

lα) = [xα]Fl

for all α < λ.

To prove this it suffices to assume that M ∈ [λ]<ω, ε ∈ M2, and∏

β∈M [xβ ]ε(β)Fl

�= 0

and show that∏

β∈M (ylβ)ε(β) �= 0. By Proposition 8.3 there is an α < λ such that

ε ⊆ glα. Now

f lα

⎛⎝∏

β∈M

(y0β , y1β)

ε(β)

⎞⎠ =

∏β∈M

(f lα(y

0β , y

1β))

ε(β) =∏β∈M

(glα(β))ε(β) = 1.

If l = 0, then

1 = f0α

⎛⎝(1, 0) ·

⎛⎝∏

β∈M

(y0β, y1β)

ε(β)

⎞⎠⎞⎠ = f0

α

⎛⎝∏

β∈M

(y0β)ε(β), 0

⎞⎠ ,

and so∏

β∈M (y0β)ε(β) �= 0. Similarly if l = 1. Hence (8) holds.

(9) Let l ∈ 2 and A ⊆ λ, and suppose that for every α ∈ A there is a homo-morphism hα : 〈[xβ ]Fl

: β ∈ A〉 → 2 such that hα([xβ ]Fl) = g1−l

α (β) for everyβ ∈ A.

Then 〈ylβ : β ∈ A〉 is irredundant in Bl.

To prove (9), first note that for any α ∈ A we have

hα([xα]Fl) = g1−l

α (α) �= glα(α) = (glα)hom([xα]Fl

),

while if β ∈ A\{α} we have

hα([xβ ]Fl) = g1−l

α (β) = glα(β) = (glα)hom([xβ ]Fl

).

It follows that 〈[xα]Fl: α ∈ A〉 is irredundant. Hence by (8), also 〈ylβ : β ∈ A〉 is

irredundant in Bl.

Now by (9) we may assume:

(10) For all A ∈ [λ]λ¬∀α ∈ A∃hα : 〈[xβ ]F0 : β ∈ A〉 → 2 such that hα([xβ ]F0) =g1α(β) for every β ∈ A.

We now construct by recursion sequences 〈αξ : ξ < λ〉 and 〈(uξ, vξ) : ξ < λ〉 sothat the following conditions hold:

(11) uξ, vξ ∈ [λ]<ω are disjoint.

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(12) (uξ ∪ vξ) ∩⋃

ζ<ξ(uζ ∪ vζ) = ∅.

(13)∏

γ∈uξ[xγ ]F0 ·

∏γ∈vξ

−[xγ ]F0 = 0.

(14)∏

γ∈uξ[xγ ]F1 ·

∏γ∈vξ

−[xγ ]F1 �= 0.

(15) ∀β ∈ uξ[g1αξ(β) = 1] and ∀β ∈ vξ[g

1αξ(β) = 0].

(16) αξ ∈ uξ ∪ vξ.

Suppose that we have constructed uζ and vζ for all ζ < ξ so that (11)–(16) hold.

Then Adef= λ\

⋃ζ<ξ(uζ ∪ vζ) has size λ. Then by (10), there is an αξ ∈ A such

that the relation

{([xβ ]F0 , g1αξ(β)) : β ∈ A}

does not extend to a homomorphism. By Sikorski’s extension criterion, this meansthat there are finite disjoint u′

ξ, v′ξ ⊆ A such that

∏γ∈u′

ξ[xγ ]F0 ·

∏γ∈v′

ξ−[xγ ]F0 = 0

while g1αξ[u′

ξ] ⊆ {1} and g1αξ[v′ξ] ⊆ {0}. Define

uξ =

{u′ξ ∪ {αξ} if g1αξ

(αξ) = 1,

u′ξ otherwise;

vξ =

{v′ξ ∪ {αξ} if g1αξ

(αξ) = 0,

v′ξ otherwise.

Clearly all conditions except possibly (14) hold; (14) follows from (7).

(17) For any ξ, ζ < λ, (∀β ∈ uξ[g1αζ(β) = 1] and ∀β ∈ vξ[g

1αζ(β) = 0]) iff ξ = ζ.

In fact, if ξ = ζ then (∀β ∈ uξ[g1αζ(β) = 1] and ∀β ∈ vξ[g

1αζ(β) = 0]) by (15). If

ξ �= ζ, then (∃β ∈ uξ[g0αζ(β) �= 1] or ∃β ∈ vξ[g

0αζ(β) �= 0]) by (13) and (7); and

αζ /∈ uξ ∪ vξ and hence by (5) g1αζ� (uξ ∪ vξ) = g0αζ

� (uξ ∪ vξ), so it follows that

(∃β ∈ uξ[g1αζ(β) �= 1] or ∃β ∈ vξ[g

1αζ(β) �= 0]). So (17) holds.

Now for each ξ < λ let

zξ =∏γ∈uξ

y1γ ·∏γ∈vξ

−y1γ .

(18) For each ξ < λ there is a homomorphism hξ : 〈{y1β : β < λ〉 → 2 such that

hξ(y1β) = g1αξ

(β) for every ξ < λ.

In fact, suppose that a, b ∈ [λ]<ω are disjoint and∏

β∈a g1αξ(β)·

∏β∈b−g1αξ

(β) �= 0.

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Then1 = f1

αξ(0, 1) ·

∏β∈a

f1αξ(y0β , y

1β) ·

∏β∈b

−f1αξ(y0β , y

1β)

= f1αξ

⎛⎝(0, 1) ·

∏β∈a

(y0β, y1β) ·

∏β∈b

−(y0β , y1β)

⎞⎠

= f1αξ

⎛⎝0,

∏β∈a

y1β ·∏β∈b

−y1β

⎞⎠ ,

and it follows that∏

β∈a y1β ·∏

β∈b−y1β �= 0, proving (18).

Now we have, for any ξ, ζ < λ, hζ(zξ) =∏

γ∈uξg1αζ

(γ) ·∏

γ∈vξ−g1αζ

(γ), and

hence by (17), hζ(zξ) = 1 iff (∀β ∈ uξ[g1αζ(β) = 1] and ∀β ∈ vξ[g

1αζ(β) = 0]) iff

ζ = ξ. So 〈zξ : ξ < λ〉 is irredundant.Case 4. {α < λ : f0

α(1, 0) = 0 and f1α(1, 0) = 1} has size λ. Let k0α = f1

α andk1α = f0

α for all α < λ. Then {α < λ : k0α(1, 0) = 1 and k1α(1, 0) = 0} has size λ,and we can apply Case 3 to get the desired result. �Corollary 8.5. If A is infinite and B is finite, then Irr(A) = Irr(A×B). �Corollary 8.6. If |B| ≤ Irr(A), then Irr(A×B) = Irr(A). �Corollary 8.7. If 〈Ai : i ∈ I〉 is a system of infinite BAs, with I infinite, then

Irr

(w∏i∈I

Ai

)= max{|I|, sup

i∈IIrr(Ai)}.


i∈I Ai. Clearly ≥ holds. Now suppose that X ⊆ B isirredundant and |X | = (max{|I|, supi∈I Irr(Ai)})+; we want to get a contradiction.For each F ∈ [I]<ω let CF be the subalgebra of B consisting of all elements withsupport contained in F . Clearly

X =⋃

F∈[I]<ω

(X ∩CF ).

Clearly each set X ∩ CF is irredundant. Now there is an F ∈ [I]<ω such that|X | = |X ∩ CF |. Clearly CF

∼=∏

i∈F Ai, so |X ∩ CF | > Irr(CF ), contradictingTheorem 8.4. �

Note that if 〈Ai : i ∈ I〉 is a system of infinite BAs and Xi ⊆ Ai is irredundantfor each i ∈ I, then

∏i∈I Ai has an irredundant subset of size |

∏i∈I Xi|. For, by

Lemma 13.9 of the Handbook, let Y ⊆∏

i∈I Xi be finitely distinguished and ofsize |

∏i∈I Xi|. Clearly Y is irredundant. This leads to the following problem.

Problem 87. For 〈Ai : i ∈ I〉 a system of infinite BAs with I infinite, determine,if possible, Irr

(∏i∈I Ai

)in terms of 〈Irr(Ai) : i ∈ I〉.

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Obviously max(Irr(A), Irr(B)) ≤ Irr(A ⊕ B). We do not know whether equalityholds here.

Problem 88. For infinite BAs A,B is it true that

max(Irr(A), Irr(B)) = Irr(A⊕B)?

Irredundance is an ultra-sup function, so Theorems 3.20–3.22 of Peterson ap-ply. By Theorem 3.22, Irr

(∏i∈I Ai/F

)≥∣∣∏

i∈I Irr(Ai)/F∣∣ for regular F ; hence

by Donder’s result it is consistent that this inequality always holds. Shelah in[03] showed that it is consistent to have Irr

(∏i∈I Ai/F

)<∣∣∏

i∈I Irr(Ai)/F∣∣.

On the other hand, Shelah in [99] shows (in 15.10) that it is consistent to haveIrr(∏

i∈I Ai/F)>∣∣∏

i∈I Irr(Ai)/F∣∣. These results answer problems 25 and 26 in

Monk [96].

Concerning the derived functions, we note just the obvious facts that IrrS+(A) =Irr(A), IrrS−(A) = ω, and dIrrS+(A) = Irr(A).

Obviously any chain is irredundant; so Length(A) ≤ Irr(A). The difference canbe large, e.g., in a free BA. By Theorem 4.25 of Part I of the BA handbook,π(A) ≤ Irr(A). In particular, if |A| is strong limit, then |A| = Irr(A), since thenπ(A) = |A|.

It is consistent that there is a BA with irredundance less than cardinality, andit is also consistent that every uncountable BA has uncountable irredundance (seeTodorcevic [93]). Now we prove the first fact, in the form that under CH there is aBA of power ω1 with countable irredundance. We give two examples for this. Thefirst example is a compact Kunen line. We say “a” since there are various Kunenlines, and we say “compact” since the standard Kunen lines are only locally com-pact. For the Kunen lines, see Juhasz, Kunen, Rudin [76]. The second constructionuses considerably less than CH, and can be found in Todorcevic [89]. For a forcingconstruction of an uncountable BA with countable irredundance, see Bell, Gins-burg, Todorcevic [82]. A generalization of the main results about irredundance (toother varieties of universal algebras) can be found in Heindorf [89a].

The history of these results is complicated. I think that the first example ofan uncountable BA with countable irredundance is due to Rubin [83] (the resultwas obtained several years before 1983). The papers with the constructions wegive do not mention irredundance; their relevance for our purposes is due to asimple theorem of Heindorf [89a]. So, modulo the simple theorem of Heindorf, thefirst example with irredundance different from cardinality is a Kunen line.

Before beginning the examples we need two topological lemmas. A topologicalspace is hereditarily separable iff every subset of it has a countable dense subset.

Lemma 8.8. Suppose that X is a non-compact, locally compact Hausdorff space,and Y is its one-point compactification. Then:

(i) If the compact-open sets of X form a base, then Y is a Boolean space.

(ii) For every integer k > 0, if kX is hereditarily separable, then so is kY .

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Proof. Recall that Y is obtained from X by adding one new point y, and declaringthe topology on Y to consist of all open sets U of X together with all sets {y}∪Usuch that U ⊆ X and X\U is compact in X . Note that if U ⊆ X and X\U iscompact in X then U is open in X .

(i): We want to show that the clopen subsets of Y form a base. Any subsetof X which is compact and open in X is clopen in Y . So it suffices to show thateach “new” basic open set contains a new basic open set which is clopen. So, letW be a new basic open set – say W = {y}∪U where U ⊆ X and X\U is compactin X . Then X\U ⊆ V for some compact open subset V of X , by the compactnessof X\U and the hypothesis of (i). Thus {y} ∪ (X\V ) ⊆ W and {y} ∪ (X\V ) isclopen in Y , as desired.

(ii): Fix x ∈ X . Assume that k is a positive integer and kX is hereditarilyseparable. Now suppose that S is a non-empty subspace of kY . For each Γ ⊆ k let

SΓ = {z ∈ S : ∀i < k(zi = y iff i ∈ Γ)};S′Γ = {w ∈ kX : ∃z ∈ SΓ∀i < k[(i ∈ Γ⇒ wi = x) and (i /∈ Γ⇒ wi = zi)]}.

Then for each Γ ⊆ k let C′Γ be a countable dense subset of S′

Γ. Next, let

CΓ = {z ∈ kY : ∃w ∈ C′Γ∀i < k[(i ∈ Γ⇒ zi = y) and (i /∈ Γ→ zi = wi)]}.

We claim that Ddef=⋃

Γ⊆k CΓ is dense in S (as desired). To this end, take an openset U such that U ∩ S �= 0. We may assume that U has the form V0 × · · · × Vk−1,where each Vi is open in Y . Say U ∩ SΓ �= 0. Define, for i < k,

V ′i =

{X, if i ∈ Γ,

Vi ∩X, if i /∈ Γ.

Set U ′ = V ′0 × · · · × V ′

k−1. Then U ′ ∩ S′Γ �= 0. In fact, if z ∈ U ∩ SΓ, define

wi =

{zi if zi �= y,

x otherwise.

Then w ∈ U ′ ∩ S′Γ. Hence U ′ ∩ C′

Γ �= 0. Take w ∈ U ′ ∩ C′Γ. Define, for i < k,

zi =

{y, if i ∈ Γ,

wi, if i /∈ Γ.

Then z ∈ U ∩ CΓ, as desired. �Lemma 8.9. If X and Y are topological spaces, X is hereditarily separable, and Yis second countable, then X × Y is hereditarily separable.

Proof. Let A be a nonempty subset of X × Y . Let A be a countable base for Y .For each U ∈ A let BU = {x ∈ X : ({x} × U) ∩ A �= ∅}. Since X is hereditarilyseparable, we can find a countable dense subset CU of each nonempty set BU .

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For each U with BU nonempty and each x ∈ CU choose yxU ∈ U such that

(x, yxU ) ∈ A. We claim now that Ddef= {(x, yxU ) : U ∈ A , BU �= ∅, x ∈ CU} is

dense in A. Clearly D is countable. To prove the claim, suppose that V and Uare open in X and Y respectively, and A ∩ (V × U) �= ∅. We want to show thatD ∩ (V × U) �= ∅. We may assume that U ∈ A . Note that V ∩ BU �= ∅. Choosex ∈ CU ∩ V . Then yxU ∈ U , and hence (x, yxU ) ∈ D. �Theorem 8.10 (CH). There is a topological space X satisfying the following con-ditions:

(i) The compact open sets form a base for the topology on X.

(ii) X is first-countable and Hausdorff.

(iii) kX is hereditarily separable for each k ∈ ω\1.(iv) X has ω1 compact open sets.

Proof. In this construction we follow de la Vega, Kunen [04]. Let r be a bijectionfrom ω1 onto R, and let ρ be the topology {U ⊆ ω1 : r[U ] open in R} on ω1.For each ξ < ω1 we denote by ρξ the subspace topology on ξ determined byρ. If n ∈ ω\1 and ξ < ω1, then Iseq(n, ξ) is the set of all f ∈ nω1 such thatf(0) < f(1) < · · · < f(n− 1) = ξ.

Now we claim

(1) There is an enumeration 〈Sμ : μ ∈ ω1\1〉 of⋃

k∈ω\1[kω1]

≤ω such that for each

μ ∈ ω1\1 there is an n(μ) ∈ ω\1 such that Sμ ⊆ n(μ)μ.

To see this, first let 〈S′μ : μ < ω1〉 be any enumeration of

⋃k∈ω\1[

kω1]≤ω. Suppose

that μ ∈ ω1\1 and Sν has been defined for all positive ν < μ. Let Sμ = S′ρ, where

ρ is minimum such that S′ρ ⊆ kμ for some k ∈ ω\1 and S′

ρ /∈ {Sν : ν < μ}. To seethat this is the desired enumeration, we just need to see that any S′

ρ appears assome Sμ. Suppose that S′

ρ is not in {Sμ : μ < ω1}, and choose ρ minimum with

this property. Say S′ρ ⊆ kτ for some k ∈ ω\1 and τ < ω1. For each σ < ρ let ν(σ)

be such that S′σ = Sν(σ). Let μ < ω1 be such that ν(σ) < μ for all σ < ρ and

τ ≤ μ. Thus S′ρ ⊆ kμ and S′

ρ /∈ {Sν : ν < μ}. If σ < ρ, then S′σ ∈ {Sν : ν < μ}.

Hence ρ is minimum such that S′ρ ⊆ kμ and S′

ρ /∈ {Sν : ν < μ}. So Sμ = S′ρ,

contradiction. This proves (1).

Now we define by recursion 〈τξ : 1 ≤ ξ ∈ ω1〉 so that the following conditionshold for each η ≤ ω1 with η �= 0:

(2) τη is a topology on η.

(3) If 0 < ξ < η, then τξ = τη ∩P(ξ).

(4) τη is first countable and Hausdorff.

(5) The compact open sets form a base for τη.

(6) ρη ⊆ τη.

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(7) If η = ξ + 1, 0 < μ < ξ, f ∈ Iseq(n(μ), ξ), and f ∈ cl(Sμ,n(μ)−1τη × ρη), then

f ∈ cl(Sμ,n(μ)τη).

For 0 < ξ ≤ ω we let τξ = P(ξ). Clearly (2)–(7) hold for ξ. For η > ω limit, defineτη = {U ⊆ η : ∀ξ < η[U ∩ ξ ∈ τξ]}. Again it is clear that (2)–(7) hold.

Now suppose that η = ξ + 1, with η infinite. We claim:

(8) If A is a countable collection of subsets of ξ, then there is a topology τ ′η on ηsuch that (2)–(6) hold for τ ′η, and also ∀E ∈ A [ξ ∈ cl(E, ρ)→ ξ ∈ cl(E, τ ′η)].

To prove (8), let A ′ = {E ∈ A : ξ ∈ cl(E, ρ)}. If A ′ = ∅, we can let τ ′η =τξ ∪ {{ξ}} ∪ {η}, and the desired conclusions follow. Now assume that A ′ �= ∅.Let 〈Un : n ∈ ω〉 be a nested countable base for ξ in the topology ρξ. Let 〈Vm :m < ω〉 enumerate A ′, with each member of A ′ appearing ω times. For eachm ∈ ω choose pm ∈ Vm ∩ Um, and by (5) and (6) let Km be compact open inτξ with pm ∈ Km ⊆ Um. Then we define τ ′η to be the topology on η with baseτξ ∪ {{ξ} ∪

⋃m>n Km : n ∈ ω}.

Clearly (2)–(3) hold, and τ ′η is first countable. To show that it is Hausdorff,it suffices to find disjoint neighborhoods of ξ and any η < ξ. Let V,W be openin ρ with η ∈ V , ξ ∈ W , V ∩W = ∅. Choose n such that ξ ∈ Un ⊆ W . ThenV and {ξ} ∪

⋃m>n Km are the desired disjoint neighborhoods. So (4) holds. To

prove (5) it suffices to show that for any n ∈ ω the set {ξ}∪⋃

m>nKm is compact.Let B be an open cover of it by base elements. Then there is a p ∈ ω such that{ξ} ∪

⋃m>p Km ∈ B. Wlog n < p. Then there is a finite B′ ⊆ B such that⋃

p<m≤n KM ⊆⋃

B′. Thus

{ξ} ∪⋃

m>n

Km ⊆ {ξ} ∪⋃m>p

Km ∪⋃

B′,

as desired; (5) holds.

For (6), suppose that V ∈ ρ. Then V ∩ξ ∈ ρξ ⊆ τα ⊆ τ ′η. Suppose that ξ ∈ V .Choose m ∈ ω such that ξ ∈ Um ⊆ V . Then

V ∩ η = (V ∩ ξ) ∪ {ξ} ∪⋃n>m

Kn ∈ τ ′η,

and (6) holds.

Now for the final condition of (8), suppose that E ∈ A and ξ ∈ cl(E, ρ).Thus E ∈ A ′. Suppose that W is a basic neighborhood of ξ in τ ′η. Say W ={ξ} ∪

⋃m>n Km. Choose m > n with E = Vm. Then pm ∈ Km ∩ Vm, and so

W ∩ E �= ∅. This proves (8).Now to actually define τη we proceed as follows. For each μ with 0 < μ < ξ

and f ∈ Iseq(n(μ), ξ) we associate a countable subset Bμf of ξ, as follows. Ifn(μ) = 1, we set

Bμ = {ρ < ω1 : {(0, ρ)} ∈ Sμ}.

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Now suppose that n(μ) > 1. By (4), let Cf be a countable neighborhood base forf � (n(μ)− 1) in n(μ)−1ξ under the topology τξ, and suppose that U ∈ Cf . Define

BμfU = {σ < ξ : ∃g ∈ Sμ[g � (n(μ)− 1) ∈ U and g(n(μ)− 1) = σ]}.

Finally, let

Cf =⋃{Bμ : μ < ξ, n(μ) = 1} ∪

⋃{BμfU : μ < ξ, n(μ) > 1, and U ∈ Cf};

A =⋃{Cf : n ∈ ω\1, f ∈ Iseq(n, ξ)}.

Thus A is a countable collection of subsets of ξ. We then apply (8) to get ourtopology τη. We now check (7) for η. Suppose that 0 < μ < ξ, f ∈ Iseq((n(μ), ξ),and f ∈ cl(Sμ,

n(μ)−1τη × ρη). To show that f ∈ cl(Sμ,n(μ)τη), let V ∈ n(μ)τη be a

system of basic open sets such that f ∈ V . Choose U ∈ Cf such that Ui ⊆ Vi forall i < n(μ)− 1.

Case 1. n(μ) = 1. Thus f={0,ξ)}, and so the assumption that f ∈cl(Sμ,n(μ)−1τη×

ρη) means that ξ ∈ cl(Bμ, ρη). Hence by (8) we get ξ ∈ cl(Bμ, τη), so that f ∈cl(Sμ, τη).

Case 2. n(μ) > 1. We claim

(9) ξ ∈ cl(BμfU , ρη).

To prove this, letW be any neighborhood of ξ in ρη. Then U×W is a neighborhoodof f in n(μ)−1τη×ρη, so we can choose g ∈ Sμ∩n(μ)−1 τη×ρη. Hence g(n(μ)−1) ∈BμfU ∩W , as desired for (9),

Now by (8) we get ξ ∈ cl(BμfU , τη). Now ξ ∈ Vn(μ)−1, so it follows that thereis a σ ∈ BμfU ∩ Vn(μ)−1. By the definition of BμfU , there is a g ∈ Sμ such thatf � (n(μ)− 1) ∈ U and g(n(μ)− 1) = σ. So g ∈ Sμ ∩ V , as desired.

This proves (7) for η, and the construction is finished.

Now we let X be ω1 with the topology τω1 . So (i) and (ii) of the lemmahold. For (iii), we first show that X itself is separable. So, let ∅ �= A ⊆ X . NowA is separable in the topology ρ, so let D be a countable dense subset of A inthe ρ sense. Fix μ < ω1 such that n(μ) = 1 and Bμ = D, where as aboveBμ = {σ : {(0, σ)} ∈ Sμ}. Suppose that μ < ξ < ω1. Let f = {(0, ξ)}. Thenf ∈ Iseq(n(μ), ξ) and f ∈ cl(Sμ, ρξ+1), so by (7), f ∈ cl(Sμ, τξ+1). This showsthat A\cl(Sμ, τξ+1) ⊆ (μ+ 1). Hence

Sμ ∪ (A\cl(Sμ, τξ+1))

is a countable dense subset of A in the τω1 sense.

Now suppose that we have shown that mX is hereditarily separable for everypositive m < n, where n > 1; we want to show that nX is hereditarily separable.Let ∅ �= A ⊆ nX . For each f ∈ A let k(f) = {(i, j) ∈ n × n : f(i) = f(j)}.

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Thus k(f) is an equivalence relation on n. For each m with 1 ≤ m ≤ n letBm = {f ∈ A : | rng(f)| = m}. So A =

⋃1≤m≤n Bm. Hence it suffices to show

that each Bm is separable. First assume that 1 ≤ m < n. For each f ∈ Bm, let〈Ki : i < m〉 enumerate the k(f)-equivalence classes, so that

(10) 0 ∈ K0, and for each i with 0 < i < m, the least element of Ki is the leastj < n such that j /∈

⋃k<i Kk.

Now for each f ∈ Bm, define gf ∈ mα1 by setting gf (i) = f(j) for some, andhence every, j ∈ Ki, for each i < m. Note that if f, h ∈ Bm with f �= h, thengf �= gh. Let Cm = {gf : f ∈ Bm}. So Cm ⊆ mX , and hence by the inductivehypothesis Cm has a countable dense subset Dm. Let Em = {f ∈ Bm : gf ∈ Dm}.Since gf �= gh for f �= h, it follows that Em is countable. We claim that it isdense in Bm. For, let U ∈ nτω1 be such that U ∩ Bm �= ∅. Say f ∈ U ∩ Bm. Foreach i < m let Vi =

⋂j∈Ki

Uj. Then V ∈ mτω1 and gf ∈ V ∩ Cm. Hence there isan h ∈ V ∩ Dm. Say h = gk with k ∈ Bm. So k ∈ Em. Clearly also k ∈ U . SoU ∩Em �= ∅, as desired. So we have shown that Bm is separable.

It remains only to show that Bn is separable. Now each permutation π ofn induces an autohomeomorphism πe of nX , defined by πe(f) = f ◦ π for everyf ∈ nX . Let P be the set of all permutations of n. Now by the inductive hypothesisand Lemma 1, n−1τω1 × ρ is hereditarily separable. For each π ∈ P let μ(π)be such that ν(μ(π)) = n and Sμ(π) is (n−1τω1 × ρ)-dense in πe[A]. Note thatπ−1e [Sμ(π)] ⊆ A.

We claim that the countable set

{f ∈ A ∩ n[0,max{μ(π) : π ∈ P}]} ∪⋃π∈P

π−1e [Sμ(π)]

is dense in A in the topology nτω1 . For, suppose that f ∈ A and max(rng(f)) >max{μ(π) : π ∈ P}. Choose π ∈ P so that f ◦ π is strictly increasing, and letξ = f(π(n− 1)). Then ξ > μ(π) and f ◦ π ∈ πe[A], and hence

f ◦ π ∈ cl(Sμ(π),n−1τω1 × ρ).

Moreover, f ◦ π ∈ Iseq(n(μ(π)), ξ). Hence by (7),

f ◦ π ∈ cl(Sμ(π),nτω1).

It follows thatf ∈ clπ−1

e (Sμ(π),nτω1).

For condition (iv) of the theorem, note that in each nontrivial step from ξ toξ + 1 countably many new compact open sets were introduced: all of the sets{ξ} ∪

⋃m>n Km. �

The theorem of Heindorf mentioned above is as follows.

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Theorem 8.11. Let X be a Boolean space, and A its BA of closed-open sets. ThenIrr(A) ≤ s(X ×X).

Proof. Suppose that I is an infinite irredundant subset of A; we will produce anideal independent subset of A⊕A of power |I| (as desired). Namely, take the set{a×−a : a ∈ I}; it is as desired, for suppose that

a×−a ⊆ (b0 ×−b0) ∪ · · · ∪ (bm−1 ×−bm−1),

where a, b0, . . . , bm−1 are distinct elements of I. Now a is not in 〈{bi : i < m}〉, soit follows that in that subalgebra, a splits some atom; this means that there is anε ∈ m2 such that, if we set d =

⋂i<m bεii then we have d∩a �= 0 �= d∩−a. Choosing

x ∈ d ∩ a and y ∈ d ∩ −a it follows that (x, y) ∈ a× −a but (x, y) /∈ bi × −bi foreach i < m, giving the desired contradiction. �

Roslanowski, Shelah [00] show that it is consistent to have Irr(A) < s(A ⊕ A),answering Problem 27 of Monk [96].

Theorem 8.12. (CH) There is a BA A of size ω1 with countable irredundance.

Proof. Recall that for any BA A, s(A) = cH+(A) ≤ dH+(A) = hd(A). �

Example 8.13. This example, which as we mentioned is from Todorcevic [89],constructs a topology on a certain subset of ωω. First, some notation: If A isa set with a linear order < on it, and if k ∈ ω, then 〈A〉k denotes the set ofall f ∈ kA such that fi < fj for all i < j < k. For f, g ∈ ωω define f <∗ gif ∃m∀n ≥ m(f(n) < g(n)). The BA we want will be constructed under theassumption that there is a subset A of ωω of power ω1 which is unbounded under<∗. This means that b = ω1, where b is a well-known cardinal. It is easy to seethat ω1 ≤ b ≤ 2ω, and it is consistent to have b = ω1 < 2ω.

Without loss of generality A has order type ω1 under <∗ and all membersof A are strictly increasing. In fact, take the A originally given, and write A ={fα : α < ω1}. Then one can inductively define fα for α < ω1 so that fβ <∗ fαfor β < α, fα <∗ fα, and fα is strictly increasing. Namely, let f0 be arbitrary. Iffβ has been constructed for all β < α, let 〈gn : n < ω〉 enumerate 〈fβ : β < α〉.Define fα(n) to be > fα(m) for all m < n, also > fα(n), and also > gm(n) for allm < n. Clearly this works. The new set {fα : α < ω1} (still denoted by A below)has the desired properties.

We will apply the above notation 〈A〉k to A under the ordering <∗. Let Tbe an Aronszajn subtree of {s ∈ <ω1ω : s is one-one}. (See Kunen [80], p. 70.)For each α < ω1 let tα be a member of T with domain α. Define e : 〈A〉2 → ω bye(fα, fβ) = tβ(α) for α < β. Then clearly

(1) For all b ∈ A, the function ebdef= e(·, b) is a one-one map from Ab

def= {a ∈ A :

a <∗ b} into ω.

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Also,

(2) For all a ∈ A, the set {eb � Aa : b ∈ A, a <∗ b} is countable.

In fact, for each e ∈ A write e = fβ(e). Then for any b, d such that a <∗ b anda <∗ d we have

(eb � Aa) �= (ed � Aa) iff 〈eb(c) : c ∈ Aa〉 �= 〈ed(c) : c ∈ Aa〉iff 〈e(c, b) : c <∗ a〉 �= 〈e(c, d) : c <∗ a〉iff 〈tβ(b)(β(c)) : c <∗ a〉 �= 〈tβ(d)(β(c)) : c <∗ a〉iff there is a γ < β(a) such that tβ(b)(γ) �= tβ(d)(γ);

hence (eb � Aa) �= (ed � Aa) implies that tβ(b) � β(a) and tβ(d) � β(a) are incompa-rable. Since level β(a) of the tree T is countable, (2) follows.

For distinct a, b ∈ A let Δ(a, b) be the least n < ω such that a(n) �= b(n).And let Δ(a, a) =∞. The following fact about this notation will be useful:

(�) If Δ(a, b) < Δ(c, b) then Δ(a, c) = Δ(a, b).

To see this, note that a � Δ(a, b) = b � Δ(a, b) = c � Δ(a, b), and a(Δ(a, b)) �=b(Δ(a, b)) = c(Δ(a, b)), so Δ(a, c) = Δ(a, b).

Now we define H : A→P(A) by

H(b) = {a ∈ A : a <∗ b and e(a, b) ≤ b(Δ(a, b))}.

Now

(3) for all l < ω and b ∈ A the set {a ∈ H(b) : Δ(a, b) = l} is finite.

In fact, if a ∈ H(b) and Δ(a, b) = l, then tβ(b)(β(a)) = e(fβ(a), fβ(b)) = e(a, b) ≤b(l). Since b(l) ∈ ω and tβ(b) is one-one, (3) follows.

Next we define C(b) for b ∈ A by recursion on b: a ∈ C(b) iff a = b or

(4) ∃c ∈ H(b)(a ∈ C(c) and ∀d ∈ H(b)(d �= a and d �= c⇒ Δ(a, d) < Δ(a, c))).

Note that

(5) H(b) ⊆ C(b)

for all b ∈ A (if a ∈ H(b), take c = a and note that Δ(a, a) =∞). Also we have

(6) If a ∈ C(b), then a <∗ b or a = b.

We prove (6) by induction on b. Assume that it is true for all c <∗ b, and supposethat a ∈ C(b) and a �= b. Then there is a c ∈ H(b) such that a ∈ C(c). We havec <∗ b since c ∈ H(b), and a <∗ c by the inductive hypothesis. So a <∗ b.

For each n ∈ ω and b ∈ A let Cn(b) = {a ∈ C(b) : Δ(a, b) ≥ n}. Then

(7) c ∈ H(b)⇒ ∃l(Cl(c) ⊆ C(b)).

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In fact, {x ∈ H(b) : Δ(x, b) = Δ(c, b)} is finite by (3). Choose l > Δ(x, c) forany x �= c which is in this set, and with l > Δ(c, b). Suppose that d ∈ Cl(c). Weclaim that d ∈ C(b), and that the element c works to show this in (4). Indeed,suppose that x ∈ H(b), x �= d, x �= c, and Δ(d, x) ≥ Δ(d, c). (In (4), replacea, b, c, d by d, b, c, x respectively.) Now Δ(c, b) < Δ(d, c), so Δ(b, d) = Δ(c, b) by(�), and Δ(b, d) < Δ(d, x), so Δ(x, b) = Δ(c, b) by (�). So x is in the indicatedfinite set, which gives Δ(x, c) < l ≤ Δ(c, d) ≤ Δ(d, x), so Δ(c, d) = Δ(x, c) by (�),contradiction.

(8) a ∈ C(b)⇒ ∃l(Cl(a) ⊆ C(b)).

For, we may assume that a /∈ H(b) by (7), and we proceed by induction on b. Theconclusion is clear if a = b, so suppose that a �= b. Choose c in accordance with(4). Then a �= c since a /∈ H(b). By the induction hypothesis, choose l such thatCl(a) ⊆ C(c). Without loss of generality, l > Δ(a, c). We claim that Cl(a) ⊆ C(b).To prove this, let d ∈ Cl(a). So, d ∈ C(c). Suppose x ∈ H(b), x �= d, and x �= c.(We are going to show that d ∈ C(b) by replacing a, b, c, d in (4) by d, b, c, xrespectively.) Then x �= a since a /∈ H(b). So Δ(a, x) < Δ(a, c) by the choice of c.Now Δ(a, c) < l ≤ Δ(a, d), so Δ(c, d) = Δ(a, c) by (�). Also, Δ(a, x) < Δ(a, d),so Δ(d, x) = Δ(a, x) < Δ(a, c) = Δ(c, d), showing that d ∈ C(b).

Now

(9) a ∈ Cm(b)⇒ ∃l(Cl(a) ⊆ Cm(b)).

In fact, suppose that a ∈ Cm(b). By (8), choose l such that Cl(a) ⊆ C(b). Letp = max{l,m}. We claim that Cp(a) ⊆ Cm(b). For, suppose that c ∈ Cp(a). Thusc ∈ C(a) and Δ(a, c) ≥ p. Since c ∈ Cp(a) ⊆ Cl(a) ⊆ C(b), we have c ∈ C(b).Also, c � m = a � m = b � m, so Δ(b, c) ≥ m, as desired.

From (9) it follows that the collection of sets {Cm(b) : b ∈ A,m ∈ ω} formsa base for a topology on A. It is Hausdorff, since, given a �= b, let l = Δ(a, b) + 1;clearly Cl(a) ∩Cl(b) = 0. Also note that each set C(b) = C0(b) is open. Next,

(10) Cl(b) is closed in C(b).

For, suppose that x ∈ C(b)\Cl(b), and let m = Δ(x, b)+1. Now Cm(x)∩Cl(b) = ∅.For suppose that d ∈ Cm(x)∩Cl(b). Then Δ(x, d) ≥ m, Δ(d, b) ≥ l, and Δ(b, x) <l. So Δ(b, x) < Δ(d, b), hence by (�) m ≤ Δ(d, x) = Δ(b, x) < m, contradiction.Then Cm(x) ∩ C(b) ⊆ C(b)\Cl(b), as desired.

(11) C(b) is compact.

We prove this by induction on b. So, assume that it is true for all c <∗ b, andsuppose that C(b) ⊆

⋃x∈X Cm(x)(x). Then choose y ∈ X such that b ∈ Cm(y)(y).

There is an l such that Cl(b) ⊆ Cm(y)(y). Now we consider two cases:

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Case 1. H(b) is finite. In this case, we can easily show that C(b) is closed: supposethat a ∈ A\C(b). Hence a �= b and

(∗) ∀c ∈ H(b)[a /∈ C(c) or ∃d ∈ H(b)[d �= a and d �= c and Δ(a, d) ≥ Δ(a, c)]].

If c ∈ H(b) and a /∈ C(c), choose an open neighborhood Uc of a with the propertythat Uc ∩ C(c) = 0, using the inductive hypothesis. For c ∈ H(b) and a ∈ C(c),choose d = d(a, c) ∈ H(b) such that d �= a, d �= c, and Δ(a, d) ≥ Δ(a, c). Let

V = CΔ(a,b)+1(a) ∩⋂

c∈H(b)a/∈C(c)

Uc ∩⋂

c∈H(b)a∈C(c)

CΔ(a,d(a,c))+1(a).

We claim that V ∩C(b) = 0 (as desired, showing that C(b) is closed). For, supposethat x ∈ V ∩C(b). Since x ∈ CΔ(a,b)+1(a), we have x �= b. Choose, then, c ∈ H(b)such that x ∈ C(c) and for all d ∈ H(b), if d �= x and d �= c then Δ(x, d) < Δ(x, c).If a /∈ C(c), then x ∈ Uc ∩ C(c), contradiction. So a ∈ C(c). Set d = d(a, c). Nowx ∈ CΔ(a,d)+1(a), so x �= d and Δ(a, x) > Δ(a, d). Hence Δ(d, x) = Δ(a, d) by (�).Now Δ(a, d) ≥ Δ(a, c), so Δ(a, c) < Δ(a, x). Hence by (�), Δ(c, x) = Δ(a, c) ≤Δ(a, d) = Δ(d, x), contradicting the choice of c.

Now for each c ∈ H(b) we have that C(c) ∩ C(b) is a closed subset of C(c).Note that H(b) = {b} ∪

⋃c∈H(b)(C(c) ∩ C(b)). Hence the inductive hypothesis

finishes this case.

Case 2. H(b) is infinite. For all c ∈ H(b) let

Dc = {a : a /∈ H(b),Δ(a, b) < l, a ∈ C(c), and

∀d ∈ H(b)(d �= a and d �= c⇒ Δ(a, d) < Δ(a, c))}.

Now

(∗∗) If c ∈ H(b) and a ∈ Dc, then Δ(c, b) < l.

For, assume otherwise. Now Δ(a, b) < Δ(c, b), so Δ(a, c) = Δ(a, b) by (�). Also,for all d ∈ H(b)\{a, c} we have Δ(d, a) < Δ(a, c) = Δ(a, b), hence by (�) we getΔ(d, b) = Δ(d, a) < Δ(a, c). So H(b) is finite by (3), contradiction. Thus (∗∗)holds.

Let Y = {c ∈ H(b) : Δ(c, b) < l}. Note that Y is finite by (3). Now

(∗∗∗) If c ∈ Y , a ∈ Dc, and m = Δ(a, c), then Cm(c) ⊆ C(b).

For, assume the hypotheses. If Δ(a, c) ≤ Δ(a, b), then d ∈ H(b)\{a, c} implies thatΔ(a, d) < Δ(a, c) ≤ Δ(a, b), so Δ(d, b) = Δ(a, d) < Δ(a, c) by (�), hence H(b) isfinite by (3), contradiction. Thus Δ(a, c) > Δ(a, b). Now suppose that u ∈ Cm(c).Thus Δ(a, c) ≤ Δ(u, c). Suppose that d ∈ H(b)\{u, c}. Then d �= a since a /∈ H(b).So Δ(a, d) < Δ(a, c) since a ∈ Dc, so Δ(d, c) = Δ(a, d) < Δ(a, c) ≤ Δ(u, c) by

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(�), hence by (�) again, Δ(d, u) = Δ(d, c) < Δ(u, c). This shows that u ∈ C(b),and it proves (∗∗∗).

For c ∈ Y with Dc �= ∅ let n(c) = min{Δ(a, c) : a ∈ Dc}. Then

(∗∗∗∗) C(b) = Cl(b) ∪ Y ∪⋃

c∈Y,Dc �=∅ Cn(c)(c).

For, ⊇ holds by (5) and (∗∗∗). For ⊆, suppose that a ∈ C(b), a /∈ Cl(b), a /∈ Y .Since a /∈ Cl(b), we have a �= b. Since a /∈ Cl(b) and a /∈ Y , we have a /∈ H(b).Since a ∈ C(b), choose c ∈ H(b) such that a ∈ C(c) and ∀d ∈ H(b)(d �= a andd �= c ⇒ Δ(a, d) < Δ(a, c)). So a ∈ Dc. By (∗∗) we get c ∈ Y . Thus a ∈ Cn(c)(c),as desired for ⊆; (∗∗∗∗) has been proved.

By the inductive hypothesis and (10), each Cn(c)(c) is compact, so it followsthat C(b) is compact in Case 2.

So, we have proved (11).

From (10) and (11) we get

(12) Cl(b) is compact; so A is locally compact.

(13) a <∗ b⇒ C(a) �= C(b).

This is true because b ∈ C(b)\C(a), using (6). So there are uncountably manycompact open sets.

A subset F ⊆ 〈A〉k is cofinal in A provided that for all a ∈ A there is anf ∈ F such that a <∗ fi for all i < k. Next we prove

(14) ∀ finite k ≥ 1 and ∀ cofinal F ⊆ 〈A〉k∃f, g ∈ F (fi ∈ H(gi) for all i < k).

This will take a while to prove, but it leads us close to the end of the proof. SinceF is cofinal in A, we may assume that there is an enumeration 〈hα : α < ω1〉 of Fsuch that hα

i <∗ hβj whenever α < β < ω1 and i, j < k. Hence every uncountable

subset of F is also cofinal in A. Let D ⊆ F be countable dense in F in the ordinarytopology (ωω has the product topology with ω having the discrete topology; k(ωω)gets the product topology too). Choose c ∈ A such that fi <

∗ c for all f ∈ D andi < k. Now

∀f ∈ F∃m ∈ ω∀n ≥ m∀i < j < k(fi(n) < fj(n)).

Hence there is an uncountable F0 ⊆ F and an m0 such that

(15) ∀f ∈ F0∀n ≥ m0∀i < j < k(fi(n) < fj(n)).

Next, let F1 = {f ∈ F0 : c <∗ f0}. Then

∀f ∈ F1∃m > m0∀n ≥ m[c(n) < f0(n)].

Hence there is an uncountable F2 ⊆ F1 and an m > m0 such that

(16) ∀f ∈ F2∀n ≥ m[c(n) < f0(n)].

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NowF2 =

⋃{{f ∈ F2 : ∀i < k(fi � m = si)} : s ∈ k(mω)}.

Hence there is an uncountable subset F3 of F2 and an s ∈ k(mω) such that

(17) ∀f ∈ F3∀i < k(fi � m = si).

Let C = {eb � Ac : c <∗ b}. Then

F3 =⋃{{f ∈ F3 : ∀i < k(efi � Ac = ui)} : u ∈ kC},

so, using (2), there is an uncountable F4 ⊆ F3 and a u ∈ kC such that

(18) ∀f ∈ F4∀i < k(efi � Ac = ui).

Now there is an n ∈ ω such that {f0(n) : f ∈ F4} is unbounded in ω, sinceotherwise, for each n ∈ ω let g(n) be greater than each f0(n) for f ∈ F4. Since F4

is cofinal in A, it follows that g is an upper bound for A, contradiction. Take m1 tobe the least such n. Let F5 be an infinite subset of F4 such that 〈f0(m1) : f ∈ F5〉is one-one. Then there is a p0 ∈ ω such that {f0 � m1 : f ∈ F5} ⊆ m1p0; sothere exist a t0 ∈ m1p0 and an infinite subset F6 of F5 such that f0 � m1 = t0for all f ∈ F6, {f0(m1) : f ∈ F6} is unbounded in ω, and 〈f0(m1) : f ∈ F6〉is one-one. Now suppose that i + 1 < k and pi, ti,mi+1, F6+i have been definedso that pi ∈ ω, mi+1 ∈ ω, ti ∈ mi+1pi, F6+i ⊆ F5, F6+i is infinite, {fi(mi+1) :f ∈ F6+i} is unbounded in ω, and 〈fi(mi+1) : f ∈ F6+i〉 is one-one. For anyj ∈ ω choose f ∈ F6+i such that j < fi(mi+1). Then j < fi+1(mi+1) also.So {fi+1(mi+1) : f ∈ F6+i} is unbounded. Let mi+2 be minimum such that{fi+1(mi+2) : f ∈ F6+i} is unbounded. Let F ′

6+i be an infinite subset of F6+i

such that 〈fi+1(mi+2) : f ∈ F ′6+i〉 is strictly increasing. Now there is a pi+1 such

that {fi+1 � mi+2 : f ∈ F ′6+i} ⊆ mi+2pi+1. So there exist ti+1 ∈ mi+2pi+1 and

F6+i+1 ⊆ F ′6+i such that F6+i+1 is infinite and ∀f ∈ F6+i+1[fi+1 � mi+2 = ti+1].

So we obtain m0, . . . ,mk+1, t0, . . . , tk−1, and F6 ⊇ · · · ⊇ F6+k−1 such that thefollowing conditions hold:

(19) ∀i < k∀f ∈ F6+k−1[ti ⊆ fi) and ti ∈ mi+1pi.

(20) ∀i < k[〈fi(mi+1) : f ∈ F6+k−1〉 is one-one ].

Now the open set in k(ωω) determined by 〈t0, . . . , tk−1〉 meets F , since F6+k−1 iscontained in it; so by the denseness of D, choose d ∈ D in this open set: ti ⊆ difor all i < k. Now dj < ∗c for all j < k by the choice of c. So dj ∈ Ac = dmn(uj)for all j < k. By (20) there is an f ∈ F6+k−1 such that

∀i < k[fi(mi+1) > max{uj(dj) : j < k}].Now for all i < k we have fi � mi+1 = ti = di � mi+1, so mi+1 ≤ Δ(di, fi).Moreover, di <

∗ c <∗ fi, and hence

e(di, fi) = efi(di) = ui(di) < fi(mi+1) ≤ fi(Δ(di, fi)),

so di is in H(fi) for all i < k, as desired; we have proved (14).

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Next,

(21) For every positive integer k, the space kA does not have an uncountablediscrete subspace.

We prove this by induction on k; suppose that it is true for all k′ < k. Supposethat F is an uncountable discrete subspace of kA. We may assume that there isan integer m such that (Cm(f0)×· · ·×Cm(fk−1))∩F = {f} for all f ∈ F . For allf ∈ F define ≡f on k by i ≡f j iff fi = fj . Without loss of generality, ≡f is thesame for all f ∈ F . Hence by the induction hypothesis, ≡ is the identity relation,so that each f ∈ F is one-one. And then by a similar argument with permutationsof k we may assume that there is a permutation π of k such that fπ(i) <∗ fπ(j)whenever f ∈ F and i < j < k. Next, we may assume that 〈rng(f) : f ∈ F 〉 isa Δ-system, say with kernel G. Say G = {ai : i < n} with a0 <∗ · · · <∗ an−1.For each f ∈ F there is a strictly increasing sequence 〈i(j, f) : j < n〉 such thatfπ(i(j,f)) = aj for each j < n. We may assume that i(j, f) does not depend on f ;so we just write i(j). Thus fπ(i(j)) = gπ(i(j)) for all f, g ∈ F and all j < n. Let

Γ = k\{i(j) : j < n}. Then F ′ def= 〈rng(f � Γ) : f ∈ F 〉 is a pairwise disjoint

system and it is cofinal in A in the sense defined before (14). Now

F ′ =⋃{{f ∈ F ′ : ∀i ∈ Γ(fi � m = si)} : s ∈ k(mω)},

so we may assume that fi � m = gi � m for all f, g ∈ F ′ and i ∈ Γ. Now we apply(14) to get distinct f, g ∈ F ′ such that gi ∈ H(fi) for all i ∈ Γ. Since H(a) ⊆ C(a)for all a, this means that g ∈ (Cm(f0) × · · · × Cm(fk−1)) ∩ F , contradiction. So(21) holds.

Now A =⋃

b∈A C(b). By (6), this cover does not have a finite subcover. SoA is not compact. By (10) and (11), it is locally compact. Take the one-pointcompactification A′ of A. Lemma 8.8(i) says that A′ is a Boolean space. Supposethat k is a positive integer and X is an uncountable discrete subset of kA′. We mayassume that the “new” point of A′ is not in X . For each x ∈ X let 〈Ux

i : i < k〉be open sets such that Ux

0 ∩ · · · ∩Uxk−1 ∩X = {x}. We may assume that each Ux

i

is open in A. This shows that X is discrete in A, contradiction.

It now follows by Theorem 8.11 that clop(A) is an uncountable BA withcountable irredundance. This finishes this example.

Problem 89. Can one construct in ZFC a BA A such that IrrA < |A|?

This is Problem 28 in Monk [96].

Concerning the min-max function Irrmm, we first note the following fromMonk [08]:

Theorem 8.14. If κ is an infinite cardinal and A is a subalgebra of P(κ) containingIntalg(κ), then Irrmm(A) = κ.

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Proof. ≥ holds by Proposition 4.23 of the handbook. Now we claim that Xdef=

{[0, α) : α < κ} is a maximal irredundant subset of A. In fact, if α < κ, thenevery element of 〈[0, β) : β ∈ κ\{α}〉 is a union of intervals of the form [a, b) witha, b ∈ (κ\{α}) ∪ {∞}, and so [0, α) /∈ 〈[0, β) : β ∈ κ\{α}〉. So X is irredundant.Now suppose that a ∈ A\〈X〉; we want to show that X ∪ {a} is redundant.We may assume that a �= ∅, κ. If 0 /∈ a, let α be the least member of a. Then[0, α + 1)\a = [0, α), so that [0, α) ∈ 〈(X ∪ {a})\{[0, α)}〉. If 0 ∈ a, let α be theleast member of κ\a. Then [0, α+1)∩a = [0, α), giving the same conclusion. �

Monk [08] gave an example of an atomless BA A such that Irrmm(A) = ω < |A| =2ω. In fact, the following conjecture seems possible.

Problem 90. Is Irrmm(A) = π(A) for every infinite BA A?

We do not know whether this assertion is true for the interval algebra on R, orfor the completion of Fr(ω).

There are reasonable finite versions of irredundance. For positive integers m,n,call a subset X of A mn-irredundant if for all x ∈ X one cannot write

x =∑i<m

∏j<n

aij ,

with aij ∈ X\{x} or −aij ∈ X\{x} for all i < m, j < n. So, X is irredundant iffit is mn-irredundant for all m,n. And define

Irrmn(A) = sup{|X | : X ⊆ A,X mn− irredundant}.

These functions have not been studied.

Since π(A) = |A| for A a tree algebra, we also have Irr(A) = |A| for A a treealgebra.

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9 Cardinality

We denote |A| also by Card(A). The behaviour of this function under algebraicoperations is for the most part obvious. Note, though, that questions about its be-haviour under ultraproducts are the same as the well-known and difficult problemsconcerning the cardinality of ultraproducts in general.

CardH− is a non-obvious function. It is related to some other “small” func-tions:

cofinality of A:

cf(A) = min{κ : κ ≥ ω and there is a strictly increasing sequence

〈Bα : α < κ〉 of subalgebras of A such that A =⋃

α<κBα

};

altitude of A:

alt(A) = min{κ : κ ≥ ω and there is a strictly increasing sequence

of length κ of filters on A whose union is an ultrafilter};pseudo-altitude of A:

p-alt(A) = min{κ : κ ≥ ω and there is an infinite homomorphic image

of A with an ultafilter generated by κ elements}.

Proposition 9.1. cf(A) ≤ alt(A) ≤ p-alt(A) ≤ CardH−(A) ≤ 2ω for any infiniteBA A.

Proof. cf(A) ≤ alt(A): If 〈Fα : α < κ〉 is a strictly increasing sequence of filterswhose union is an ultrafilter G, then 〈〈Fα〉 : α < κ〉 is a strictly increasing sequenceof proper subalgebras of A with union A; the inequality follows.

alt(A) ≤ p-alt(A): Let f be a homomorphism from A onto a BA B, andlet G be an ultrafilter on B generated by a set {xα : α < κ}, with κ minimum.Then 〈〈f−1[{xβ : β < α}]〉fi : α < κ〉 is an increasing sequence of proper filterswhose union is the ultrafilter f−1[G]; a strictly increasing subsequence proves thisinequality.

p-alt(A) ≤ CardH−(A): Obvious.

CardH−(A) ≤ 2ω: take a denumerable subalgebra B of A, and extend theidentity on B to a homomorphism from A onto a subalgebra of B. �

, ,OI 10.1007/978-3- - - _ ,


014


Basel 210

311

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312 Chapter 9. Cardinality

We now give several results concerning cf and Card due to Sabine Koppelberg [77].

Lemma 9.2. Suppose that f is a homomorphism from A onto a denumerable algebraB, and 〈Cn : n ∈ ω〉 is an increasing sequence of finite BAs with union B. (Notnecessarily strictly increasing.) Let F be a nonprincipal ultrafilter on B. Then〈f−1[F ∩ Cn]〉fiA �= f−1[F ], for all n ∈ ω.

Proof. Take any n ∈ ω. Then F∩Cn is an ultrafilter on Cn, and so there is an atomb of Cn with b ∈ F . Take c ∈ B with 0 < c < b; c exists since F is nonprincipal.Take a ∈ A with f(a) = c. Now 〈f−1[F ∩Cn]〉fiA ∪ {a} has fip, as otherwise we getsome e ∈ f−1[F ∩ Cn] such that e · a = 0, hence f(e) · c = 0. Since f(e) ∈ F ∩Cn

we have b ≤ f(e), so b ·c = 0, contradiction. Also 〈f−1[F ∩Cn]〉fiA∪{−a} has fip, asotherwise we get some e ∈ f−1[F ∩Cn] such that e · −a = 0, hence f(e) · −c = 0.Since f(e) ∈ F we have b ≤ f(e), so b · −c = 0, contradiction. It follows that〈f−1[F ∩Cn]〉fiA is not an ultrafilter. Since it is obviously contained in f−1[F ], thelemma is proved. �

Theorem 9.3. For any infinite BA A the following conditions are equivalent:

(i) CardH−(A) = ω.

(ii) There exist a strictly increasing sequence 〈Bn : n ∈ ω〉 of subalgebras of Aand an ultrafilter F on A such that A =

⋃n∈ω Bn and 〈F ∩ Bn〉fiA �= F , for

all n ∈ ω.

Proof. (i)⇒(ii): Assume (i), and let f be a homomorphism from A onto a de-numerable BA B. Let 〈Cn : n ∈ ω〉 be a strictly increasing sequence of finitesubalgebras of B with union B. Let F be any nonprincipal ultrafilter on B. Then〈f−1[Cn] : n ∈ ω〉 is a strictly increasing sequence of subalgebras of A with unionA, f−1[F ] is an ultrafilter on A, and for any n ∈ ω, 〈f−1[F ∩ Cn]〉fiA �= f−1[F ] byLemma 9.2. This proves (ii).

Now assume (ii). Then

(1) For each n ∈ ω there is an ultrafilter Gn on A such that F �= Gn but Gn∩Bn =F ∩Bn.

In fact, let n ∈ ω and choose b ∈ F\〈F ∩Bn〉fiA. We claim that 〈F ∩Bn〉fiA ∪ {−b}has fip. Otherwise, we get c ∈ 〈F ∩Bn〉fiA such that c · −b = 0, hence c ≤ b, henceb ∈ 〈F ∩Bn〉fiA, contradiction. Let Gn be an ultrafilter containing this set. Clearly(1) holds.

Now let X = {F} ∪ {Gn : n ∈ ω}. For each a ∈ A let f(a) = {H ∈ X : a ∈H}. Clearly f is a homomorphism from A into P(X).

(2) If n ∈ ω and a ∈ Bn ∩ F , then {Gm : m ≥ n} ⊆ f(a).

For, assume the hypothesis and suppose that m ≥ n. Then a ∈ Bn∩F ⊆ Bm∩F =Gm ∩Bm, so a ∈ Gm.

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Chapter 9. Cardinality 313

By (2), f maps into Finco(X). Suppose that f [F ] is finite. Each member off [F ] is a cofinite subset of X , so there is a Gn ∈

⋂f [F ]. Choose a ∈ F\Gn. Then

Gn /∈ f(a), contradiction. Hence f [F ] is infinite, and so f maps onto an algebraof size ω. �

Proposition 9.4. alt(A) = ω iff p-alt(A) = ω iff CardH−(A) = ω, for any infiniteBA A.

Proof. By Proposition 9.1 it suffices to show that CardH−(A) = ω under theassumption that alt(A) = ω. So, suppose that 〈Fn : n ∈ ω〉 is a strictly increasingsequence of filters on A whose union is an ultrafilter G. For each n ∈ ω let Bn =〈Fn〉. Note that Bn = Fn ∪ {a : −a ∈ Fn}. Hence if a ∈ Fn+1\Fn then alsoa ∈ Bn+1\Bn. For each n ∈ ω we have G ∩Bn = Fn, and hence 〈G ∩ Bn〉fiA �= G.Hence CardH−(A) = ω by Theorem 9.3. �

The following theorem is a generalization of a result of Koppelberg; it is due inde-pendently to Balcar and Simon, Kunen, and Shelah (all evidently unpublished);we follow van Douwen [89].

Theorem 9.5. If A is a BA such that Ind(A) ≥ ω1, then p-alt(A) ≤ ω1.

Proof. Since Ind(A) ≥ ω1, there is a homomorphism f from A onto a BA B suchthat Fr(ω1) ≤ B ≤ Fr(ω1). Now we claim that it suffices to find ultrafilters F andGη on B for η < ω1 such that the following two conditions hold.

(1) ∀ξ < ω1∃a ∈ F∀η < ω1[a ∈ Gη ↔ ξ ≤ η].

(2) ∀a ∈ F∃ξ < ω1∀η < ω1[ξ ≤ η → a ∈ Gη].

In fact, suppose that (1) and (2) hold. Let X = {F} ∪ {Gξ : ξ < ω1}. Defineg : B → P(X) by setting g(a) = {H ∈ X : a ∈ H}. Thus g is a homomorphism.For each ξ < ω1 choose aξ ∈ F by (1).

(3) g[F ] is an ultrafilter on rng(g).

In fact, clearly g[F ] is closed under multiplication. Suppose that a ∈ F and g(a) ≤g(b). Then g(a · −b) = 0, so in particular F /∈ g(a · −b), i.e., a · −b /∈ F , hence−b /∈ F , hence b ∈ F . So g[F ] is closed upwards. Clearly 0 /∈ g[F ]. For any a ∈ Awe have a ∈ F or −a ∈ F , hence g(a) ∈ g[F ] or −g(a) = g(−a) ∈ g[F ]. Thus (3)holds.

Now {g(aξ) : ξ < ω1} generates g[F ]. In fact let b ∈ F . By (2), choose ξ < ω1

such that

(4) ∀η < ω1[ξ ≤ η → b ∈ Gη].

We claim that g(aξ) ≤ g(b). In fact, suppose that Gη ∈ g(aξ). Thus aξ ∈ Gη, soby (1), ξ ≤ η. Hence b ∈ Gη by (4), so Gη ∈ g(b).

Hence {g(aξ) : ξ < ω1} generates g[F ].

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Finally, rng(g) is infinite, since if ξ < η we have aξ ∈ Gξ and aη �∈ Gξ by (1),so Gξ ∈ g(aξ)\g(aη).

Now to construct ultrafilters satisfying (1) and (2), let 〈xξ : ξ < ω1〉 be a

system of free generators of Fr(ω1). Let H be an ultrafilter on Fr(ω1) containingthe set {xξ : ξ < ω1}. For each η < ω1 let hη be the automorphism of Fr(ω1) suchthat

hη(xξ) =

{xξ if ξ ≤ η,

−xξ if η < ξ.

Then hη extends to an automorphism of Fr(ω1) which we still denote by hη. For

each η < ω1 let Kη = hη[H ]. So Kη is an ultrafilter on Fr(ω1). Define F = H ∩Band Gη = Kη ∩B.

To check (1), suppose that ξ < ω1. We try a = xξ. Suppose that η < ω1.Then

xξ ∈ Gη iff h−1η (xξ) ∈ H iff ξ ≤ η,

as desired.

To check (2), suppose that a ∈ F . Write a =∑

M , where M is a countablesubset of Fr(ω1). Choose ξ < ω1 such that ∀b ∈ M∀xη ∈ supp(b)[η < ξ]. Nowsuppose that ξ ≤ η < ω1. Then hη(a) = a ∈ F , so a ∈ Gη, as desired. �

Example 9.6. There is a BA A such that cf(A) = ω, alt(A) = p-alt(A) = ω1, andCardH−(A) = 2ω.

Proof. Let Z′ = Z\{0}. For each n ∈ ω let Bn = {a ⊆ Z′ : ∀i ∈ ω\(n + 1)[i ∈a iff − i ∈ a]}. Thus each Bn is a subalgebra of P(Z′), and Bm ≤ Bn if m < n.Let A =

⋃m∈ω Bm. So cf(A) = ω. Note that each Bm is isomorphic to P(ω). In

fact, Bm is complete and has as its set of atoms the set

{{i} : 1 ≤ i ≤ m} ∪ {{−i} : 1 ≤ i ≤ m} ∪ {{i,−i} : m < i < ω}.

Now suppose that f is a homomorphism from A onto an infinite BA C. Then wecan write C =

⋃m∈ω f [Bm]. If some f [Bm] is infinite, then |f [Bm]| = 2ω since

f [Bm] satisfies CSP; hence |C| = 2ω in this case. So assume that each f [Bm] isfinite; we will get a contradiction in this case. Let F be an ultrafilter on C. Now〈f [Bm] : m ∈ ω〉 is an increasing sequence of finite BAs with union C. Hence byLemma 9.2,

(∗) 〈f−1[F ∩ f [Bm]]〉fiA �= f−1[F ], for all m ∈ ω.

Now we have two possibilities for the ultrafilter B0 ∩ f−1[F ] on B0:

Case 1. There is an atom {−m,m} ∈ B0 ∩ f−1[F ], where m is a positive integer.Thus f−1[F ] is a principal ideal of A; say {m} ∈ f−1[F ]. By (∗), choose a ∈f−1[F ]\〈f−1[F ∩f [Bm+1]]

fiA. Then {m} ⊆ a. But {m} ∈ f−1[F ]∩Bm+1 ⊆ f−1[F ∩

f [Bm+1]], contradiction.

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Case 2. B0 ∩ f−1[F ] is a nonprincipal ultrafilter on B0. By (∗), choose an elementa ∈ f−1[F ]\〈f−1[F ∩ f [B0]]

fiA. Say a ∈ Bm. Then

a ⊇ a\⋃

0<i<m

{i,−i} ∈ B0 ∩ f−1[F ],

contradiction.

From Theorem 9.4 it follows that alt(A) ≥ ω1.

Since, as observed above, each Bn is isomorphic to P(ω), it follows thatInd(A) ≥ ω1. So p-alt(A) ≤ ω1 by Theorem 9.5. �

The following is an old problem, implicit in Koppelberg [77]:

Problem 91. Can one prove in ZFC that cf(A) ≤ ω1 for every infinite BA A?

The following is a problem stated in van Douwen [89]

Problem 92. Is alt(A) = p-alt(A) for every infinite BA A?

Theorem 9.7. If A is an infinite superatomic BA, then CardH−(A) = ω.

Proof. Let a ∈ A be such that [a] is an atom, where [a] is the equivalence class ofa with respect to the ideal of atoms. Let M be a denumerable set of atoms of Abelow a. For each x ∈ A let f(x) = {b ∈M : b ≤ x}. Clearly f is a homomorphismof A into P(M), and the range of f is infinite. We claim that in fact f maps intoFinco(M). For, take any x ∈ A.

Case 1. [x · a] = 0. Then f(x) is finite.

Case 2. [x · a] = [a]. Then [a · −x] = 0, and so f(−x) is finite. �

For our final result on CardH− we need a standard fact about Martin’s axiom.

Theorem 9.8 (MA). Suppose that A is an infinite ccc BA, and A ≤ B ≤ A with|B| < 2ω. Then there is an ultrafilter F on B such that for every b ∈ F ∩B thereis an a ∈ F ∩ A such that a ≤ b.

Proof. We consider the following collection of dense subsets of B+:

{{a ∈ A+ : a ≤ b or a ≤ −b} : b ∈ B}.

Let F be a generic filter on B intersecting all of these dense subsets. Clearly F isas desired. �Theorem 9.9 (MA). If A is an infinite BA of size less than 2ω, then CardH−(A)=ω.

Proof. By Theorem 9.7 we may assume that A is not superatomic. Hence it has asubalgebra isomorphic to Fr(ω), and hence by Sikorski’s extension theorem it hasa homomorphic image B such that Fr(ω) ≤ B ≤ Fr(ω). By Theorem 9.8 let F bean ultrafilter on B such that

(∗) For every b ∈ F ∩B there is an a such that a ∈ F ∩ Fr(ω) and a ≤ b.

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Now F ∩Fr(ω) is an ultrafilter in a denumerable algebra, and hence it has a strictlydecreasing sequence 〈bn : n ∈ ω〉 of generators. For each n ∈ ω let Cn = {c ∈ B :bn ≤ c or bn ≤ −c}. Clearly Cn is a subalgebra of B. Moreover, bn+1 ∈ Cn+1\Cn

for each n ∈ ω.

(1)⋃

n∈ω Cn = B.

In fact, take any c ∈ B. Say c ∈ F . By (∗), choose d ∈ F ∩ Fr(ω) such that d ≤ c.Choose n ∈ ω such that bn ≤ d. Then bn ≤ c, so that c ∈ Cn. This proves (1).

Let G be an ultrafilter on B containing F .

(2) 〈G ∩ Cn〉fiB �= G, for all n ∈ ω.

In fact, take any n ∈ ω. Clearly bn+1 ∈ G\〈G ∩ Cn〉fiB .Now Theorem 9.3 gives the desired conclusion. �

W. Just and P. Koszmider [91] have shown that there is a model M in which2ω = κ arbitrarily large, in M there is a BA A such that cf(A) = ω1, and forevery λ ≤ κ of uncountable cofinality there is a BA B such that CardH−(B) = λ.Koszmider [90] shows that it is consistent to have 2ω > ω1 while every BA haspseudo-altitude ≤ ω1.

The cardinal function Cardh− coincides with CardH−: obviously Cardh−(A) ≤CardH−(A), and if X is any infinite subset of Ult(A), then the function f suchthat f(a) = S(a) ∩ X is a homomorphism from A onto an algebra B such that|B| ≤ Clop(X); so CardH−(A) ≤ Cardh−(A). To see that rng(f) is infinite, givenm ∈ ω choose distinct Fi ∈ X for all i < m. For each i < m there is an elementai ∈ Fi\

⋃j �=i Fj , and it follows that f(ai) �= f(aj) for i �= j.

The cardinal function Cardh+ is defined as follows:

Cardh+(A) = sup{|Clop(X)| : X ⊆ UltA}.

It is possible to have Cardh+(A) > |Ult(A)| : this is true, for example, with Athe finite-cofinite algebra on an infinite cardinal κ, taking X to be the set of allprincipal ultrafilters on A, so that X is discrete and hence Clop(X) = P(X) andCardh+(A) = 2κ. Clearly dCardS+(A) = |A|, and dCardS−(A) = π(A).

We shall now go into some detail concerning the spectrum function CardHs, whichseems to be another interesting derived function associated with cardinality. Firstwe note some more-or-less obvious facts: (1) If A is an infinite free BA, thenCardHs(A) = [ω, |A|]; (2) If A is infinite and complete, then CardHs(A) = [ω, |A|]∩{κ : κω = κ} (using in an essential way the Balcar, Franek theorem); (3) ifω ≤ κ ≤ |A|, then CardHs(A)∩ [κ, 2κ] �= 0; (4) if A has a free subalgebra of powerκ ≥ ω, then CardHs(A) ∩ [κ, κω] �= 0. Now we prove a few more involved things.

Lemma 9.10. If κ is an infinite cardinal, L is a linear ordering, the sequence〈aα : α < κ〉 is strictly increasing in L, and A is the interval algebra on L, then[ω, κ] ⊆ CardHs(A).

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Proof. It suffices to show that κ ∈ CardHs(A). Define x ≡ y iff x, y ∈ L and∀α < κ[(aα < x iff aα < y) and (x < aα iff y < aα)]. Then ≡ is a convexequivalence relation on L with the equivalence classes of order type κ or κ + 1,and the desired homomorphism is easy to define. �Corollary 9.11. If κ is an infinite cardinal and A is the interval algebra on κ, thenCardHs(A) = [ω, κ]. �Corollary 9.12. If κ is an infinite cardinal, L is a linear ordering of power ≥ (2κ)+,and A is the interval algebra on L, then [ω, κ+] ⊆ CardHs(A).

Proof. One can apply the Erdos–Rado theorem(2κ)+ → (κ+)2κ to get a chain in Lof order type κ+ or (κ+)∗. �Theorem 9.13. Let A be the interval algebra on R. Then CardHs(A) = {ω, 2ω}.

Proof. The inclusion ⊇ is obvious. Now suppose that f is a homomorphism ofA onto an uncountable BA B; we want to show that |B| = 2ω. Notice that fis determined by a convex equivalence relation E on R, where the number of

E-equivalence classes is |B|. Now L′ def=⋃{k : k is an E-equivalence class with

|k| > 1} is Borel, so L′′ def= R\L′ is also. There are only countably many E-

equivalence classes k such that |k| > 1, so clearly |L′′| = |B|. Hence |B| = 2ω bythe Alexsandroff–Hausdorff theorem (see Kuratowski [58] Theorem 3, p. 355). �Theorem 9.14. CardHs(A×B) = CardHs(A) ∪ CardHs(B).

Proof. The inclusion ⊇ is obvious. For ⊆, use Proposition 1.1. �Theorem 9.15. CardHs(A⊕B) = CardHs(A) ∪ CardHs(B).

Proof. The inclusion ⊇ is obvious. If f is a homomorphism from A ⊕ B onto C,then f [A] ∪ f [B] generates C, so the other inclusion follows. �Corollary 9.16. If ω ≤ κ ≤ 2ω, then there is a BA A such that CardHs(A) =[ω, κ] ∪ {2ω}.

Proof. Apply Theorem 9.14 to A ×P(ω), where A is the free BA on κ free gen-erators. �

The strongest result known about CardHs is a special case of the following theoremof Juhasz [93]:

Let κ be a regular uncountable cardinal and let X be a compact Hausdorff space ofweight at least κ. Then there is a closed subspace F ⊆ X such that the weight ofF is in [κ, 2<κ] and

|F | ≤∑λ<κ

22λ

.

As a corollary, under GCH for every BA A the set CardHs(A) contains all regularuncountable cardinals≤ |A|. As a special case, this solves Problem 22 of Monk [90].

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We now give the proof of this theorem, for BAs. It depends on a resultabout free sequences which is independently interesting, and we first prove thisresult. Both proofs are highly topological, but we will try to make our treatmentself-contained, appealing to Engelking [89] for simple topological facts.

We introduce some notation and terminology from duality theory.

A subset D of a topological space is regular closed iff D = int(D). Engelkinguses the term closed domain.

Suppose that X ⊆ Ult(A) and M ⊆ A. We defne

Xfi = {a ∈ A : X ⊆ S(a)};X id = {a ∈ A : S(a) ⊆ X};

M c =⋂a∈M

S(a);

Mo =⋃a∈M

S(a);

M r = {b : ∀a[∀x ∈M(a ≤ x)⇒ a ≤ b]}.

Note that M r is a filter, and M ⊆M r. A filter F is regular iff F = F r.

Let κ be an infinite cardinal and S ⊆ κ, X ⊆ κ2. We say that X is S-determined iff for all x, y ∈ κ2, if x � S = y � S and x ∈ X , then y ∈ X .

For κ an infinite cardinal, let 〈xκα : α < κ〉 be a system of free generators of

Fr(κ).

Given Y ⊆ Ult(Fr(κ)) and S ⊆ κ, we say that Y is determined by S iff forall F,G ∈ Ult(Fr(κ)), if ∀α ∈ S[xκ

α ∈ F iff xκα ∈ G], then F ∈ Y iff G ∈ Y .

Suppose that F ∈ Ult(Fr(κ)), and S ⊆ κ. Then we define Φ(F, S) = {G ∈Ult(Fr(κ)) : ∀α ∈ S[xκ

α ∈ G iff xκα ∈ F ]}. Suppose that B

def= Fr(κ) ≤ A, F ∈

Ult(A), and S ⊆ κ. Then we define A (F, S) = {a ∈ F : {G : G ∈ Ult(B) andG∪{a} has fip} is determined by S}. Suppose that κ is an infinite cardinal. Witheach f ∈ κ2 we associate the ultrafilter ultrf on Fr(κ) generated by {xκ

α : f(α) =1} ∪ {−xκ

α : f(α) = 0}. Conversely, with each U ∈ Ult(Fr(κ)) we associate thefunction kU ∈ κ2 defined by

kU (α) =

{1 if xκ

α ∈ U ,

0 otherwise.

Lemma 9.17. For any X ⊆ Ult(A), the set Xfi is a filter in A, and X =⋂{S(a) :

a ∈ Xfi}.

Proof. Clearly Xfi is a filter in A. Now we work on the equality.

⊆: Suppose that F ∈ X and a ∈ Xfi. If F /∈ S(a), then a /∈ F , hence −a ∈ F ,so there is a G ∈ X such that −a ∈ G. But X ⊆ S(a), contradiction.

⊇: suppose that F ∈⋂{S(a) : a ∈ Xfi}. Let a ∈ F , and suppose that S(a)∩

X = ∅. Then X ⊆ S(−a), and hence −a ∈ Xfi. So F ∈ S(−a), contradiction. �

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Proposition 9.18. For any X ⊆ Ult(A), int(X) =⋃{S(a) : a ∈ X id} and X id is

an ideal in A. �

Proposition 9.19. For any X ⊆ Ult(A), int(X)fi= {a ∈ A : ∀b ∈ A[S(b) ⊆ X ⇒

b ≤ a]}.

Proof. First suppose that a ∈ int(X)fi. Thus int(X) ⊆ S(a). Suppose that S(b) ⊆

X . Then S(b) ⊆ int(X) ⊆ int(X) ⊆ S(a), and hence b ≤ a.

Second, suppose that ∀b ∈ A[S(b) ⊆ X ⇒ b ≤ a]; we want to show that

a ∈ int(X))fi. Thus by definition we want to show that int(X) ⊆ S(a). It suffices

to show that int(X)) ⊆ S(a). By Proposition 9.18 we take any b ∈ X id and try toshow that S(b) ⊆ S(a), i.e., that b ≤ a. Now by definition, S(b) ⊆ X . Hence bysupposition, b ≤ a, as desired. �

Proposition 9.20. For any filter F on A we have

int(F c)fi= {a ∈ A : ∀b ∈ A[∀c ∈ F (b ≤ c)⇒ b ≤ a]}.

Proof. First suppose that a is in the left-hand side. Suppose that b ∈ A and∀c ∈ F (b ≤ c). Thus S(b) ⊆

⋂c∈F S(c) = F c. Hence by Proposition 9.19, b ≤ a.

Second, suppose that a is in the right-hand side. To apply Proposition 9.19again, suppose that b ∈ A and S(b) ⊆

⋂c∈F S(c). Then ∀c ∈ F (b ≤ c). Hence

b ≤ a, as desired. �

Lemma 9.21. If B is a dense subalgebra of A and F is a regular filter on A, thenF ∩B is a regular filter on B.

Proof. Clearly F ∩B is a filter on B. Now suppose that b ∈ B, and

(1) ∀a ∈ B[∀x ∈ F ∩B(a ≤ x)⇒ a ≤ b].

We want to show that b ∈ F . To do this, assume that a ∈ A and

(2) ∀x ∈ F (a ≤ x).

If we show that a ≤ b, then b ∈ F from the regularity of F . Suppose that a �≤ b.Thus a · −b �= 0, so we can choose 0 �= c ≤ a · −b with c ∈ B. If x ∈ F ∩ B, thena ≤ x by (2). Hence c ≤ x for all x ∈ F ∩B. Hence by (1) we get c ≤ b. But c �= 0and c ≤ −b, contradiction. �

Lemma 9.22. If F is a regular filter on a BA A, then F c is regular closed.

Proof. By Proposition 9.20 we have int(F c)fi= F r = F . Hence by Proposition

9.17,

int(F c) =⋂{S(a) : a ∈ int(F c)

fi} =⋂{S(a) : a ∈ F} = F c. �

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Lemma 9.23. Let A be any BA, and let a ∈ A. Then {x ∈ A : a ≤ x} is a regularfilter on A.

Proof. Let F = {x ∈ A : a ≤ x}. Clearly F is a filter on A. Suppose that b ∈ A and∀c[∀x ∈ F (c ≤ x)⇒ c ≤ b]; we want to show that b ∈ F . Obviously ∀x ∈ F (a ≤ x),so a ≤ b, as desired. �Lemma 9.24. Suppose that B ≤ A and a ∈ A. Then

{G ∈ Ult(B) : {x ∈ B : a ≤ x} ⊆ G} = {G ∈ Ult(B) : G ∪ {a} has fip}.

Proof. Let G be in the left side, and suppose that x ∈ G and x · a = 0. Thena ≤ −x, so −x ∈ G, contradiction.

Now suppose that G is in the right side and a ≤ x ∈ B. If x /∈ G, then−x ∈ G and so G ∪ {a} does not have fip, contradiction. �Lemma 9.25. Suppose that B is a dense subalgebra of A, and a ∈ A. Then {G ∈Ult(B) : G ∪ {a} has fip} is regular closed.

Proof. Let F = {x ∈ A : a ≤ x}. Thus F is a regular filter on A by Lemma9.23. Hence by Lemma 9.21, {x ∈ B : a ≤ x} is a regular filter on B. Now{x ∈ B : a ≤ x}c =

⋂{S(x) : x ∈ B, a ≤ x} = {G ∈ Ult(B) : {x ∈ B, a ≤ x} ⊆

G} = {G ∈ Ult(B) : G∪{a} has fip} by Lemma 9.24. Thus the desired conclusionfollows by Lemma 9.22. �Lemma 9.26. Suppose that κ is an infinite cardinal, S ⊆ κ, and X ⊆ κ2 is S-determined. Then also X is S-determined.

Proof. Suppose that x, y ∈ κ2, x � S = y � S, and x ∈ X; we want to showthat y ∈ X. To this end, let {z : z � F = y � F}, F ⊆ κ finite, be a basic openneighborhood of y. Then x ∈ {z : z � (F ∩ S) = x � (F ∩ S)}, so we can choosew ∈ X such that w � (F ∩S) = x � (F ∩S). Now let w′ be such that w′ � S = w � Swhile w′ � (F\S) = y � (F\S). Then also w′ ∈ X , since X is S-determined. Also,w′ ∈ {z : z � F = y � F}, as desired. �Lemma 9.27. Suppose that B is a dense subalgebra of A, and a, b ∈ A. Then

int({G ∈ Ult(B) : G ∪ {a} has fip }) ∩ int({G ∈ Ult(B) : G ∪ {b} has fip }) =int({G ∈ Ult(B) : G ∪ {a · b} has fip }).

Proof. Clearly rhs⊆lhs. Suppose that

Cdef= (int({G ∈ Ult(B) : G ∪ {a} has fip}) ∩ int({G ∈ Ult(B) : G ∪ {b} has fip}))\{G ∈ Ult(B) : G ∪ {a · b} has fip} �= ∅.

Note that C is open by Lemma 9.25. Then Ddef= S(a) ∩ {G ∈ Ult(A) : G ∩ B ∈

C} �= ∅. In fact, take any H ∈ C. Then H ∪ {a} has fip, and so is contained in

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an ultrafilter G; so G ∈ D. Now {G ∈ Ult(A) : G ∩B ∈ C} is open since it is theinverse image of the open set C under a continuous mapping. It follows that wecan choose 0 �= x ∈ B such that S(x) ⊆ D. Thus x ≤ a. Let x ∈ G ∈ Ult(A). ThenG ∩B ∈ C, and hence (G ∩B) ∪ {b} has fip; so it is contained in an ultrafilter Hon A. Since x ∈ G∩B ⊆ H , we also have x ∈ H , hence a ∈ H . So a · b ∈ H . Thus(G ∩B) ∪ {a · b} has fip, contradicting G ∩B ∈ C. �

Lemma 9.28. Suppose that M ⊆ κ2, M is S-determined, f, g ∈ κ2, f � S = g � S,and f ∈ int(M). Then g ∈ int(M). Thus int(M) is S-determined.

Proof. LetN be a finite subset of κ such that Pdef= {h ∈ κ2 : h � N = f � N} ⊆M .

Let k = g � N . Let Q = {l ∈ κ2 : k ⊆ l}. So g ∈ Q. We claim that Q ⊆M . (Henceq ∈ int(M), as desired.) For, suppose that l ∈ Q. We define l′ ∈ κ2:

l′(α) ={f(α) if α ∈ N ,

l(α) otherwise.

Then l′ ∈ P , so l′ ∈ M . We claim that l � S = l′ � S. (Hence l ∈ M , as desired.)If fact, suppose that α ∈ S. If α ∈ N , then l′(α) = f(α) = g(α) = l(α). If α /∈ N ,then l′(α) = l(α) by definition. �

Lemma 9.29.

(i) For any ultrafilter U on Fr(κ) we have U = ultrkU .

(ii) For any f ∈ κ2 we have f = kultrf .

Proof. (i): ultrkU is generated by

{xκα : kU (α) = 1} ∪ {−xκ

α : kU (α) = 0} = {xκα : xκ

α ∈ U} ∪ {−xκα : xκ

α /∈ U},

so (i) follows.

(ii): For any α < κ,

kultrf (α) =

{1 if xκ

α ∈ ultrf ,

0 otherwise

= f(α).

�

Lemma 9.30. k is a homeomorphism from Ult(Fr(κ)) onto κ2, and ultr is a home-omorphism from κ2 onto Ult(Fr(κ)).

Proof. Let M be a finite subset of κ and ε ∈ M2. Then

k−1[{f : ε ⊆ f}] = {G : ε ⊆ kG} = S( ∏

α∈M

xε(α)α

).

Also, k is one-one and onto by Lemma 9.29, so it is a homeomorphism.

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Similarly, if M be a finite subset of κ, ε ∈ M2, and a =∏

α∈M (xκα)

ε(α), then

ultr−1[S(a)] = {f ∈ κ2 : ultrf ∈ S(a)} ={f ∈ κ2 :

∏α∈M

(xκα)

ε(α) ∈ ultrf

}

= {f ∈ κ2 : ε ⊆ f}.

Again ultr is one-one and onto by Lemma 9.29, and so it is a homeomorphism. �

Lemma 9.31. Suppose that κ is an infinite cardinal and S ⊆ κ. Also suppose thatX ⊆ κ2. Then X is S-determined iff {ultrf : f ∈ X} is determined by S.

Proof. ⇒: Suppose thatX is S-determined, F,G ∈ Ult(Fr(κ)), and ∀α ∈ S[xκα ∈ F

iff xκα ∈ G]. Let α ∈ S. Then

kF (α) = 1 iff xκα ∈ F iff xκ

α ∈ G iff kG(α) = 1,

so kF � S = kG � S. Hence kF ∈ X iff kG ∈ X . So

F ∈ {ultrf : f ∈ X} iff ultrkF ∈ {ultrf : f ∈ X}iff kF ∈ X

iff kG ∈ X

iff ultrkG ∈ {ultrf : f ∈ X}iff G ∈ {ultrf : f ∈ X}.

⇐: Suppose that {ultrf : f ∈ X} is determined by S, f, g ∈ κ2, f � S = g � S,and f ∈ X . For α ∈ S we have xκ

α ∈ ultrf iff f(α) = 1 iff g(a) = 1 iff xκα ∈ ultrg.

Moreover, ultrf ∈ {ultrh : h ∈ X}. Hence ultrg ∈ {ultrh : h ∈ X}, so g ∈ X . Theconverse is similar. �

Lemma 9.32. Suppose that κ is an infinite cardinal and S ⊆ κ. Also suppose thatY ⊆ Ult(Fr(κ)). Then Y is determined by S iff {kU : U ∈ Y } is S-determined.

Proof.

Y is determined by S iff {ultrkU : U ∈ Y } is determined by S

iff {kU : U ∈ Y } is S-determined, by Lemma 9.31. �

Lemma 9.33. Suppose that Bdef= Fr(κ) is a dense subalgebra of A, F ∈ Ult(A),

and S ⊆ κ. Then A (F, S) is closed under ·.

Proof. Suppose a, b ∈ A (F, S). Thus a, b ∈ F and:

{G ∈ Ult(B) : G ∪ {a} has fip} is determined by S;

{G ∈ Ult(B) : G ∪ {b} has fip} is determined by S;

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hence by Lemma 9.32,

{kG : G ∈ Ult(B) : G ∪ {a} has fip} is S-determined;

{kG : G ∈ Ult(B) : G ∪ {b} has fip} is S-determined;

hence by Lemma 9.28,

int({kG : G ∈ Ult(B) : G ∪ {a} has fip}) is S-determined;

int({kG : G ∈ Ult(B) : G ∪ {b} has fip}) is S-determined;

hence clearly

int({kG : G ∈ Ult(B) : G∪{a} has fip})∩ int({kG : G ∈ Ult(B) : G∪{b} has fip})

is S-determined, and so by Lemma 9.27,

int({kG : G ∈ Ult(B) : G ∪ {a · b} has fip})

is S-determined. Hence by Lemma 9.26, also

int({kG : G ∈ Ult(B) : G ∪ {a · b} has fip})

is S-determined. By Lemma 9.31,

ultr[int({kG : G ∈ Ult(B) : G ∪ {a · b} has fip})

]is determined by S; and by Lemma 9.30,

ultr[int({kG : G ∈ Ult(B) : G ∪ {a · b} has fip})

]= int({ultr(kG) : G ∈ Ult(B) : G ∪ {a · b} has fip})= int({G : G ∈ Ult(B) : G ∪ {a · b} has fip}).

But by Lemma 9.25, {G ∈ Ult(B) : G ∪ {a · b} has fip} is regular closed, so weconclude that it is determined by S. �

Lemma 9.34. Suppose that I is an ideal in Fr(κ). Then there is a countably gen-erated ideal J ⊆ I such that (Io)fi = (Jo)fi.

Proof. Let X be maximal pairwise disjoint with X ⊆ I, and let J = 〈X〉id. ThusX is countable, and so J is countably generated. We claim that (Io)fi = (Jo)fi.Clearly J ⊆ I, and so Jo ⊆ Io, hence (Io)fi ⊆ (Jo)fi. Now suppose that b ∈ (Jo)fi.Thus Jo ⊆ S(b), so a ≤ b for all a ∈ J . Suppose that a ∈ I and a �≤ b. Then0 �= a · −b ∈ I, so there is an x ∈ X such that a · −b · x �= 0. But x ∈ J , so x ≤ b,contradiction. Hence ∀a ∈ I(a ≤ b). So Io ⊆ S(b), and hence b ∈ (Io)fi. �

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Lemma 9.35. If I is an ideal in Fr(κ) and S ⊆ κ, then the following conditionsare equivalent:

(i) I is generated by some set K ⊆ 〈{xκα : α ∈ S}〉.

(ii) Io is determined by S.

Proof. (i)⇒(ii): Assume (i), and suppose that F,G ∈ Ult(Fr(κ)) and for all α ∈ S,xκα ∈ F iff xκ

α ∈ G. Then for all a ∈ K, a ∈ F iff a ∈ G. Suppose that F ∈ Io.Say F ∈ S(a) with a ∈ I. Let M be a finite subset of K such that a ≤

∑M . Now

a ∈ F , so there is a b ∈ M such that b ∈ F . Since b ∈ K, we also have b ∈ G. SoG ∈ S(b) ⊆ Io.

(ii)⇒(i): Assume (ii). Let K = I∩〈{xκα : α ∈ S}〉. We claim that K generates

I. Suppose that b ∈ I, but b �≤ a for all a ∈ K. Then {b · −a : a ∈ K} has the fip,and so is contained in an ultrafilter F . We claim that also

(∗) {(xκα)

ε(α) : α ∈ S and (xκα)

ε(α) ∈ F} ∪ {−a : a ∈ I}

has fip. Otherwise there is a finite M ⊆ S such that∏

α∈M (xκα)

ε(α) ∈ F ∩ I; it

follows that∏

α∈M (xκα)

ε(α) ∈ K, contradicting −∏

α∈M (xκα)

ε(α) ∈ F . So (∗) hasfip, and we extend it to an ultrafilter G. Clearly ∀α ∈ S[xκ

α ∈ F iff xκα ∈ G],

F ∈ Io, and G /∈ Io, contradicting (ii). �


and a ∈ F . Then there is a countable S ⊆ κ such that a ∈ A (F, S).

Proof. Let I = {b ∈ Fr(κ) : b ≤ a}. Thus I is an ideal in Fr(κ). By Lemma 9.34,there is a countably generated ideal J ⊆ I in Fr(κ) such that (Io)fi = (Jo)fi. SayJ is generated by K ⊆ 〈{xκ

α : α ∈ S}〉 with S countable. By Lemma 9.35, Jo isdetermined by S. Hence by Lemma 9.32, {kU : U ∈ Jo} is S-determined. Thus by

Lemma 9.26, {kU :U ∈Jo} is S-determined, so by Lemma 9.31, ultr[{kU :U ∈Jo}

]is determined by S. Now by Lemma 9.30, ultr

[{kU : U ∈ Jo}

]= ultr[k[Jo]] = Jo,

so Jo is determined by S. Now by Proposition 9.17,

Jo =⋂

a∈(Jo)fi

S(a) =⋂

a∈(Io)fi

S(a) = Io.

So Io is determined by S. Hence it suffices to prove

(∗) Io = {G : G ∈ Ult(B) and G ∪ {a} has fip}.To do this, first suppose that G ∈ Io. Suppose that G ∪ {a} does not have fip.Then there is a b ∈ G such that b · a = 0. Then a ≤ −b, S(a) ⊆ S(−b), and so−b ∈ (Io)fi. Hence G ∈ S(−b) by Lemma 9.17, so −b ∈ G, contradiction.

Second, suppose that G ∈ Ult(B) and G ∪ {a} has fip. In order to applyLemma 9.17, suppose that x ∈ (Io)fi; we want to show that G ∈ S(x), i.e., thatx ∈ G. Now S(a) ⊆ Io, so S(a) ⊆ S(x) and hence a ≤ x. If −x ∈ G, then G∪ {a}does not have fip, contradiction. So x ∈ G. �

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Lemma 9.37. Suppose that Fr(κ) is a dense subalgebra of A, F ∈ Ult(A), andS ⊆ T ⊆ κ. Then (A (F, T ))c ⊆ (A (F, S))c.

Proof. Clearly A (F, S) ⊆ A (F, T ), so the conclusion is clear. �


and S ⊆ κ. Then Φ(F ∩B,S) = {G ∩B : G ∈ (A (F, S))c}.

Proof. First suppose that G ∈ (A (F, S))c. Suppose that α ∈ S, ε ∈ 2, and(xκ

α)ε ∈ F ∩ B. Clearly (xκ

α)ε ∈ A (F, S). It follows that (xκ

α)ε ∈ G. This shows

that G ∩B ∈ Φ(F ∩B,S).

Second, suppose that H ∈ Φ(F ∩ B,S). We claim that H ∪ A (F, S) hasfip. Otherwise, by Lemma 9.33 there exist b ∈ H and a ∈ A (F, S) such thatb · a = 0. Since a ∈ A (F, S), we have a ∈ F and {G ∈ Ult(B) : G∪ {a} has fip} isdetermined by S. Now F ∩B is in this set, so from H ∈ Φ(F ∩B,S) we infer thatH is in the set also. So H ∪ {a} has fip. But b ∈ H and b · a = 0, contradiction.This proves the claim. Let G be an ultrafilter on A such that H ∪A (F, S) ⊆ G.Then G ∈ (A (F, S))c and G ∩B = H , as desired. �

Theorem 9.39. Suppose that κ is an uncountable regular cardinal, Bdef= Fr(κ) is a

dense subalgebra of A, and F ∈ Ult(A). Then there is a free sequence 〈Hα : α < κ〉in Ult(A) which converges to F .

Proof. Choose ε ∈ κ2 such that xε(α)α ∈ F for each α ∈ κ. For each α < κ let Gα

be an ultrafilter on Fr(κ) which contains the set

{xε(β)β : β ≤ α} ∪ {x1−ε(β)

β : α < β < κ}.

Then

(1) 〈Gα : α < κ〉 is a free sequence in Ult(Fr(κ)).

In fact, suppose that β < κ. Then {Gα : α < β} ⊆ S(x1−ε(β)β ) and {Gα : β ≤ α <

κ} ⊆ S(xε(ββ ), as desired.

Now clearly Gα ∈ Φ(F ∩ B,α + 1), so by Lemma 9.38 we can choose Hα ∈(A (F, α + 1))c such that Hα ∩B = Gα. Hence by the proof of (1),

(2) 〈Hα : α < κ〉 is a free sequence in Ult(A).

(3)⋂

α<κ(A (F, α + 1))c = {F}.

In fact, if a ∈ F , choose by Lemma 9.36 a countable S ⊆ κ such that a ∈ A (F, S).Then choose β < κ such that S ⊆ β + 1. Then by Lemma 9.37,⋂

a<κ

(A (F, α + 1))c ⊆ (A (F, β + 1))c ⊆ (A (F, S))c ⊆ S(a);

hence (3) follows.

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(4) 〈Hα : α < κ〉 converges to F .

In fact, suppose that a ∈ F . Then F ∈ S(a), so⋂

α<κ(A (F, α + 1))c ⊆ S(a) by(3). Hence by compactness there is an α < κ such that (A (F, α + 1))c ⊆ S(a).By Lemma 9.37 it follows that (A (F, β + 1))c ⊆ S(a) for all β > α, and henceHβ ∈ S(a) for all β > α. �

Proposition 9.40. If f : A→ B is a surjective homomorphism and 〈Fα : α < κ〉 isa free sequence in Ult(B), then 〈f−1[Fα] : α < κ〉 is a free sequence in Ult(A).

Proof. Suppose that β < κ. Choose bβ ∈ B such that {Fα : α < β} ⊆ S(bβ)and {Fα : β ≤ α < κ} ⊆ S(−bβ). Choose aβ ∈ A such that f(aβ) = bβ. Then{f−1[Fα] : α < β} ⊆ S(aβ) and {f−1[Fα] : β ≤ α < κ} ⊆ S(−aβ). �

Proposition 9.41. If f : A→ B is a surjective homomorphism and 〈Fα : α < κ〉 isa sequence in Ult(B) which converges to G, then 〈f−1[Fα] : α < κ〉 converges tof−1[G].

Proof. Let a ∈ f−1[G]. Then f(a) ∈ G, so there is a β < κ such that f(a) ∈ Fα

for all α ∈ [β, κ). So a ∈ f−1[Fα] for all α ∈ [β, κ). �

Theorem 9.42. (Juhasz, Szentmiklossy [92]) Suppose that κ is an uncountable reg-ular cardinal and Ult(A) has a free sequence of length κ. Then Ult(A) has a con-vergent free sequence of length κ.

Proof. By the proof of Theorem 4.26, there is a homomorphism f of A onto aBA B such that B has a strictly increasing sequence of order type κ. Then usingSikorski’s extension theorem, there is a homomorphism g of B onto a BA C suchthat Intalg(κ) is a dense subalgebra of C.

(1) We may assume that C does not have an independent subset of size κ.

In fact, suppose that it does have such a subset. Then by Sikorski’s extensiontheorem, there is a homomorphism h of C onto a BA D such that Fr(κ) is a densesubalgebra of D. By Theorem 9.39, Ult(D) has a convergent free sequence of ordertype κ. Hence by Propositions 9.40 and 9.41, so does A, as desired. Thus (1) holds.

Now let I = 〈{[0, α) : α < κ}〉idC , and let h : C → C/I be the naturalhomomorphism. Let k = h ◦ g ◦ f .

(2) If F ∈ Ult(C/I) and α < κ, then [α, κ) ∈ h−1[F ].

In fact, h([α, κ)) = 1, so (2) follows.

By (1), C/I does not have an independent subset of size κ. Hence by Hand-book 10.21, πχinf(C/I) < κ. Let F be an ultrafilter on C/I such that πχ(F ) < κ.Choose D ⊆ (C/I)+ dense in F with |D| < κ.

Now suppose that G ∈ Ult(C); we define ϕ(G) ∈ κ + 1 as follows. If G isthe principal ultrafilter generated by {α} with α < κ, we set ϕ(G) = α + 1. If Gis nonprincipal and [α, κ) ∈ G for all α < κ, we set ϕ(G) = κ. Now suppose that

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G is nonprincipal and [0, α) ∈ G for some α < κ. Then there is a limit ordinal λsuch that [0, λ) ∈ G, and we let ϕ(G) be the least such λ. Then we have

(3) For any G ∈ Ult(C) and any γ < ϕ(G) we have [γ, ϕ(G)) ∈ G.

In fact, (3) is clear if G is principal, or nonprincipal with [α, κ) ∈ G for all α < κ.So suppose that λ is the least limit ordinal such that [0, λ) ∈ G. Suppose thatγ < λ and [γ, λ) /∈ G. Then [0, γ) ∈ G or [λ, κ) ∈ G. The latter is not possible,as otherwise ∅ = [0, λ) ∩ [λ, κ) ∈ G. So [0, γ) ∈ G. Since G is nonprincipal, γ isinfinite. If δ is the greatest limit ordinal ≤ γ, then [0, δ) ∈ G since {α} /∈ G for allα ∈ [δ, γ). This contradicts the choice of λ.

(4) If a ∈ C\I, then {α < κ : ∃G ∈ Ult(C)[a ∈ G and ϕ(G) = α]} is club in κ.

In fact, let E = {α < κ : ∃G ∈ Ult(C)[a ∈ G and ϕ(G) = α]}. If E is bounded in κ,then there is a β < κ such that ∀G ∈ Ult(C)[a ∈ G⇒ ϕ(G) < β]. Now a �≤ [0, β),

so a · [β, κ) �= 0. Note that [β, κ) =∑C

γ<κ[β, γ). Hence there is a γ < κ such thata · [β, γ) �= 0. Let G is an ultrafilter such that a · [β, γ) ∈ G, then ϕ(G) ≥ β,contradiction. So E is unbounded in κ. To show that E is closed, suppose thatβ < κ is a limit ordinal and E ∩ β is unbounded in β. Then a ∪ {[γ, β) : γ < β}has fip, and so is contained in an ultrafilter G. Clearly ϕ(G) = β, as desired. Thisproves (4).

Now for each x ∈ D choose ax ∈ C such that [ax] = x, and let

Cx = {α < κ : ∃G ∈ Ult(C)[ax ∈ G and ϕ(G) = α]},

and let E =⋂

x∈D Cx. Thus E is club in κ.

Now we define

Vα = {b ∈ h−1[F ] : ∀x ∈ D[x ≤ h(b)⇒ ax · [α, κ) ≤ b]}.

Note that Vα is closed under ·. Also, let Fα =⋂

b∈VαS(b).

(5) ϕ[Fα] =⋂

b∈Vαϕ[S(b)].

For, first suppose that β ∈ ϕ[Fα], and suppose that b ∈ Vα. Choose G ∈ Fα suchthat ϕ(G) = β. Then b ∈ G, and so β ∈ ϕ[S(b)], as desired.

Second, suppose that β ∈⋂

b∈Vαϕ[S(b)]. We claim that Vα ∪{[γ, β) : γ < β}

has fip. Otherwise, there exist b ∈ Vα and γ < β such that b · [γ, β) = 0. Nowβ ∈ ϕ[S(b)], so we can choose an ultrafilter G on C such that ϕ(G) = β and b ∈ G.Then by (3) we get [γ, β) ∈ G. Since also b ∈ G, this is a contradiction. So ourclaim holds. Let G be an ultrafilter on C such that Vα ∪ {[γ, β) : γ < β} ⊆ G.Then G ∈ Fα and ϕ(G) = β. This proves (5).

(6) E\(α+ 1) ⊆ ϕ[Fα].

For, suppose that b ∈ Vα; we show that E\(α+1) ⊆ ϕ[S(b)], and so (6) will followby (5).

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Choose x ∈ D such that x ≤ h(b). Then by the definition of Vα, ax ·[α, κ) ≤ b.Then we claim

(7) Cx\(α+ 1) ⊆ ϕ[S(ax · [α, κ))].

In fact, suppose that β ∈ Cx\(α + 1). Choose G ∈ Ult(C) such that ax ∈ G andϕ(G) = β. Then ∀γ < β[[γ, β) ∈ G], so [α, β) ∈ G. Hence ax · [α, κ) ∈ G, so thatβ ∈ ϕ[S(ax · [α, κ))], as desired in (7).

By (7)

E\(α+ 1) ⊆ Cx\(α+ 1) ⊆ ϕ[S(ax · [α, κ))] ⊆ ϕ[S(b)].


(8) h−1[F ] ∈ Fα but Fα �= {h−1[F ]}.

Clearly Vα ⊆ h−1[F ], so h−1[F ] ∈ Fα. Since h([β, κ)) = 1 for each β < κ, we haveϕ(h−1[F ]) = κ. Now choose β ∈ E\(α+ 1). Then by (6), β ∈ ϕ[Fα], so there is aG ∈ Fα such that ϕ(G) = β. Hence G �= h−1[F ], and (8) holds.

Now for each α < κ let Zα = ϕ−1[(κ+ 1)\(α+ 1)]. Then

(9) ϕ−1[{κ}] =⋂

α<κ Zα.

In fact,

⋂α<κ

Zα =⋂α<κ

ϕ−1[(κ+ 1)\(α+ 1)] = ϕ−1

[ ⋂α<κ

((κ+ 1)\(α+ 1))

]= ϕ−1[{κ}].

Now for α limit, Zα is closed. For, suppose that G ∈ Zα. If [0, α) ∈ G, then thereis an H ∈ Zα such that [0, α) ∈ H , which is clearly impossible. So [α, κ) ∈ G, andso ϕ(G) ≥ α+ 1, as desired.

(10) For every b ∈ h−1[F ] there is an α < κ such that b ∈ Vα.

In fact, fix b ∈ h−1[F ]. Then we claim

(11) For all x ∈ D, if x ≤ h(b) then S(ax) ∩ ϕ−1[{κ}] ⊆ S(b).

In fact, take any x ∈ D such that x ≤ h(b). Suppose that G ∈ S(ax) ∩ ϕ−1[{κ}].Now ϕ(G) = κ implies that G ∩ I = ∅, and so h[G] is an ultrafilter on C/I. Sincex ∈ h[G], it follows that h(b) ∈ h[G], and hence b ∈ G, as otherwise −b ∈ G andso −h(b) ∈ h[G]. This proves (11).

Now note that α < β < κ implies that Zβ ⊆ Zα. Hence from (9), (11)and compactness, for each x ∈ D with x ≤ h(b) we get an αx < κ such thatS(ax) ∩ Zαx ⊆ S(b). Since |D| < κ and κ is regular, choose β < κ such thatαx < β for all x ∈ D with x ≤ h(b). We claim now that b ∈ Vβ , as desired in (10).For, suppose that x ∈ D and x ≤ h(b). If ax · [β, κ) · −b �= 0, we can choose an

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ultrafilter G on C such that ax · [β, κ) · −b ∈ G. Thus ϕ(G) > β > αx, so b ∈ G,contradiction. This proves (10).

(12) {h−1[F ]} =⋂

α<κ Fα.

For, suppose that G is an ultrafilter on C different from h−1[F ]. Choose b ∈h−1[F ]\G. By (10), choose α ∈ κ such that b ∈ Vα. Since b /∈ G, we have G /∈⋂

c∈VαS(c) = Fα, and (12) follows, using (8).

Now if α < β then Vα ⊆ Vβ and hence Fβ ⊆ Fα. By (12),⋂

α<κ Fα ={h−1[F ]}, and by (8), Fα �= {h−1[F ]}. Hence there is a strictly increasing sequence〈β(γ) : γ < κ〉 of ordinals less than κ such that for all γ, δ, if γ < δ < κ thenFβ(δ) ⊂ Fβ(γ). Now choose Gα ∈ Fβ(α)\Fβ(α+1) for all α < κ.

(13) The sequence 〈Gα : α < κ〉 converges to h−1[F ].

In fact, let b ∈ h−1[F ]. Then h−1[F ] ∈ S(b), so⋂

α<κ Fα = {h−1[F ]} ⊆ S(b). Bycompactness, there is an α < κ such that Fβ(α) ⊆ S(b). Hence Gγ ∈ S(b) for allγ ≥ α, proving (13).

(14) ∀γ < κ∃α < κ([γ, κ) ∈ Gα).

In fact, given γ < κ, by (10) there is an α < κ such that [γ, κ) ∈ Vα ⊆ Vβ(α) andso Fβ(α) ⊆ S([γ, κ)) and hence [γ, κ) ∈ Gα.

(15) ∀γ < κ∃α < κ[ϕ(Gα) > γ].

This is immediate from (14).

By (15), there is a strictly increasing sequence 〈γ(α) : α < κ〉 such that〈ϕ(Gγ(α)) : α < κ〉 is also strictly increasing. We claim, finally, that 〈Gγ(2α+1) :α < κ〉 is a free sequence converging to h−1[F ]. Clearly it converges to h−1[F ].Now suppose that ξ < κ. Let ϕ(Gγ(2ξ)) = η. Then [0, η) ∈ Gγ(2α+1) for all α < ξ,and [η, κ) ∈ Gγ(2α+1) for all α ≥ ξ. �

For the next two lemmas, see Engelking [89], page 17.

Lemma 9.43. If κ is an infinite cardinal, X is a topological space, w(X) ≤ κ, andif A is a collection of open subsets of X, then there is an A ′ ⊆ A such that|A ′| ≤ κ and

⋃A =

⋃A ′.

Proof. Let B be a base for X such that |B| ≤ κ. Define B′ = {U ∈ B : ∃V ∈A [U ⊆ V ]}. For each U ∈ B′ choose VU ∈ A such that U ⊆ VU . Let A ′ ={VU : U ∈ B′}. Thus |A ′| ≤ κ. Now suppose that x ∈

⋃A . Choose W ∈ A such

that x ∈ W , and then choose U ∈ B such that x ∈ U ⊆ W . Thus U ∈ B′. Sox ∈ U ⊆ VU ∈ A ′, hence x ∈

⋃A ′. So

⋃A =

⋃A ′. �

Lemma 9.44. If κ is an infinite cardinal, X is a topological space, w(X) ≤ κ, andif A is a base for X, then there is a base A ′ ⊆ A such that |A ′| ≤ κ.

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Proof. Let B be a base for X such that |B| ≤ κ. For each U ∈ B let S(U) = {V ∈A : V ⊆ U}. Since A is a base for X , we have U =

⋃S(U) for every U ∈ B; by

Lemma 9.43 there is an S′(U) ∈ [S(U)]≤κ such that⋃S′(U) =

⋃S(U) = U . Let

A ′ =⋃

U∈B S′(U). Then |A ′| ≤ κ. Since B is a base and U =⋃S′(U) for every

U ∈ B, it follows that A ′ is a base. �

For any topological space X , w(X) is the smallest size of a base for X .

Lemma 9.45. If 〈xα : α < κ〉 is a left separated sequence in a space X, thenw({xα : α < κ}) = κ.

Proof. For each α < κ let Uα be an open set such that Uα ∩ {xβ : β < κ} = {xβ :α ≤ β}. Let B be a basis for {xα : α < κ} of size w({xα : α < κ}). For eachα < κ let Vα ∈ B be such that xα ∈ Vα ⊆ Uα. If α < γ, then xα ∈ Vα\Vγ . Hence|B| = κ. �

For the next two results, see Juhasz [80].

Lemma 9.46. Suppose that κ is regular, X is a space, X does not have a leftseparated sequence of length κ, 〈Yα : α < κ〉 is an increasing sequence of subsets ofX, Z =

⋃α<κ Yα, and B is a family of open sets of X such that for each α < κ,

the set {B ∩Yα : B ∈ B} is a base for Yα. Then {B ∩Z : B ∈ B} is a base for Z.

Proof. Suppose not. Then there is a z ∈ Z and an open set U of X such thatz ∈ U and for all B ∈ B, if z ∈ B then B ∩Z �⊆ U . We now define by induction asequence 〈να : α < κ〉 of ordinals less than κ, a sequence 〈yα : α < κ〉 of membersof Z, and a sequence 〈Bα : α < κ〉 of members of B such that yα ∈ Yνα for allα < κ. Let ν0 be such that z ∈ Yν0 . Now z ∈ U ∩ Yν0 and {B ∩ Yν0 : B ∈ B} is abase for Yν0 , so we can choose B0 ∈ B such that z ∈ B0 ∩Yν0 ⊆ U ∩Yν0 . Then byassumption B0 ∩ Z\U �= ∅, so we choose y0 ∈ B0 ∩ Z\U .

Now suppose that α < κ and we have defined νβ , yβ and Bβ for all β < α.Choose να < κ such that νβ < να and {yβ : β < α} ⊆ Yνα for all β < α. This ispossible since κ is regular. Now z ∈ U ∩ Yνα , so by hypothesis there is a Bα ∈ Bsuch that z ∈ Bα ∩ Yνα ⊆ U ∩ Yνα . Then choose yα ∈ Bα ∩Z\U . This finishes theconstruction.

We claim that 〈yα : α < κ〉 is left-separated. (Contradiction.) For, supposethat α < κ. Then, we claim,

(∗) {yβ : β < κ} ∩⋃α≤γ

Bγ = {yβ : α ≤ β}.

For, if β < α ≤ γ and yβ ∈ Bγ , then yβ ∈ Bγ ∩ Yνγ ⊆ U , contradiction. Thus ⊆holds in (∗). If α ≤ β, then yβ ∈ Bβ; so ⊇ holds too. �

Theorem 9.47. If κ is an infinite cardinal, X is a topological space, and w(Y ) < κfor all Y ∈ [X ]≤κ, then w(X) < κ.

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Proof. First we assume that κ is regular. Suppose that the conclusion fails, sothat there is no base for X of size less than κ. We now define by recursion twosequences 〈Yα : α < κ〉 and 〈Bα : α < κ〉. Let Y0 = ∅ and B0 = ∅. Suppose thatYβ and Bβ have been defined for all β < α, where α < κ so that |Yβ |, |Bβ | < κ forall β < α. Let Zα =

⋃β<α Yβ . Then |Zα| < κ since κ is regular. By assumption,

there is a family Bα of open sets such that⋃

β<α Bβ ⊆ Bα, |Bα| < κ, and{U ∩ Zα : U ∈ Bα} is a base for Zα. Now by assumption Bα is not a basefor X . Hence there is a pα ∈ X and an open set Uα such that pα ∈ Uα, and ifpα ∈ V ∈ Bα then V \Uα �= ∅. Let Cα = {V ∈ Bα : pα ∈ V } and for each V ∈ Cα

pick q(V ) ∈ V \Uα. Let

Yα = Zα ∪ {pα} ∪ {q(V ) : V ∈ Cα}.

Clearly |Yα| < κ. This finishes the construction.

Let C =⋃

α<κ Bα. Now by Lemma 9.45, X does not have a left-separatedsequence of length κ. Let W =

⋃α<κ Yα. We apply Lemma 9.46 to conclude that

{D ∩W : D ∈ C } is a base for W . Clearly |C | ≤ κ, so w(W ) < κ. By Lemma9.44 there is a C ′ ∈ [C ]<κ such that {D ∩W : D ∈ C ′} is a base for W . By theregularity of κ, there is an α < κ such that C ′ ⊆ Bα, so {D ∩W : D ∈ Bα} is abase forW . Now pα ∈ Uα, so there is a D ∈ Bα such that pα ∈ D∩W ⊆ Uα. HenceD ∈ Cα, so q(D) ∈ D\Uα, contradiction. This finishes the proof for κ regular.

Now assume that κ is singular. Then we claim

(∗) There is a λ < κ such that w(Y ) < λ whenever Y ∈ [X ]≤κ.

In fact, if there is no such κ, then for every λ < κ there is a Yλ ∈ [C]≤κ suchthat w(Yλ) ≥ λ. Let Z =

⋃λ<κ Yλ. Then Z ∈ [X ]≤κ and for each λ < κ we have

w(Z) ≥ w(Yλ) ≥ λ, hence w(Z) ≥ κ, contradiction. So (∗) holds.Choose a λ as in (∗). Then w(Y ) < λ+ for all Y ∈ [X ]≤λ+

, so by the firstpart of this proof, w(X) < λ+ < κ. �

Lemma 9.48. If X ⊆ Ult(A), then w(X) ≤ 2|X|.

Proof. The set {S(a)∩X : a ∈ A} is a base for the topology on X. Now if a, b ∈ A,then

S(a) ∩X �= S(b) ∩X iff S(a�b) ∩X �= ∅iff S(a�b) ∩X �= ∅iff S(a) ∩X �= S(b) ∩X ;

hence there is a one-one map from {S(a) ∩X : a ∈ A} into P(X). �

Theorem 9.49. If X is a Hausdorff space and Y ⊆ X, then |Y | ≤ 22|Y |

.

Proof. For any p ∈ Y let f(p) = {U ∩ Y : p ∈ U,U open}. Thus f(p) ⊆ P(Y ),so f(p) ∈ P(P(Y )). Hence it suffices to show that f is one-one. Let p, q ∈ Y ,p �= q. Choose U, V open and disjoint with p ∈ U and q ∈ V . Thus U ∩ Y ∈ f(p).

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Suppose that U ∩Y ∈ f(q). Let W be open with q ∈W and U ∩Y = W ∩Y . Thenq ∈ V ∩W , hence V ∩W �= ∅, hence V ∩W ∩Y �= ∅. But V ∩W∩Y = U∩V ∩Y = ∅,contradiction. �

For any topological space X , the net weight of X is

nw(X) = min{|B| : B is a family of subsets of X and

∀x ∈ X∀U [U is open and x ∈ U ⇒ ∃B ∈ B[x ∈ B ⊆ U ]]}.

A set B as in this definition is called a network for X .

Theorem 9.50. If X is an infinite compact Hausdorff space, then nw(X) = w(X).

Proof. Obviously nw(X) ≤ w(X). Now let B be a network for X . For any pair(C,D) of elements of B such that there are disjoint open U, V with C ⊆ U andD ⊆ V , let UCD and VCD be such a pair U, V . We claim that {UCD : C,D ∈ Band such U, V exist} is a subbase for X . This will give the equality of the theorem.

Suppose that x ∈ X and x ∈ W , W open. For any y ∈ X\W let S, T bedisjoint open sets with x ∈ S and y ∈ T . Then choose Exy, Fxy ∈ B such thatx ∈ Exy ⊆ S and y ∈ Fxy ⊆ T . Then x ∈ UExyFxy , y ∈ VExyFxy , Exy ⊆ UExyFxy ,and Fxy ⊆ VExyFxy . Thus {VExyFxy : y ∈ X\W} covers X\W , so there is a finitesubcover {VExy(0)Fxy(0)

, . . . , VExy(n−1)Fxy(n−1)}. It follows that

x ∈⋂i<n

UExy(i)Fxy(i)⊆W,

as desired. �Theorem 9.51 (Juhasz [93). ] If A is a BA and ω < κ ≤ |A| with κ regular, then

A has a homomorphic image B such that κ ≤ |B| ≤ 2<κ and |Ult(B)| ≤∑{22λ :

λ < κ}.

Proof. We consider two cases.

Case 1. There exist y ∈ Ult(A) and Y ⊆ Ult(A) such that y ∈ Y and ∀Z ∈[Y ]<κ[y /∈ Z]]. Then by Theorem 4.25, Ult(A) has a free sequence 〈Fξ : ξ < κ〉 oflength κ. By Theorem 9.42 we may assume that this sequence converges to someultrafilter G. For each α < κ let Yα = {Fξ : ξ < α}. If there is an α < κ such thatw(Y α) ≥ κ, then by Lemma 9.48, w(Y α) ≤ 2|α| ≤ 2<κ. Moreover, |Yα| < κ, so by

Theorem 9.49, |Yα| ≤∑{22λ : λ < κ}.

Thus assume that w(Y α) < κ for all α < κ. Let M = {Fξ : ξ < κ}.

(1) M = {G} ∪⋃

α<κ Y α.

In fact, ⊇ is obvious. Now assume that H ∈ M\⋃

α<κ Y α and G �= H . Choosea ∈ H\G. Then −a ∈ G, so there is a ξ < κ such that ∀η ∈ [ξ, κ)[−a ∈ Fη]. Takeany b ∈ H . Then a · b ∈ H , so there is a ξ < κ such that a · b ∈ Fξ. Clearly ξ < η.This shows that H ∈ Y η, contradiction. So (1) holds.

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(2) nw(M ) ≤ κ.

To prove this, choose for each α < κ a subset Bα of A of size less than κ such that{S(b) ∩ Y α : b ∈ Bα} is a base for Y α. Let

P =⋃α<κ

{S(b) ∩ Y α : b ∈ Bα} ∪ {M\Y α : α < κ}.

We claim that P is a network for M . Clearly |P | ≤ κ, so this will prove (2). So,suppose that H ∈ S(a) ∩M .

Subcase 1.1. H ∈ Y α for some α < κ. Choose b ∈ Bα such that H ∈ S(b) ∩ Y α ⊆S(a) ∩ Y α; this is as desired.

Subcase 1.2. H = G. Since a ∈ H = G, there is a ξ < κ such that ∀η ∈ [ξ, κ)[a ∈Fη]. Now G ∈ {Fη : ξ ≤ η < κ}, so by the freeness of 〈Fρ : ρ < κ〉 it follows thatG /∈ Fξ, i.e., G ∈ M\Fξ. We claim that M\Yξ ⊆ S(a) (as desired). For, supposethat K ∈ M\Yξ. We may assume that K �= G. Say K ∈ Y η; so ξ ≤ η. SinceK /∈ Y ξ, we can choose b ∈ K such that S(b) ∩ Y ξ = ∅. Suppose that −a ∈ K.Choose Fρ ∈ Yη so that b · −a ∈ Fρ. Since b ∈ Fρ, it follows that ξ ≤ ρ. But thena ∈ Fρ, contradiction. Hence a ∈ K, as desired.

This finishes the proof of (2). Hence by Theorem 9.50 we get w(M) ≤ κ.

(3) M is discrete.

See Theorem 5.18 and its proof.

Thus κ = w(M) ≤ w(M ), so by the above, w(M ) = κ. Moreover, |M | ≤∑α<κ |Y α| ≤

∑λ<κ 2

λ ≤∑

λ<κ 22λ .

Case 2. ∀y ∈ Ult(A)∀Y ⊆ Ult(A)[y ∈ Y implies that ∃Z ∈ [Y ]<κ[y ∈ Z]]. Noww(Ult(A)) = |A| ≥ κ, so by Theorem 9.47 there is an M ∈ [Ult(A)]≤κ such thatw(M) ≥ κ. If |M | < κ, define f(a) = S(a) ∩ M for all a ∈ A. Then f is ahomomorphism, and κ ≤ |f [A]| ≤ 2|M| ≤ 2<κ; by Theorem 5.1, d(rng(f)) ≤ |M |and hence |Ult(rng(f))| ≤ 2|M| ≤

∑λ<κ 2

2λ .

So assume that |M | = κ; write M = {Fξ : ξ < κ} with F one-one. DefineYα = {Fξ : ξ < α} for every α < κ. Now since κ is regular, the case assumptionyields M =

⋃α<κ Yα.

Subcase 2.1. There is an α < κ such that w(Y α) ≥ κ. Define f(a) = S(a) ∩ Y α

for any a ∈ A. Then f is a homomorphism, and κ ≤ |f [A]| ≤ 2|α| ≤ 2<κ; and by

Theorem 9.49, |Y α| ≤ 22|Yα| ≤

∑λ<κ 2

2λ .

Subcase 2.2. w(Y α) < κ for all α < κ. For each α < κ let Bα ⊆ A be such that|Bα| < κ and {S(b) ∩ Y α} is a base for Y α.

(4) nw(M ) ≤ κ.

In fact, clearly⋃

α<κ{S(a) ∩ Y α : a ∈ Bα} is a network for M of size at most κ.

By Theorem 9.50 we have w(M) ≤ κ. Since w(M ≥ w(M) ≥ κ, it follows

that w(M ) = κ. Finally, |M | ≤∑

α<κ |Y α| ≤∑

λ<κ 2λ ≤

∑λ<κ 2

2λ . �

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As remarked above, Theorem 9.51 implies under GCH that for every BA A andevery uncountable regular cardinal κ ≤ |A|, A has a homomorphic image of size κ.This does not extend to κ = ω, as any CSP algebra has only uncountable infinitehomomorphic image. It also does not extend to all singular cardinals, by a resultof Juhasz, Shelah [98]. That result, for BAs, is a consequence of Corollary 4.78.For example, under CH we have

CardHs(〈[ℵω]≤ωP(ℵω)) = {ℵn : 1 ≤ n < ω} ∪ {ℵω+1}.

It is an open question to what extent Theorem 9.51 is best possible. Recall fromChapter 3, (20) in the treatment of cHr, that Koszmider [99] has shown the con-sistency of the existence of a BA A such that CardHr(A) = {ω1, λ}, where λ = 2ω

can be arbitrarily large.

Problem 93. Is Theorem 9.51 best possible? In particular, is it consistent that thereis a BA A such that CardHs(A) = {ω2, ω4}?

In comparing cardinality with the cardinal functions so far introduced, we nownote explicitly that 2d(A) ≥ |A| for any infinite BA A. Finally, recall from Part Iof the BA Handbook, Theorem 12.2, that |A|ω = |A| for any infinite CSP algebraA, in particular for any (countably) complete infinite BA A.

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10 Independence

There is a lot of information about independence in Part I of the Handbook. Aneven more extensive account is in Monk [83].

If A ≤ B, then Ind(A) ≤ Ind(B), and the independence can increase arbi-trarily. Now we consider our special subalgebras.

• A ≤free B: the independence obviously can increase arbitrarily. Hence by thegeneral results of Chapter 2, the same applies to ≤proj, ≤u, ≤rc, ≤reg, and≤σ.

• A ≤π B: the independence can increase; for example, in Fr(ω) ≤π Fr(ω),using the Balcar, Franek theorem. For an even larger increase, take A =Finco(κ) and B = A.

Proposition 10.1. If A and B are infinite BAs and A ≤s B, then Ind(A) = Ind(B).

Proof. By Proposition 2.29 there are ideals I0 and I1 such that B ∼= (A/I0) ×(A/I1). By Corollary 10.3 (easily proved directly), Ind(C × D) = max(Ind(C),Ind(D)) for any infinite BAs C,D. Clearly Ind(C/J) ≤ Ind(C) for any infinite BAC and any ideal J of C. So our proposition follows. �

Thus also if A ≤m B then Ind(A) = Ind(B).

Problem 94. Do there exist infinite BAs A,B such that A ≤mg B and Ind(A) <Ind(B)?

IfB is a homomorphic image ofA, then Ind(B) ≤ Ind(A); and clearly the differencehere can be arbitrary.

To treat the attainment problem, it is again convenient to first talk aboutindependence in products.

Theorem 10.2. If neither A nor B has an independent set of power κ ≥ ω thenA×B also does not.

Proof. Let 〈(aα, bα) : α < κ〉 be a system of elements of A× B; we want to showthat this system is dependent. Choose a finite subset Γ of κ and ε ∈ Γ2 so that∏

α∈Γ aε(α)α = 0, and then choose a finite subset Δ of κ\Γ and δ ∈ Δ2 so that∏

α∈Δ bδ(α)α = 0. Let Θ = Γ ∪ Δ and θ = δ ∪ ε. Then

∏α∈Θ(aα, bα)

θ(α) = 0, asdesired. �

, ,OI 10.1007/978-3- - - _ ,


014


Basel 211

335

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336 Chapter 10. Independence

Corollary 10.3. Ind(A×B) = max(Ind(A), Ind(B)) for infinite BAs A,B. �

Corollary 10.4. If 〈Ai : i ∈ I〉 is a system of BAs, κ is an infinite cardinal, andfor every i ∈ I, the set Ai does not have an independent subset of power κ, then∏w

i∈I Ai also has no such subset.

Proof. Suppose that X is an independent subset of∏w

i∈I Ai of power κ. Fix x ∈ X .We may assume x is of type 1, with 1-support F . Then⟨

y �∏i∈F

Ai : y ∈ X\{x}⟩

gives κ independent elements of∏

i∈F Ai, contradicting Theorem 10.2. �

Corollary 10.5. Ind(∏w

i∈I Ai) = supi∈IInd(Ai). �

Corollary 10.5 enables us to take care of the attainment problem for independence.For each limit cardinal κ there is a BA A with independence κ not attained. Forκ = ω we simply take for A any infinite superatomic BA. Now assume that κ isan uncountable limit cardinal. Let I be the set of all infinite cardinals < κ, and

for each λ ∈ I let Bλ be the free BA with λ free generators. Then Adef=∏w

λ∈I Bλ

is as desired, by Corollary 10.5.

It is perhaps surprising that the analog of Corollary 10.5 for arbitrary prod-ucts is false. This follows from a theorem of L. Heindorf (unpublished); it wasknown earlier – see Cramer [74], but the construction there is rather ad hoc. Hein-dorf proves that |A| ≤ Ind(

∏n∈ω\1 A

∗n) for any infinite BA A, where A∗n is thefree product of A with itself n times.

Theorem 10.6. If A is an infinite BA, then |A| ≤ Ind(∏

n∈ω\1 A∗n).

Proof. Let F be the set of all functions f such that f maps m2 into 2 for somem ∈ ω\1; m is denoted by ρ(f). Let B be a free BA with free generators xa fora ∈ A. It suffices to isomorphically embed B into

∏f∈F A∗ρ(f). For each f ∈ F

and each i < ρ(f) let gfi be the natural embedding of A into the ith free factor ofA∗ρ(f). We define G : B →

∏f∈F A∗ρ(f) by setting, for each a ∈ A and f ∈ F

(G(xa))f =∑

ε∈ρ(f)2,f(ε)=1

⎛⎝ ∏

j<ρ(f)

(gfj (a))ε(j)

⎞⎠ ,

extending G to a homomorphism. We want to show that G is one-one. To thisend, let a0, . . . , an−1 be distinct elements of A and suppose that ε ∈ n2; we wantto show that

ydef= (G(xa0))

ε(0) · . . . · (G(xan−1 ))ε(n−1) �= 0.

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Chapter 10. Independence 337

Let Γ = {(i, j) : i < j < n}, and choose m and h so that h is a one-one functionfrom m onto Γ. For each k < m write h(k) = (i, j), and let Fk be any ultrafilteron A such that ai�aj ∈ Fk. For each l < n we define δl ∈ m2 by setting, for anyk < m,

δl(k) =

{1 if al ∈ Fk,

0 otherwise.

Note that if i < j < n then δi �= δj , since if k = h−1(i, j) we have δi(k) �= δj(k).Hence there is an f : m2 → 2 such that f(δi) = ε(i) for all i < n. Now weclaim that yf �= 0, as desired. If l < n, then for ε(l) = 1 we have f(δl) = 1, and

so∏

k<m(gfk (al))δl(k) ≤ (G(xal

))f ; and for ε(l) = 0 we have f(εl) = 0 and so∏k<m(gfk (al))

δl(k) ≤ ((G(xal))f )

0; so in either case we have∏

k<m(gfk (al))δl(k) ≤

(G(xal))f )

ε(l). It follows that∏

l<n

∏k<m(gfk (al))

δl(k) ≤ yf . Suppose that l < n

and k < m. Then aδl(k)i ∈ Fk, so

∏l<n a

δl(k)l ∈ Fk, so

∏l<n a

δl(k)l �= 0. Hence

∏l<n

∏k<m

(gfk (al))δl(k) =

∏k<m

gfk

(∏l<n

aδl(k)l

)�= 0,

as desired. �

For the next remark we need the following result from Day [67].

Theorem 10.7. If A and B are superatomic BAs, A,B ≤ C, and C = 〈A ∪ B〉,then C is superatomic.

Proof. Let f be a homomorphism from C onto a BA D. Assume that D is non-trivial. It suffices to show that D has an atom. Note that the elements of C arefinite joins of elements of the form a · b with a ∈ A and b ∈ B. Hence there existelements u ∈ A and v ∈ B such that f(u · v) �= 0. Hence f [A] is nontrivial andso there is an a ∈ A such that f(a) is an atom of f [A]. Now for each b ∈ B letg(b) = f(a) · f(b). Then g is a homomorphism, so there is a b ∈ B such that g(b)is an atom of g[B]. We claim that g(b) is an atom of D. To prove this, take u ∈ Aand v ∈ B; we show that f(u · v) · g(b) = 0 or f(u · v) · g(b) = g(b).

Case 1. f(u) · f(a) = 0. Then f(u · v) · g(b) = f(u) · f(v) · f(a) · f(b) = 0.

Case 2. f(a) ≤ f(u). Since g(v) ∈ f [B] we have two cases.

Subcase 2.1. g(v) · g(b) = 0. Then f(u · v) · g(b) = f(u) · f(v) · f(a) · f(b) =f(u) · g(v) · g(b) = 0.

Subcase 2.2. g(b) ≤ g(v). Then g(b) ≤ f(a) · f(v) ≤ f(u) · f(v) = f(u · v). �

Now, given an infinite cardinal κ, let A be the finite-cofinite algebra on κ. Theneach algebra A∗n is superatomic, hence has no infinite independent set, but theproduct

∏n∈ω\1A

∗n has independence at least κ. This shows a total failure ofCorollary 10.5 for full direct products. Although this example takes care of themost obvious question about independence in products, there is another related

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question, namely whether an example of this sort can be done with an intervalalgebra (they always have independence ω too, just like superatomic algebras,although independence is attained for some interval algebras). The answer is no,and after several partial results by several mathematicians a complete solutionwas given by Shelah in December 1992; see Shelah [94a]:

If Ai is a non-trivial interval algebra for each i ∈ I, where I is infinite, thenInd(

∏i∈I Ai) = 2|I|.

This answers Problem 23 in Monk [90]. We give this result here. The followingelementary lemma will be useful.

Lemma 10.8. Suppose that 〈ai : i < 24〉 is a system of elements of a BA A, ε ∈ 32,

and aε(0)3i · aε(1)3i+1 · a

ε(2)3i+2 = 0 for all i < 8. Let 〈δi : i < 8〉 enumerate 32. Then∏

i<8

aδi(0)3i · aδi(1)3i+1 · a

δi(2)3i+2 = 0. �

Now we make some general definitions concerning interval algebras. Let L be alinear order with first element 0. An element ∞ not in L is assumed to be greaterthan each element of L. Then the interval algebra of L, denoted by Intalg(L),is the algebra of subsets of L generated by all the half-open intervals [a, b) witha, b ∈ L ∪ {∞} and a < b. Every element a of Intalg(L) can be written uniquelyin the form a = [sa0 , s

a1) ∪ · · · ∪ [sa2n(a)−2, s

a2n(a)−1) for some n(a) ∈ ω, with sa0 <

sa1 < · · · < sa2n(a)−1 ≤ ∞. We define

ends(a) = {sai : i < 2n(a)};ends′(a) = {sai : i < 2n(a)} ∪ {0,∞};

n′(a) = |ends′(a)|;

and we write ends′(a) = {s′ai : i < n′(a)} with 0 = s′a0 < · · · < s′an′(a)−1 = ∞.Note:

(1) ∀i < n′(a)− 1[[0 ∈ ends(a)→ [sai , sai+1) ⊆ a iff i is even]], and

[0 /∈ ends(a)→ [sai , sai+1) ⊆ a iff i is odd]].

If A ⊆ L and a ∈ Intalg(L), an A-partition of a is a subset C of A ∪ {0,∞} suchthat the following conditions hold:

(A) {0,∞} ⊆ C.

(B) ends′(a) ∩A ⊆ C.

(C) For every i < n′(a)− 1, if (s′ai , s′ai+1) ∩ A �= ∅, then (s′ai , s′ai+1) ∩ C �= ∅.If L is a linear order with first element 0 and a < b ≤ ∞, then Intalg(L) � [a, b) isthe interval algebra on the linear order [a, b).

Now if S is a set of ordinals and a = 〈aα : α ∈ S〉 is a system of elements ofIntalg(L), we say that

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(a) a is homogeneous if the following conditions hold:

(i) There is a k ∈ ω such that for every α ∈ S, n′(aα) = k.

(ii) For all α, β ∈ S, ends(aα) ∩ {0,∞} = ends(aβ) ∩ {0,∞}.(iii) For all α < β in S there is an lαβ < n′(α)−1 such that s′aα

lαβ< t < s′aα

lαβ+1

for all t ∈ ends′(aβ)\{0,∞}.(b) a is weakly homogeneous iff there are 0 = t0 < t1 < · · · < tk = ∞ such

that for every l < k the sequence 〈aα ∩ [tl, tl+1) : α ∈ S〉 is homogeneousin Intalg(L) � [tl, tl+1). The sequence 〈t0, t1, . . . , tk〉 is called a partitioningsequence for a.

A 3-signature is a sequence 〈A,m, s01, s02, s12, ε01, ε02, ε12〉 such that A ⊆ {0,∞},m is an integer greater than 1, s01, s02, s12 < m − 1, and ε01, ε02, ε12 ∈ 2. If〈a0, a1, a2〉 is homogeneous with notation as above, then its 3-signature is thefollowing sequence: 〈a0 ∩ {0,∞}, n′(a0), l01, l02, l12, ε0, ε1, ε2〉, where s′a0

l01< t <

s′a0

l01+1 for all t ∈ ends′(a1)\{0,∞}, and similarly for l02 and l12, and ε01 = 1 iff

[s′a0

l01, s′ai

l01+1) ⊆ a0, and similarly for ε02 and ε12.

Lemma 10.9. With each 3-signature σ as above we can associate εσ ∈ 32 such thatif A is an interval algebra and 〈ai : i < 3〉 is homogeneous with 3-signature σ, and

with the integer k of (a)(i) at least 2, then aεσ(0)0 · aεσ(1)1 · aεσ(2)2 = 0.

Proof. Note that k ≥ 2 implies that {a0, a1, a2} ∩ {0, 1} = ∅. Note:Case 1. A = ∅, ε01 = 0. Let εσ be 〈1, 1, 1〉.Case 2. A = ∅, ε01 = 1. Let εσ be 〈0, 1, 1〉.Case 3. A = {0,∞}, ε12 = 0. Let εσ be 〈1, 1, 0〉.Case 4. A = {0,∞}, ε12 = 1. Let εσ be 〈1, 0, 0〉.

Now in cases with A = {0}, note:

(1) a1 = [0, s′a11 ) ∪ · · · s′a1

n′(a1)−2);

(2) −a2 = [s′a21 · · · ∪ [s′a2

n′(a2)−2,∞);

Case 5. A = {0}, ε01 = 0, 0 < l12. Let εσ be 〈1, 1, 0〉. By (1) and (2), a1 · −a2 ⊆[s′a1

l12, sa1

n′(a1)−2).

Case 6. A = {0} and l12 = 0. Let εσ = 〈1, 0, 1〉. Clearly a2 ⊆ a1 in this case.

Case 7. A = {0}, ε01 = 1, 0 < l12. Let εσ be 〈0, 1, 0〉. See Case 5.

In cases with A = {∞} the following hold.

(3) a1 = [s′a11 · · · ∪ [s′a1

n(a1)−2,∞).

(4) −a2 = [0, s′a21 ) ∪ · · · s′a2

n(a2)−2).

Case 8. A = {∞}, ε01 = 0, 0 < l12. Let εσ be 〈1, 0, 1〉. By (3) and (4) we havea1 · −a2 ⊆ [s′a1

1 · · · s′a1

n(a1)−2).

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Case 9. A = {∞}, ε01 = 1, 0 < l12. Let εσ be 〈0, 0, 1〉. See Case 8.

Case 10. A = {∞}, l12 = 0. εσ be 〈1, 1, 0〉. Clearly a1 · −a2 = 0.

Case 11. in any other case let εσ = 〈1, 1, 1〉.Clearly if 〈ai : i < 3〉 is homogeneous with 3-signature σ, notation as before thislemma, then one of Cases 1–10 holds. �Theorem 10.10. If 〈Aζ : ζ < κ〉 is a system of infinite interval algebras, κ an

infinite cardinal, then Ind(∏

ζ<κ Aζ

)= 2κ.

Proof. Note that κ2 is a subalgebra of∏

ζ<κ Aζ , andκ2 is isomorphic to P(κ).

Hence∏

ζ<κAζ has an independent subset of size 2κ by the theorem of Fichtenholz,Kantorovich, Hausdorff. So our task is to show that there does not exist a largerindependent subset.

Suppose that Aζ = Intalg(Lζ) for each ζ < κ, each Lζ a linear order withfirst element. Suppose that a : (2κ)+ →

∏ζ<κ Aζ . For brevity let λ = (2κ)+. We

want to show that a is not independent. For each ζ < κ let

L′ζ = {s(a(α))ζi : α < λ, i < 2n(a(α)), s

(a(α))ζi �=∞}.

Thus L′ζ ⊆ Lζ and |L′

ζ | ≤ λ. Let A′ζ = Intalg(L′

ζ). If F is a finite subset of λ and

ε ∈ F 2, then∏α∈F

(a(α))ε(α) �= 0 in∏ζ<κ

Aζ iff∏α∈F

(a(α))ε(α) �= 0 in∏ζ<κ

A′ζ .

Hence it suffices to work with 〈A′ζ : ζ < κ〉. Now for each ζ < κ isomorphically

embed L′ζ into a linear order L′′

ζ of size λ, and let A′′ζ = Intalg(L′′

ζ ). Clearly if F

is a finite subset of λ and ε ∈ F 2, then∏α∈F

(a(α))ε(α) �= 0 in∏ζ<κ

A′ζ iff

∏α∈F

(a(α))ε(α) �= 0 in∏ζ<κ

A′′ζ .

Now each L′′ζ is isomorphic to a linear algebra (L′′′, <ζ) whose universe is λ, with

least element the ordinal 0. Let A′′′ζ = Intalg(L′′′

ζ ). Now by a change of notationwe may assume that we are given linear orders Lζ = (λ,<ζ) with least elementthe ordinal 0, Aζ = Intalg(Lζ), and a : λ→

∏ζ<κ Aζ .

The following claim gives the main part of the proof.

Claim 1. There is a stationary subset S of λ such that 〈(a(α))ζ : α ∈ S〉 is weaklyhomogeneous for every ζ < κ.

Proof of Claim 1. Let S0 = {α < λ : cf(α) = κ+}. Thus S0 is stationary in λ. Nowfor each α ∈ S0 define fα : κ→ 4 by setting, for any ζ ∈ κ,

fα(ζ) =

⎧⎪⎪⎨⎪⎪⎩0 if 0,∞ ∈ ends((a(α))ζ ),

1 if 0 ∈ ends((a(α))ζ ) and ∞ /∈ ends((a(α))ζ ),

2 if 0 /∈ ends((a(α))ζ ) and ∞ ∈ ends((a(α))ζ ),

3 if 0,∞ /∈ ends((a(α))ζ ).

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Then there exist a g ∈ κ4 and a stationary S1 ⊆ S0 such that fα = g for allα ∈ S1. Thus

(2) ∀α, β ∈ S1∀ζ ∈ κ[ends((a(α))ζ ) ∩ {0,∞} = ends((a(β))ζ) ∩ {0,∞}].

Now for each α < λ let gα ∈ κω be defined by gα(ζ) = n′((a(α))ζ ). Then thereexist a stationary S2 ⊆ S1 and an h ∈ κω such that gα = h for all α ∈ S2. Thus

(3) ∀α, β ∈ S2∀ζ ∈ κ[n′((a(α))ζ ) = n′((a(β))ζ )]

Now for each α ∈ S2 let

Cα = {0,∞}∪

⎛⎝α ∩

⋃ζ∈κ

ends′((a(α))ζ)

⎞⎠ ∪ {min((s

′(a(α))ζi , s

′(a(α))ζi+1 ) ∩ α) :

ζ < κ, i < n′((a(α))ζ)− 1, (s′(a(α))ζi , s

′(a(α))ζi+1 ) ∩ α �= ∅}.

Clearly |Cα| ≤ κ and Cα is an α-partition of (a(α))ζ for each ζ < κ. Sincecf(α) = κ+ for each α ∈ S2, it follows that sup(Cα\{∞}) < α. Hence by Fodor’stheorem there exist a stationary S3 ⊆ S2 and a β < λ such that sup(Cα\{∞}) = βfor all α ∈ S3.

Now |[β]≤κ| ≤ (2κ)κ = 2κ < λ, so there exist a stationary S4 ⊆ S3 and aD ∈ [β]≤κ such that Cα = D ∪ {∞} for all α ∈ S4.

Note that for any α ∈ S4 and ζ < κ we have

α∩ends′((a(α))ζ ) ⊆ D∩ends′((a(α))ζ ) = Cα∩ends′((a(α))ζ) ⊆ α∩ends′((a(α))ζ);

so α ∩ ends′((a(α))ζ ) = D ∩ ends′((a(α))ζ).Now for any α ∈ S4 and ζ < κ, let Tαζ be a finite subset of D such that if

i < n′(a(α))ζ−1 and (s′(a(α))ζi , s

′(a(α))ζi+1 )∩D �= ∅, then (s

′(a(α))ζi , s

′(a(α))ζi+1 )∩Tαζ �= ∅.

Define

Fα(ζ) = (D ∩ ends′((a(α))ζ ), 〈(s′(a(α))ζi , s′(a(α))ζi+1 ) ∩ Tαζ : i < n′(a(α))ζ − 1〉).

Let S5 ⊆ S4 be stationary such that Fα = Fβ for all α, β ∈ S5. For any ζ < κ,take any α ∈ S5 and let

Tζ = (D ∩ ends′((a(α))ζ )) ∪⋃{(s′(a(α))ζi , s

′(a(α))ζi+1 ) ∩ Tαζ : i < n′(a(α))ζ}.

Then for any α, β ∈ S5 and any ζ < κ we have

(4) (a) Tζ is an α-partition of (a(α))ζ .

(b) For every l < n′((a(a))ζ , if s′((a(α))ζl ∈ Tζ, then s′((a(α))ζl = s

′((a(β))ζl .

(c) For every l < n′((a(a))ζ − 1,

Tζ ∩ (s′((a(α))ζi , s

′((a(α))ζi+1 ) = Tζ ∩ (s

′((a(β))ζi , s

′((a(α))ζi+1 ).

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For each ζ < κ let Tζ = {tζ0, . . . , tζmζ} with 0 = tζ0 < · · · < tζmζ

= ∞. Now letF = {γ < λ : for all α < γ and all ζ < κ, ends′((a(α))ζ )\{∞} ⊆ γ}. Clearly F isclub in λ. Let S6 = S5 ∩ F . Thus S6 is stationary in λ. The following claim willfinish the proof of Claim 1.

Claim 2. For each ζ < κ, the sequence 〈(a(α))ζ : α ∈ S6〉 is weakly homogeneous,with partitioning sequence 〈tζ0, . . . , tζmζ

〉.

To prove this claim, fix ζ < κ and l < mζ . We want to show that 〈(a(α))ζ ∩[tζl, tζ(l+1)) : α ∈ S6〉 is homogeneous in Intalg(Lζ) � [tζl, tζ(l+1)). For brevity leta(α))ζ ∩ [tζl, tζ(l+1)) = b(α)ζ for each α ∈ S6.

Case 1. tζ(l+1) �= ∞. For any α ∈ S6 choose m(α), p(α) < n′(a(α))ζ − 1 such that

tζl ∈[s′(a(α))ζm(α) , s

′(a(α))ζm(α)+1

)and tζ(l+1) ∈

[s′(a(α))ζp(α) , s

′(a(α))ζp(α)+1

).

Subcase 1.1. m(α) = p(α) for some α ∈ S6. Then (b(α))ζ is either [tζl, tζ(l+1)) or

∅. In either case, n′(b(α))ζ = 2. Moreover, by (4)(b) and (4)(c), m(β) = p(β) for allβ ∈ S6, so n′(b(β))ζ = 2 for all β ∈ S6. Thus (a)(i) holds.

Now if 0 ∈ ends((a(α))ζ) and m(α) is even, then (b(α))ζ is [tζl, tζ(l+1)); itfollows that ends((b(α))ζ) = {0,∞} in the sense of Intalg(Aζ) � [tζl, tζ(l+1)). By(4)(c) the same is true for (b(β))ζ for any β ∈ S6. If 0 ∈ ends((a(α))ζ) and m(α)is odd, then (b(α))ζ = ∅, and this is also true for any (b(β))ζ for β ∈ S6. Similararguments work for 0 /∈ ends((a(α))ζ ). This shows that (a)(ii) holds.

(a)(iii) holds vacuously, since ends′((b(β))ζ) = {0,∞} in the sense of

Intalg(Aζ) � [tζl, tζ(l+1)).

Subcase 1.2. p(α) = m(α)+ 1. If 0 ∈ ends((a(α))ζ ) and m(α) is even, then (b(α))ζ

is [tζl, s′(a(α))ζm(α)+1). Hence n′(b(α))ζ = 3. Similarly, n′(b(β))ζ = 3 for any β ∈ S6. So

(a)(i) holds. Clearly ends((b(β))ζ ) ∩ {0,∞} = {0} for any β ∈ S6. Thus (a)(ii)

holds. Now assume that α < β in S6. Then s′(a(α))ζm(α)+1 �= s

′(a(β))ζm(α)+1, as otherwise

s′(a(α))ζm(α)+1 = s

′(a(β))ζm(α)+1 ∈ β ∩ ends((a(β))ζ ) ⊆ Tζ . Hence (a)(iii) holds.

On the other hand, if 0 ∈ ends((a(α))ζ ) and m(α) is odd, then (b(α))ζ is the

interval [s′(a(α))ζm(α)+1, tζ(l+1)), and a similar argument works.

The case 0 /∈ ends((a(α))ζ ) is also similar.

Subcase 1.3. p(α) > m(α) + 1. Now the n′ sequence for (b(a))ζ as a member ofIntalg(Aζ) � [tζl, tζ(l+1)) is

〈tζl, s′(b(α))ζm(α)+1, . . . , s′(b(α))ζp(α) , tζ(l+1)〉,

where there are at least two s terms. Thus n′((b(α))ζ ) = p(α)−m(α)+2. By (4)(b)and (4)(c), this is true for any β ∈ S6. So (a)(i) holds. Clearly ends((b(α))ζ) ∩{0,∞} = {0,∞} for any α ∈ S6, so (a)(ii) holds. To prove (a)(iii) it suffices to

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note that for α < β there do not exist i, j such that s′(b(β))ζi < s

′(b(α))ζj < s

′(b(β))ζi+1 .

This is true since s′(b(α))ζj < β, using (4)(a). Thus (a)(iii) holds.

Case 2. tζ(l+1) =∞. This is similar to Case 1.

This finishes the proof of Claims 1 and 2.

Now let S6 =⋃

β<λ Tβ, where |Tβ | = 3 for all β < λ and Tβ < Tγ for β < γ.For each ζ < κ choose mζ and 0 = tζ0 < · · · < tαmζ

= ∞ so that for all n < mζ

the sequence 〈(a(α))ζ ∩ [tζn, tζ(n+1)) : α ∈ S6〉 is homogeneous. For each β < λ letσβ be the function with domain κ such that for each ζ < κ,

σβ(ζ) = 〈(n, τ) : n < mζ and τ is the 3-signature of

〈(a(αi))ζ ∩ [tζn, tζ(n+1)) : i < 3〉〉,

where Tβ = {α0, α1, α2} with α < 0 < α1 < α2. Then there is an infinite U ⊆ λsuch that σβ = σγ for all β, γ ∈ U . Let β0, . . . , β7 be members of U such that Tβ0 <· · · < Tβ7 , and let 〈α0, . . . , α23〉 be the strictly increasing sequence of ordinals suchthat Tβi = {α3i, α3i+1, a3i+2} for each i < 8.

Claim 3.∏

i<8,k<3[a(α3i+k)]δi(k) = 0, where 〈δi : i < 8〉 enumerates 32.

This claim finishes the proof. To prove the claim, suppose that ζ < κ and n < mζ ;it suffices to show that

(∗)∏

i<8,k<3

[(a(α3i+k))ζ ∩ [tζn, tζ(n+1))]δi(k) = 0

Now for i, j < 8, the 3-signature of

〈(a(α3i+k))ζ ∩ [tζn, tζ(n+1)) : k < 3〉

is the same as the 3-signature of

〈(a(α3j+k))ζ ∩ [tζn, tζ(n+1)) : k < 3〉.

Thus by Lemma 10.9 there is an ε ∈ 32 such that∏k<3

[(a(α3i+k))ζ ∩ [tζn, tζ(n+1))]ε(k) = 0

for all i < 8. Then by Lemma 10.8 we obtain (∗). �

Now we discuss independence in free products.

Theorem 10.11. Ind(A ⊕ B) = max(Ind(A), Ind(B)) for any BAs A,B, at leastone of which is infinite.

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Proof. If both A and B are superatomic, then so is A⊕B by Theorem 10.8 or theHandbook 10.19. If both are not superatomic, then Ind(A ⊕ B) = max(Ind(A),Ind(B)) by the Handbook 10.16 and 11.15. Finally, suppose for example that Ais superatomic but B is not. By the Handbook 11.17 we have (A⊕ B)/p ∼= A forevery ultrafilter p on B. Then by the Handbook 10.16, 10.17,10.19 we have

Ind(A⊕B) = Ind∗(A⊕B)

= max(Ind∗(B), sup{Ind∗(A⊕B)/p : p ∈ Ult(B)})= Ind∗(B) = Ind(B) �

Theorem 10.12. If I is infinite and 〈Ai : i ∈ I〉 is a system of BAs each with atleast 4 elements, then Ind(⊕i∈IAi) = max(|I|, sup{Ind(Ai) : i ∈ I}).

Proof. This is immediate from the Handbook 11.15, 10.16, and 10.19. �

We turn to independence in ultraproducts. As in the case of cellularity, it is easyto see that if F is a countably complete ultrafilter on an index set I and eachAi has countable independence, then so does

∏i∈I Ai/F . Namely, suppose that

〈fα/F : α < ω1〉 is a system of independent elements of∏

i∈I Ai/F . Now for everyi ∈ I there exist finite disjoint subsets M(i), N(i) of ω1 such that∏

α∈M(i)

fα(i) ·∏

α∈N(i)

−fα(i) = 0.

HenceI =

⋃M,N

{i ∈ I : M = M(i) and N = N(i)},

with M and N ranging over finite subsets of ω1, so, since F is ω2-complete, thereexist finite disjoint M,N ⊆ ω1 such that {i ∈ I : M = M(i) and N = N(i)} ∈ F .But then

∏α∈M fα/F ·

∏α∈N −fα/F = 0, contradiction.

Further, if F is countably incomplete and each algebra Ai is infinite, then∏i∈I Ai/F is ω1-saturated, hence is CSP, from which it follows that

∏i∈I Ai/F

has independence ≥ 2ω. (See the Handbook, Theorem 13.20.) Like with cellularity,if I is infinite, F is a |I|-regular ultrafilter on I, and Ai is an infinite BA for eachi ∈ I, then Ind(

∏i∈I Ai/I) ≥ 2|I|. The proof is similar to that for cellularity: let

E be a subset of F such that |E| = |I| and each i ∈ I belongs to only finitelymany members of E; let G(i) be the set of all e ∈ E such that i ∈ e. With eachg ∈ E2 we associate g′ ∈

∏i∈I Ai as follows. Let 〈xh : h ∈ G(i)2〉 be a system of

independent elements of Ai. Then for any i ∈ I we set g′(i) = xg�G(i). We claimthat 〈[g′] : g ∈ E2〉 is an independent system of elements of

∏i∈I Ai/I. To see this,

let [(g(0))′], . . . , [(g(m−1))′] be distinct elements of∏

i∈I Ai/I and let ε ∈ m2. LetH be a finite subset of E such that (g(0)) � H, . . . , (g(m− 1)) � H are all distinct.Let i ∈

⋂H be arbitrary. Now H ⊆ G(i), so (g(0)) � G(i), . . . , (g(m− 1)) � G(i)

are all distinct. Hence

((g(0))′(i))ε(0) · . . . · ((g(m− 1))′(i))ε(m−1) = xε(0)(g(0))�G(i) · . . . · x

ε(m−1)(g(m−1))�G(i) �= 0,

as desired.

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Independence is an ultra-sup function, so Theorems 3.20–3.22 of Petersonapply, Theorem 3.22 saying that Ind

(∏i∈I Ai/F

)≥∣∣∏

i∈I IndAi/F∣∣ for F regular.

So by Donder’s theorem it is consistent that ≥ always holds. The inequality canbe strict, as is seen by the following adaptation of Theorem 10.6.

Theorem 10.13. If A is an infinite BA and X is an infinite disjoint subset of A+,then for any nonprincipal ultrafilter F on ω\1 we have |X | ≤ Ind(

∏n∈ω\1 A

∗n/F ),where A∗n is the free nth power of A.

Proof. For each a ∈ X we define fa ∈∏

n∈ω\1 A∗n by setting fa(n) = g0(−a) ·

. . . · gn−1(−a) for each n ∈ ω\1, where gi is the natural isomorphism of A ontothe ith free factor of A∗n for each i < n. Given finite disjoint subsets M and Nof X , let m = |N | + 1. We claim that (

∏a∈M fa(n) ·

∏a∈N −fa(n) �= 0 for all

n ≥ m. For, extend N to a subset N ′ of X , still disjoint from M , of size n. WriteN ′ = {a0, . . . , an−1}. Then(∏

a∈M

fa ·∏a∈N ′

−fa

)(n) =

∏a∈M

g0(−a) · . . . ·∏a∈M

gn−1(−a)·

∏a∈N ′

(g0(a) + · · ·+ gn−1(a))

≥∏a∈M

g0(−a) · . . . ·∏a∈M

gn−1(−a)·

g0(a0) · . . . · gn−1(an−1) �= 0,

as desired. �

On the other hand, Magidor and Shelah have shown that it is consistent thatthere is an infinite set I, a system 〈Ai : i ∈ I〉 of infinite BAs, and an ultrafilter Fon I such that Ind

(∏i∈I Ai/F

)<∣∣∏

i∈I IndAi/F∣∣. See Roslanowski, Shelah [98],

Proposition 1.14.

Next, note that independence is an ordinary sup-function, and so its be-haviour with respect to unions of well-ordered chains is given by Theorem 3.16.

The following theorem is due to Kevin Selker.

Let I, B, 〈Ji : i ∈ I〉, and 〈Ai : i ∈ I〉 be as in the definition of moderateproducts. Then

Ind

(B∏i∈I

Ai

)= max{Ind(B), sup

i∈IInd(Ai)}.

Problem 95. Describe the independence of the one-point gluing of a system of BAs.In particular, does the analog of Theorem 10.6 hold?

Proposition 10.14. If A is an infinite BA, then Ind(Dup(A)) = Ind(A).

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Proof. Clearly ≥ holds. Now suppose that X ⊆ Dup(A) and |X | > Ind(A); wewant to show that X is dependent. For each x ∈ X write x = (ax, Yx). Let F,Gbe finite disjoint subsets of X such that

∏x∈F ax ·

∏y∈G−ay = 0. Then there is

a finite Z ⊆ Ult(A) such that∏

x∈F x ·∏

y∈G−y = (0, Z). Let |Z| < 2k. Let W ⊆X\(F ∪G) with |W | = k. Then there is an ε ∈ W 2 such that Z∩

⋂w∈W Y

ε(w)w = ∅.

Hence∏

w∈W wε(w) ·∏

x∈F x ·∏

y∈G−y = (0, 0), as desired. �

Concerning the exponential, the following result is of some interest.

Proposition 10.15. If A is an infinite BA, then c(A) ≤ Ind(Exp(A)).

Proof. Let X be an infinite disjoint subset of A; we produce an independent sub-set of Exp(A) of size |X |. Namely, for each x ∈ X let ax = V (S(x),S(−x)).By Lemma 1.22(iv) we have ax = −(V (S(x)) ∪ V (S(−x))). Now suppose thatx0, . . . , xm−1, y0, . . . , yn−1 are distinct members of X . Then

(1) ax0 ∩ · · · ∩ axm−1 ∩ −ay0 ∩ · · · ∩ −ayn−1

= −(⋃

i<m

V (S(xi)) ∪⋃i<m

V (S(−xi))

)

∩⋂j<n

(V (S(yj)) ∪ V (S(−yj)).

Here we may assume that n > 0. The nonempty closed set S(−y0 · . . . · −yn−1) isa member of (1). �

We turn to the functions derived from independence. IndH+, IndS+, and dIndS+all coincide with Ind itself. IndH− appears to be a new function. Fedorchuk [75]has constructed, using ♦, a BA A such that IndH−(A) = Ind(A) = ω andCardH−(A) = ω1.

Problem 96. Can one construct in ZFC a BA A with the property that IndH−A <CardH−A?

This is Problem 29 in Monk [96]. Clearly IndS−(A) = ω for any infinite BA A. Wedefine

Indh+A = sup{|X | : X ⊆ ClopY , X is independent, Y ⊆ UltA}.

Then it is possible to have A superatomic, hence with IndA = ω, while Indh+A >|UltA|; see the argument for Card. In fact, maybe it is always true that Indh+A =Cardh+A:

Problem 97. Is Indh+A = Cardh+A for every infinite BA A?

This is Problem 30 in Monk [96]. Indh− is defined analogously. Again we do notknow anything about this cardinal function; for example:

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Problem 98. Is Indh−A = IndH−A for every infinite BA A?

This is Problem 26 in Monk [90]. The function dIndS− appears to be interesting; ifA is P(X) for some infinite X , then we have dIndS−A = ω since the BA of finiteand cofinite subsets of X is dense in A. On the other hand, if A is an infinite freeBA of regular cardinality, then dIndS−A = |A|; see the Handbook, Theorem 9.16.

We now go into the notions Indmm and Indspect. Because of usage in thespecial case P(ω)/fin, we use the notation i and isp instead:

isp(A) = {|X | : X is an infinite maximal independent subset of A};i(A) = min(isp(A)).

The following theorem is from Monk [04].

Theorem 10.16. For any set X of infinite cardinals there is a BA A such thatisp(A) = X.

Proof. For X empty we just take any infinite superatomic BA. So suppose thatX is nonempty. Let κ be the least member of X . The desired algebra is then

Adef=

(w∏

λ∈X

Fr(λ)

)⊕ Fr(κ).

To prove this, first note that each element of A can be written in the form

(1) z =∑i<mz

(uzi · vzi ) with each uz

i ∈w∏

μ∈X

Fr(μ),

each vzi ∈ Fr(κ), and vzi · vzj = 0 for i �= j,

uzi �= 0 �= vzi for all i.

Now take any λ ∈ X ; we show that λ ∈ isp(A). Let 〈xα : α < λ〉 be a one-oneenumeration of free generators of Fr(λ). For each α < λ define yα ∈

∏wμ∈X Fr(μ)

by setting, for each μ ∈ X ,

yα(μ) =

{xα if μ = λ,

0 otherwise.

Clearly {yα : α < λ} is an independent system of elements of A; extend it to amaximal independent family Y . To show that λ ∈ isp(A) it suffices to show:

(2) |Y | = λ.

Suppose not; so |Y | > λ. Since |Y | > λ, we can choose Y ′ ∈ [Y ]λ+

such that thereis a positive integer n such that mz = n for all z ∈ Y ′ and the two sequences

〈uzi (λ) : i < n〉 and 〈vzi : i < n〉

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do not depend on the particular z ∈ Y ′. Take any distinct w, z ∈ Y ′, and chooseα < λ such that yα �= w, z. Then w · −z · yα = 0, contradiction. in fact,

w · −z · yα =

(∑i<n

(uwi · vwi )

)·∏i<n

((−uz)i +−vzi ) · yα

=

⎛⎜⎝∑

i<n,J⊆n

⎛⎝uw

i · vwi ·∏j∈J

−uzj ·

∏j∈n\J

−vzj

⎞⎠⎞⎟⎠ · yα.

Now take any i < n and J ⊆ n. If i ∈ J , then uwi ·∏

j∈J −uzi · yα = 0, as desired.

If i /∈ J , then vwi ·∏

j∈n\J −vzi = 0, as desired.

So (2) holds.

Now suppose that μ /∈ X but Y is a maximal independent subset of A of sizeμ; we want to get a contradiction. Let 〈xα : α < κ〉 be a one-one system of freegenerators of Fr(κ).

(3) κ < |Y |.

For, suppose that |Y | < κ. Choose α < κ with xα not in the support of any vzi forz ∈ Y . Clearly each element of 〈Y 〉 can be written in the form

∑i<n(si · ti) with

si ∈∏w

λ∈X Fr(λ) and ti in the subalgebra of Fr(κ) generated by {vzi : z ∈ Y, i < n}.Hence Y ∪ {xα} is independent, contradiction. So (3) holds.

Note by (3) that Y is uncountable.

(4) Suppose that 〈y(i) : i < s〉 and 〈z(j) : j < t〉 are systems of distinct elements

of Y with s, t > 0, my(i) = mz(j) = n and vy(i)k = v

z(j)k = wk for all i < s, j < t,

and k < n. Then

∏i<s

y(i) ·∏j<t

−z(j) =∑k<n

⎛⎝∏

i<s

uy(i)k ·

∏j<t

−uz(j)k · wk

⎞⎠ .

To prove (4), first note that

∏i<s

y(i) =∏i<s

∑k<n

(uy(i)k · wk) =

∑k<n

(∏i<s

uy(i)k · wk

)

since the wk’s are pairwise disjoint. Hence

∏i<s

y(i) ·∏j<t

−z(j)=(∑

k<n

(∏i<s

uy(i)k ·wk

))·∏j<t

∏k<n

(−uz(j)k +−wk)

=∑k<n

⎛⎝∏

i<s

uy(i)k ·wk ·

∏j<t

∏k<n

(−uz(j)k +−wk)

⎞⎠=

∑k<n

(∏a<s

uy(a)k ·

∏b<t

−uz(b)k ·wk

),

proving (4).

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Now for each y ∈ Y and i < mi let

δiy =

{0 if {λ ∈ X : uy

i (λ) �= 0} is finite,

1 if {λ ∈ X : uyi (λ) �= 1} is finite,

Fiy = {λ ∈ X : uyi (λ) �= δiy}.

Now clearly there is an uncountable subset Z of Y such that the following condi-tions hold:

(5) There is an n ∈ ω such that my = n for all y ∈ Z.

(6) δiy = δi for all y ∈ Z.

(7) vyi = vi for all y ∈ Z.

(8) For each i < n, 〈Fiy : y ∈ Z〉 is a Δ-system, say with kernel Gi.

Now let H =⋃

i<n Gi. Then

(9) If z, s, t, w are distinct members of Z, i < n, and λ ∈ X\H , then (uzi ·us

i ·−uti ·

−uwi )(λ) = 0.

For, if δi = 0, then since Fiz∩Fis = Gi, we get λ /∈ Fiz or λ /∈ Fis, hence uzi (λ) = 0

or usi (λ) = 0; so (uz

i · usi )(λ) = 0, and similarly if δi = 1 then (−ut

i · −uwi )(λ) = 0,

so (9) holds.

Note that by (4) we have

(10) If z, s, t, w are distinct members of Z, then

z · s · −t · −w =∑i<n

(uzi · us

i · −uti · −uw

i · vi).

Now let L = {λ ∈ X : λ < μ} and K = {λ ∈ X : μ < λ}.

(11) Suppose that a, b ∈ Y , a �= b, ma = mb, uai (λ) = ub

i(λ) for all i < ma and allλ ∈ H ∩ L, and vai = vbi for all i < ma. Then (ua

i · −ubi)(λ) = 0 for all i < ma and

λ ∈ H ∩ L, and

a · −b =∑i<ma

(uai · −ub

i · vai ).

This follows from (4). Hence by (1) we get

(12) If z, s, t, w are distinct members of Z, a, b ∈ Y , a �= b, {a, b}∩ {z, s, t, w} = ∅,ma = mb, u

ai (λ) = ub

i(λ) for all i < ma and all λ ∈ H ∩ L, and vai = vbi for alli < ma, then

z · s · −t · −w · a · −b =∑i<n

j<ma

(uzi · us


i · uaj · −ub

j · vi · vaj ).

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Case 1. H ∩K = ∅. Let z, s, t, w be distinct members of Z. Let

M =

{(f, p, g) : p ∈ ω, f ∈ p

∏λ∈H∩L

Fr(λ), g ∈ pFr(κ)

}.

Clearly |M | < μ. Now with each a ∈ Y we can associate the triple

(〈uai � (H ∩ L) : i < ma〉,ma, 〈vai : i < ma〉)

which is a member of M . It follows that there are distinct a, b ∈ Y \{z, s, t, w}such that ma = mb, u

ai (λ) = ub

i(λ) for all i < ma and all λ ∈ H ∩ L, and vai = vbifor all i < ma. Then we obtain the equation in (12). But by (9) and (11) we haveuzi · us


i · uaj · −ub

j = 0 for all i < n and j < ma, contradiction.

Case 2. H ∩K �= ∅. For every λ ∈ H ∩K let w(λ) be a free generator of Fr(λ)not in the support of any element uy

i (λ) for y ∈ Y and i < my. Clearly w /∈ Y , soby the maximality of Y there exist a finite N ⊆ Y , an ε ∈ N2, and a δ ∈ 2, suchthat wδ ·

∏y∈N yε(y) = 0. Now take distinct z, s, t, w ∈ Z\N . By the argument

in Case 1 there are distinct a, b ∈ Y \(N ∪ H ∪ {z, s, t, w}) such that ma = mb,vai = vbi for all i < ma, and ua

i (λ) = ubi(λ) for all i < ma and all λ ∈ H ∩ L. Now

if λ ∈ X\(H ∩K), then either λ ∈ X\H or λ ∈ H ∩ L. Hence by (9) and (11),

(13) (uzi · us

i · −uti · −u

pi · ua

j · −ubj) � (X\(H ∩K)) = 0 for all i < n and j < ma.

Let∏

y∈N yε(y) =∑

k<p ck · dk, with ck ∈∏w

ν∈X Fr(ν) and dk ∈ Fr(κ). Then by(12), ∏

y∈N

yε(y) · z · s · −t · −w · a · −b

=∑{uz

i · usi · −ut

i · −uwi · ua

j · −ubj · ck · vi · vaj · dk :

i < n, j < ma, kp}.

Hence by (13) there is a λ ∈ H ∩K such that

(uzi · us


i · uaj · −ub

j · ck)(λ) �= 0.

By the choice of w it follows that

w(λ) · (uzi · us


i · uaj · −ub

j · ck)(λ) �= 0.

This contradicts w ·∏

y∈N yε(y) = 0. �

For our next result concerning i we need the following lemma, which is of generalinterest.

Lemma 10.17. If A is a countable BA and F is an infinite free BA, then F ∼= A⊕F .

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Proof. Let X be a set of free generators of F and write X = Y0 ∪ Y1, whereY0 ∩ Y1 = ∅ and |Y0| = ω. Then A ⊕ 〈Y0〉 is denumerable and atomless, and so isisomorphic to 〈Y0〉. Hence A⊕ F ∼= A⊕ 〈Y0〉 ⊕ 〈Y1〉 ∼= 〈Y0〉 ⊕ 〈Y1〉 ∼= F . �

The following result is due to McKenzie, Monk [04].

Theorem 10.18. If A0 and A1 are atomless BAs, then isp(A0 × A1) = isp(A0) ∪isp(A1).

Proof. First suppose that κ ∈ isp(A0); say X ⊆ A0 is maximal independent with|X | = κ. Let Y = {(x, 0) : x ∈ X}. Clearly Y is an independent subset of A0×A1.Suppose that (a, b) ∈ A0×A1 with a /∈ X or b �= 0. Choose a finite subset F of X ,an ε ∈ F 2, and a δ ∈ 2 such that

∏x∈F xε(x) · aδ = 0. Take any y ∈ X\F . Then∏

x∈F (x, 0)ε(x) · (y, 0) · (a, b)δ = (0, 0). So Y is maximal independent. This shows

that isp(A0) ⊆ isp(A0 ×A1). Similarly, isp(A1) ⊆ isp(A0 ×A1).

Now suppose that κ ∈ isp(A0 × A1)\(isp(A0) ∪ isp(A1); we want to get acontradiction. Let X be a maximal independent subset of A0 ×A1 of size κ.

(1) κ > ω.

In fact, suppose that κ = ω. Then π0[X ] is contained in a denumerable atomlesssubalgebra B of A0. Let Y be a set of free generators of B. Since i(A0) > ω, thereis a y0 ∈ A0\Y such that Y ∪ {y0} is still independent. It follows that if x ∈ Xand π0(x) �= 0, then π0(x) · y0 �= 0 �= π0(x) · −y0. Similarly we get an elementy1 ∈ A1 such that if x ∈ X and π1(x) �= 0, then π1(x) · y1 �= 0 �= π1(x) · −y1. Then(y0, y1) /∈ X and X ∪ {(y0, y1)} is still independent, contradiction. So (1) holds.

Temporarily fix j ∈ 2. We define

Dj = {w : w is an X-monomial, and πj(v) �= 0 for every X-monomial v ≤ w}.

Thus if w ∈ Dj and v ≤ w is an X-monomial, then v ∈ Dj.

(2) There is an X-monomial w such that πj(w) = 0.

In fact, suppose not. Then 〈πj(a) : a ∈ X〉 is an independent system in Aj . SinceAj does not have a maximal independent set of size κ, it follows that there isa c ∈ Aj\{πj(a) : a ∈ X} such that 〈πj(a) : a ∈ X〉�〈c〉 is independent. Takeany b ∈ A0 × A1 such that πj(b) = c. Then 〈a : a ∈ X〉�〈b〉 is independent,contradicting the maximality of X . Thus (2) holds.

(3) If w is an X-monomial such that πj(w) = 0, then w ∈ D1−j .

For, suppose that v is an X-monomial such that v ≤ w and π1−j(v) = 0. Thenv = 0, contradiction. So (3) holds.

(4) Dj �= ∅

This is true by (2) and (3), since j is arbitrary.

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(5) If w is an X-monomial and w /∈ Dj , then there is an X-monomial v ≤ w suchthat v ∈ D1−j .

For, choose an X monomial v ≤ w such that πj(v) = 0. Then v ∈ D1−j by (3).

Now let Mj be a maximal set of pairwise disjoint members of Dj . Thus Mj

is countable. Let Xj be a denumerable subset of X such that Mj ⊆ 〈Xj〉, and letYj = X\Xj.

(6) There exist an element bj of 〈X〉 and a subalgebra Cj of 〈X〉 � bj such thatCj is free of size κ and the following conditions hold:

(a) For all w ∈ Dj there is a nonzero v ∈ Cj such that v ≤ w.

(b) For all nonzero v ∈ Cj there is a w ∈ Dj such that w ≤ v.

To prove this, we consider two cases.

Case 1. Mj is infinite. Let bj = 1. Let Jj be the ideal of 〈Xj〉 generated by Mj ,and let Bj be the subalgebra of 〈Xj〉 generated by Jj . Note that Bj = Jj ∪ {a ∈〈Xj〉 : −a ∈ Jj}. Let Cj = 〈Bj ∪ Yj〉. Since Bj ≤ 〈Xj〉 and Yj = X\Xj, it followsthat b · c �= 0 for all b ∈ B+

j and c ∈ 〈Yj〉+. So Cj = Bj ⊕ 〈Yj〉. By Lemma 10.17,Cj is free. Clearly |Cj | = κ.

To check (a), suppose that w ∈ Dj . By the maximality of Mj, there is amember r of Mj such that w · r �= 0. As a product of X-monomials, w · r is also anX-monomial. Write w · r = s0 · s1 with s0 an Xj-monomial and s1 a Yj-monomial.So s0 is the product of r with a part of w, and hence s0 ≤ r. It follows thats0 ∈ Jj ⊆ Bj . This shows that w · r = s0 · s1 ∈ Cj . Thus v = w · r is as desired.

For (b), suppose that v ∈ C+j . Then there is an Xj-monomial d and a Yj-

monomial e such that d · e ≤ v. First suppose that d ∈ Jj . Then there existr0, . . . , rm−1 ∈ Mj such that d ≤ r0 + · · · + rm−1. Wlog d · r0 �= 0. So d · r0 · eis an X-monomial and d · r0 · e ≤ r0 ∈ Mj ⊆ Dj. Hence d · r0 · e ∈ Dj . Sinced · r0 · e ≤ v, this is as desired. Second, suppose that d /∈ Jj . Then −d ∈ Jj , and sowe get members r0, . . . , rm−1 of Mj such that −d ≤ r0 + · · ·+ rm−1. Since Mj isinfinite, there is a member s of Mj different from each ri, and hence s ≤ d. Clearlys ∈ Dj , so s · e ∈ Dj , and s · e ≤ v, as desired.

Case 2.Mj is finite. Let Bj =∑

Mj and Bj = 〈Xj〉 � bj. Thus Bj is a denumerableatomless BA. Let Cj be the subalgebra of 〈X〉 � bj generated by Bj ∪ {bj · y :y ∈ Yj}. Now for each c ∈ Mj let Dc be the subalgebra of 〈X〉 generated by(〈Xj〉 � c) ∪ {c · y : y ∈ Yj}. Then C ∼=

∏c∈Mj

Dc. Clearly for each c ∈ Mj we

have Dc = (〈Xj〉 � c) ⊕ 〈{c · y : y ∈ Yj}〉. Moreover, 〈Xj〉 � c is denumerable and〈{c · y : y ∈ Yj}〉 is free, so again by Lemma 10.12, Dc is free of size |X |. Hencealso C is free of size |X | by the Handbook, Proposition 9.14.

To check (a), suppose that w ∈ Dj . By the maximality of Mj, there is amember r of Mj such that w · r �= 0. As a product of X-monomials, w · r is also anX-monomial. Write w · r = s0 · s1 with s0 an Xj-monomial and s1 a Yj-monomial.So s0 is the product of r with a part of w, and hence s0 ≤ r. It follows thats0 ∈ Bj . This shows that w · r = s0 · s1 ∈ Cj . Thus v = w · r is as desired.

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For (b), suppose that v ∈ C+j . Then there is an Xj-monomial d ≤ bj and

a Yj-monomial e with bj · e �= 0 such that d · e ≤ v. Since d ≤ bj, say wlog thatd · r0 �= 0. So d · r0 · e is an X-monomial and d · r0 · e ≤ r0 ∈ Mj ⊆ Dj . Henced · r0 · e ∈ Dj . Since d · r0 · e ≤ v, this is as desired.


(7) πj is injective on Cj .

For, suppose that v ∈ C+j . By (6)(b) choose w ∈ D+

j such that w ≤ v. Then bythe definition of Dj we have πj(w) �= 0, and so also πj(v) � −0, proving (7).

Now since κ /∈ isp(Aj), it follows from the easy direction in our theoremthat κ /∈ isp(Aj � πj(bj)). Now Cj is free of size κ, so also πj [Cj ] is free ofsize κ. A set of free generators of πj [Cj ] is not maximal, and so we can choosewj ∈ (Aj � πj(bj))\πj [Cj ] such that wj is free over πj [Cj ].

(8) wj · πj(d) �= 0 �= −wj · πj(d) for all d ∈ Dj .

For, by (6)(a) choose v ∈ C+j such that v ≤ d. Then 0 �= wj · πj(v) ≤ wj · πj(d),

so wj · πj(d) �= 0. Similarly, −wj · πj(d) �= 0. so (8) holds.

Now unfix j. Let w = (wo, w10. Suppose that v is anyX-monomial. If v ∈ D0,then w · v �= 0 �= −w · v by (8). Suppose that v /∈ D0. By (5), let s ≤ v be anX-monomial such that s ∈ D1. Then by (8) we get w ·s �= 0, so w ·v �= 0. Similarly−w · v �= 0.

This contradicts the maximality of X . �Corollary 10.19. If A and B are atomless BAs, then i(A×B) = min{i(A), i(B)}.

�Proposition 10.20. If I is an infinite set and 〈Ai : i ∈ I〉 is a system of atomlessBAs, then

isp

(w∏i∈I

Ai

)= {ω} ∪

⋃i∈I

isp(Ai).

Proof. Let B =∏w

i∈I Ai. Clearly isp(Ai) ⊆ isp(B) for each i ∈ I.

Now to show that ω ∈ isp(B), for convenience suppose that I = κ, an infinitecardinal. For each α < κ let 〈xα,i : i ∈ ω〉 be a system of independent elements ofAα. We define yn ∈ B for each n ∈ ω by setting, for each α < κ

yn(α) =

⎧⎨⎩xα,n−α−1 if α < n,

1 if α = n,

0 if n < α.

We claim that 〈yn : n ∈ ω〉 is a maximal independent system of elements of B.To show that it is independent, it suffices to take any positive integer m and any

ε ∈ m2 and show that zdef=∏

n<m yε(n)n �= 0. If ε(n) = 0 for all n < m, then

z(m) =∏n<m

−yn(m) = 1.

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So suppose that ε(i) = 1 for some i < m, and take the least such i. Then

z(i) =∏n<m

yn(i)

=∏n<i

−yn(i) · yi(i) ·∏

i<n<m

yε(n)n (i)

=∏

i<n<m

xε(n)i,n−i−1 �= 0.

So 〈yn : n ∈ ω〉 is independent.Now suppose that w ∈ B+\{yn : n ∈ ω}; we want to show that {yn : n ∈

ω}∪ {w} is no longer independent. Wlog {α < κ : w(α) �= 0} is finite. If w(α) = 0for all α < ω, then y0 · w = 0, as desired. So assume that there is an α < ωsuch that w(α) �= 0, and take the greatest such. Let v = w ·

∏n≤α−yn · yn+1.

We claim that v = 0 (as desired). To show this, take any β, κ. If β ≤ α, thenv(β) ≤ −yβ(β) = 0. If β = α + 1, then v(β) ≤ w(β) = 0. Finally, if α + 1 < β,then v(β) ≤ yα+1(β) = 0.

This finishes the proof that ω ∈ isp(B), and hence ⊇ in our proposition holds.

For the other inclusion, suppose that

κ ∈ isp(B)\({ω} ∪

⋃i∈I

isp(Ai)

);

we want to get a contradiction. Let X be a maximal independent subset of B of

size κ. Wlog for all x ∈ X the set Fxdef= {i ∈ I : x(i) �= 0} is finite. Let Y be

an uncountable subset of X such that 〈Fx : x ∈ Y 〉 forms a Δ-system, say withkernel G. Obviously G �= ∅. Now

(1) 〈x � G : x ∈ X〉 is independent in∏

i∈GAi.

In fact, suppose that K is a finite subset of X and ε ∈ K2. Choose x, z ∈ Y \K.Then x · z ·

∏y∈K yε(y) �= 0, so

∏y∈K(y � G)ε(y) �= 0, as desired for (1).

Now by Theorem 10.18 there is a w ∈∏

i∈G Ai such that 〈x � G : x ∈ X〉�〈w〉is independent. Let v be the member ofB whose restriction to G is w, with v(i) = 0for all i /∈ G. For any finite subset K of X and any ε ∈ K2, choose x, z ∈ Y \K;then for any δ ∈ 2,

x · z ·∏y∈K

yε(y) · vδ �= 0.

But this contradicts the maximality of X . �

A corollary of Proposition 10.20 is that i(∏w

i∈I Ai) = ω. This has been generalizedto moderate products by Kevin Selker.

Proposition 10.21. r(A) ≤ i(A) for every atomless BA A.

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Proof. Let X be maximal independent with |X | = i(A). Clearly the set{∏x∈F

xε(x) : F ∈ [X ]<ω, ε ∈ F 2

}

is weakly dense. �

The results above leave the following questions open.

Problem 99. Determine i(∏

i∈I Ai

)in terms of I and 〈i(Ai) : i ∈ I〉, where I is

infinite and each Ai is atomless.

Problem 100. Determine i(∏B

i∈I Ai

)in terms of B and 〈i(Ai) : i ∈ I〉, where I is

infinite, Finco(I) ≤ B ≤P(I), each Ai is atomless (moderate products).

Problem 101. Is πχinf(A) ≤ i(A) for every atomless BA A?

Concerning the spectrum function IndHs, note that if IndH−(A) ≤ μ ≤ Ind(A),then A has a homomorphic image B such that μ ≤ Ind(B) ≤ μω. Moreover, if Ahas CSP, then this cannot be improved:

IndHs(A) = {λ : 2ω ≤ λ ≤ Ind(A), λω = λ} for CSP A.

(These remarks are due to S. Koppelberg.)

The spectrum function IndSs is trivial: IndSs(A) = [ω, IndA] for every infiniteBA.

A BA A has free caliber κ if κ ≤ |A| and ∀X ∈ [A]κ∃Y ∈ [X ]κ(Y is independent).Freecal(A) is the set of all κ ≤ |A| such that A has free caliber κ.

Proposition 10.22. For any BA A, if cf(κ) = ω, then κ /∈ Freecal(A).

Proof. If κ = ω, take a disjointX ∈ [A]ω; clearly thisX shows that κ /∈ Freecal(A).

Now suppose that κ > ω and cf(κ) = ω. We can assume that A has anindependent subset of size κ. Let 〈μm : m ∈ ω〉 be a strictly increasing sequenceof infinite cardinals with supremum κ, and suppose that 〈xα : α < κ〉 is a systemof independent elements of A. Then⎧⎨

⎩xα · xμm+1 ·∏n≤m

−xμn : μm < α < μm+1

⎫⎬⎭

is a subset of A of size κ, but it does not have any independent subset of size κ. �

We give just a few results about free caliber; a more extensive treatment is inMonk [83]. The notion of free caliber is motivated by the central example of afree BA itself. By Theorem 9.16 of the Handbook, λ ∈ Freecal(Fr(κ)) for everyuncountable regular cardinal λ ≤ κ. This result is generalized below.

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Proposition 10.23. κ ∈ Freecal(A×B) iff one of the following conditions holds:

(i) κ ∈ Freecal(A) and |B| < κ.

(ii) κ ∈ Freecal(B) and |A| < κ.

(iii) κ ∈ Freecal(A) ∩ Freecal(B).

Proof. First suppose that κ ∈ Freecal(A × B). If X ∈ [A]κ, then {(x, 0) : x ∈X} ∈ [A × B]κ, and so there is an independent Z ∈ [{(x, 0) : x ∈ X}]κ. WritingZ = W ×{0}, it is clear that W is independent and W ∈ [X ]κ. A similar argumentworks for B, and this shows that one of (i)–(iii) holds.

Second, suppose that (i) holds. Suppose that X ∈ [A×B]κ. Then Y = {a ∈A : ∃b ∈ B[(a, b) ∈ X ]} has size κ, and hence there is an independent Z ∈ [Y ]κ.Clearly this gives rise to an independent W ∈ [X ]κ.

(ii) is similar. Now suppose that (iii) holds, and X ∈ [A×B]κ. For each i < 2let

x ≡i y iff x, y ∈ X and πi(x) = πi(y).

This gives two equivalence relations on X . Now for each x ∈ X let f(x) = (x/ ≡0,x/ ≡1). Clearly f is one-one. Hence there are κ many equivalence classes under≡0 or under ≡1. This easily yields the desired conclusion. �Proposition 10.24. Let I be an infinite set, and 〈Ai : i ∈ I〉 a system of infiniteBAs. Let κ be an infinite cardinal, and assume that |I| < cf(κ). Then the followingconditions are equivalent:

(i) κ ∈ Freecal(∏w

i∈I Ai

).

(ii) ∃i ∈ I[κ ≤ |Ai|], and κ ∈⋂{Freecal(Ai) : κ ≤ |Ai|}.

Proof. Let B =∏w

i∈I Ai.

Assume (i). In particular, κ ≤ |B|. Suppose that ∀i ∈ I[|Ai| < κ]. Since|I| < cf(κ), this is a contradiction. Clearly κ ∈

⋂{Freecal(Ai) : κ ≤ |Ai|}.

Now assume (ii), and suppose that X ∈[∏w

i∈I Ai

]κ. For each x ∈ X let

Fx = {i ∈ I : xi �= 0}. We may assume that each set Fx is finite. Now

X =⋃

G∈[I]<ω

{x ∈ X : Fx = G}.

Since |[I]<ω| = |I| < cf(κ), there exist a Y ∈ [X ]κ and a G ∈ [I]<ω such thatFx = G for all x ∈ Y . Let Z = {f � G : f ∈ Y }. Then Z is a subset of

∏i∈G Ai of

size κ. By Proposition 10.23 there is an independent W ∈ [Z]κ. Clearly this givesan independent U ∈ [Y ]κ. �Proposition 10.25. Let I be an infinite set, and 〈Ai : i ∈ I〉 a system of infiniteBAs. Let κ be an infinite cardinal, and assume that cf(κ) ≤ |I|. Then the followingconditions are equivalent:

(i) κ ∈ Freecal(∏w

i∈I Ai

).

(ii) The following conditions hold:

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(a) |I| < κ.

(b) sup{|Ai| : i ∈ I, |Ai| < κ} < κ.

(c) |{i ∈ I : κ ≤ |Ai|}| < cf(κ).

(d) ∃i ∈ I[κ ≤ |Ai|].(e) κ ∈

⋂{Freecal(Ai) : i ∈ I, κ ≤ Ai}.

Proof. Again let B =∏w

i∈I Ai. Assume (i). In particular, κ ≤ |B|. (a): If κ ≤ |I|,let 〈iα : α < κ〉 be a one-one sequence of elements of I. Let X = {χ{iα} : α < κ,where in general χF is the member of B such that χF (j) = 1 for j ∈ F andχF (j) = 0 otherwise. Clearly X does not have an independent subset even of size2, contradiction.

(b): Suppose that (b) fails. Then there is a sequence 〈i(α) : α < κ} ofmembers of I such that 〈|Ai(α)| : α < κ〉 is strictly increasing with supremum κ.Define

X ={χai(α) : α < κ, a ∈ Ai(α)

},

where χai(α) is the member of B which is equal to a at i(α) and is 0 otherwise.

Let Y ∈ [X ]κ be independent. Then there are distinct α, β < κ and a ∈ Ai(α),

b ∈ Ai(β), such that χai(α), χ

bi(β) ∈ Y . Since χa

i(α) ·χbi(β) = 0, this is a contradiction.

(c): Suppose that (c) fails. Then there is a one-one sequence 〈i(α) : α < cf(κ)}of members of I such that |Ai(α)| = κ for all α < cf(κ). Let 〈Xα : a < cf(κ) besuch that ∀α < cf(κ)[Xα ⊆ Ai(α)] and 〈|Xα| : α < cf(κ)〉 is strictly increasingwith supremum κ. Define

Y ={χai(α) : α < cf(κ), a ∈ Xα

};

this gives a contradiction as in (b).

(d): Suppose that ∀i ∈ I[|Ai| < κ]. Then by (b) there is a cardinal λ < κsuch that |Ai| ≤ λ for all i ∈ I. Hence using (a),

|B| ≤∑

F∈[I]<ω

∣∣∣∣∣∏i∈F

Ai

∣∣∣∣∣ ≤ |I| · λ < κ,

contradiction.

(e): Clear.

Now assume (ii), and suppose that X ∈ [B]κ. Let J = {i ∈ I : |Ai| ≥ κ},K = {i ∈ I : |Ai| < κ}, and C =

∏wi∈K Ai. Then

(1) |C| < κ.

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In fact, by (b) we have λdef= supi∈K |Ai| < κ, and, using (a),

|C| ≤

∣∣∣∣∣∣⋃

G∈[K]<ω

{x ∈ C : supp(x) = G}

∣∣∣∣∣∣≤

∑G∈[K]<ω

∣∣∣∣∣∏i∈G

Ai

∣∣∣∣∣≤ |I| · λ < κ,

proving (1).

Now again let Fx = {i ∈ I : xi �= 0}. Wlog each Fx is finite. Now

X =⋃{{x ∈ X : Fx ∩ J = G} : G ∈ [J ]<ω}.

Since |[J ]<ω| < cf(κ) by (c), it follows that there exist Y ∈ [X ]κ and G ∈ [I]<ω

such that ∀x ∈ Y [Fx ∩ J = G}. Let D =∏

i∈G Ai × C.

(2) G �= ∅, and {x � (G ∪K) : x ∈ Y } ∈ [D]κ.

If G = ∅, then Fx ⊆ K for each x ∈ Y . Then 〈x � K : x ∈ Y 〉 is a one-one function, so |Y | = κ contradicts (1). For the second part of (2), supposethat x, y ∈ Y with x �= y. Choose i ∈ I such that xi �= yi. If i ∈ K, then(x � (G ∪K))i �= (y � (G ∪K))i. If i ∈ J , then i ∈ G since i ∈ Fx or i ∈ Fy, soagain (x � (G ∪K))i �= (y � (G ∪K))i. This proves (2).

Now by (e), κ ∈ Freecal(Ai) for all i ∈ G (since G ⊆ J). Hence by Proposition10.24, κ ∈ Freecal

(∏i∈G Ai

). By (1) and Proposition 10.23, κ ∈ Freecal(D). It

follows from (2) that there is a W ∈ [Y ]κ such that {x � (G ∪ K) : x ∈ W} isindependent. Hence W itself is independent. �

Proposition 10.26. Suppose that I is an infinite set, κ ≤∣∣∏

i∈I Ai

∣∣, and for everysystem 〈λi : i ∈ I〉 of cardinals less that κ we have

∏i∈I λi < κ. Then the following

hold:

(i) There is an i ∈ I such that κ ≤ |Ai|.(ii) κ ∈ Freecal

(∏i∈I Ai

)iff κ ∈

⋂{Freecal(Ai) : κ ≤ |Ai|}.

Proof. Clearly the hypotheses imply (i). The direction ⇒ in (ii) is clear. Nowsuppose that κ ∈

⋂{Freecal(Ai) : κ ≤ |Ai|}, and X ∈

[∏i∈I Ai

]κ. Then there is

an i ∈ I such that |πi[X ]| ≥ κ, as otherwise |X | ≤∏

i∈I |πi[X ]| < κ, contradiction.Hence the desired conclusion follows. �

Corollary 10.27. If κ ≤∣∣∏

i∈I Ai

∣∣, κ = λ+, and λ|I| = λ, then (i) and (ii) ofProposition 10.26 hold. �

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Of course Proposition 10.26 does not take care of all possibilities concerning freecaliber and arbitrary products; the following problem is open.

Problem 102. Describe Freecal(∏

i∈I Ai

)in terms of 〈Freecal(Ai) : i ∈ I〉.

A particular infinite product to which Proposition 10.26 does not apply is∏n∈ω Fr(�n); it is natural to conjecture that this product has free caliber �+

ω .But Shelah [99], 6.11, shows that consistently the conjecture can be true or false.This answers Problem 35 of Monk [96].

We now turn to free products with respect to free caliber. The followinggeneral lemma will be useful.

Lemma 10.28. If A and B are subalgebras of C and A∪B generates C, then everyelement x ∈ C can be written in the form

x =∑i<mx

(axi · bxi ),

with each axi ∈ A, bxi ∈ B. With x written in this form, let 〈cxj : j < nx〉 be aone-one enumeration of the atoms of the finite subalgebra 〈{axi : i < mx}〉 of A,and let

dxj =∑{bxk : k < mx, c

xj ≤ axk}

for each j < nx. Then

(∗) x =∑j<nx

(cxj · dxj );

−x =∑j<nx

(cxj · −dxj ).

Proof. We only give the derivation of (∗):∑i<mx

(axi · bxi ) =∑i<mx

∑{cxj · bxi : j < nx, c

xj ≤ axi }

=∑j<nx

(cxj ·

∑{bxi : i < mx, c

xj ≤ axi }

)

=∑j<mx

(cxj · dxj ). �

Note that by symmetry there is a similar result using atoms of 〈{bxj : j < mx}〉.Now we take first the case of a free product of two BAs, and κ regular.

Lemma 10.29. Let κ be an infinite cardinal. Suppose that A and B are BAs,X ∈ [A⊕B]κ, n ∈ ω, and for each x ∈ X we have

x =∑i<n

(axi · bxi ),

[email protected]

with each axi ∈ A, bxi ∈ B, 〈bxi : i < n〉 a partition of B. Suppose that i < n,〈axi : x ∈ X〉 is independent, and {bxi : x ∈ X} has fip. Then X is independent.

Proof. Suppose that F,G ∈ [X ]<ω with F ∩G = ∅. Then∏x∈F

x ·∏x∈G

−x =∏x∈F

∑k<n

(axk · bxk) ·∏x∈G

∑k<n

(−axk · bxk)

≥∏x∈F

(axi · bxi ) ·∏x∈G

(−axi · bxi )

�= 0. �

Lemma 10.30. Let κ be an infinite cardinal. Suppose that A and B are BAs, andκ ∈ Freecal(A). Suppose that X ∈ [A⊕B]κ, n ∈ ω, and for each x ∈ X we have

x =∑i<n

(axi · bxi ),

with each axi ∈ A, bxi ∈ B, 〈bxi : i < n〉 a partition of B. Suppose that i < n and|{axi : x ∈ X}| = κ while {bxi : x ∈ X} has fip. Then there is a Y ∈ [X ]κ such thatY is independent.

Proof. There is a Y ∈ [X ]κ such that 〈axi : x ∈ Y 〉 is independent. Y is independentby Lemma 10.29. �

Proposition 10.31. Let κ be regular and uncountable. Then κ ∈ Freecal(A0 ⊕ A1)iff ∃i < 2[κ ∈ Freecal(Ai)] and κ ∈

⋂{Freecal(Ai) : i < 2, κ ≤ |Ai|}.

Proof. Clearly⇒ holds. Now assume the above conditions, and suppose that X ∈[A0 ⊕A1]

κ. By Lemma 10.28, write each x ∈ X in the form

x =∑i<mx

(axi · bxi ),

where each axi ∈ A0, and 〈bxi : i < mx〉 is a partition of A1. Since cf(κ) > ω, thereexist an X0 ∈ [X ]κ and an n ∈ ω such that mx = n for all x ∈ X0.

Case 1. ∀i < n[|{axi : x ∈ X0}| < κ]. Then

X0 =⋃{

{x ∈ X0 : 〈axi : i < n〉 = y} : y ∈∏i<n

{axi : x ∈ X0}},

so by the regularity of κ there exist an X1 ∈ [X0]κ and a system 〈yi : i < n〉 of

elements of A0 such that axi = yi for all x ∈ X1 and i < n. By Lemma 10.28 againwe can write each x ∈ X1 as

x =∑i<u

(ci · dxi )

[email protected]

with 〈ci : i < u〉 a partition of A0 and each dxi ∈ A1. Now there is an i < u suchthat |{dxi : x ∈ X1}| = κ, as otherwise |X1| < κ. By Lemma 10.30 there is anindependent Y ∈ [X1]

κ.

Case 2. ∃i < n[|{axi : x ∈ X0}| = κ]. Then there is an X1 ∈ [X0]κ such that

〈axi : x ∈ X1〉 is independent.Subcase 2.1. |{bxi : x ∈ X0}| < κ. Then there exist an X1 ∈ [X0]

κ and an elementc of B such that bxi = c for all x ∈ X1. By Lemma 10.29 X1 is independent.

Subcase 2.2. |{bxi : x ∈ X0}| = κ. Then there is an X1 ∈ [X0]κ such that 〈axi :

x ∈ X1〉 and 〈bxi : x ∈ X1〉 are independent. Then X1 is independent by Lemma10.29. �Corollary 10.32. Let κ be regular and uncountable. Suppose that I is finite withat least two elements. Then κ ∈ Freecal(⊕i∈IAi) iff ∃i ∈ I[κ ∈ Freecal(Ai)] andκ ∈

⋂{Freecal(Ai) : i ∈ I, κ ≤ |Ai|}. �

To treat the case of singular κ, first recall Proposition 10.22, which implies thatwe only need to consider singular κ with cf(κ) > ω. We also need a new notion.A BA A has precaliber κ iff

∀X ∈ [A]κ∃Y ∈ [X ]κ[Y has fip].

We let precal(A) = {κ : A has precaliber κ}. Note that there is no assumptionhere that κ ≤ |A|; in fact, A vacuously has precaliber κ for every cardinal κ greaterthan |A|.Proposition 10.33. Suppose that κ is singular, cf(κ) ∈ precal(A), X ∈ [A]κ, anda ∈ XA with | rng(a)| < κ. Then there is a Y ∈ [X ]κ such that {ax : x ∈ Y } has fip.

Proof. We claim

(∗) ∃d ∈ cf(κ) rng(a)∀P ∈ [cf(κ)]cf(κ)[|{x ∈ X : ∃α ∈ P (ax = dα)}| = κ].

For, define x ≡ y iff x, y ∈ X and ax = ay. Thus ≡ is an equivalence relationon X with less than κ equivalence classes. If some class K has κ elements, letd : cf(κ)→ rng(a) be constant with value a member of K. Then if P ∈ [cf(κ)]cf(κ)

we have {x ∈ X : ∃α ∈ P (ax = dα)} = K, as desired. If no such class exists, thenlet 〈Zα : α < cf(κ)〉 enumerate some of the classes so that 〈|Zα| : α < cf(κ)〉 isstrictly increasing with supremum κ. Let dα ∈ Zα for all α < cf(κ). Clearly thedesired conclusion follows again.

Now by the assumption cf(κ) ∈ precal(A), choose P ∈ [cf(κ)]cf(κ) such that{dα : α ∈ P} has fip. Let Y = {x ∈ X : ∃α ∈ P (ax = dα)}. Clearly {ax : x ∈ Y }has fip. �Proposition 10.34. Suppose that κ is a singular cardinal, κ ∈ precal(B), X ∈[A⊕B]κ, n ∈ ω, and for every x ∈ X

x =∑i<n

axi · bxi ,

[email protected]

with axi ∈ A and bxi ∈ B. Suppose that i < n, |{axi : x ∈ X}| = κ, and |{bxi : x ∈X}| < κ.

Then there is a Y ∈ [X ]κ such that |{axi : x ∈ Y }| = κ and {bxi : x ∈ Y }has fip.

Proof. Choose Z ∈ [X ]κ such that 〈axi : x ∈ Z〉 is one-one. By Proposition 10.33choose Y ∈ [Z]κ such that {bxi : x ∈ Z} has fip. �Proposition 10.35. If κ is regular and uncountable, then κ ∈ precal(A ⊕ B) iffκ ∈ precal(A) ∩ precal(B).

Proof. ⇒ is clear. Now suppose that κ ∈ precal(A)∩precal(B), and X ∈ [A⊕B]κ.For each x ∈ X write

x =∑i<mx

(axi · bxi ),

where each axi ∈ A+ and bxi ∈ B+. Then there exist n and Y ∈ [X ]κ such thatmx = n for all x ∈ Y . Choose i < n such that |{axi · bxi : x ∈ Y }| = κ. Say bysymmetry |{axi : x ∈ Y }| = κ. Choose Z ∈ [Y ]κ such that {axi : x ∈ Z} has fip.

Case 1. |{bxi : x ∈ Z}| < κ. Then there exist a V ∈ [Z]κ and a c ∈ B+ such thatbxi = c for all x ∈ V . Clearly then V has fip.

Case 2. |{bxi : x ∈ Z}| = κ. Then there is a V ∈ [Z]κ such that {bxi : x ∈ V } hasfip. Hence V has fip. �Lemma 10.36. Let κ be a singular cardinal with cf(κ) > ω. Then κ ∈ Freecal(A⊕B)iff one of the following conditions holds:

(i) κ ≤ |A|, |B| < κ, κ ∈ Freecal(A), and cf(κ) ∈ precal(B).

(ii) κ ≤ |B|, |A| < κ, κ ∈ Freecal(B), and cf(κ) ∈ precal(A).

(iii) κ ≤ |A|, |B|, κ ∈ Freecal(A)∩Freecal(B), and cf(κ) ∈ precal(A)∩precal(B).

Proof. Let 〈λα : α < cf(κ)〉 be a strictly increasing sequence of infinite cardinalswith supremum κ.

For ⇒, suppose that κ ∈ Freecal(A ⊕ B). The assertions in (i)–(iii) con-cerning free caliber are clear. Suppose that κ ≤ |A| and |B| < κ. To show thatcf(κ) ∈ precal(B), suppose that Y ∈ [B]cf(κ). Let 〈yα : α < cf(κ)〉 be a one-oneenumeration of Y . For each α < cf(κ), let Zα ∈ [A]λα . Let X = {a · yα : α <cf(κ), a ∈ Zα}. Then X ∈ [A ⊕ B]κ, and so X has an independent subset W ofsize κ. Let V = {α : (a · yα) ∈ W for some a ∈ Zα}. Then |V | = cf(κ) since|X | = κ. If F is a finite subset of V , for each α ∈ F choose a(α) ∈ Zα suchthat (a(α) · yα) ∈ W . Then

∏α∈F (a(α) · yα) �= 0, and hence

∏α∈F yα �= 0. So

{yα : α ∈ V } has fip. It follows that B has precaliber cf(κ). The other precaliberassertions in (i)–(iii) are checked similarly.

Now for⇐, assume that one of (i)–(iii) holds, and suppose that X ∈ [A⊕B]κ.For each x ∈ X write

x =∑j<mx

(axj · bxj ),

[email protected]

with axj ∈ A, bxj ∈ B, and 〈bxj : j < mx〉 a partition of B. Then there existX0 ∈ [X ]κ and n ∈ ω such that mx = n for all x ∈ X0. We now consider severalcases.

Case 1. ∃X1 ∈ [X0]κ∃i < n[|{bxi : x ∈ X1}| < κ and |{axi : x ∈ X1}| = κ].

By Proposition 10.34 choose X2 ∈ [X1]κ such that |{axi : x ∈ X2}| = κ and

{bxi : x ∈ X2} has fip. By Proposition 10.30 there is a Z ∈ [X2]κ such that Z is

independent.

Case 2. ∀X1 ∈ [X0]κ∀i < n[|{bxi : x ∈ X1}| < κ → |{axi : x ∈ X1}| < κ], and

there is an i < n such that |{axi : x ∈ X1}| = κ. Choose X2 ∈ [X1]κ such that

〈axi : x ∈ X2〉 is independent. Applying our case condition to X2 in place of X1

we infer that |{bxi : x ∈ X2}| = κ. Choose X3 ∈ [X2]κ such that 〈bxi : x ∈ X3〉 is

independent. Hence X3 is independent by Lemma 10.29.

Case 3. ∀X1 ∈ [X0]κ∀i < n[|{bxi : x ∈ X1}| < κ → |{axi : x ∈ X1}| < κ], and

∀i < n[|{axi : x ∈ X1}| < κ. For each x ∈ X1 let 〈cxj : j < px〉 be a one-oneenumeration of the atoms of 〈{axi : i < n}〉. Then by Lemma 10.28 we can write

x =∑j<px

(cxj · dxj )

for each x ∈ X1, where each dxj is in B. Choose X2 ∈ [X1]κ and q so that px = q

for all x ∈ X2. Now |{cxj : x ∈ X2}| < κ for all j < q, so there is a j < q suchthat |{dxj : x ∈ X2}| = κ. Choose X3 ∈ [X2]

κ so that 〈dxj : x ∈ X3〉 is one-one. ByProposition 10.34, there is an X4 ∈ [X3]

κ such that {cxj : x ∈ X4} has fip. Nowthe desired result follows from Lemma 10.30. �Proposition 10.37. Let I be a finite set with at least two elements, and let κ be asingular cardinal with cf(κ) > ω. Then κ ∈ Freecal(⊕i∈IAi) iff one of the followingconditions hold:

(i) There is exactly one i ∈ I such that κ ≤ |Ai|; for this i we have κ ∈Freecal(Ai) and cf(κ) ∈

⋂j∈I\{i} precal(Aj).

(ii) There are at least two i ∈ I such that κ ≤ |Ai|, κ ∈⋂{Freecal(Ai) : i ∈

I, κ ≤ |Ai|}, and cf(κ) ∈⋂

i∈I precal(Ai).

Proof. We prove this by induction on |I|. The case |I| = 2 is given by Lemma10.36. Now suppose that the proposition holds for I, with |I| ≥ 2, and we aregiven I ∪ {i} with i /∈ I. Let B = ⊕j∈IAj . Suppose that κ ∈ Freecal(B ⊕Ai). ByLemma 10.36 we have three cases.

Case 1. κ≤ |⊕j∈IAj |, |Ai|<κ, κ∈Freecal(⊕j∈IAj), and cf(κ)∈precal(Ai). It isstraightforward to check the desired conditions for I ∪ {i}.Case 2. κ≤|Ai|, |⊕j∈IAj | < κ, κ∈Freecal(Ai), and cf(κ)∈precal(⊕j∈IAj). Thisgives condition (i) for I ∪ {j}, using Proposition 10.35.

Case 3. κ ≤ |⊕j∈IAj |, κ ≤ |Ai|, κ ∈ Freecal (⊕j∈IAj) ∩ Freecal(Ai), and cf(κ) ∈precal (⊕j∈IAj) ∩ precal(Ai). By the inductive hypothesis, (ii) holds.

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For the converse we have three possibilities.

Case 1. κ≤|Ai|, ∀j ∈ I[|Aj | < κ]; κ∈Freecal(Ai) and cf(κ)∈⋂

j∈Iprecal(Aj). ByPropositions 10.35 and 10.36 we get κ ∈ Freecal(⊕j∈IAi ⊕Ai).

Case 2. There is exactly one j ∈ I such that κ ≤ |Aj |; also |Ai| < κ; andκ ∈ Freecal(Aj) and cf(κ) ∈ precal(Ai) ∩

⋂k∈I\{j} precal(Aj). By the inductive

hypothesis we get κ ∈ Freecal(⊕k∈IAk), and hence κ ∈ Freecal(⊕j∈IAi ⊕ Ai) by10.36.

Case 3. (ii) holds for I ∪ {i}. The desired conclusion is clear. �

Turning to arbitrary free products, the case of regular κ is simple:

Proposition 10.38. Suppose that |I| ≥ 2 and κ is uncountable and regular. Thenκ ∈ Freecal (⊕i∈IAi) iff κ ≤ |⊕i∈IAi| and ∀i ∈ I[κ ≤ |Ai| → κ ∈ Freecal(Ai)].

Proof. For I finite Corollary 10.32 gives the desired result. Now suppose that I isinfinite. For brevity let B = ⊕i∈IAi. The implication ⇒ is obvious. Now assumethe two conditions, and suppose that X ∈ [B]κ. For all x ∈ X choose Fx ∈ [I]<ω

such that x ∈ ⊕i∈FxAi. Then there exist an X0 ∈ [X ]κ and an integer m such that〈Fx : x ∈ X0〉 forms a Δ-system, say with kernel G, and |Fx| = m for all x ∈ X0.If Fx = G for all x ∈ X0, then the desired conclusion follows from Corollary 10.32.Hence assume that Fx �= G for all x ∈ X0. For each x ∈ X0 write

x =∑i<nx

(axi · bxi )

with each axi ∈ ⊕j∈GAj , bxi ∈ ⊕j∈Fx\GAj , 〈bxi : i < nx〉 a partition of ⊕j∈Fx\GAj .

Then there exist X1 ∈ [X0]κ and p such that nx = p for all x ∈ X1. Note by

disjointness of supports that 〈bxi : x ∈ X1〉 is independent for every i < p.

Case 1. ∃i < p[|{axi : x ∈ X1}| = κ]. Then there is an X2 ∈ [X1]κ such that

〈axi : x ∈ X2〉 is independent. Clearly also X2 itself is independent.

Case 2. ∀i < p[{axi : x ∈ X1}| < κ]. For each x ∈ X1 let 〈cxj : j < qx〉 be thesystem of atoms of the algebra 〈{axi : i < p}〉. By Lemma 10.28 write

x =∑j<qx

(cxj · dxj )

with each dxj ∈ ⊕k∈Fx\GAk. TakeX2 ∈ [X1]κ and r such that qx = r for all x ∈ X2.

There is a j < r such that |{dxj : x ∈ X2}| = κ. Hence X2 is independent. �Corollary 10.39. If κ is uncountable and regular, |Ai| < κ for all i ∈ I, with Iinfinite, and κ ≤ |⊕i∈IAi|, then κ ∈ Freecal (⊕i∈IAi). �

Taking eachAi of size 4, we obtain another proof of Theorem 9.16 in the Handbook.

Corollary 10.40. If κ is uncountable and regular and I is a set with at least twoelements, then κ ∈ precal(⊕i∈IAi) iff ∀i ∈ I[κ ∈ precal(Ai)].

[email protected]

Proof. Repeat the proof of Proposition 10.38, replacing “independent” by “fip”.�

Now we take the case of singular κ. Here we need the well-known double deltasystem lemma. We give a proof of it which is due to Richard Laver First we havean elementary lemma about singular cardinals.

Lemma 10.41. Let κ be a singular cardinal. Then there is a strictly increas-ing sequence 〈λα : α < cf(κ)〉 of successor cardinals such that cf(κ) < λ0,supα<cf(κ) λα = κ, and ∀α < cf(κ)[supβ<α λβ < λα].

Proof. First let 〈μα : α < cf(κ)〉 be any sequence of cardinals less than κ but withsupremum κ. Then set λ0 = (cf(κ))+ and, for 0 < α < cf(κ),

λα =

(max

{μα, sup

β<αλβ

})+

. �

Lemma 10.42. (Double Δ-system lemma) Suppose that κ is singular with cf(κ) >ω. Take 〈λα : α < cf(κ)〉 as in Lemma 10.41. Suppose that 〈Aξ : ξ < κ〉 is asystem of finite sets. Then there exist M ⊆ cf(κ), G, and sequences 〈Nα : α ∈M〉,〈Fα : α ∈M〉 such that the following conditions hold:

(i) 〈Nα : α ∈ M〉 is a system of pairwise disjoint subsets of κ, with |Nα| = λα

for all α ∈M .

(ii) |M | = cf(κ).

(iii) 〈Aξ : ξ ∈ Nα〉 is a Δ system with kernel Fα, for every α ∈M .

(iv) For distinct α, β ∈M we have:

(a) Fα ∩ Fβ = G.

(b) Aξ ∩Aη = G for all ξ ∈ Nα and η ∈ Nβ.

Proof. Write κ =⋃

α<cf(κ) N′α, where the N

′α’s are pairwise disjoint and |N ′

α| = λα

for all α < cf(κ). For each α < cf(κ) choose N ′′α ∈ [N ′

α]λα so that 〈Aξ : ξ ∈ N ′′

α〉 isa Δ system, say with kernel Fα. Choose M ∈ [cf(κ)]cf(κ) so that 〈Fα : α ∈ M〉 isa Δ system, say with kernel G.

For each α ∈M let

Bα =⋃⎧⎨⎩⋃

ξ∈N ′′β

Aξ : α ∈M,β < α

⎫⎬⎭ .

Thus∣∣∣⋃ξ∈N ′′

βAξ

∣∣∣ = λβ for each β < α, and hence |Bα| < λα by the choice of the

sequence 〈λα : α < cf(κ)〉. Now

N ′′α =

⋃{{ξ ∈ N ′′

α : Aξ ∩Bα = D} : D ∈ [Bα]<ω},

[email protected]

so there exist an N ′′′α ∈ [N ′′

α ]λα and a finite Cα such that Aξ ∩ Bα = Cα for all

ξ ∈ N ′′′α . Note that Cα ⊆ Fα. In fact, take distinct ξ, η ∈ N ′′′

α . Then Fα ∩ Bα =Aξ ∩Aη ∩Bα = Aξ ∩Bα ∩ Cα = Cα, so that Cα ⊆ Fα. Finally, let

(∗) Nα = N ′′′α \{ξ ∈ N ′′′

α : (Aξ\Fα) ∩ Fβ �= ∅ for some β ∈M}.

Note that if α, β, γ ∈M with β �= γ, then Fβ ∩ Fγ = G ⊆ Fα; so (Aξ\Fα) ∩ Fβ ∩(Aξ\Fα) ∩ Fγ = ∅. From this fact and the fact that |M | = cf(κ) < λα it followsthat |Nα| = λα.

Now we check the conditions of the lemma. All of them are clear except for(iv)(b). For it, suppose by symmetry that β < α, ξ ∈ Nα, and η ∈ Nβ . ThenAη ⊆ Bα, and hence Aξ ∩ Aη ⊆ Aξ ∩Bα = Cα ⊆ Fα. Therefore

Aξ ∩ Aη = Aξ ∩Aη ∩ Fα

⊆ Aη ∩ Fα

= (Aη ∩ Fα ∩ Fβ) ∪ ((Aη ∩ Fα)\Fβ)

= Aη ∩ Fα ∩ Fβ

⊆ G,

using (∗). To show equality, pick ξ′ ∈ Nα\{ξ} and η′ ∈ Nβ\{η}. Then

Aξ ∩Aξ′ ∩ Aη ∩ Aη′ = Fα ∩ Fβ = G,

and so G ⊆ Aξ ∩ Aη. �Theorem 10.43. Let κ be singular with cf(κ) > ω, and suppose that |I| ≥ 2 andκ ≤ |⊕i∈IAi|. Then κ ∈ Freecal (⊕i∈IAi) iff (i) below holds, and exactly one of(ii), (iii) holds.

(i) If I is infinite and G ∈ cf(κ)([I]<ω) consists of pairwise disjoint sets, and if〈Xα : α < cf(κ)〉 is such that

∀α < cf(κ) [Xα ⊆ ⊕i∈GαAi] and supα<cf(κ)

|Xα| = κ,

then there is a system 〈Yα : α < cf(κ)〉 such that

∀α < cf(κ)[Yα ⊆ Xα and Yα is independent] and supα<cf(κ)

|Yα| = κ.

(ii) There is an i ∈ I such that

κ ≤ |Ai| and κ ∈ Freecal(Ai) and |⊕j∈I,j �=iAj | < κ

and cf(κ) ∈⋂

j∈I,j �=iprecalAi.

(iii) For all i ∈ I we have

|⊕j∈I,j �=iAj | ≥ κ and cf(κ) ∈ precal(Ai)

and if κ ≤ |Ai| then κ ∈ Freecal(Ai).

[email protected]

Proof. The case I finite is given by Proposition 10.37. So suppose that I is infinite.

For ⇒, suppose that κ ∈ Freecal (⊕i∈IAi). Then (i) is clear. Clearly at mostone of (ii), (iii) holds.

Case 1. ∀i ∈ I[|⊕j∈I,j �=iAj | ≥ κ]. Then (iii) holds, using Proposition 10.36 for theprecaliber assertion.

Case 2. ∃i∈I[|⊕j∈I,j �=iAj |<κ]. By Proposition 10.36 we have κ∈Freecal(Ai) andcf(κ) ∈ precal (⊕j∈I,j �=iAj). Clearly then cf(κ) ∈

⋂j∈I,j �=i precal(Aj).

Turning to⇐, we assume (i) and one of (ii), (iii). We now note that it sufficesto treat these two cases:

(1) ∀i ∈ I[κ ≤ |Ai|, κ ∈ Freecal(Ai), and cf(κ) ∈ precal(Ai)].

(2) ∀i ∈ I[|Ai| < κ and cf(κ) ∈ precal(Ai)].

For, suppose that we have proved that κ ∈ Freecal (⊕i∈IAi) in these two cases. If(ii) holds, then this conclusion follows immediately from Propositions 10.40 and10.36(i), without using these cases. Assume that (iii) holds. Let J = {i ∈ I : |Ai| ≥κ}. If J = ∅, then (2) gives the desired result. So suppose that J �= ∅. Then case(1) applies to J , and so by assumption κ ∈ Freecal (⊕j∈JAj). Now Propositions10.36 and 10.40 give the desired result.

We treat the cases (1) and (2) simultaneously. Take 〈λα : α < cf(κ)〉 as inLemma 10.41. Suppose that X ∈ [⊕i∈IAi]

κ. For each x ∈ X let Hx ∈ [I]<ω be

such that x ∈ ⊕i∈HxAi. Let f : κ → X be a bijection, and define Kα = Hf(α)

for all α < κ. Now we apply the double Δ-system lemma 10.42 to the system〈Kα : α < κ〉 to obtain M ⊆ cf(κ), G, and sequences 〈Nα : α ∈M〉, 〈Fα : α ∈M〉with the following properties:

(3) 〈Nα : α ∈M〉 is a system of pairwise disjoint subsets of κ, with |Nα| = λα forall α ∈M .

(4) |M | = cf(κ).

(5) 〈Kξ : ξ ∈ Nα〉 is a Δ system with kernel Fα, for every α ∈M .

(6) For distinct α, β ∈M we have:

(a) Fα ∩ Fβ = G.

(b) Kξ ∩Kη = G for all ξ ∈ Nα and η ∈ Nβ .

NowM =

⋃{{α ∈M : |Fα| = n} : n ∈ ω},

so there exist an M0 ∈ [M ]cf(κ) and n ∈ ω such that |Fα| = n for all α ∈M0. LetX0 = f

[⋃α∈M0

Nα

]. So X0 ∈ [X ]κ. For each x ∈ X0 let α(x) ∈ M0 be such that

x ∈ f [Nα(x)]. Now

X0 =⋃{{x ∈ X0 : |Hx| = m} : m ∈ ω},

so there exist an X1 ∈ [X0]κ and m ∈ ω such that |Hx| = m for all x ∈ X1.

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Note that for any x ∈ X1 we have Hx = Kf−1(x) ⊆ Fα(x) ⊆ G. Now thereare several possibilities, and we first treat the most complicated one.

Case 1. 0 < |G| < n < m. For each x ∈ X1 write

x =∑i<px

(axi · bxi · cxi ),

where each axi ∈ ⊕j∈GAj , 〈axi : i < px〉 a partition, bxi ∈ ⊕j∈Fα(x)\GAj with

f−1(x) ∈ Nα(x), and cxi ∈ ⊕j∈Kf−1(x)\Fα(x)Aj . Then there exist X2 ∈ [X1]

κ and q

such that px = q for all x ∈ X2.

Now by Proposition 10.33 we have

(7) If i < q, Y ∈ [X2]κ, and |{axi : x ∈ Y }| < κ, then there is a Z ∈ [Y ]κ such that

〈axi : x ∈ Z〉 has fip.We now consider some subcases.

Subcase 1.1. There exist an X3 ∈ [X2]κ and an i < q such that 〈axi : x ∈ X3〉

has fip and |{bxi · cxi : x ∈ X3}| = κ. Then there is an X4 ∈ [X3]κ such that

〈bxi · cxi : x ∈ X4〉 is one-one.Now we consider two subsubcases.

Subsubcase 1.1.1. |{bxi : x ∈ X4}| < κ. By Proposition 10.33 there is an X5 ∈ [X4]κ

such that 〈bxi : x ∈ X5〉 has fip. Since 〈bxi ·cxi : x ∈ X4〉 is one-one, {bxi ·cxi : x ∈ X5}still has size κ, and so |{cxi : x ∈ X5}| = κ. Hence there is an X6 ∈ [X5]

κ suchthat 〈cxi : x ∈ X6〉 is one-one. Now(8) for distinct x, y ∈ X6 we have (Kf−1(x)\Fα(x)) ∩ (Kf−1(y)\Fα(y)) = ∅.In fact, if α(x) = α(y), then Kf−1(x) ∩Kf−1(y) = Fα(x) by (5), and the conclusionfollows. If α(x) �= α(y), (6) gives the conclusion.

It follows from (8) that 〈cxi : x ∈ X6〉 is independent. ThenX6 is independent.For, suppose that Y, Z ∈ [X6]

<ω and Y ∩ Z = ∅. Then∏x∈Y

x ·∏x∈Z

−x =∏x∈Y

∑j<q

(axj · bxj · cxj ) ·∏x∈Z

∑j<q

(axj · (−bxj +−cxj )

≥∏x∈Y

(axi · bxi · cxi ) ·∏x∈Z

(axi · (−bxi +−cxi )

≥∏x∈Y

(axi · bxi · cxi ) ·∏x∈Z

(axi · −cxi )

�= 0.

Subsubcase 1.1.2. |{bxi : x ∈ X4}| = κ. Then there is an X5 ∈ [X4]κ such that

〈bxi : x ∈ X5〉 is one-one. Now for each α < cf(κ) let Zα = {bxi : x ∈ X5 andα(x) = α}, and

G′α =

{Fα(x)\G if α = α(x) for some x ∈ X5,

∅ otherwise.

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Now {bxi : x ∈ X5} =⋃

α<cf(κ) Zα, and Zα ∩ Zβ = ∅ if α �= β. Hence

supα<cf(κ) |Zα| = κ.

Thus the hypothesis of (i) holds with G′ in place of G and 〈Zα : α < cf(κ)〉 inplace of 〈Xα : α < cf(κ)〉. Hence there is a system 〈Yα : α < cf(κ)〉 such that

∀ξ < cf(κ)[Yα ⊆ Zα and Yα is independent] and supα<cf(κ)

|Yα| = κ.

Now by (6)(a) we have G′α ∩ G′

β = ∅ if α �= β. It follows that⋃

α<cf(κ) Yα is

independent. Now there is an X6 ∈ [X5]κ such that {bxi : x ∈ X6} =

⋃α<cf(κ) Yα.

Hence {bxi : x ∈ X6} is independent.

Now if |{cxi : x ∈ X6}| < κ, then by Proposition 10.33 there is an X7 ∈ [X6]κ

such that 〈cxi : x ∈ X7〉 has fip. If |{cxi : x ∈ X6}| = κ, then there is anX7 ∈ [X6]<ω

such that 〈cxi : x ∈ X7〉 is one-one; by (8) we get that 〈cxi : x ∈ X7〉 is independent,hence satisfies fip. It follows now that X7 is independent. The argument here isvery similar to that in Subsubcase 1.1.1.

Subcase 1.2. For all i < q and X3 ∈ [X2]κ, if 〈axi : x ∈ X3〉 has fip then |{bxi · cxi :

x ∈ X3}| < κ. Now

(9) ∀i < q∃X3 ∈ [X2]κ[〈axi : x ∈ X3〉 has fip].

In fact, take any i < q. If |{axi : x ∈ X2}| = κ, we can apply (1) to get the desiredX3. Otherwise apply (7).

By (9) there is an X3 ∈ [X2]κ such that for all i < q the sequence 〈axi : x ∈

X3〉 has fip. By the subcase we are in, ∀i < q[|{bxi · cxi : x ∈ X4}| < κ. For eachx ∈ X3 let 〈dxk : k < rx〉 be the system of atoms of 〈{bxj · cxj : j < q〉. ApplyingLemma 10.28 we get 〈exk : k < rx〉 in ⊕s∈GAs such that x =

∑k<rx

(exk · dxk)for all x ∈ X3. Since |{dxk : k < rx, x ∈ X3}| < κ, there is a k < rx such that|{exk : x ∈ X3}| = κ. Then by (1) and Proposition 10.37 there is anX4 ∈ [X3]

κ suchthat 〈exk : x ∈ X4〉 is independent. By Proposition 10.33 there is an X5 ∈ [X4]

κ

such that 〈dxk : x ∈ X4〉 satisfies fip. Hence X5 is independent.

Case 2. G = ∅, n = 0. Clearly then m > 0. Since Hx ∩Hy = ∅ for x �= y, it followsthat X2 is independent.

Case 3. G = ∅, 0 < n < m. For all x ∈ X1 write

x =∑i<px

(bxi · cxi )

with each bxi ∈ ⊕j∈Fα(x)Aj , f

−1(x) ∈ Nα(x), cxi ∈ ⊕j∈Kf−1(x)\Fα(x)

Aj , 〈bxi : i < px〉a partition. Choose X2 ∈ [X1]

κ and q so that px = q for all x ∈ X2.

(10) For all i < q and Y ∈ [X2]κ there is a Z ∈ [Y ]κ such that 〈bxi : x ∈ Z〉 has fip.

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To prove (10), let i < q and Y ∈ [X2]κ. If |{bxi : x ∈ Y }| < κ, then by Proposition

10.33 there is a Z satisfying the desired condition. If |{bxi : x ∈ Y }| = κ, then by(1) there is an Z ∈ [Y ]κ such that 〈bxi : x ∈ Z〉 is independent, hence has fip.

Now we apply (10) q times and obtain X3 ∈ [X2]κ such that for all i < q,

〈bxi : x ∈ X3〉 has fip.Subcase 3.1. There is an i < q such that |{cxi : x ∈ X3}| = κ. By (1), let X4 ∈[X3]

κ be such that 〈cxi : x ∈ X4〉 is independent. By Proposition 10.29, X4 isindependent.

Subcase 3.2. |{cxi : x ∈ X3}| < κ for all i < q. Rewrite each x ∈ X3 as∑i<rx

dxi · exi

with dxi ∈ ⊕j∈Fα(x)Aj , e

xi ∈ ⊕j∈Kf−1(x)\Fα(x)

Aj , 〈exi : i < rx〉 a partition. Let

X4 ∈ [X3]κ be such that rx = s for all x ∈ X4. Now |{exi : x ∈ X4}| < κ for all

i < s, so there is an i < s such that |{dxi : x ∈ X4}| = κ. By Proposition 10.33there is an X5 ∈ [X4]

κ such that {exi : x ∈ X5} has fip. Then by Lemma 10.30there is an X6 ∈ [X5]

κ such that X6 is independent.

Case 4. G = ∅, 0 < n = m. Then x ∈ ⊕i∈FαAi for every x ∈ X1 such thatf−1(x) = α. Let g : cf(α) → M0 be a bijection, and for each α < cf(κ) letYα = {x ∈ X1 : f−1(x) = g(α)}. Then (i) applies to give the desired result, sincethe Fα’s are pairwise disjoint.

Case 5. 0 < |G| < n = m. For each x ∈ X1 write

x =∑i<px

axi · bxi

where axi ∈ ⊕j∈GAj , 〈axi : i < px〉 is a partition, and bxi ∈ ⊕j∈Fα(x)\GAj . TakeX2 ∈ [X1]

κ such that px = q for all x ∈ X2.

Subcase 5.1. There is an i < q such that |{bxi : x ∈ X2}| = κ. Let X3 ∈ [X2]κ

be such that 〈bxi : x ∈ X3〉 is independent. Then take X4 ∈ [X3]κ such that

{axi : x ∈ X4} has fip. By Lemma 10.29 X4 is independent.

Subcase 5.2. |{bxi : x ∈ X2}| < κ for all i < q. Rewrite each x ∈ X2 in the form

x =∑i<rx

cxi · dxi

with cxi ∈ ⊕j∈GAj , dxi ∈ ⊕j∈Fα(x)\GAj , with 〈dxi : i < rx〉 a partition. Then

proceed as in Subcase 5.1.

Case 6. 0 < |G| = n < m. Then for x, y ∈ X1 with x �= y we have Kf−1(x) ∩Kf−1(y) = G. This is very similar to Case 5.

Case 7. 0 < |G| = n = m. The result follows from Proposition 10.37. �

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Corollary 10.44. For any cardinals κ, λ, if ω < cf(κ) and κ ≤ λ, then κ ∈Freecal(Fr(λ)). �

The following problem concerns trying to characterize the classes Freecal(A). Thisis Problem 38 in Monk [96].

Problem 103. For K any nonempty set of regular cardinals, are the followingconditions equivalent?

(i) There is a BA A such that K = {κ : κ ∈ Freecal(A) and κ is regular}.(ii) The following conditions hold, where μ = minK and ν = supK:

(a) μ is uncountable.

(b) For all λ ∈ (μ, ν], if λ is regular and is not the successor of a singularcardinal, then λ ∈ K.

(c) For all λ ∈ (μ, ν], if λ = σ+ for some singular σ with μ ≤ cf σ, thenλ ∈ K.

For background and evidence for this problem, see Monk [83].

The comparison of independence with the cardinal functions already introducedis simple: Ind(A) ≤ Irr(A) for every infinite BA A, and the difference can be arbi-trarily large, for example in an interval algebra; it is possible to have Ind(A) biggerthan π(A), for example in P(κ). Depth(A) can be much larger than Ind(A), forexample in the interval algebra on κ. Note that there are some close relationshipsbetween independence and cellularity, though. The main fact here is Theorem10.1 in the Handbook, a result due to Shelah. A simple application of that the-orem is that if (2c(A))+ ≤ |A|, then (2c(A))+ ≤ Ind(A) by Corollary 10.9 of theHandbook. In particular, |A| ≤ 2max(c(A),Ind(A)). And if |A| is strong limit, then|A| = max(c(A), Ind(A)).

The following is Problem 32 in Monk [96].

Problem 104. Assume that ρ < ν < κ ≤ 2ρ < λ ≤ 2ν with κ and λ regular. Isthere a κ-cc BA A of power λ with no independent subset of power λ?

Shelah [99] has several partial results concerning this problem. A partial positivesolution follows from 6.8 of Shelah [99], which says

If κ is weakly inaccessible and 〈2μ : μ < κ〉 is not eventually constant, then thereis a κ-cc BA A of size 2<κ with no independent subset of size κ+.

This gives, consistently, several examples solving Problem 106 positively. For ex-ample, take a model in which κ = ℵα is weakly inaccessible and 2ℵβ = ℵα+β+1 foreach regular ℵβ < κ. Then 2<κ = ℵα+α. For Problem 104 one can take ρ = ℵ1,ν = ℵ3, and λ = ℵα+3.

The above result also solves Problem 34 of Monk [96] positively.

Another result from Shelah [99] is as follows:

[email protected]

(14.24 Conclusion) If μ = μ<μ < θ = θ<θ, then for some μ-complete, μ+-cc forcingP , in V P : If B is a κ-cc BA of size ≥ λ, μ<κ = μ, and λ is a regular cardinal in(μ, θ], then λ is a free caliber of B.

This gives a negative solution of Problem 33 of Monk [96].

Problem 36 of Monk [96] asks whether �ω+1 ∈ Freecal(Fr(�ω+1)). By 6.11 ofShelah [99] either answer to this question is consistent.

Concerning the possibility of a complete BA with empty free caliber set, 8.1of Shelah goes as follows:

Assume GCH in the ground model, and let P be the partial order for adding ℵω1

Cohen reals. Then in the generic extension we have 2ℵ0 = ℵω1 , 2ℵ1 = ℵω1+1, and:

There is no complete BA A of size 2ℵ1 such that Freecal(A) = ∅; in fact,every complete BA of that size has free caliber 2ℵ1 .

This solves Problem 37 of Monk [96].

As mentioned in the introduction, there are bounded versions of independence. Aset X ⊆ A is m-independent (where m is a positive integer) if for every Y ∈ [X ]m

and every ε ∈ Y 2 we have∏

y∈Y yεy �= 0. Then we set

Indn(A) = sup{|X | : X ⊆ A and X is n-independent}.

This notion is briefly studied in Monk [83], where the following problem is statedwhich is somewhat relevant to the notion; this is Problem 39 in Monk [96].

Problem 105. Can one prove the following in ZFC? For every m ∈ ω with m ≥ 2there is an interval algebra having a subset P of size ω1 such that for all Q ∈ [P ]ω1 ,Q has m pairwise comparable elements and also m independent elements.

The condition in this problem is shown to be consistent in Monk [83].

Roslanowski, Shelah [00] consider the finite version of independence moreextensively, proving the following results (and more):

(1) If n ≥ 2 and λ is an infinite cardinal, then there is a BA A such thatIndnA = λ = |B| and Indn+1A = ω.

(2) If λ is an infinite cardinal and n is an even integer > 2, then there is a BAA such that IndnA = λ and Ind(A×A) = ω.

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11 π-Character

First of all, note that if F is a non-principal ultrafilter on a BA A, then πχ(F ) ≥ ω.To see this, suppose that X is a finite set of non-zero elements of A which is densein F . Choose y ∈ F such that y <

∏(X ∩ F ). Then choose x ∈ X such that

x ≤ y ·∏{z ∈ F : −z ∈ X}. This clearly gives a contradiction, whether x ∈ F or

not.

The following equivalent definition of πχ(A) is frequently useful.

Proposition 11.1. For any infinite BA A, πχ(A) is the smallest cardinal κ suchthat for every ultrafilter F on A there is an X ∈ [A]≤κ such that X is dense in F .

Proof. If F is an ultrafilter on A, then πχ(F ) ≤ πχ(A), and there is a subsetX of A of size πχ(F ) which is dense in F . So πχ(A) is a cardinal κ of the typeindicated. Now suppose that λ < κ. Then there is an ultrafilter F on A such thatλ < πχ(F ); so there is no X ∈ [A]≤λ which is dense in F . This shows that πχ(A)is the least κ with the indicated property. �

Note that if A is atomic, then the set of atoms of A is dense in any ultrafilter.Hence it is possible for A ≤ B with πχ(A) > πχ(B). For example, let A be anuncountable free subalgebra of P(ω).

Turning to our special kinds of subalgebras, first we have:

Proposition 11.2. If A and B are infinite BAs such that A ≤π B, then πχ(A) ≤πχ(B).

Proof. Let F be any ultrafilter on A, and extend it to an ultrafilter on B. ChooseX ∈ B+ which is dense in G, with |X | = πχ(G). For each x ∈ X choose yx ∈ A+

such that yx ≤ x. Then {yx : x ∈ X} is dense in F . It follows that πχ(F ) ≤ πχ(G).Since F is arbitrary, this shows that πχ(A) ≤ πχ(B). �

An example of A,B with A ≤π B and πχ(A) < πχ(B) is given by A = Finco(ω1)and B = P(ω1); see below, where we prove that πχ(Finco(κ)) = ω and thatπχ(P(κ)) ≥ κ for any infinite cardinal κ.

Proposition 11.3. If A ≤rc B then πχ(A) ≤ πχ(B).

, ,OI 10.1007/978-3- - - _ ,


014


Basel 212

373

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374 Chapter 11. π-Character

Proof. For each b ∈ B let C(b) be the smallest element of A which is above b. LetF be any ultrafilter on A, and extend it to an ultrafilter G on B. Let X ⊆ B+ bedense in G with |X | = πχ(G). We claim that {C(x) : x ∈ X} is dense in F . For,take any a ∈ F . Then also a ∈ G, so we can choose x ∈ X such that x ≤ a. ThenC(x) ≤ a, as desired. Thus πχ(F ) ≤ πχ(G). �

Strict inequality is possible; for example, we show below that πχ(Fr(κ)) = κfor every infinite cardinal κ. Hence Proposition 11.8 of the Handbook gives anexample.

By the diagram at the end of Chapter 2, the conclusion πχ(A) ≤ πχ(B) alsoholds if A ≤free B or A ≤proj B.

For the remaining special subalgebra notions we do not know what happens.

Problem 106. Describe the relationship between πχ(A) and πχ(B) for the variousspecial notions of A a subalgebra of B.

Turning to homomorphic images, let B = P(ω). Using the fact that B has anindependent set of size ω1, we obtain a homomorphism f from B onto an algebraA such that A is a subalgebra of the completion of the free algebra C on ω1 freegenerators {xα : α < ω1}, and C is a subalgebra of A. Then, we claim, πχA = ω1.For, suppose that F is an ultrafilter on A, and X is a countable subset of A.Then each element of X is a countable sum of monomials in the xα’s. If we takesome α with xα not in any of these monomials, then xα (or −xα) is an elementof F with no element of X below it. Thus A is a homomorphic image of B andπχ(B) = ω < ω1 = πχ(A).

We turn to products. Clearly πχ(A × B) = max(πχ(A), πχ(B)) for anyinfinite BAs A and B. More generally, we have:

Theorem 11.4. πχ(∏w

i∈I Ai

)= supi∈Iπχ(Ai) for any system 〈Ai : i ∈ I〉 of BAs.

Proof. We may assume that I is infinite. Since Ult(∏w

i∈I Ai

)is the one-point

compactification of the disjoint union of all of the spaces Ult(Ai), it suffices toprove the following:

(1) Let F be the ultrafilter on∏w

i∈I Ai consisting of all x ∈∏w

i∈I Ai such that{i ∈ I : xi �= 1} is finite. Then πχ(F ) = ω.

To prove (1), let J be any denumerable subset of I. For each j ∈ J we define anelement xj of

∏wi∈I Ai by setting, for each i ∈ I,

xji =

{0 if j �= i,

1 if j = i.

We claim that {xj : j ∈ J} is dense in F . To see this, take any y ∈ F . Then thereis a j ∈ J such that yj = 1. So xj ≤ y, as desired. �

Taking each Ai in Proposition 11.4 to be the two-element BA, we get:

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Chapter 11. π-Character 375

Corollary 11.5. πχ(Finco(κ)) = ω for any infinite cardinal κ.

Note that the proof of Theorem 11.4 shows that π-character is attained in∏w

i∈I Ai

iff there is an i ∈ I such that πχ(∏w

i∈I Ai

)= πχ(Ai) and πχ(Ai) is attained. Using

this remark, we can describe the attainment property of π-character: for eachuncountable limit cardinal κ there is a BA A with π-character κ not attained: wetake the weak product of free algebras of the obvious sizes. On the other hand,if πχA = ω, then it is attained, since any non-principal ultrafilter has infiniteπ-character by our initial remark.

Turning to arbitrary products, we have:

Theorem 11.6. If 〈Ai : i ∈ I〉 is a system of non-trivial BAs with∏

i∈I Ai infinite,

then πχ(∏

i∈I Ai

)≥ max(|I|, supi∈Iπχ(Ai)).

Proof. If i ∈ I and G is an ultrafilter on Ai, then the set Fdef= {y ∈

∏i∈I Ai :

yi ∈ G} is an ultrafilter on∏

i∈I Ai, and a subset of∏

i∈I Ai dense in F clearlygives rise to a subset of Ai with no more elements which is dense in G. Henceπχ(Ai) ≤ πχ

(∏i∈I Ai

).

Next, assume that I is infinite; we show that |I| ≤ πχ(∏

i∈I Ai

). For each

subset J of I let xJ be the characteristic function of J , considered as a memberof∏

i∈I Ai. Let F be any ultrafilter on∏

i∈I Ai containing all elements xI\J suchthat |J | < |I|. Then, we claim, πχ(F ) ≥ |I|. In fact, suppose that X ⊆ A+, X isdense in F , and |X | < |I|. For each y ∈ X choose i(y) ∈ I such that yi(y) �= 0.Let J = {i(y) : y ∈ X}. Then the element xI\J of F is not ≥ any element of X ,contradiction. �

Taking each Ai to be the two-element BA, we get:

Corollary 11.7. πχ(P(κ)) = κ for every infinite cardinal κ.

Actually, πχ can jump tremendously in a product. This follows in an obvious wayfrom the following theorem, which is an observation of Douglas Peterson based onTheorem 10.13 and its proof.

Theorem 11.8. If A is an infinite atomless BA, then c(A) ≤ πχ(∏

i∈ω\1 A∗i)

(with notation as in Theorem 10.13).

Proof. Wlog c(A) > ω. Let X and f be as in Theorem 10.13 and its proof, withX uncountable. Then by that proof, {−fx : x ∈ X} generates a proper filterin∏

i∈ω\1 A∗i, and we extend it to an ultrafilter F . Since A is atomless, F is

nonprincipal. We claim that πχ(F ) ≥ |X |; this will prove the Theorem. Suppose

that Y ⊆(∏

i∈ω\1 A∗i)+

, |Y | < |X |, and Y is dense in F ; we want to get a

contradiction. There exist a y ∈ Y and an uncountable Z ⊆ X such that y ≤ −fzfor all z ∈ Z. Say yi �= 0. Wlog yi has the form a0 · a1 · . . . · ai−1, where aj is inthe jth free factor of A∗i. Then a0 · a1 · . . . · ai−1 ≤ g0(z) + · · · + gi−1(z) for all

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z ∈ Z, so there is a j < i such that aj ≤ gj(z). This being true for all z ∈ Z,and Z being infinite, it follows that there exist a j < i and two distinct z, w ∈ Zsuch that aj ≤ gj(z) and aj ≤ gj(w). Since z · w = 0, it follows that aj = 0,contradiction. �

The possibility of doing the above with interval algebras, which naturally arosein Chapter 10, is not so interesting here, since interval algebras can have highπ-character (see the end of this chapter).

We turn to ultraproducts, giving some results of Douglas Peterson. Since πχ is asup-min function, Theorems 6.5–6.7 hold. An additional result of the sort describedin these theorems, with a proof using independent matrices, is the following theo-rem of Peterson: If 〈Ai : i ∈ I〉 is a system of infinite BAs, with I infinite, F is aregular ultrafilter on I, and ess.sup F

i∈I |Ai| ≤ 2|I|, then πχ(∏

i∈I Ai/F)≥ cf(2|I|).

From 6.5–6.7 the following theorem follows, with a proof similar to that of Theo-rem 4.18:

Theorem 11.9. (GCH) Suppose that 〈Ai : i ∈ I〉 is a system of infinite BAs,with I infinite, and F is a regular ultrafilter on I. Then πχ

(∏i∈I Ai/F

)≥∣∣∏

i∈I πχAi/F∣∣.

As usual, the result of Donder shows that it is consistent to always have≥. Petersonhas shown that it is consistent to have < in Theorem 11.9 in the absence of GCH.See also Chapter 4 for an independent solution by Shelah. For > we have thefollowing extension of Theorem 11.8, which shows that πχ can jump very much inan ultraproduct.

Theorem 11.10. If A is an infinite atomless BA, X ⊆ A+ is disjoint, and F isa nonprincipal ultrafilter on ω, then, with notation as in Theorem 10.13, |X | ≤πχ(∏

i∈ω\1 A∗i/F

).

Proof. By Theorem 10.13, if N is a finite subset of X then {n :∏

x∈N −fx(n) = 0}is finite, and hence

∏x∈N −fx/F �= 0. Thus {−fx/F : x ∈ X} has the finite inter-

section property, and we can let G be an ultrafilter on∏

i∈ω\1 A∗i/F containing

this set. We claim that πχ(G) ≥ |X |, which will prove the theorem. To get acontradiction, suppose that Y ⊆ (

∏i∈ω\1 A

∗i/F )+, |Y | < |X |, and Y is dense in

G. Then there is a y/F ∈ Y and an uncountable X ′ ⊆ X such that y/F ≤ −fx/Ffor all x ∈ X ′. We may assume that yi �= 0 for all i ∈ ω, and further that each yihas the form a0i · a1i · . . . · ai−1

i with aji from the jth factor. Now for any x ∈ X ′ wehave y/F ≤ −fx/F , and so there is an i ∈ ω such that yi ≤ −fxi. Hence there isan i ∈ ω and an uncountable X ′′ ⊆ X ′ such that yi ≤ −fxi for all x ∈ X ′′. Nowwe proceed to a contradiction as in the proof of Theorem 11.8. �

Next we describe π-character for free products:

Theorem 11.11. If 〈Ai : i ∈ I〉 is a system of BAs each with at least 4 elements,then πχ(⊕i∈IAi) = max(|I|, supi∈Iπχ(Ai)).

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Proof. For brevity let B = ⊕i∈IAi. First take any i ∈ I; we show that πχ(Ai) ≤πχ(B). Let F be any ultrafilter on Ai, and extend F to an ultrafilter G on B.Suppose X ⊆ B is dense in G. We may assume that each x ∈ X has the form

(1) x =∏

j∈Mx

yxj

for some finite subset Mx of I, where yxj ∈ Aj for every j ∈ Mx. Now defineY = {yxi : x ∈ X, i ∈ Mx}. Then clearly Y is dense in F and |Y | ≤ |X |. Thisproves that πχ(Ai) ≤ πχ(B).

Next, we show that |I| ≤ πχ(B), where we assume that I is infinite. For eachi ∈ I choose ai ∈ Ai such that 0 < ai < 1. Let F be an ultrafilter on B such thatai ∈ F for each i ∈ I; clearly such an ultrafilter exists. Suppose that X ⊆ B isdense in F ; we may assume that each x ∈ X has the form (1) indicated above.Clearly then, by the free product property, we must have |X | ≥ |I|.

Now let F be an ultrafilter on B. Then for each i ∈ I, F ∩Ai is an ultrafilteron Ai, and so there is an Xi ⊆ Ai of cardinality ≤ πχ(Ai) which is dense in F ∩Ai.Let

Y =

⎧⎨⎩y : there is a finite J ⊆ I and a b in

∏j∈J Xj such that y =

∏j∈J

bj

⎫⎬⎭ .

Clearly |Y | ≤ max(|I|, supi∈I(Ai)) and Y is dense in F , as desired. �Corollary 11.12. πχ(Fr(κ)) = κ for every infinite cardinal κ. �

Next we discuss the behaviour of πχ under unions.

Theorem 11.13. Suppose that 〈Aα : α < κ〉 is a strictly increasing sequence ofBAs with union B, where κ is regular. Let λ = supα<κ πχ(Aα). Then πχ(B) ≤∑

α<κ πχ(Aα) ≤ max(κ, λ). Assume in addition that Aα =⋃

β<αAβ for all limit

α < κ. Then πχ(B) ≤ λ+.

Proof. Let F be an ultrafilter on B. Choose Xα ⊆ Aα which is dense in F ∩ Aα,with |Xα| = πχ(F ∩ Aα), for each α < κ. Then

⋃α<κ Xα is dense in F , and

|⋃

α<κXα| ≤∑

α<κ πχ(Aα). So πχ(B) ≤∑

α<κ πχ(Aα) ≤ max(κ, λ).

Now we make the additional assumption indicated, and suppose that πχ(B)>λ+. Let F be an ultrafilter on B such that πχ(F ) > λ+. Thus κ > λ+ by the firstpart of this proof. Let S = {α < κ : cf(α) = λ+}. So, S is stationary in κ. Foreach α < κ let Xα ⊆ Aα be dense in F ∩ Aα with |Xα| ≤ λ. For α ∈ S we thenhave Xα ⊆ Af(α) for some fα < α. Therefore f is constant, say equal to β, onsome stationary subset of S. So Xβ is dense in F , contradicting πχ(F ) > λ+. �

In contrast to Theorem 6.4, we did not assert in 11.13 that κ ≤ 2λ. In fact, for anyinfinite cardinal κ there is a strictly increasing continuous sequence 〈Aα : α < κ〉of BAs such that πχ(Aα) = ω for all α < κ. Namely, take a strictly increasing

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continuous sequence of subalgebras of Finco(κ) with union Finco(κ); recall that ifA ≤ Finco(κ), then A is isomorphic to Finco(λ) for some λ ≤ κ. (In Monk [90],κ ≤ 2λ was mistakenedly asserted.)

The upper bound λ+ mentioned in Theorem 11.13 can be attained – take asequence of free algebras.

Problem 107. What is the relationship between πχ(A) and πχ(B) for A the one-point gluing of B?

Proposition 11.14. πχ(Dup(A)) ≤ πχ(A) for any infinite BA A.

Proof. Suppose that F is a nonprincipal ultrafilter on Dup(A). By Proposition1.19 there is an ultrafilter G on A such that F = {(a,X) ∈ Dup(A) : a ∈ G}. LetM be dense in G with |M | ≤ πχ(A). We claim that {(c,S(c)\{G}) : c ∈ M} isdense in F . For, let (a,X) ∈ F , with a ∈ G. Then S(a)\(X∪{G}) is finite. For eachH ∈ S(a)\(X ∪{G}) choose bH ∈ G\H . Let d = a ·

∏{bH : H ∈ S(a)\(X ∪{G})}.

Then d ∈ G. Choose c ∈ M such that c ≤ d. Then (c,S(c)\{G}) ≤ (a,X), asdesired. �

Problem 108. Characterize πχ(Dup(A)) in terms of A.

Next, we consider moderate products.

Proposition 11.15. Suppose that 〈Ai : i ∈ I〉 is a system of infinite BAs, each Ai

being a field of subsets of Ji. Assume that I is infinite and Ji∩Jj = ∅ for all i �= j.Let B be a field of subsets of I containing all the finite sets. Then

πχ

(B∏i∈I

Ai

)= max{πχ(B), sup

i∈Iπχ(Ai)}.

Proof. We use some of the notation in the treatment of moderate products inChapter 1. For brevity let C =

∏Bi∈I Ai. Let H be a nonprincipal ultrafilter on B.

Set G = 〈{b : b ∈ H}〉fiC .

(1) G is an ultrafilter on C.

For, suppose that h(b, F, a) is any element of C, with h(b, F, a) normal. If b ∈ H ,then clearly h(b, F, a) ∈ G. Suppose that I\b ∈ H . Now K\h(b, F, a) = h(I\(b ∪F ), F, a′) for a certain a′. SinceH is nonprincipal, we have I\F ∈ H . So I\(b∪F ) ∈H too, and consequently K\h(b, F, a) ∈ G. This proves (1).

Let X ⊆ C+ be dense in G, with |X | = πχ(G). For each x ∈ X writex = h(bx, Fx, a

x). Let Y = {bx ∪ Fx : x ∈ X}\{0}. We claim that Y is densein H . For, suppose that c ∈ H . Choose x ∈ X such that x ⊆ c. Clearly thenbx ∪ Fx ⊆ c. From this it follows that πχ(H) ≤ πχ(G). Since H is arbitrary, weget πχ(B) ≤ πχ(C).

Next, suppose that i ∈ I and H is a nonprincipal ultrafilter on Ai. LetG = {c ∈ C : c ∩Ai ∈ H}. Clearly G is an ultrafilter on C. Let X ⊆ C+ be dense

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in G, with |X | = πχ(G). Let Y = {x∩Ai : x ∈ X}\{0}. We claim that Y is densein H . For, let d ∈ H . So also d ∈ G, so there is an x ∈ X such that x ≤ d. Nowd ⊆ Ai, so x = x∩Ai. So, indeed, Y is dense in H . It follows that πχ(H) ≤ πχ(G).Hence πχ(Ai) ≤ πχ(C).

This proves ≥ in the proposition. Suppose that actually > holds. Let G bean ultrafilter on C such that πχ(G) > max{πχ(B), supi∈I πχ(Ai)}.Case 1. ∀i ∈ I[Ji /∈ G]. Let H = {b ∈ B : b ∈ G}. Thus H is a nonprincipalultrafilter on B. Let X ⊆ B+ be dense in H with |X | = πχ(H) < πχ(G). Weclaim that {b : b ∈ X} is dense in G. For, suppose that h(b, F, a) ∈ G. Thenb ∈ G. Otherwise we get h(b, F, a)\b ∈ G, so that

∑i∈F ai ∈ G. This implies that

Ji ∈ G for some i ∈ F , contradiction. Now choose c ∈ X such that c ≤ b. Thenc ≤ b ≤ h(b, F, a). This proves the claim, which gives a contradiction.

Case 2. There is a (unique) i ∈ I such that Ji ∈ G. Let H = {a ∈ Ai : a ∈ G}.Then H is a nonprincipal ultrafilter on Ai. Let X ⊆ A+

i be dense in H with|X | = πχ(H) < πχ(G). We claim that X is also dense in G (contradiction). For,suppose that h(b, F, a) ∈ G. Then also Ji∩h(b, F, a) ∈ G, and so Ji∩h(b, F, a) ∈ H .Choose x ∈ X with x ⊆ Ji ∩ h(b, F, a). So x ⊆ h(b, F, a), as desired. �

For the exponential we have:

Proposition 11.16. πχ(A) ≤ πχ(Exp(A)) for any infinite BA A.

Proof. From Proposition 1.22(vi) and Sikorski’s extension criterion it follows thatthere is a homomorphism f from Exp(A) onto A such that f(V (S(a))) = a forall nonzero a ∈ A. Now suppose that D is an ultrafilter on A. Then f−1[D] is anultrafilter on Exp(A), and we claim that πχ(D) ≤ πχ(f−1[D]). This will provethe proposition.

Let X ⊆ Exp(A)+ be dense in f−1[D], with |X | = πχ(f−1[D]). By Proposi-tion 1.21 we may assume that each x ∈ X has the form

x = V (S(ax0 )) ∩ · · · ∩ V (S(axmx−1)) ∩−V (S(bx0 )) ∩ · · · ∩ −V (S(bxnx−1)),

with clear assumptions. Since each x ∈ X is nonzero, from Proposition 1.22(iii) itfollows that ax0 · . . . · axmx−1 is nonzero. Let

Y = {ax0 · . . . · axmx−1 : x ∈ X}.

We claim that Y is dense in D; this will finish the proof.

Take any c ∈ D. So V (S(c)) ∈ f−1[D]. Choose x ∈ X such that x ≤ V (S(c)).Thus

V (S(ax0 ))∩ · · · ∩V (S(axmx−1))∩−V (S(bx0 ))∩ · · · ∩−V (S(bxnx−1)) ·−V (S(c)) = 0.

Since x �= 0, it follows from Proposition 1.21(vi) that ax0 · . . . · axmx−1 ≤ c, asdesired. �

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An example with < in Proposition 11.16 is given by the example worked out inChapter 1, with κ an uncountable cardinal. We assume the notation there. ByCorollary 11.5 we get πχ(A) = ω. We claim that πχ(Exp(A)) ≥ κ. Let D be theultrafilter on Exp(A) containing the set

{a : −a is an atom} ∪ {xα : α < κ}.

Suppose that X ⊆ Exp(A)+ is dense in D and |X | < κ. Each y ∈ X can be writtenin the form

y = Fy +my · −Gy,

where my is a monomial in {xα : α < κ} and Fy and Gy are finite sums of atoms.Take β < κ such that xβ is not in the support of any monomial my with y ∈ X ,and also

β /∈⋃{Γ : {zΓ} ∈ Fy for some y ∈ X}.

Take y ∈ X such that y ≤ xβ . By the last condition on β we must have Fy = ∅.But then this implies that the monomial my · −xβ is finite, contradiction.

Concerning the derived functions of π-character, the first result is that t(A) =πχH+(A) = πχh+(A), where πχh+(A) = sup{πχ(F, Y ) : F ∈ Y, Y ⊆ Ult(A)},and for where πχh+(A) = sup{πχ(F, Y ) : F ∈ Y, Y ⊆ Ult(A)}, and for any pointx of any space X , πχ(x,X) is defined to be min{|M | : M is a collection of non-empty open subsets of X and for every neighborhood U of x there is a V ∈ Msuch that V ⊆ U}. Such a set M is called a local π-base for x.

It is also convenient for this proof to have an algebraic version of free se-quences. Let A be a BA. A free sequence in A is a sequence 〈xξ : ξ < α〉 of elementsof A such that if F and G are finite subsets of α, and ∀ξ ∈ F∀η ∈ G[ξ < η], then∏

η∈F xη ·∏

η∈G−xη �= 0. Note that by taking G = ∅ it follows that 〈xξ : ξ < α〉has fip. Taking F = ∅ we see that 〈−xξ : ξ < α〉 has fip.

Proposition 11.17. For any infinite BA and any infinite ordinal α, the followingare equivalent:

(i) A has a free sequence of length α.

(ii) Ult(A) has a free sequence of length α.

Proof. (i)⇒(ii): Suppose that 〈xξ : ξ < α〉 is a free sequence in A. For each ξ < αlet Fξ be an ultrafilter containing {xη : η ≤ ξ} ∪ {−xη : ξ < η < α}. This ispossible by the definition of free sequence. To show that 〈Fξ : ξ < α〉 is a free

sequence in Ult(A), suppose to the contrary that ξ < α and G ∈ {Fη : η < ξ} ∩{Fη : ξ ≤ η < α}. Now −xξ ∈ Fη for all η < ξ, so also −xξ ∈ G. Hence there is anη with ξ ≤ η < α and −xξ ∈ Fη. This contradicts the definition of Fη.

Conversely, let 〈Fξ : ξ < α〉 be a free sequence in Ult(A). Then by thedefinition of free sequences in spaces, for each ξ < α there is a yξ ∈ A such that{Fη : η < ξ} ⊆ S(−yξ) and {Fη : ξ ≤ η} ⊆ S(yξ). Let xξ = y1+ξ for every ξ < α.

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Then 〈xξ : ξ < α〉 is a free sequence in A. In fact, suppose that M and N arefinite subsets of α and ∀ξ ∈M∀η ∈ N [ξ < η].

Case 1. M = ∅. Now for any η ∈ N we have 0 < 1 + η, so −xη = −y1+η ∈ F0.Hence

∏η∈N −xη ∈ F0 and consequently

∏η∈N −xη �= 0.

Case 2. M �= ∅. Let ξ be the greatest member of M . For any ρ ∈M we have ρ ≤ ξ,hence 1 + ρ ≤ 1 + ξ, and hence xρ = y1+ρ ∈ F1+ξ. For any η ∈ N we have ξ < η,hence 1 + ξ < 1 + η, and hence −xη = −y1+η ∈ F1+ξ. Thus

∏ρ∈M xρ ·

∏η∈N −xη

is a member of F1+ξ, and hence is nonzero. �

This equivalence shows, in particular, that Ind(A) ≤ t(A). Note that tightness inthese two free sequence senses has the same attainment properties: one is attainediff the other is.

Theorem 11.18. (Shapirovskiı) For any infinite BA A we have t(A) = πχH+(A) =πχh+(A).

Proof. First we show t(A) ≤ πχh+(A). For brevity let κ = πχh+(A). Supposethat F ⊆

⋃Y with Y ∪ {F} ⊆ Ult(A); we want to find Z ⊆ Y with |Z| ≤ κ and

F ⊆⋃Z. By the definition of πχ, let M be a collection of nonempty open subsets

of Y ∪ {F} such that M is a local π-base for F in Y ∪ {F}. We may assume thatF /∈ Y .

(1) ∀a ∈ F [S(a) ∩ Y �= ∅].

In fact, assume that a ∈ F . Now F ⊆⋃Y , so there is a G ∈ Y such that a ∈ G.

Thus G ∈ S(a) ∩ Y , proving (1).

(2) ∀V ∈M [V ∩ Y �= ∅].

For, take any V ∈M . Then V is a nonempty open subset of Y ∪ {F}, so there isan a ∈ A such that ∅ �= S(a) ∩ (Y ∪ {F}) ⊆ V . By (1) it follows that V ∩ Y �= ∅.

Now let Z be a subset of Y of size at most κ such that V ∩ Z �= ∅ for allV ∈ M ; this is possible by (2). We claim that F ⊆

⋃Z, showing that t(A) ≤ κ.

For, take any a ∈ F . Choose V ∈ M such that V ⊆ S(a). Then take G ∈ Z ∩ V .Then G ∈ S(a), so a ∈ G, as desired.

Next we show that πχh+(A) ≤ πχH+(A). Given Y ⊆ Ult(A), let Y be theclosure of Y , and recall from the duality theory that Y corresponds to a homo-morphic image of A. So, we just need to show that πχ(Y ) ≤ πχ(Y ). Let y ∈ Y ,and let M be a local π-base for y in Y with |M | ≤ πχ(y). It suffices now to showthat {U ∩ Y : U ∈ M} is a local π-base for y in Y . Suppose that W is a neigh-borhood of y in Y . Choose a ∈ A so that y ∈ S(a) ∩ Y ⊆ W . Then S(a) ∩ Y isa neighborhood of y in Y , so there is a V ∈ M such that V ⊆ S(a) ∩ Y . ThenV ∩ Y ⊆ S(a) ∩ Y ⊆W , a desired.

Finally, we show that πχH+(A) ≤ t(A). Note that if f is a homomorphismfrom A onto a BA B, 〈bξ : ξ < κ〉 is a free sequence in B, and f(aξ) = bξ for allξ < κ, then 〈aξ : ξ < κ〉 is a free sequence in A. Hence it suffices to show that if

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B is a homomorphic image of A, F is an ultrafilter on B, and κ = πχ(F,B), thenthere is a free sequence of length κ in B. We may assume that κ is infinite, andhence F is not isolated in Ult(B). We construct 〈bξ : ξ < κ〉 by recursion. Supposethat bη has been constructed for all η < ξ so that the following conditions hold:

(3)ξ bη ∈ F for all η < ξ.

(4)ξ If M,N ∈ [ξ]<ω and ∀η ∈M∀ρ ∈ N [η < ρ], then∏

η∈M bη ·∏

ρ∈N −bρ �= 0.

Let A be the collection of all products described in (4)ξ. Then |A | < κ. It followsthat {S(u) : u ∈ A } is not a local π-base for F . Hence there is a neighborhoodU of F such that ∀u ∈ A [S(u) �⊆ U ]. Say F ∈ S(bξ) ⊆ U . Then ∀u ∈ A [u �≤ bξ].Now we verify (3)ξ+1 and (4)ξ+1. Actually (3)ξ+1 is obvious. For (4)ξ+1, supposethat M,N ∈ [ξ + 1]<ω and ∀η ∈ M∀ρ ∈ N [η < ρ]. If ξ /∈ M ∪N , then (4)ξ givesthe desired conclusion. Suppose that ξ ∈ N . Then the desired conclusion is clear.Finally, suppose that ξ ∈ M . It follows that N = ∅, and again the conclusion isclear. �

Note from the proof of Theorem 11.18 that one of πχh+ and πχH+ is attained iffthe other is; and if πχh+ is attained, then so is t, in the free sequence sense.

It is possible to have πχS+(A) > πχ(A); this is true, for example, for A =P(ω), using the fact that P(ω) has a free subalgebra of size 2ω.

Clearly πχS−(A) = πχH−(A) = ω. On the other hand, πχh−(A) = 1 for anyinfinite BA A, since Ult(A) has a denumerable discrete subspace. If B is densein A, then πχ(B) ≤ πχ(A). In fact, if F is an ultrafilter on A, let X ⊆ A bedense in F with |X | = πχ(F ). Wlog X ⊆ B. Hence X is dense in F ∩ B, soπχ(F ∩B) ≤ πχ(F ). This shows that, indeed, πχ(B) ≤ πχ(A). It is possible thatπχ(B) < πχ(A) when B is dense in A. For example, let A be the interval algebraon an uncountable cardinal κ and let B be Finco(κ); see the description of πχ forinterval algebras below. These comments show that dπχS+(A) = πχ(A), but thereis an example with dπχS−(A) < πχ(A) (contradicting a statement in Monk [90]).

The function πχinf and related functions were discussed in Chapter 6.

We now consider the relationships of πχ with the other cardinal functions so farconsidered. Clearly πχ(A) ≤ π(A) for any infinite BA A. The difference betweenπχ and π can be large, for example in a finite-cofinite algebra: see Corollary 11.5.πχ(A) > d(A) for some free algebras A; a free algebra also shows that πχ(A)can be greater than LengthA. It is easy to construct an example where πχ ismuch smaller than Ind. In fact, let A be a free BA on κ free generators. Thenwe construct a sequence 〈Bn : n ∈ ω〉 of algebras by recursion. Let B0 = A.Having constructed Bn, let Bn+1 be an extension of Bn obtained by adding foreach ultrafilter F on Bn an element 0 �= ynF such that ynF ≤ b for all b ∈ F ; itis easy to see that this is possible. Let C =

⋃n∈ω Bn. Then Ind(C) ≥ κ, while

πχ(C) = ω. For, let G be any ultrafilter on C. Then {ynG∩Bn: n ∈ ω} is dense in

G, showing that πχ(G) ≤ ω.

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If A is the interval algebra on an uncountable cardinal, then πχ(A) > Ind(A),while Depth(A) > πχ(A) for A the interval algebra on 1 + ω∗ · (κ + 1); both ofthese results are clear on the basis of the description of πχ for interval algebrasgiven at the end of this chapter.

There are two interesting positive results concerning the relationship of πχwith our earlier cardinal functions. The first of these is true for arbitrary non-discrete regular Hausdorff spaces, with no complications in the proof from the BAcase. A π-basis for a point x in a space X is a collection B of nonempty opensets such that for every neighborhood U of x there is a V ∈ B such that V ⊆ U .πχ(x,X) is the least size of a π-basis for x, and πχ(X) = supx∈X πχ(x,X).

Theorem 11.19. d(X)≤πχ(X)c(X) for any non-discrete regular Hausdorff spaceX.

Proof. For each x ∈ X let Ox be a family of non-empty open subsets of X suchthat |Ox| ≤ πχ(X) and for every neighborhood U of x there is a V ∈ O suchthat V ⊆ U . Now we define subsets Yα ⊆ X and collections Pα of open sets forα < (c(X))+ by induction so that the following conditions hold:

(1) |Yα| ≤ (πχ(X))c(X);

(2) |Pα| ≤ (πχ(X))c(X).

Fix x0 ∈ X . Set Y0 = {x0} and P0 = Ox0 . Suppose that Yβ and Pβ have beendefined for all β < α. If α is a limit ordinal, set Yα =

⋃β<α Yβ and Pα =

⋃β<α Pβ .

Now suppose that α is a successor ordinal β + 1. Set

Qα = {R : R ⊆ Pβ , |R| ≤ c(X),⋃R �= X}

Clearly |Qα| ≤ πχ(X)c(X). For every R ∈ Qα choose ϕR ∈ X\⋃R and put

Yα = Yβ ∪ {ϕR : R ∈ Qα}, Pα =⋃

x∈Yα

Ox.

This finishes the definition. Now we claim

(3) Ldef=⋃

α<(c(X))+ Yα is dense in X .

Since |L| ≤ (πχ(X))c(X), (3) finishes the proof. To prove (3), suppose that it isnot true. Then by regularity, there is an open U such that L ⊆ U ⊆ U �= X . SetP∗ =

⋃x∈LOx, and T = {V ∈ P∗ : V ⊆ U}. Let R be a maximal disjoint subset

of T . Then L ⊆⋃R; for, if x ∈ L\

⋃R, then x ∈ U\

⋃R, which is open, so there

is a V ∈ Ox such that V ⊆ U\⋃R, and R ∪ {V } contradicts the maximality of

R. Also,⋃R ⊆

⋃T ⊆ U �= X . Since R ⊆ Pβ for some β < (c(X))+, it follows

that R ∈ Qβ for some β < (c(X))+, and hence we get ϕR ∈ X\⋃R ⊆ X\L,

contradiction. �Theorem 11.20. d(A) · πχ(A) = π(A) for any infinite BA A.

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Proof. We already know that d(A) ≤ π(A) and πχ(A) ≤ π(A). Now let D be adense subset of Ult(A) with |D| = d(A), and for each F ∈ D let XF be a localπ-base for F of size ≤ πχ(A). Clearly

⋃F∈D XF is dense in A, as desired. �

There are many problems concerning the functions πχSr and πχHr, so we restrictourselves to the following vague questions.

Problem 109. Give a purely cardinal number characterization of πχSr.

Problem 110. Give a purely cardinal number characterization of πχHr.

Concerning πχ for special classes of algebras, we first give a description of whathappens for interval algebras. Let L be a linearly ordered set with first element0, and let A be the interval algebra on L. The ultrafilters on A are in one-onecorrespondence with the end segments of L different from L itself; correspondingto the ultrafilter F is the segment {a ∈ L : [0, a) ∈ F}. Given an end segment Tof L, let κ be the type of a shortest cofinal sequence in L\T and λ the type of ashortest coinitial sequence in T . If both κ and λ are infinite, then πχ(F ) is theminimum of κ and λ. If one is infinite and the other is 1, then πχ(F ) is the infiniteone. If both are 1, then πχ(F ) is 1. From this description it is easy to constructa linear order L such that if A is the interval algebra on L then πχ(A) < χ(A),with the difference arbitrarily large: for example, let κ be any infinite cardinal,and let L be 0+ω∗ ·κ+ω∗. The above description implies that πχ(A) = ω, whileif F is the ultrafilter corresponding to the end segment ω∗, then χ(F ) = κ. In thisexample we also have πχ(A) < Depth(A). The description of πχ also shows thatπχ(A) ≤ Depth(A) for an interval algebra A.

If A is complete, then c(A) ≤ πχ(A): in fact, suppose that πχ(A) < c(A). LetX be disjoint in A with

∑X = 1 and |X | = (πχ(A))+. Let F be an ultrafilter on

A such that∑

(X\Y ) ∈ F for each Y ⊂ X such that |Y | < |X |. Let Y be a π-basefor F with |Y | < |X |. For each y ∈ Y choose xy ∈ X such that y · xy �= 0. Then{xy : y ∈ Y } is a π-base for F ∩ 〈X〉cm (where 〈X〉cm is the complete subalgebraof A generated by X). But −

∑y∈Y xy ∈ F ∩ 〈X〉cm, contradiction.

Also recall the result of K. Bozeman [91] given in Chapter 6 that under GCHwe have π(A) = πχ(A) for A complete.

In Chapter 6 we gave an example of a complete algebra A with the propertythat d(A) < π(A); hence by Theorem 11.20 we have d(A) < πχ(A) also.

πχ is characterized for tree algebras by the following theorem.

Theorem 11.21. Let T be an infinite tree. Then πχ(Treealg(T )) = sup{cf(C) : Cis an initial chain of T with finitely many immediate successors}.

Proof. We describe πχ(F ) for each ultrafilter F on Treealg(T ). Recall that theultrafilters on Treealg(T ) are in one-one correspondence with the initial chains ofT , where if T has finitely many roots we exclude the empty chain (a correction

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of the description in the Handbook). Given an initial chain C, we let FC be thefilter generated by

{T ↑ t : t ∈ C} ∪ {T \(T ↑ t) : t ∈ T \C}.

This is the ultrafilter associated with C. We now consider several cases.

Case 1. C has a maximal element t, and t has finitely many immediate successors.Then {t} ∈ FC , which is thereby principal, so that πχ(F ) = 1.

Case 2. C has infinitely many immediate successors. Let M be a countable set ofsuch immediate successors, and let X = {T ↑ t : t ∈ M}. Then X is dense in FC .So πχ(FC) ≤ ω in this case.

Case 3. C has no maximal element, but has finitely many immediate successors.Let M be the set of all immediate successors of C, and let N be a cofinal subsetof C of size cf(C). Then {(T ↑ t)\

⋃s∈M (T ↑ s) : t ∈ N} is dense in FC . Suppose

that X is dense in FC but |X | < cf(C). Wlog each element x ∈ X has the form(T ↑ tx)\

⋃s∈Px

(T ↑ s). Choose u ∈ N such that tx < u for all x ∈ X . Then(T ↑ u)\

⋃s∈M (T ↑ s) ∈ FC , and no element of X is below it, contradiction. Thus

πχ(FC) = cf(C) in this case. �

For tree algebras we have πχ(A) ≤ Depth(A), since Depth(A) = t(A) for them.The difference can be arbitrarily large; this is an observation of Douglas Peterson.Namely, given κ, consider the tree

Tdef= {f : f : α+ 1→ ω for some α ≤ κ} ∪ {0}

under ⊆. Since every initial chain of T has countably many immediate successors,it follows that πχ(Treealg(T )) = ω by Theorem 11.21; but Depth(Treealg(T )) = κ.

In Dow, Monk [94] the relationship between depth and π-character for super-atomic BAs is described. There is a superatomic BA A such that Depth(A) = ωand πχ(A) = ω1. If πχ(A) ≥ ω2, then πχ(A) = Depth(A). See also Shelah,Spinas [99].

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12 Tightness

Again we note first of all that if F is a non-principal ultrafilter in a BA A, thent(F ) ≥ ω. To see this, note that for each x ∈ F there is a y /∈ F such that0 < y < x; hence there is an ultrafilter Gx such that x ∈ Gx but Gx �= F . LetY = {Gx : x ∈ F}. Thus F ⊆

⋃Y . Suppose that Z is a finite subset of Y such

that F ⊆⋃Z. But it is a very elementary exercise to show that no ultrafilter is

included in a finite union of other, different, ultrafilters. So, t(F ) ≥ ω, and hencet(A) ≥ ω for every infinite BA A.

From the algebraic form of the free sequence equivalent of tightness (seeChapter 11), it is clear that Ind(A) ≤ t(A).

Also, using the free sequence equivalent to the definition of tightness it isclear that if A is a subalgebra of B, then t(A) ≤ t(B). Clearly if A ≤free B,then tightness can increase arbitrarily from A to B; so the same applies for ≤u,≤proj, ≤rc, ≤σ, and ≤reg. If A ≤π B clearly tightness can increase in going fromA to B. The increase can be arbitrary, as one sees by taking A = Finco(κ) andB = P(κ) for κ a cardinal. We have t(A) = ω by the argument for weak productswhich follows next, while t(B) = 2κ since 2κ = Ind(B) ≤ t(B). If A ≤s B, thent(A) = t(B). This is true since clearly t(C) ≤ t(D) whenever C is a homomorphicimage of D; then use Proposition 2.29. t can increase arbitrarily from A to B whenA ≤mg B, as one sees by considering interval algebras.

From the definition of tightness it is clear that t(A×B) = max{t(A), t(B)}.Furthermore, t

(∏wi∈I Ai

)= max(ω, supi∈It(Ai)) for any system 〈Ai : i ∈ I〉 of

non-trivial BAs with I infinite. By the topological description of weak products,

to prove this it suffices to show that t(F ) = ω for the “new” ultrafilter Fdef= {x ∈∏w

i∈I Ai : there is a finite subset G of I such that xi = 1 for all i ∈ I\G}. To see

this, first note that if G ∈ Ult(∏w

i∈I Ai

)and G �= F , then there is an iG ∈ I and

an ultrafilter KG on Ai such that G ={x ∈

∏wi∈I Ai : xiG ∈ KG

}. Next, for H a

finite subset of I let

xH(i) =

{1 if i ∈ I\H0 if i ∈ H .

Now suppose that F ⊆⋃Y with Y ⊆ Ult

(∏wi∈I Ai

). The case F ∈ Y is easy, so

suppose that F /∈ Y . Now Hdef= {iG : G ∈ Y } is infinite; otherwise xH ∈ F gives

a contradiction. (If xH ∈ G ∈ Y , then xH(iG) = 0, contradiction.) Let Z be a

, ,OI 10.1007/978-3- - - _ ,


014


Basel 213

387

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388 Chapter 12. Tightness

countable subset of Y such that {iG : G ∈ Z} is infinite. Suppose that x ∈ F . Sayxi = 1 for all i ∈ I\L, L finite. Choose G ∈ Z such that iG /∈ L. Then x ∈ G, asdesired.

Note that this argument again shows that tightness is attained in∏w

i∈I Ai

iff there is an i ∈ I such that t(∏w

i∈I Ai

)= t(Ai) and tightness is attained in Ai

(for infinite Ai’s). From this, the attainment property of tightness follows: for eachlimit cardinal κ > ω there is a BA A with tightness κ not attained: take the weakproduct of 〈Aλ : ω < λ < κ, λ a cardinal〉, where Aλ is the free BA of size λ.

By Theorem 10.6, t can increase arbitrarily from algebras to their (full)product.

For the free sequence equivalents of tightness see Chapters 4 and 11. Con-cerning attainment in the free sequence sense, we first show

Theorem 12.1. If κ is an infinite cardinal with cf(κ) > ω and 〈Ai : i ∈ I〉 is asystem of BAs none of which has a free sequence of type κ, then also

∏wi∈I Ai does

not have a free sequence of type κ.

Proof. Suppose that 〈Fα : α < κ〉 is a free sequence in Ult(∏w

i∈I Ai

). We think

of Ult(∏w

i∈I Ai

)as the one-point compactification of the disjoint union of all of

the spaces Ult(Ai). We may assume that the “new” ultrafilter G is not among theFα’s. For each α < κ let i(Fα) be the unique i ∈ I such that {a · χ{i} : a ∈ F} isan ultrafilter on Ai, where

χ{i}(j) ={1 if i = j,

0 otherwise.

Set J = {i(Fα) : α < κ}. Now for each j ∈ J the set {α < κ : i(Fα) = j} has sizeless than κ, since Aj does not have a free sequence of length κ. Hence |J | ≥ cf(κ),since κ =

⋃j∈J{η < κ : i(Fη) = j}. It follows that there is a ξ < κ such that

|{i(Fη : η < ξ}| = ω, and hence |{i(Fη) : ξ ≤ η}| = cf(κ). Therefore

G ∈ {Fη : η < ξ} ∩ {Fη : ξ ≤ η < κ},

which contradicts the free sequence property. �

It follows from Theorem 12.1 that for every limit cardinal κ with cf(κ) > ω thereis a BA with tightness κ not attained in the free sequence sense.

Now we turn to the case of cofinality ω:

Theorem 12.2. Let t(A) = κ, where κ is a singular cardinal of cofinality ω. ThenA has a free sequence of length κ.

Proof. This will be a modification of the proof of Theorem 4.8; see also Theorem7.3. An element a ∈ A is called a μ-element if for some ideal I of A � a, thealgebra (A � a)/I has a strictly increasing sequence of type μ. Let 〈λi : i < ω〉 bea strictly increasing sequence of infinite regular cardinals with supremum κ. Wecall an element a ∈ A an ∞-element if it is a λi-element for all i < ω.

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Chapter 12. Tightness 389

(1) If a is an ∞-element and a = b+ c with b · c = 0, then b is an ∞-element or cis an ∞-element.

For, it is enough to show that for every i < ω, either b is a λi-element or c is aλi-element. Suppose that for some i < ω, neither b nor c is a λi-element. Let I bean ideal in A � a and 〈[xα] : α < λi〉 a strictly increasing sequence of elements in(A � a)/I. Now if α < β < λi, then

xα · b · −(xβ · b) = xα · −xβ · b ∈ I ∩ (A � b),

and hence in A � b we have [xα · b] ≤ [xβ · b]. Hence there is an α < λi such that ifα < β < γ < λi then xγ · −xβ · b ∈ I. Similarly for c: there is an α′ < λi such thatif α′ < β < γ < λi, then xγ · −xβ · c ∈ I. But then if max(α, α′) < β < γ < λi weget xγ · −xβ ∈ I, contradiction. This proves (1).

Now we construct disjoint elements a0, a1, . . . such that ai is a λi-elementfor all i < ω. Note from Theorem 4.26 that 1 is an ∞ element. Suppose thatai has been constructed for all i < n so that

∏i<n−ai is an ∞-element; so this

holds for n = 0. Now there exists an ideal I in A �∏

i<n−ai with a sequence〈[xα] : α < λn+1〉 strictly increasing in (A �

∏i<n−ai)/I. Then clearly

(2) xλn is a λn-element.

Now by (1) and (2) there is a λn-element an such that∏

i≤n−ai is an∞-element.

Now for each i < ω choose an ideal Ii in A � ai such that (A � ai)/Ii has achain of type λi. Let J = 〈

⋃i<ω Ii〉Id. Then J ∩ (A � ai) = Ii for each i < ω, and

hence A/J has a chain of type λi for all i < ω. Hence as in the proof of 4.8, A/Jhas a chain of type κ, as desired. �

We also recall from Theorem 11.18 that t(A) = πχH+(A) = πχh+(A). And, asmentioned after the proof of Theorem 11.18, πχH+ and πχh+ have the same at-tainment properties, while πχH+ attained implies that t is attained in the freesequence sense. Another of the attainment problems is answered by the followingtheorem.

Theorem 12.3. Suppose that κ is a singular cardinal. Then tightness is not attainedin Intalg(κ).

Proof. Let 〈λα : α < cf(κ)〉 be a strictly increasing continuous sequence of cardi-nals with supremum κ. By the Handbook, each ultrafilter on Intalg(κ) is deter-mined by an end segment of κ not containing 0 (this last restriction is not foundin the Handbook, but it is clearly necessary). If C is such an end segment of κ,then its associated ultrafilter FC is generated by

{[0, c) : c ∈ C} ∪ {[c,∞) : c ∈ κ\C}.

So, take any end segment C; we want to show that t(FC) < κ.

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Case 1. C = ∅. In this case we claim that t(FC) ≤ cf(κ). In fact, suppose that FC ⊆⋃Y , where Y ⊆ Ult(Intalg(κ)). For each α < cfκ we have [λα,∞) ∈ FC , so we can

choose Gα ∈ Y such that [λα,∞) ∈ Gα. We claim that FC ⊆⋃{Gα : α < cfκ}

(as desired). In fact, let x ∈ FC . Without loss of generality x has the form [c,∞)for some c ∈ κ. Choose α < cfκ such that c < λα. Then [c,∞) ⊇ [λα,∞) ∈ Gα,as desired.

Case 2. C �= ∅. Let c be the least element of C. Then we claim that t(FC) ≤max(ω, |c|). For, again suppose that FC ⊆

⋃Y , where Y ⊆ Ult(Intalg(κ)). For

each d < c we have [d, c) ∈ FC , and so we can choose Gd ∈ Y such that [d, c) ∈ Gd.Now we claim that FC ∈

⋃{Gd : d < c}, as desired. For, let x ∈ FC . Wlog x has

the form [d, e) with d ∈ κ\C and e ∈ C. Then x ∈ Gd, as desired. �

Corollary 12.4. For every singular cardinal κ there is a BA A such that t(A) = κnot attained but A has a free sequence of type κ. �

This corollary answers Problem 29 of Monk [90].

The description of πχ for interval algebras given at the end of Chapter 11shows that if κ is singular, then πχ(Intalg(κ)) = κ not attained. Thus attainmentin the free sequence sense does not imply attainment in the πχH+ sense, answeringProblem 30 in Monk [90] negatively. But if t(A) is regular limit and it is attained inthe free sequence sense then it is attained in the πχh+ sense. The argument here is alittle lengthy, but will be useful in discussing character too. Let t(A) = κ, κ regularlimit, and suppose that 〈Fα : α < κ〉. is a free sequence in Ult(A). For each ξ < κchoose aξ ∈ A such that {Fα : α < ξ} ⊆ S(aξ) and S(aξ) ∩ {Fα : ξ ≤ α < κ} = ∅.Then

{−aξ : ξ < κ} ∪ {x ∈ A : {Fα : α < κ} ⊆ S(x)}

has the finite intersection property. In fact, otherwise we would get −aξ1 · . . . ·−aξn · x = 0, where {Fα : α < κ} ⊆ S(x). Choose α < κ with ξi < α for alli = 1, . . . , n. Then x ∈ Fα, so aξi ∈ Fα for some i, contradiction. So, let G bean ultrafilter containing the given set. Let Y = {Fα : α < κ} ∪ {G}. We claimthat πχ(G, Y ) = κ. For, suppose that M ∈ [A]<κ and {S(x) ∩ Y : x ∈ M} isa π-base for G, where S(x) ∩ Y �= ∅ for all x ∈ M . Then by the regularity ofκ, there is an x ∈ M and a Γ ∈ [κ]κ such that S(x) ∩ Y ⊆ S(−aξ) ∩ Y for allξ ∈ Γ. Then {Fα : α < κ} ⊆ S(−x). In fact, let α < κ. Choose ξ ∈ Γ suchthat α < ξ. Then Fα ∈ S(aξ), so Fα /∈ S(x), hence Fα ∈ S(−x), proving that{Fα : α < κ} ⊆ S(−x). It follows that −x ∈ G too. So S(x)∩Y = 0, contradiction.

We now give an example, due to J.C. Martınez [02], of a BA A in which fora certain singular cardinal κ, t(A,F ) = κ for some ultrafilter F , but A does nothave a free sequence of order type κ. This answers Problems 41 and 42 of Monk[96]. In fact, let κ be any singular cardinal with cf(κ) > ω, and let 〈λα : α < cf(κ)〉be a strictly increasing sequence of infinite successor cardinals with supremum κ.For each α < cf(κ) let Gα be the ultrafilter on Intalg(λα) generated by the set

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{[ξ, λα) : ξ < λα}. Define

A =

⎧⎨⎩x ∈

w∏α<cf(κ)

Intalg(λα) : ∀α, β < cf(κ)[xα ∈ Gα iff xβ ∈ Gβ ]

⎫⎬⎭ .

Clearly A is a subalgebra of∏w

α<cf(κ) Intalg(λα). Then let

F = {x ∈ A : ∀α < cf(κ)[xα ∈ Gα]}.Clearly F is an ultrafilter onA. Clearly |A| = κ. We claim that t(F ) = κ. Obviouslyt(F ) ≤ κ. Suppose that t(F ) < κ. Choose α < cf(κ) such that t(F ) < λα. For eachξ < λα let Hξ be an ultrafilter on A containing the set {x ∈ A : ξ ∈ xα}. ThenF ⊆

⋃ξ<λα

Hξ. In fact, if x ∈ F , then xα ∈ Gα, and so there is an η < λα such

that [η, λα) ⊆ xα; hence η ∈ xα and so x ∈ Hη. Choose M ∈ [λα]≤t(F ) such that

F ⊆⋃

ξ∈M Hξ. Since λα is regular, choose η ∈ λα such that ξ < η for all ξ ∈ M .Let x ∈ A be such that xα = [η, λα). Then x ∈ F\

⋃ξ∈M Hξ, contradiction. This

proves our claim that t(F ) = κ.

Now suppose that 〈Hξ : ξ < κ〉 is a free sequence of ultrafilters on A; we wantto get a contradiction. We may assume that Hξ �= F for all ξ < κ. Now for eachultrafilter K �= F on A there is a unique α(K) < cf(κ) such that {xα(K) : x ∈ K}is an ultrafilter on Intalg(λα(K)).

(∗) If M ⊆ Ult(A)\{F} and {α(K) : K ∈M} is infinite, then F ∈M .

In fact, let F ∈ S(a) with a ∈ A. Then {β < cf(κ) : aβ �= 1} is finite, and so thereis a K ∈M such that aα(K) = 1. Thus K ∈ S(a), as desired in (∗).

Now we define a strictly increasing sequence 〈β(n) : n ∈ ω〉 of ordinals lessthan cf(κ) by recursion. Let β(0) = 0. Now suppose that β(n) has been defined.Then

|{ξ < κ : α(Hξ) ≤ α(Hβ(n))}| ≤∑

η≤α(Hβ(n))

λη ≤ λη · λη < κ;

hence there is a β(n + 1) < κ with β(n) < β(n + 1) and α(Hβ(n+1)) > α(Hβ(n)).

Let M = {β(n) : n ∈ ω}. Then F ∈M by (∗). Let γ = supn∈ω β(n). By the aboveargument we get a strictly increasing sequence 〈δn : n ∈ ω〉 of ordinals less thancf(κ) such that γ = δ0 and α(Hδn) < α(Hδ(m)) for n < m. Let N = {δn : n ∈ ω}.By (∗) again, F ∈ N . Hence

F ∈ {Hξ : ξ < γ} ∩ {Hξ : γ ≤ ξ < κ},contradicting the fact that 〈Hξ : ξ < κ〉 is a free sequence.

The following problem about attainment is still open; this is Problem 43 inMonk [96].

Problem 111. Does attainment of tightness in the πχH+ sense imply attainmentin the sense of the definition?

We return to the discussion of products.

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Theorem 12.5. If 〈Ai : i ∈ I〉 is a system of non-trivial BAs, with I infinite, thent(∏

i∈I Ai) ≥ max(2|I|, supi∈It(Ai)).

Proof. If j ∈ I, then Aj is isomorphic to a subalgebra of∏

i∈I Ai; so t(Ai) ≤t(∏

i∈I Ai). Since independence is less than or equal to tightness, it also follows

that 2|I| ≤ t(∏

i∈I Ai). �

Now we consider ultraproducts, giving some results of Douglas Peterson. Recallfrom the introduction that tightness is an order-independence function. For suchfunctions we have the following theorem, which uses the notion of depth of a linearordering, which is the supremum of cardinalities of well-ordered subsets of theordering.

Theorem 12.6. Suppose that k is an order-independence function, 〈Ai : i ∈ I〉 is asequence of infinite BAs, with I infinite, F is an ultrafilter on I, and 〈κi : i ∈ I〉 isa sequence of cardinals such that κi < k′(Ai) for all i ∈ I. Then k

(∏i∈I Ai/F

)≥

Depth(∏

i∈I κi/F).

Proof. For each i ∈ I let 〈aiα : α < κi〉 be a sequence of elements of Ai such thatfor all finite G,H ⊆ κi, if 〈κi, <,G,H〉 |= ϕ then

∏α∈F aiα ·

∏α∈H −aiα �= 0. Let

λ = Depth(∏

i∈I κi/F). We consider two cases.

Case 1. λ is a successor cardinal. Let 〈fα/F : α < λ〉 be a sequence of elementsof∏

i∈I κi/F such that fα/F < fβ/F if α < β. Define gα(i) = aifα(i) for allα < λ and i ∈ I. Now suppose that G and H are finite subsets of λ such that(λ,<,G,H) |= ϕ. Let

K = {i ∈ I : ∀α, β ∈ G ∪H(α < β ⇒ fα(i) < fβ(i))}.

ThenK ∈ F . By (2) in the definition of order-independence function we have (κi, <, {fα(i) : α ∈ G}, {fα(i) : α ∈ H}) |= ϕ for each i ∈ K, and hence

∏α∈G aifα(i) ·∏

α∈H −aifα(i) �= 0. Therefore∏

α∈G fα/F ·∏

α∈H −gα/F �= 0, as desired.

Case 2. λ is a limit cardinal. Then k(∏

i∈I Ai/F)≥ κ for each successor κ < λ,

by the above argument; hence k(∏

i∈I Ai/F)≥ λ. �

Theorem 12.7. (GCH) Suppose that 〈Ai : i ∈ I〉 is a system of infinite BAs, with Iinfinite, and F is a regular ultrafilter on I. Then t

(∏i∈I Ai/F

)≥∣∣∏

i∈I t(Ai)/F∣∣.

Proof. Let κ = ess.sup Fi∈It(Ai).

Case 1. κ ≤ |I|. The ultraproduct has an independent subset of size 2|I|, and thedesired result follows.

Case 2. cf(κ) > |I|. Then if κ is a successor cardinal, we may assume thatt′(Ai) = κ+ for all i ∈ I, and hence by Theorem 12.6 we have t

(∏i∈I Ai/F

)≥

Depth(∏

i∈I κ/F)= κ. The limit case clearly follows from this case.

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Case 3. cf(κ) ≤ |I| < κ. Using Lemma 3.17 or Lemma 3.18, we obtain a sys-tem 〈λi : i ∈ I〉 of infinite cardinals such that λi < t(Ai) for each i ∈ I, andess.sup F

i∈Iλi = κ. Hence by Theorem 12.6 again, and by the proof of Theorem4.18, t

(∏i∈I Ai/F

)≥ Depth

(∏i∈I λi/F

)≥ κ+. �

As usual, Donder’s theorem then says that ≥ holds for any uniform ultrafilter,assuming V = L. The inequality can be strict, from the discussion of independence.A consistent example exists for the other direction by Roslanowski, Shelah [98].

The tightness of free products is described by a theorem of Malyhim [72]; we givethe result here. The proof we give is due to Todorcevic. (private communication);he uses the idea of this proof to strengthen Malyhin’s result.

Theorem 12.8. t(A⊕B) = max(t(A), t(B)).

Proof. The inequality ≥ is clear. For the other inequality it suffices to show thatif 〈cα : α < θ〉 is a free sequence in A ⊕ B with θ regular and uncountable, theneither A or B has a free sequence of that length too. We use free sequence here inthe algebraic sense described in Chapter 11. First we claim:

(1) We may assume that each cα has the form aα · bα with aα ∈ A and bα ∈ B.

To see this, first write cα =∑

i<mα(aαi · bαi) with each aαi ∈ A and each bαi ∈ B.

Since θ is regular and uncountable, we may assume that mα = m does not dependon α. Now for each α < θ let Fα be an ultrafilter on A ⊕ B such that {cξ : ξ ≤α} ∪ {−cξ : α < ξ < θ} ⊆ Fα. Then by the first part of the proof of Theorem 4.23we get an ultrafilter G on A⊕B such that

(2) |{α < θ : a ∈ Fα}| = θ for all a ∈ G.

Then cα ∈ G for all α < θ; for if −cα ∈ G we would get −cα ∈ Fβ for some β ≥ αby (2), and this is impossible. It follows that for all α < θ there is an i < m suchthat aαi ·bαi ∈ G. Hence there exist an i < m and a Γ ∈ [θ]θ such that aαi ·bαi ∈ Gfor all α ∈ Γ. Now let

K = {δ ∈ Γ : ∀H ∈ [Γ ∩ δ]<ω∃α ∈ (maxH, δ)∀ξ ∈ H(aξi · bξi ∈ Fα)}.

We claim that K is unbounded in θ. For, let δ0 < θ. For every finite H ⊆ Γ ∩ δ0we have

∏ξ∈H(aξi · bξi) ∈ G, and hence by (2) there is an αH > maxH such that∏

ξ∈H(aξi · bξi) ∈ FαH . Choose δ1 ∈ Γ greater than δ0 and all ordinals αH for

H ∈ [Γ∩δ0]<ω. Then repeat the construction for δ1, obtaining δ2 ∈ Γ, etc. Finally,let δω be the least member of Γ greater than all δi, i < ω. Clearly δω ∈ K, provingthe claim about K.

Let 〈δξ : ξ < θ〉 enumerate K in increasing order. We claim, then, that〈aδξi · bδξi : ξ < θ〉 is a free sequence in A ⊕ B; this will prove the claim (1). Toprove this, let M and N be finite subsets of θ such that each member of M isless than each member of N . We may assume that N is nonempty. Let ξ be theleast member of N . We then apply the definition of K to its member δξ to get an

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α ∈ (max{δη : η ∈ M}, δξ) such that aδηi · bδηi ∈ Fα for all η ∈ M . Note that wealso have −cδη ∈ Fα for all η ∈ N . Now∏

η∈M

(aδηi · bδηi) ·∏η∈N

−cδη ≤∏η∈M

(aδηi · bδηi) ·∏η∈N

−(aδηi · bδηi),

and the left side is in Fα and hence is nonzero, so the right side is nonzero too,and this proves that 〈aδξi · bδξi : ξ < θ〉 is a free sequence in A⊕B.

So now we assume (1). We consider two cases.

Case 1. ∀α < θ∃β ≥ α∀K ∈ [α]<ω∀L ∈ [θ\β]<ω(aKLdef=∏

ξ∈K aξ ·∏

ξ∈L−aξ �= 0). Define 〈αξ : ξ < θ〉 as follows. If αη has been defined for all η < ξ, letβξ = supη<ξ αη and choose αξ > βξ such that ∀K ∈ [βξ]

<ω∀L ∈ [θ\αξ]<ω(aKL �=

0). Then 〈aαξ: X < θ〉 is a free sequence in A. For, assume that M andN are finite

subsets of θ, each member ofM less than each member ofN . Let ξ = supη∈M (η+1)(ξ = 0 if M = 0). Then {αη : η ∈ M} ∈ [βξ]

<ω and {αη : η ∈ N} ∈ [θ\αξ]<ω, so∏

η∈M aαη ·∏

η∈N −aαη �= 0, as desired.

Case 2. Case 1 fails: ∃α0 < θ∀β ≥ α0∃Kβ ∈ [α0]<ω∃L ∈ [θ\β]<ω(aKβL = 0).

So ∃K ∈ [α0]<ω∃Γ ∈ [θ\α0]

θ∀β ∈ Γ∃L ∈ [θ\β]<ω(aKL = 0). Hence we get〈Lα : α < θ〉 such that ∀α∀β(α < β < θ ⇒ ∀ξ ∈ Lα∀η ∈ Lβ(ξ < η)) and∀α < θ[∀ξ ∈ K∀η ∈ Lα(ξ < η)) and aKLα = 0]. Let cα =

∏ξ∈Lα

bξ for allα < θ. Then 〈cα : α < θ〉 is a free sequence in B. For, suppose that M and N arefinite subsets of θ, each member of M less than each member of N . Then, withP =

⋃α∈N Lα,

0 �=∏ξ∈K

(aξ · bξ) ·∏

α∈M,ξ∈Lα

(aξ · bξ) ·∏

α∈N,ξ∈Lα

−(aξ · bξ)

=∏ξ∈K

(aξ · bξ) ·∏

α∈M,ξ∈Lα

(aξ · bξ) ·∑Γ⊆P

⎛⎝∏

ξ∈Γ

−aξ ·∏

ξ∈P\Γ−bξ

⎞⎠ ,

so choose Γ ⊆ P so that

0 �=∏ξ∈K

(aξ · bξ) ·∏

α∈M,ξ∈Lα

(aξ · bξ) ·∏ξ∈Γ

−aξ ·∏

ξ∈P\Γ−bξ.

Now if Lα ⊆ Γ for some α ∈ N , then∏ξ∈K

(aξ · bξ) ·∏

α∈M,ξ∈Lα

(aξ · bξ) ·∏ξ∈Γ

−aξ ·∏

ξ∈P\Γ−bξ ≤ aKLα = 0,

contradiction. So for all α ∈ N there is a ξ ∈ Lα\Γ. Thus

0 �=∏ξ∈K

bξ ·∏

α∈M,ξ∈Lα

bξ ·∏

ξ∈P\Γ−bξ ≤

∏α∈M

cα ·∏α∈N

∑ξ∈Lα

−bξ =∏α∈M

cα ·∏α∈N

−cα,

as desired. �

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Theorem 12.9. If 〈Ai : i ∈ I〉 is a system of BAs each with at least four elements,then t(⊕i∈IAi) = max(|I|, supi∈I t(Ai)).

Proof. Obviously t(Aj) ≤ t(⊕i∈IAi) for each j ∈ I; and |I| ≤ ⊕i∈IAi sinceInd ≤ t. Thus ≥ holds. To prove ≤, let κ = max(|I|, supi∈I t(Ai)), and supposethat 〈cα : α < κ+〉 is a free sequence in ⊕i∈IAi; we shall get a contradiction. Foreach α < κ+ there is a finite Sα ⊆ I such that cα ∈ ⊕i∈SαAi. We may assumethat S = Sα does not depend on α. But then κ+ ≤ supi∈S tAi by Theorem 12.8,contradiction. �

The behaviour of tightness in the free sequence sense under unions of chains ofBAs is similar to the case of cellularity (Theorem 3.16). The definition of ordinarysup-function does not quite fit, but essentially the same proof can be used:

Theorem 12.10. Let κ and λ be infinite cardinals, with λ regular. Then the follow-ing conditions are equivalent:

(i) cf(κ) = λ.

(ii) There is a strictly increasing sequence 〈Aα : α < λ〉 of infinite Booleanalgebras each with no free sequence of type κ such that

⋃α<λ Aα has a free

sequence of type κ. �

In view of the equivalence of tightness with its free sequence variant, 12.10 alsoapplies to tightness when κ is a successor cardinal. And actually 12.10 extendsin the following form to tightness itself; this answers, negatively, Problem 31 inMonk [90].

Theorem 12.11. Let κ and λ be infinite cardinals, with λ regular. Then the follow-ing conditions are equivalent:

(i) cf(κ) = λ.

(ii) There is a strictly increasing sequence 〈Aα : α < λ〉 of Boolean algebras eachwith tightness less than κ such that

⋃α<λ Aα has tightness κ.

Proof. By the comment before the theorem, we assume that κ is a limit cardinal.(i)⇒(ii): Take a free BA of size κ and write it as an increasing union of smalleralgebras in the obvious way.

(ii)⇒(i): Assume that (ii) holds and (i) fails. Let μ = supα<λ t(Aα); the firstpart of the proof will consist in showing that μ = κ; to this end, suppose that μ < κ.Fix ν such that μ < ν < κ. Let 〈aα : α < ν〉 be a free sequence in B. For each β < λlet Sν

β = {α < ν : aα ∈ Aβ}. Thus Sνβ ⊆ Sν

γ for β < γ < λ, and ν =⋃

β<λ Sνβ .

If ∃β < λ∀γ ∈ (β, λ)[Sνβ = Sν

γ ], then ν = Sνβ and so {aα : α < ν} ⊆ Aβ , hence

t(Aβ) ≥ ν, contradiction. Thus ∀β < λ∃γ ∈ (β, λ)[Sνβ ⊂ Sν

γ ]. Applying this to

ν = μ+ we get λ ≤ μ+; then applying it to ν = μ++ we get μ++ =⋃

β<λ Sμ++

β , so

there is a β < λ such that |Sμ++

β | = μ++, so Aβ has a free sequence of type μ++,which contradicts t(Aβ) ≤ μ. This contradiction proves that μ = κ.

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Since each t(Aα) is less than κ, from μ = κ it follows that cf(κ) ≤ λ; since(i) fails, we have in fact that cf(κ) < λ. Now ∀α < λ∃β ∈ (α, λ)[t(Aα) < t(Aβ)],since otherwise we would have μ < κ. Hence λ ≤ supα<λ t(Aα) = κ. Since λ isregular, λ < κ. So κ is singular. Let 〈να : α < cfκ〉 be a strictly increasing sequenceof cardinals with supremum κ. For each α < cf(κ) there is a βα < λ such thatt(Aβα) ≥ να, since μ = κ. Let γ = supα<cfκ βα; then γ < λ since cfκ < λ. Butthen t(Aγ) = κ, contradiction. �

A more natural version of Theorem 12.10 for tightness itself would be the equiva-lence expressed in the following problem.

Problem 112. Is the following true? Let κ and λ be infinite cardinals, with λ regular.Then the following conditions are equivalent:

(i) cf(κ) = λ.

(ii) There is a strictly increasing sequence 〈Aα : α < λ〉 of Boolean algebras eachhaving no ultrafilter with tightness κ such that

⋃α<λ Aα has an ultrafilter

with tightness κ.

This was Problem 44 in Monk [96].

Theorem 12.12. t(∏B

i∈I Ai

)= max(t(B), supi∈I t(Ai)).

Proof. For brevity let C =∏B

i∈I Ai. B and each Ai are isomorphic to subalgebrasof C, so ≥ holds. For the other direction we use the description of the ultrafilterson C given in Theorem 1.6(x). Let U be an ultrafilter on C, and suppose thatA ⊆ Ult(C) with U ⊆

⋃A .

Case 1. There exist i ∈ I and an ultrafilter V on Ai such that U = {h(b, F, a) :(b, F, a) is normal, and i ∈ b or (i /∈ b and i ∈ F and ai ∈ V )}. Supposethat x ∈ V . Then h(∅, {i}, {(i, x)}) ∈ U , and so there is a Wx ∈ A such thath(∅, {i}, {(i, x)}) ∈ Wx. By Theorem 1.6(x) there is an ultrafilter Yx on Ai suchthat

Wx = {h(b, F, a) : (b, F, a) is normal, and i ∈ b or (i /∈ b and i ∈ F and ai ∈ Yx)}.

It follows that x ∈ Yx. Thus V ⊆⋃

x∈V Yx. So there is a M ∈ [V ]≤t(V ) suchthat V ⊆

⋃x∈M Yx. We claim that U ⊆

⋃x∈M Wx. In fact, take any normal

h(b, F, a) ∈ U . If i ∈ b, then h(b, f, a) ∈ Wx for all x ∈ M . Suppose that i /∈ b.Then ai ∈ V , so there is an x ∈ M such that ai ∈ Yx. Then h(b, f, a) ∈ Wx. Thisproves the claim. Thus t(U) ≤ t(V ).

Case 2. There is a nonprincipal ideal W on B such that U = {h(b, F, a) : h(b, F, a)is normal and b ∈ W}. Suppose that x ∈ W . Then h(x, ∅, ∅) ∈ U , so there isa Vx ∈ A such that h(x, ∅, ∅) ∈ Vx. By Theorem 1.6(x) there is a nonprincipalultrafilter Yx on B such that Vx = {h(b, F, a) : h(b, F, a) is normal and b ∈ Yx}.Thus x ∈ Yx. Hence W ⊆

⋃x∈W Yx. Choose M ∈ [W ]≤t(W ) such that W ⊆⋃

x∈M Yx. We claim that U ⊆⋃

x∈M Vx. For, suppose that h(b, F, a) ∈ U , with

[email protected]

h(b, F, a) normal. Thus b ∈ W , so there is an x ∈ W such that x ∈ Yx. Thenh(b, F, a) ∈ Vx, as desired. �

Problem 113. Do there exist a system 〈Ai : i ∈ I〉 of BAs and a system 〈Fi : i ∈ I〉of ultrafilters, each Fi an ultrafilter on Ai, such that, with B the one-point gluingusing these inputs, t(B) < t

(∏i∈I Ai

)?

If 〈aα : α < κ〉 is a free sequence in A, then clearly 〈(aα,S(aα)) : α < κ〉 is a freesequence in Dup(A). Thus t(A) ≤ t(Dup(A)).

Problem 114. Is there a BA A such that t(A) < t(Dup(A))?

From Proposition 2.6 it follows that t(A) ≤ t(Exp(A)) for any infinite BA A. Theconcrete example of the exponential described at the end of Chapter 1 shows that< is possible here.

Lemma 12.13. If X is a topological space and X ∈ Z in Exp(X), then⋃Z is

dense in X.

Proof. Let 0 �= U be open in X . Then X ∈ V (X,U), so there is an F ∈ Z ∩V (X,U). Thus F ∩ U �= 0, so

⋃Z ∩ U �= 0. �

Lemma 12.14. For any Hausdorff space X we have d(X) ≤ t(Exp (X)).

Proof. Let Z be the collection of all finite non-empty subsets ofX . Then Z is a sub-set of Exp (X). Moreover, X ∈ Z, since if V (U0, . . . , Um−1) is any neighborhoodof X , choose ai ∈ Ui for all i < m; then {ai : i < m} ∈ Z ∩V (U0, . . . , Um−1). Nowchoose a subset Y of Z of size ≤ t(Exp (X)) such that X ∈ Y . Then by Lemma12.13,

⋃Y is dense in X . Clearly |

⋃Y | ≤ t(Exp (X)). �

We turn to derived functions for tightness. By Theorem 11.18 we have thattH+(A) = th+(A) = t(A). Clearly tS+(A) = t(A), tS−(A) = ω, and dtS+(A) =t(A). In the algebra of Fedorchuk [75] we have tH−(A) ≤ t(A) < CardH−(A); soProblem 32 of Monk [90] was solved long ago. See Chapter 16 for Fedorchuk’salgebra. Note that dtS−(A) �= t(A) in general; this can be seen by consideringP(ω) and its dense subalgebra consisting of the finite and cofinite subsets of ω.

Recall also our earlier results that DepthH+(A) = t(A) = πχH+(A); seeTheorems 4.26, 11.18. Next we mention more about the relationships betweentightness and our previously introduced functions. By Theorem 4.26 we haveDepth(A) ≤ t(A) for any BA A; the difference can be big, for example in a freealgebra. πχ(A) ≤ t(A) by Theorem 11.18. We observed in Chapter 11 that onecan have πχ(A) < Depth(A) in an interval algebra with the difference arbitrarilylarge. This solves Problem 33 in Monk [90]. In particular, it is possible to haveπχ(A) < t(A) with the difference arbitrarily large.

It is obvious that Ind(A) ≤ t(A); the difference is large in some intervalalgebras. Obviously t(A) ≤ |A|. Note that t(A) ≤ s(A) ≤ Irr(A) by Theorem 3.30and a later remark. Thus Problem 34 in Monk [90] has the obvious answer “no”.t(A) > π(A) for A = P(ω). t(A) > Length(A) for A an uncountable free BA.

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Length(A) > t(A) for A the interval algebra on the reals. c(A) > t(A) for A anuncountable finite-cofinite algebra.

We also give the following result relating π with t; it is from Todorcevic [90a].

Theorem 12.15. For every infinite BA A there is a sequence 〈aα : α < β〉 ofnonzero elements of A such that {aα : α < β} is dense in A and for every subsetΓ of β with no maximum element, the sequence 〈aα : α ∈ Γ〉 is free iff {aα : α ∈Γ} has the finite intersection property. (Since Γ has a natural order from β, themeaning of “free” in this extended sense is clear.)

Proof. Let P be a maximal disjoint subset of A+ such that A � b is π-homogeneousfor every b ∈ P , that is, π(A � c) = π � b) for every nonzero c ≤ b. Temporarilyfix b ∈ P . Let π(A � b) = κb, and let 〈cbα : α < κb〉 enumerate a dense subset of(A � b)+ of size κb. Now we define 〈abα : α < κb〉 by induction. Suppose that abαhas been defined for all α < β. Let F be the collection of all nonzero elements ofthe form cbβ ·

∏α∈F (a

bα)

ε(α) for F a finite subset of β and ε ∈ F 2. Then F is not

dense in A � cbβ , so there is an abβ ∈ (A � cbβ)+ such that x �≤ abβ for all x ∈ F .

This finishes the construction.

Concatenating the so obtained sequences 〈abα : α < κb〉 in any order, weobtain a sequence 〈aα : α < β〉 as desired in the theorem. In fact, first we checkthat {aα : α < β} is dense in A. Suppose that a ∈ A+. Choose b ∈ P such thata·b �= 0. There is a γ < κb such that cbγ ≤ a·b. By construction, abγ ≤ cbγ , as desired.Next we check that for any subset Γ of β with no maximum element, 〈aα : α ∈ Γ〉is free iff {aα : α ∈ Γ} has the finite intersection property.⇒: obvious.⇐: Assumethat {aα : α ∈ Γ} has the finite intersection property. Then there is a b ∈ P suchthat each of the aα’s for α ∈ Γ is of the form abγ . So without loss of generality

we assume that {abα : α ∈ Γ} has the finite intersection property, and we want toshow that 〈abα : α ∈ Γ〉 is free. We prove

(∗) If F and G are finite subsets of γ and F < G, then∏

α∈F abα ·∏

α∈G−abα �= 0.

This we do by induction on |G|. The case G = 0 is given. Assume that (∗) istrue for G, and G < γ ∈ Γ. If

∏α∈F abα ·

∏α∈G−abα · cbγ = 0, then also

∏α∈F abα ·∏

α∈G−abα ·abγ = 0, and so 0 �=∏

α∈F abα ·∏

α∈G−abα =∏

α∈F abα ·∏

α∈G−abα ·−abγ ,as desired. If

∏α∈F abα ·

∏α∈G−abα · cbγ �= 0, then

∏α∈F abα ·

∏α∈G−abα · −abγ �= 0

by construction. �

A min-max version of tightness is defined as follows. A free sequence 〈aξ : ξ < α〉is maximal iff there is no b ∈ A such that 〈aξ : ξ < α〉�〈b〉 is a free sequence,where 〈aξ : ξ < α〉�〈b〉 is the result of adjoining b at the end of the sequence〈aξ : ξ < α〉. Now we define

fsp(A) = {|α| : A has an infinite maximal free sequence of length α};f(A) = min(fsp(A)).

These notions are studied in Monk [11].

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Proposition 12.16. If 〈an : n ∈ ω〉 is a strictly decreasing sequence of elements ofa BA A, with 1 > a0, then it is a free sequence.

Proof. Suppose that F,G ⊆ ω with F < G. If F = ∅ �= G, then∏

ξ∈F aξ ·∏η∈G−aη = −aν �= 0, where ν is the least member of G. If F �= ∅ = G, then∏ξ∈F aξ ·

∏η∈G−aη = aν �= 0, where ν is the greatest member of F . If F �= ∅ �= G,

then∏

ξ∈F aξ ·∏

η∈G−aη = aν · −aμ �= 0, where ν is the greatest member of Fand μ is the least member of G. �

Proposition 12.17. A free sequence 〈aξ : ξ < α〉 of elements of A is maximal ifffor every b ∈ A one of the following conditions holds:

(i) There is a finite F ⊆ α such that∏

ξ∈F aξ · b = 0.

(ii) There exist finite F,G⊆α such that F <G and∏

ξ∈F aξ ·∏

η∈G−aη ·−b=0.�

Proposition 12.18. Suppose that 〈aξ : ξ < κ〉 is a strictly decreasing sequence ofelements of a BA A such that {aξ : ξ < κ} generates an ultrafilter on A and1 > a0. Then 〈aξ : ξ < κ〉 is a maximal free sequence.

Proof. 〈aξ : ξ < κ〉 is a free sequence by Proposition 12.16. Clearly it is maximal.�

The following proposition gives a connection between maximal free sequences ina BA A and towers in homomorphic images of A.

Proposition 12.19. Suppose that 〈aξ : ξ < α〉 is a maximal free sequence in anatomless BA A. For each ξ ≤ α let Fξ be an ultrafilter containing the set {aη : η <ξ} ∪ {−aη : ξ ≤ η < α}. Let I = {x ∈ A : ∀ξ ≤ α[−x ∈ Fξ]}. Then I is an ideal inA and 0 < [aη]I < [aξ]I < 1 if ξ < η < α. Moreover, if α is a limit ordinal, then∏

ξ<α[aξ]I = 0, while if α = β + 1 then [aβ]I is an atom of A/I.

Proof. Clearly I is an ideal on A. Now suppose that ξ < η < α. If ν ≤ α andaη · −aξ ∈ Fν , then η < ν, hence also ξ < ν and so aξ ∈ Fν , contradiction. Hence∀ν ≤ α[−(aη · −aξ) ∈ Fν ], and so aη · −aξ ∈ I and consequently [aη]I ≤ [aξ]I .Suppose that [aη]I = [aξ]I . Then aξ · −aη ∈ I, and so −aξ + aη ∈ Fξ+1. Alsoaξ ∈ Fξ+1, so aη ∈ Fξ+1. Since ξ + 1 ≤ η, this is a contradiction.

Thus we have shown that [aη]I < [aξ]I if ξ < η < α. If [aη]I = 0, thenaη ∈ I. But aη ∈ Fα, contradiction. If [aξ]I = 1, then −aξ ∈ I. But −aξ ∈ F0,contradiction.

Now suppose that 0 < [b]I < [aξ]I for all ξ < α. By the maximality of〈aξ : ξ < α〉 there are then two possibilities. If

∏ξ∈F aξ ·b = 0 for some finite subset

F of α, then [b]I = 0, contradiction. Suppose that∏

ξ∈F aξ ·∏

η∈G−aη · −b = 0,where F < G are finite subsets of α. If ξ is the greatest member of F and η is thesmallest member of G, then [aξ]I · −[aη]I ≤ [b]I ≤ [aη]I , so that [aξ]I · −[aη]I = 0,contradiction. If ξ is the greatest member of F and G = ∅, then [aξ]I · −[b]I = 0,hence [aξ]I ≤ [b]I < [aξ]i, contradiction. If η is the smallest element of G and

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F = ∅, then −[aη]I · −[b]I = 0, so −[aη]I ≤ [b]I ≤ [aη]I , so that −[aη]I = 0,contradiction. �Proposition 12.20. πχinf(A) ≤ f(A) for any atomless BA A.

Proof. Suppose that 〈aξ : ξ < α〉 is a maximal free sequence. Suppose that 2 ≤m < ω. We claim that⎧⎨

⎩∏ξ∈F

aξ : F ∈ [α]<ω

⎫⎬⎭ ∪

⎧⎨⎩∏ξ∈F

aξ ·∏ξ∈G

−aξ : F,G ∈ [α]<ω, F < G

⎫⎬⎭

is m-dense. To see this, let 〈bi : i < m〉 be a weak partition of A. If there is ani < m such that

∏ξ∈F aξ ·

∏ξ∈G−aξ · −bi = 0 for some finite F < G, this is as

desired. If for every i < m there is a finite Fi such that∏

ξ∈Fiaξ · bi = 0, then

with G =⋃

i<m Fi we have

∏ξ∈G

aξ =

⎛⎝∏

ξ∈G

aξ

⎞⎠ · (b0 + · · ·+ bm−1) = 0,

contradiction. �Theorem 12.21. p(A) ≤ f(A) for any atomless BA A.

Proof. Let 〈aξ : ξ < α〉 be a maximal free sequence, with α an infinite ordinal.Clearly

∏ξ∈F aξ �= 0, for every finite F ⊆ α. Suppose that 0 �= b ≤ aξ for every

ξ < α. Choose u with 0 < u < b. First suppose that∏

ξ∈G aξ · u = 0 for somefinite G ⊆ α. Now u < b ≤

∏ξ∈G aξ, so u = 0, contradiction. Suppose that∏

ξ∈G aξ ·∏

η∈H −aη · −u = 0 with finite G < H . If H �= ∅, choose η ∈ H .Then u < b ≤ aη, so −aη < −u, and it follows that

∏ξ∈G aξ ·

∏η∈H −aη =∏

ξ∈G aξ ·∏

η∈H −aη ·−u = 0, contradiction. Hence H = ∅. Hence∏

ξ∈G aξ ≤ u <b ≤

∏ξ∈G aξ, contradiction. �

Example 12.22. There is an atomless BA A such that f(A) < i(A). This is analgebra A of McKenzie, Monk [04]: with ω < κ < λ both regular, A has a strictlydecreasing sequence of length κ which generates an ultrafilter, while i(A) = λ.

Kevin Selker has constructed under CH an atomless BA A such that f(A) < u(A),where u(A) is the smallest size of a set generating a nonprincipal ultrafilter.

Problem 115. Describe the possibilities for tHs.

Clearly tSs(A) = [ω, t(A)] for any infinite BA A.

Problem 116. Describe tSr(A) in cardinal number terms.

Even for small cardinals there are many specific problems related to Problem 117.

Problem 117. Describe tHr(A) in cardinal number terms.

Again, for small cardinals there are many specific problems related to Problem 118.

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There are several natural finite versions of tightness, using the free sequence equiv-alent. For m,n ∈ ω, an m,n-free sequence is a sequence 〈aα : α < κ〉 such that ifΓ,Δ ⊆ α with |Γ| = m, |Δ| = n, and Γ < Δ, then

∏α∈Γ aα ·

∏β∈Δ−aβ �= 0. Then

we set

tmn(A) = sup{κ : there is an m,n-free sequence of length κ}.

Similarly we get three more notions:

An m-free sequence is a sequence 〈aα : α < κ〉 such that if Γ,Δ ⊆ α with |Γ| = m,Δ finite and Γ < Δ, then

∏α∈Γ aα ·

∏β∈Δ−aβ �= 0;

tm(A) = sup{κ : there is an m-free sequence of length κ}.

utmn(A) = sup{|X | : ∀Y ∈ [X ]m and ∀Z ∈ [X ]n(Y ∩Z = 0⇒∏y∈Y

y·∏z∈Z

−z �= 0)};

utm(A) = sup{|X | : ∀Y ∈ [X ]m and ∀ finite Z(Y ∩Z = 0⇒∏y∈Y

y ·∏z∈Z

−z �= 0)}.

These notions are studied in Roslanowski, Shelah [94].

Concerning tightness for special classes of algebras, note first of all that t(A) = |A|whenever A is complete. The description of t for interval algebras is similar to thatfor πχ. Since t coincides with DepthH+, t(A) = Depth(A) for A an interval algebra,by retractiveness. But it is of some interest to describe t(F ) for each ultrafilter Fon an interval algebra; this is a correction of the description in Monk [96]. Let Abe the interval algebra on a linearly ordered set L with first element 0. Let C bea terminal segment of L not containing 0. The ultrafilter FC determined by C isgenerated by

{[a,∞) : a /∈ C} ∪ {[0, c) : c ∈ C}.

Case 1. C = ∅, and L has a greatest element a. Then FC is the principal ultrafiltergenerated by {a}, and t(FC) = 1.

Case 2. C = ∅, and L does not have a greatest element. Let κ be the cofinality ofL. We claim that t(FC) = κ. To prove this, let 〈aξ : ξ < κ〉 be a strictly increasingcofinal sequence of elements of L. For each ξ < κ let Gξ be an ultrafilter such that[aξ, aξ+1) ∈ Gξ. Clearly FC ⊆

⋃ξ<κ Gξ. If M ∈ [κ]<κ, choose ξ < κ such that

η < ξ for all η ∈M . Then [aξ,∞) ∈ FC\⋃

η∈M Gη. This shows that t(FC) ≥ κ.

Now suppose that FC ⊆⋃Y , with Y ⊆ Ult(A). For each ξ < κ choose Hξ ∈

Y such that [aξ,∞) ∈ Hξ. Clearly FC ⊆⋃

ξ<κ Hξ. This shows that t(FC) = κ.

Case 3. C has a smallest element b and L\C has a greatest element a. Then FC

is the principal ultrafilter generated by {a}, and t(FC) = 1.

Case 4. C has a smallest element b, but L\C does not have a greatest element.Then t(FC) = κ, where κ is the cofinality of L\C; see Case 2.

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Case 5. C does not have a smallest element, but L\C has a greatest element a.Then t(FC) = κ, where κ is the coinitiality of C; see Case 2.

Case 6. C does not have a smallest element, and L\C does not have a greatestelement. Let κ be the cofinality of L\C and λ the coinitiality of C. Then t(FC) =max(κ, λ). We prove this under the assumption that κ ≤ λ, by symmetry. Let〈aξ : ξ < κ〉 be strictly increasing and cofinal in L\C, and 〈bξ : ξ < λ〉 strictlydecreasing and coinitial in C. For any ξ < κ and η < λ let Gξη be an ultrafiltercontaining [aξ, bη). Clearly FC ⊆

⋃(ξ,η)∈κ×λ Gξη. Suppose that M ∈ [κ × λ]<λ.

Then there is a ρ < λ such that η < ρ whenever (ξ, η) ∈ M . Then [a0, bρ) is amember of FC\

⋃(ξ,η)∈M Gξη. This shows that t(FC) ≥ λ.

Now suppose that FC ⊆⋃Y with Y ⊆ Ult(A). For each ξ, η) ∈ κ × λ

let Hξη ∈ Y be such that [aξ, bη) ∈ Hξη. Clearly F ⊆⋃

(ξ,η)∈κ×λ Hξη. Hence

t(FC) = λ.

Concerning f and interval algebras we have the following result.

Proposition 12.23. Let L be a linear ordering.

(i) If 〈aξ : ξ < α〉 is a strictly increasing sequence with lub b, with α a limitordinal, then 〈[aξ, b) : ξ < α〉 is a maximal free sequence in Intalg(L).

(ii) If 〈aξ : ξ < α〉 is a strictly decreasing sequence with glb b, with α a limitordinal, then 〈[b, aξ) : ξ < α〉 is a maximal free sequence in Intalg(L).

(iii) Suppose that 〈aξ : ξ < α〉 is strictly increasing, 〈bξ : ξ < α〉 is strictlydecreasing, ∀ξ < α[aξ < bξ], and there is no element c ∈ L such that∀ξ < α[aξ < c < bξ]. Then 〈[aξ, bξ) : ξ < α〉 is a maximal free sequencein Intalg(L).

(iv) Suppose that κ and λ are distinct infinite cardinals, 〈aξ : ξ < κ〉 is strictlyincreasing, 〈bξ : ξ < λ〉 is strictly decreasing, ∀ξ < κ∀η < λ[aξ < bη], andthere is no element c ∈ L such that ∀ξ < κ∀η < λ[aξ < c < bη]. Then thereis a maximal free sequence of length max(κ, λ) in Intalg(L).

Proof. (i): Let x be any nonzero element of Intalg(L). We consider two cases.

Case 1. For every component [c, d) of x we have b ≤ c or d < b. Clearly then thereis a ξ < α such that [aξ, b) ∩ [c, d) = ∅ for every component [c, d) of x. Hence[aξ, b) ∩ x = ∅, as desired in 12.17.

Case 2. There is a component [c, d) of x such that c < b ≤ d. Then there is a ξ < αsuch that [aξ, b) ⊆ [c, d) ⊆ x. So [aξ, b) · −x = ∅, again as desired in 12.17.

The proof of (ii) is similar, but (iii) is more complicated. Clearly 〈[aξ, bξ) :ξ < α〉 is a free sequence in Intalg(L). Now suppose that x is a nonzero elementof Intalg(L). If for every component [c, d) of x there is a ξ < α such that bξ < cor d < aξ, then there is a ξ < α such that x ∩ [aξ, bξ) = ∅, as desired. So, supposethat there is a component [c, d) of x such that for every ξ < α we have c ≤ bξ andaξ ≤ d. Then by the hypothesis of (iii) there is a ξ < α such that [aξ, bξ) ⊆ [c, d).Hence [aξ, bξ)\x = ∅, as desired.

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Now assume the hypothesis of (iv). By symmetry say κ < λ. For each ϕ < λwrite ϕ = κ · ρ + ξ with ξ < κ, and set cϕ = [aξ, bϕ). Suppose that F,G ∈ [λ]<ω

and F < G. We may assume that G �= ∅. First suppose also that F �= ∅. Let ϕ bethe greatest element of F and ψ the least element of G. Then

⋂F has the form

[aτ , bϕ) and ∩e∈G − e has the form [0, aσ) ∪ [bψ,∞). So⋂

F ∩⋂

e∈G−e contains[bψ, bϕ) and hence is nonempty.

The case F = ∅ is also clear by this argument.

Thus we have a free sequence. To show that it is maximal, it suffices toshow that {cϕ : ϕ < λ} generates an ultrafilter. Let x be any member of A, with0 < x < 1. First suppose that x has a component [u, v) with u /∈ L and v ∈ L.Choose ξ < κ and η < λ such that u < aξ and bη < v. Let ϕ = κ · η + ξ. Thenϕ ≥ η, and hence bϕ ≤ bη. So cϕ ⊆ x. Second, if x has no such component, then−x does have such a component and so we get a ϕ < λ such that cϕ ⊆ −x. �

Corollary 12.24. f(A) ≤ tow(A) for any atomless interval algebras A.

Proof. By Theorem 4.82 and Proposition 12.23. �

For tree algebras we have t(Treealg(T )) = Depth(Treealg(T )) by retractiveness.Now take any ultrafilter F on Treealg(T ). It corresponds to an initial chain C ofT ; see the Handbook. A description of t(F ), due to Brenner [82], is as follows:

Theorem 12.25. Let T be a tree with a single root and F an ultrafilter on Treealg(T).Let C = {t ∈ T : (T ↑ t) ∈ F}. Then one of the following holds:

(i) C has a greatest element t, and t has only finitely many immediate successors.Then F is principal, and t(F ) = 1.

(ii) C has a greatest element t, and t has infinitely many immediate successors.Then t(F ) = ω.

(iii) C has no greatest element. Then t(F ) = cf(C).

Proof. (i) is obvious. For (ii), suppose that C has a greatest element t and thas infinitely many immediate successors. Suppose that F ⊆

⋃Y , where Y ⊆

Ult(Treealg(T )). Without loss of generality F /∈ Y . We claim

(1) Sdef= {s : s is an immediate successor of t and (T ↑ s) ∈ G for some G ∈ Y } is

infinite.

For, suppose that S is finite. Now (T ↑ t)\⋃

s∈S(T ↑ s) ∈ F , so choose G ∈ Ysuch that (T ↑ t)\

⋃s∈S(T ↑ s) ∈ G. For every immediate successor s of t we have

(T ↑ s) /∈ G. So F = G, contradiction. So (1) holds.

Let U ∈ [S]ω. For each u ∈ U choose Gu ∈ Y such that (T ↑ u) ∈ Gu. Nowsuppose that x ∈ F . Without loss of generality x has the form (T ↑ t)\

⋃v∈V (T ↑

v) where V is a finite set of immediate successors of t. Choose u ∈ U\V . Clearly(T ↑ t)\

⋃v∈V (T ↑ v) ∈ Gu, as desired.

In the present case it is clear that F is nonprincipal, so t(F ) = ω.

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For (iii), suppose that C has no greatest element. Let 〈sα : α < cf C〉 be astrictly increasing cofinal sequence of elements of C.

First we show that t(F ) ≥ cf(C). For each α < cf(C) the set

{T ↑ sα}∪{T \(T ↑ u) : u is an immediate successor of sα}∪{T \(T ↑ v) : v and sα are incomparable}

has the fip, as is easily seen; let Gα be an ultrafilter containing this set. We claimthat F ⊆

⋃α<cf(C) Gα. For, suppose that x ∈ F . We may assume that

(2) x = (T ↑ r)\⋃

u∈U (T ↑ u) where r ∈ C, U is a set of immediate successors ofr, and U ∩C = 0.

Choose α < cf(C) such that r ≤ sα. Clearly x ∈ Gα, as desired.

Now suppose that Γ ⊆ cf(C) and |Γ| < cf(C). We claim that F �⊆⋃

α∈Γ Gα.For, choose β < cf(C) such that Γ < β. Clearly (T ↑ sβ) ∈ F but (T ↑ sβ) /∈⋃

α∈Γ Gα.

So, we have shown that t(F ) ≥ cf(C).

Now suppose that F ⊆⋃Y , Y ⊆ Ult(Treealg (T )). We want to find Z ∈

[Y ]≤cf(C) such that F ⊆⋃Z. We may assume that F /∈ Y . For each α < cf(C)

let yα = (T ↑ sα) and let Gα ∈ Y be such that yα ∈ Gα. We claim that F ⊆⋃α<cf(C) Gα. Let x ∈ F . We may assume that x is as in (2). Choose α < cf(C)

such that r < α. Then yα ⊆ x, and so x ∈ Gα, as desired. �

Concerning superatomic BAs, Shelah and Spinas [99] proved the following, im-proving a result of Dow, Monk [94]:

Suppose that 0� exists. Let B be a superatomic Boolean algebra in the constructibleuniverse L, and let λ be an uncountable cardinal in V . Then in L it is true thatt′(B) ≥ λ+ implies that Depth′(B) ≥ λ.

This solves Problem 45 of Monk [96]. But the following is still open.

Problem 118. Can one prove in ZFC that for every uncountable cardinal λ, t′(B) ≥λ+ implies that Depth′(B) ≥ λ?

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13 Spread

First note that any infinite BA A has an infinite disjoint subset D, which gives riseto an infinite discrete subspace of Ult(A). So s(A) is always infinite, for A infinite.

The following theorem gives some equivalent definitions of spread.

Theorem 13.1. For any infinite BA A, s(A) is equal to each of the following car-dinals:

sup{|X | : X is a minimal set of generators of 〈X〉Id};sup{|X | : X is ideal-independent};sup{|X | : X is the set of all atoms in some homomorphic image of A};sup{|At(B)| : B is an atomic homomorphic image of A};sup{c(B) : B is a homomorphic image of A}.

Proof. Six cardinals are mentioned in this theorem; let them be denoted byκ0, . . . , κ5 in the order that they are mentioned. In Theorem 3.29 we proved thatκ0 = κ2, and in Theorem 3.30 that κ2 = κ5. It is obvious that κ1 = κ2. Toshow that κ3 ≤ κ4, suppose that B is a homomorphic image of A with an infinitenumber of atoms. Let I be the ideal 〈{x : x · a = 0 for every atom a of B}〉Id ofB. Clearly B/I is atomic with the same number of atoms as B. This shows thatκ3 ≤ κ4. Obviously κ4 ≤ κ5 Finally, for κ5 ≤ κ3, let B be a homomorphic imageof A, and let D be an infinite disjoint subset of B. We show how to find an atomichomomorphic image C of B with exactly |D| atoms. Let M be the subalgebra ofB generated by D. Let f be an extension of the identity on D to a homomorphismof B into D; the image of B under f is as desired. �

From these characterizations it follows that if A is a subalgebra or homomorphicimage of B, then s(A) ≤ s(B). Clearly the difference can be arbitrarily large.

Clearly one can have s(B) much larger than s(A) when A ≤free B; so thesame applies to A ≤proj B, A ≤u B, A ≤rc B, A ≤reg B, and A ≤σ B. One canhave s(A) < s(B) also when A ≤π B; for example, take A = Fr(κ) and B = A.But when A ≤π B, we have s(B) ≤ |B| ≤ 2|A|. If A ≤s B, in particular if A ≤m B,then s(A) = s(B); this follows easily from Proposition 2.29. From Proposition 2.54it is clear that one can have A ≤mg B with s(B) much larger than s(A).

Note that all of the equivalents of spread given in Theorem 13.1 have the sameattainment properties. The attainment question for spread has been thoroughly

, ,OI 10.1007/978-3- - - _ ,


014


Basel 214

405

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406 Chapter 13. Spread

studied in topology, mainly by Juhasz. We present these results, specialized toBoolean algebras.

For the first result we will use the general canonization lemma, 28.1 in Erdos,Hajnal, Mate, Rado [84]. We describe a special case of that lemma which we willuse:

Define �0(κ) = κ and �m+1(κ) = �m(2κ) for every m ∈ ω. Suppose that κ is astrong limit singular cardinal, and 〈λξ : ξ < cf(κ)〉 is a strictly increasing sequenceof cardinals with supremum κ satisfying the following conditions:

(i) 2cf(k) ≤ λ0.

(ii) �3(λξ) < λη for ξ < η < cf(κ).

Suppose also that 〈Aξ : ξ < cf(κ)〉 is a system of pairwise disjoint sets, and |Aξ| ≥(�3(λξ))

+ for all ξ < cf(κ). Let B =⋃

ξ<cf(κ)Aξ. Suppose that f : [B]3 → 2× 2.

Then there is a system 〈Cξ : ξ < cf(κ)〉 satisfying the following conditions:

(iii) Cξ ⊆ Aξ and |Cξ| ≥ λ+ξ for each ξ < cf(κ).

(iv) Let D =⋃

ξ<cf(κ) Cξ. Then for any u, v ∈ [D]3, if |u ∩ Cξ| = |v ∩ Cξ| for all

ξ < cf(κ), then f(u) = f(v).

Theorem 13.2. If κ is strong limit singular and κ ≤ |Ult(E)|, then Ult(E) has adiscrete subset of size κ.

Proof. Let 〈λα : α < cf(κ)〉 be sequence of cardinals as in the above statementof the canonization lemma. Let ≺ be a well-order of Ult(E). 〈Aξ : ξ < cf(κ)〉be a system of pairwise disjoint subsets of Ult(E), with |Aξ| = (�3(λξ))

+ for allξ < cf(κ), and with Aξ ≺ Aη for ξ < η < cf(κ). Let B =

⋃ξ<cf(κ) Aξ. We will find

a discrete subset of B of size λ. For distinct F,G ∈ B choose aFG ∈ F\G. Let ≺be a well-order of B. We now define a function f : [B]3 → 2× 2. For F ≺ G ≺ H ,let f({F,G,H}) = (ε1, ε2), where

ε1 =

{0 if aGH ∈ F ,

1 otherwise;ε2 =

{0 if −aFG ∈ H ,

1 otherwise.

We now apply the above canonization lemma to get a system 〈Cξ : ξ < cf(κ)〉satisfying the following conditions:

(1) Cξ ⊆ Aξ and |Cξ| ≥ λ+ξ for each ξ < cf(κ).

(2) Let D =⋃

ξ<cf(κ)Cξ. Then for any u, v ∈ [D]3, if |u ∩ Cξ| = |v ∩ Cξ| for all

ξ < cf(κ), then f(u) = f(v).

(3) Cξ ≺ Cη for ξ < η < cf(κ).

It suffices to find a discrete subset of D of size λ. In fact, let

Z = {G ∈ D : G has an immediate predecessor F , an immediate successor H ,

and there is a ξ < cf(κ) such that{F,G,H} ⊆ Cξ}.

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Chapter 13. Spread 407

By conditions (1)–(3) it is clear that |Z| = λ. We claim that it is discrete. For, takeany G ∈ Z; say that G has an immediate predecessor F an immediate successorH , ξ < cf(κ), and {F,G,H} ⊆ Cξ. We want to show that G is isolated in Z. Infact, let N = S(−aFG)∩S(aGH)∩Z. Clearly G ∈ N and F,H /∈ N . Suppose thatK ∈ Z and K ≺ F . Clearly ∀η < cf(κ)[|{K,F,G} ∩ Cη}| = |{K,G,H} ∩ Cη|], sof({K,F,G}) = f({K,G,H}).Case 1. aFG ∈ K. Obviously then K /∈ N .

Case 2. −aFG ∈ K. Then f({K,F,G}) has the form (1, ε2), so also

f({K,G,H}) = (1, ε2).

So −aGH ∈ K and hence K /∈ N .

The case H ≺ K is similar: Clearly

∀η < cf(κ)[|{K,F,G} ∩ Cη}| = |{K,G,H} ∩ Cη|],

so f({K,F,G}) = f({K,G,H}).Case a. −aGH ∈ K. Obviously then K /∈ N .

Case b. aGH ∈ K. Then f({G,H,K}) has the form (ε1, 1), so also

f({F,G,K}) = (ε1, 1).

So aFG ∈ K and hence K /∈ N . �Corollary 13.3. If s(A) is singular strong limit, then it is attained. �

Our second attainment theorem for spread uses an argument (in Juhasz [80]) easilyadapted for hL and hd; so we treat all three simultaneously. To this end we needa definition, whose connection with hL will come later.

A sequence 〈xξ : ξ < κ〉 of distinct elements of a topological space X isright-separated provided that for every ξ < κ the set {xη : η ≤ ξ} is open in{xξ : ξ < κ};

Also recall from just before Theorem 6.12 the definition of left-separated.

Proposition 13.4. Suppose that λ is a singular cardinal of cofinality ω, and supposethat A is an infinite BA. A subset X of Ult(A) is of type t, where t < 3, if t = 0and X is discrete; or t = 1 and X is the range of a left-separated sequence; ort = 2 and X is the range of a right-separated sequence. Then for any t < 3, if λis the supremum of sizes of type t subsets of Ult(A), then Ult(A) has a subset oftype t and cardinality λ.

Proof. Let 〈μn : n ∈ ω〉 be a strictly increasing sequence of uncountable regularcardinals with supremum λ. Let t < 3. Let 〈Dn : n ∈ ω〉 be a system of subspacesof Ult(A) of type t such that |Dn| = μn for all n ∈ ω, and let Y =

⋃n∈ω Dn. It

suffices to find a subspace of Y of type t and size λ. Let Z =⋃{U : U is an open

subset of Y of size less than λ}.

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(1) If W ⊆ Y has size κ with ω < κ < λ and κ is regular, then W has a subset ofsize κ and of type t.

In fact, W =⋃

n∈ω(W ∩Dn), so by the regularity of κ there is an n ∈ ω such that|W ∩Dn| = κ. So (1) holds.

Now to prove the Proposition we consider two cases.

Case 1. |Z| < λ. Thus |Y \Z| = λ.

(2) Every nonempty open subspace of Y \Z has size λ.

In fact if U is a nonempty open subspace of Y \Z, then there is an open subset Vof Y such that U = V ∩ (Y \Z). Then |V | = λ, and hence |U | = λ.

(3) There is a system 〈an : n ∈ ω〉 of pairwise disjoint nonzero elements of A suchthat S(an) ∩ (Y \Z) �= ∅ for all n ∈ ω.

In fact, suppose that we have constructed ai for all i < n so that they are nonzero,pairwise disjoint, S(ai)∩(Y \Z) �= ∅ for all i < n, and Y \

(Z ∪

⋃i<n S(ai)

)is open

in Y \Z and of size λ. Fix distinct F,G ∈ Y \(Z ∪

⋃i<n S(ai)

), and let an ∈ F ,

bn ∈ G, with an ·bn = 0 and S(an)∩(Y \Z),S(bn)∩(Y \Z) ⊆ Y \(Z ∪

⋃i<n S(ai)

).

This finishes the construction.

Now by (1) each set S(an) ∩ (Y \Z) contains a set En of size μn and type t,and then

⋃n∈ω En is a subset of Y \Z of size λ and type t.

Case 2. |Z| = λ. It suffices to find a subset of Z of size λ and type t. For eachκ < λ let

Xκ = {F ∈ Z : ∃aF ∈ F [|S(aF ) ∩ Y | < κ]}.Note that Z =

⋃κ<λ Xκ.

Subcase 2.1. ∀κ < λ[|Xκ| < λ]. We define 〈Fm : m ∈ ω〉, 〈n(i) : i < ω〉, and〈ai : i ∈ ω〉 by recursion. Let n(0) = 0. Choose F0 ∈ Z\Xμ0 . Choose n(1) > n(0)so that F0 ∈ Xμn(1)

. Let a0 ∈ F0 with |S(a0) ∩ Y | < μn(1). Now F0 /∈ Xμ0 , so|S(a0) ∩ Y | ≥ μ0. By (1), S(a0) ∩ Y has a subset of size μ0 and type t.

Now suppose that Fi, ai, n(i), n(i + 1) have been defined for all i ≤ m sothat for all i ≤ m, Fi ∈ Z\Xμn(i)

, Fi ∈ Xμn(i+1), S(ai) ∩ Y ) has a subset of

size μn(i) and type t, and |S(ai) ∩ Y | < μn(i+1). Clearly these conditions hold fori = 0. Choose Fm+1 ∈ Z\Xμn(m+1)

. Say Fm+1 ∈ Xμn(m+2)with m + 1 < m + 2.

Choose c ∈ Fm+1 with |S(c) ∩ Y | < μn(m+2). Note that if i ≤ m then ai /∈ Fm+1.In fact, if ai ∈ Fm+1, then, since |S(ai) ∩ Y | < μn(i+1), we would get Fm+1 ∈Xn(i+1) ⊆ Xn(m+1), contradiction. Let am+1 = c ·

∏i≤m−ai. Then am+1 ∈ Fm+1;

since Fm+1 /∈ Xμn(m+1), it follows that S(am+1) ∩ Y has a subset of size μn(m+1)

and type t.

This finishes the construction, and shows that Y has a subset of size λ andtype t.

Subcase 2.2. There is a κ < λ such that |Xκ| = λ. For each F ∈ Xκ let F (F ) =S(aF ) ∩Xκ. Thus F : Xκ → [Xκ]

<κ, and κ < λ. It follows from Hajnal’s free set

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theorem that there is a set S ⊆ Xκ of size λ which is free for F , i.e., ∀F,G ∈ S[F �=G → F /∈ F (G)]. Thus {F} = S(aF ) ∩ Xκ for all F ∈ S, so Xκ is discrete. Let〈Fξ : ξ < κ〉 enumerate S without repetitions. Take any ξ < κ. Then

⋃η≤ξ S(xFη )

is an open set whose intersection with S is {Fη : η < ξ}; so 〈Fξ : ξ < κ〉 isright-separated. Similarly, it is left-separated. �

Assuming V=L, if κ is inaccessible but not weakly compact, then there is a BAA with spread κ not attained. Namely, one can take a κ-Suslin tree T in whichevery element has infinitely many immediate successors, and let A = Treealg(T ).By the retractiveness of A, Theorem 13.1, and Theorem 3.59, A is as desired.

Theorem 13.5. If κ is uncountable and weakly compact, and if s(A) = κ, then s(A)is attained.

Proof. We modify the proof of Theorem 13.2. Let 〈μα : α < κ〉 be a strictlyincreasing sequence of uncountable regular cardinals with supremum κ. For eachα < κ let Dα be a discrete subset of Ult(A) of size μα. Let X =

⋃α<cf(κ) Dα.

It suffices to find a discrete subset of X of size κ. For distinct F,G ∈ X chooseaFG ∈ F\G. Let ≺ be a well-order ofX . We now define a function f : [X ]3 → 2×2.For F ≺ G ≺ H , let f({F,G,H}) = (ε1, ε2), where

ε1 =

{0 if aGH ∈ F ,

1 otherwise;ε2 =

{0 if −aFG ∈ H ,

1 otherwise.

Now κ → (κ)34; see Erdos, Hajnal, Mate, Rado [84], Theorem 29.6. Hence chooseY ∈ [X ]κ and (ε0, ε1) ∈ 2 × 2 such that f � [Y ]3 takes on the constant value(ε0, ε1). Let Z = {L ∈ Y : L has an immediate predecessor and an immediatesuccessor}. We claim that Z is discrete. To prove this, take any G ∈ Z. Let F bethe immediate predecessor of G and H its immediate successor. Note that possiblyF /∈ Z or H /∈ Z. Let b = aGH · −aFG. Thus b ∈ G, b /∈ F , and b /∈ H . Take anyK ∈ Z\{F,G,H}.Case 1. aGH ∈ F . Then f(F,G,H) = 0. If K < F < G, then f(K,F,G) = 0 andso aFG ∈ H and b /∈ H . If F < G < K, then f(G,H,K) = 0 and so aGH /∈ K andb /∈ K.

Case 2. aGH /∈ F . Then f(F,G,H) = 1. If K < G < H , then f(K,G,H) = 1 andso aGH /∈ K and b /∈ K. If F < G < K, then f(F,G,K) = 1 and so aFG ∈ K andb /∈ K. �

The following two results are clear from topological duality.

Proposition 13.6. If A and B are infinite BAs, then s(A×B) = max(s(A), s(B)).�

Proposition 13.7. Suppose that 〈Ai : i ∈ I〉 is a system of BAs each with at leasttwo elements. Then s

(∏wi∈I Ai

)= max(|I|, supi∈Is(Ai)). �

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Clearly s(∏

i∈I Ai

)≥ max(2|I|, supi∈Is(Ai)). Shelah and Peterson independently

observed that strict inequality is possible, thus answering Problem 35 of Monk [90].Namely, let κ be the first limit cardinal bigger than 2ω (thus κ has cofinality ω), letA be the finite-cofinite algebra on κ, and consider ωA. Then for any non-principalultrafilter on ω we have

κω = |ωA| ≥ s(ωA) ≥ c(ωA/F ) = κω

by the discussion of ultraproducts for cellularity. Thus ωA gives a product wherethis inequality is strict.

Turning to ultraproducts, note that spread is an ultra-sup function, so Theo-rems 3.20–3.22 apply; Theorem 3.22 says that s

(∏i∈I Ai/F

)≥∣∣∏

i∈I s(Ai)/F∣∣ for

F regular, and Donder’s theorem says that under V = L the regularity assump-tion can be removed. A consistent example with > is described in Roslanowski,Shelah [98]. Two further consistency results are as follows. Shelah [99] proved thefollowing result, part of Conclusion 15.13:

If D is a uniform ultrafilter on κ, then there is a class of cardinals χ with thefollowing properties:

(i) χκ = χ.

(ii) There are BAs Bi for i < κ such that:

(a) s(Bi) ≤ χ for each i < κ, and hence |∏

i<κ s(Bi)/D| ≤ χ.

(b) s(∏

i<κ Bi) = χ+.


The following is a result of Shelah, Spinas [00], part of Corollary 2.4:

There is a model in which there exist cardinals κ, μ, a system 〈Bi : i < κ〉 of BAs,and an ultrafilter D on κ such that |

∏i<κ s(Bi)/D| = μ++ and s(

∏i<κ Bi) ≤ μ+.


We turn to free products. .

Theorem 13.8. If 〈Ai : i ∈ I〉 is a system of BAs each with at least 4 elements,then s(⊕i∈IAi) ≥ max(|I|, supi∈Is(Ai)). �

Equality does not hold in Theorem 13.8, in general. For example, let A be theinterval algebra on the reals. We observed in Corollary 3.48 that s(A) = ω. Hereis a system of 2ω ideal independent elements in A⊕A: for each real number r, letar = [r,∞) × [−∞, r) (considered as an element of A ⊕ A). Suppose that F is afinite subset of R, r ∈ R\F , and ar ∈ 〈as : s ∈ F 〉Id. Thus

[r,∞)× [−∞, r) ·∏s∈F

([−∞, s) + [s,∞)) = 0.

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But if Tdef= {s ∈ F : r < s} and U

def= F\T , then

[r,∞)× [−∞, r) ·∏s∈F

([−∞, s) + [s,∞)) ≥

[r,∞)× [−∞, r) ·∏s∈T

[−∞, s) ·∏s∈U

[s,∞) �= 0,

contradiction.

We give now the proof that |B| ≤ 2s(B) for any BA B. It depends on several otherresults which are of interest. We will use the notion of a network introduced inChapter 9.

Lemma 13.9. If X is a Hausdorff space and 2κ < |X |, then there is a sequence〈Fα : α < κ+〉 of closed subsets of X such that α < β implies Fβ ⊂ Fα.

Proof. For each f ∈⋃

α<κ+α2 we define a closed subset Xf of X . Let X0 = X .

For dmn(f) limit, let Xf =⋂

α<dmn(f) Xf�α. Now suppose that Xf has been

constructed. If |Xf | ≤ 1, let Xf�〈0〉 = Xf�〈1〉 = Xf . Otherwise, let Xf�〈0〉 andXf�〈1〉 be two proper closed subsets of Xf whose union is Xf . This finishes theconstruction. Clearly

⋃dmn(f)=α Xf = X for all α < κ+. Now

(∗) there is an f ∈ κ+

2 such that |Xf�α| ≥ 2 for all α < κ+.

For, otherwise, for all x ∈ X there is an f ∈⋃

α<κ+α2 such that Xf = {x}, and so

|X | ≤∣∣∣∣∣⋃

α<κ+

α2

∣∣∣∣∣ = 2κ,

contradiction. So (∗) holds, and it clearly gives the desired result. �Theorem 13.10. |A| ≤ 2s(A) for any BA A.

Proof. To start with, we prove:

(1) d(A) ≤ 2s(A).

In fact, suppose that (1) fails. Note that for every Y ⊆ Ult(A) of power < d(A)we have Y �= Ult(A). Hence one can construct two sequences 〈Fα : α < (2s(A))+〉and 〈aα : α < (2s(A))+〉 such that aα ∈ Fα ∈ Ult(A) and S(aα)∩{Fβ : β < α} = 0for all α < (2s(A))+. Let X = {Fα : α < (2s(A))+}. Clearly F is one-one, so|X | > 2s(A). By Lemma 13.9, let 〈Kα : α < (sA)+〉 be a system of closed subsetsof X such that α < β implies that Kβ ⊂ Kα. Say Fβα ∈ Kα\Kα+1 for allα < (s(A))+, and choose bα ∈ A so that Fβα ∈ S(bα) ∩X ⊆ X\Kα+1. Then

(2) S(bα) ∩ {Fβγ : γ > α} = 0.

For, suppose γ > α and Fβγ ∈ S(bα). But Fβγ ∈ Kγ ⊆ Kα+1, contradiction.

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Define f : [(s(A))+]2 → 2 as follows: f{γ, δ} = 0 iff when γ < δ we haveβγ > βδ. We now use the partition relation μ+ → (ω, μ+). Since there is no infinitedecreasing sequence of ordinals, we get a subset Γ of (s(A))+ of size (s(A))+ suchthat if γ, δ ∈ Γ and γ < δ, then βγ < βδ. Hence for any α ∈ Γ we have

S(aβα · bα) ∩ {Fβγ : γ ∈ Γ} = {Fβα},

and {Fβγ : γ ∈ Γ} is discrete, contradiction. So, we have finally proved (1).

Let Y be a subset of Ult(A) which is dense in Ult(A) and of cardinality d(A).Let

N = {Z : Z ⊆ Y, |Z| ≤ t(A)}.

From (1), Theorem 5.18, and Lemma 13.9 we see that |N | ≤ 2s(A). So, we will befinished, by Theorem 13.9, after we show that N is a network for A. Let F ∈ U ,with U open. Say F ∈ S(a) ⊆ U . Now F ∈ Ult(A) = Y , so we can choose Z ⊆ Ywith |Z| ≤ t(A) such that F ∈ Z. Let Z ′ = V ∩Z. Then F ∈ Z ′ ⊆ U and Z ′ ∈ N ,as desired. �

By Theorem 13.1, spread can be considered to be an ordinary sup-function, andso its behaviour under unions is given by Theorem 3.16.

Concerning moderate products we have the following.

Proposition 13.11. Suppose that I is infinite, κ is an infinite successor cardinal,s(B) < κ, |I| < κ, and s(Ai) < κ for every i ∈ I. Then s(

∏Bi∈I Ai) < κ.

Proof. We write any element x of∏B

i∈I Ai as h(bx, Fx, ax), in normal form as

described in Chapter 1. Now suppose that A is a discrete subset of Ult(∏B

i∈I Ai

)such that |A | = κ; we want to get a contradiction. For each D ∈ A chooseh(bD, FD, aD) such that S(h(bD, FD, aD)) ∩A = {D}.

Case A. Xdef= {D ∈ A : Theorem 1.6(x)(b) holds for D} has size > s(B).

For each D ∈ X choose an ultrafilter ED on B in accordance with Theorem1.6(x)(b). Suppose that D �= D′. Then ED �= ED′ . In fact, choose x ∈ D\D′. Thenx = h(bx, Fx, ax). By Proposition 1.5(ii), −x = h(−(bx ∪ Fx), Fx, a

′x) for some a′.

Since −x ∈ D′, it follows that bx ∈ ED\ED′ . Let Y = {ED : D ∈ X}. Now weclaim that S(bD) ∩ Y = {ED} for each D ∈ X . Suppose that H ∈ S(bD) ∩ Y .Say H = EK with K ∈ X . Now bD ∈ H , so h(bD, FD, aD) ∈ K. Hence K = Dand so H = ED. Conversely, clearly ED ∈ S(bD) ∩ Y . This proves the claim, andcontradicts |Y | = |X | ≥ s(B).

Case B. Case A fails. It follows that Xdef= {D ∈ A : Theorem 1.6(x)(a) holds

for D} has size κ. Hence for each D ∈ X there exist iD ∈ I and an ultrafilterHD on Ai such that D = {h(b, F, a) : iD ∈ b or iD ∈ F and a(iD) ∈ HD}. Since|X | > |I|, there exist a j ∈ I and a Y ⊆ X such that iD = j for all D ∈ Y ,with |Y | = κ. Clearly HD �= HD′ for D �= D′. Let Z = {HD : D ∈ Y }. We claimthat Z is discrete. For, take any D ∈ Y . S(aD(j)) ∩ Z = {HD}. In fact, if HE ∈

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S(aD(j)) with E ∈ Y , then aD(j) ∈ HE and hence h(bD, FD, aD) ∈ E. So E ∈S(h(bD, FD, aD))∩Y and so E = D. Conversely, clearly HD ∈ S(aD(j))∩Z. �

Concerning one-point gluing, it is clear that s(C) = max(s(A), s(B)) if C is ob-tained from A and B by one-point gluing. Not having a good description for thespread of an infinite product, we also do not know what happens under one-pointgluing in general:

Problem 119. Completely describe the behaviour of spread under one-point gluing.

If B is the Alexsandroff duplicate of A, clearly s(B) = |B| = 2|ult(A)|.By Proposition 2.6 we have s(Exp(A)) ≥ s(A) for any infinite BA A.

Problem 120. Is there a BA A such that s(A) < s(Exp(A))?

We turn to the derived functions for spread. The following facts are clear: sH+(A) =s(A); sS+(A) = s(A); sS−(A) = ω; sh−(A) = ω; dsS+(A) = s(A). The algebra Aof Fedorchuk [75] (constructed under ♦ and presented in Chapter 16) is such thatsH−(A) ≤ s(A) < CardH−(A). Thus Problem 36 of Monk [90] was solved long ago.It is also easy to see that sh+(A) = s(A). The status of the derived function dsS−is not clear; note that dsS−(A) < s(A) for A = Pκ.

Problem 121. Completely describe dsS− in terms of the other cardinal functions.

We turn to smm and sspect, which were discussed in Monk [08]. They are definedas follows.

sspect(A) = {|X | : X is an infinite maximal ideal independent subset of A};smm(A) = min(sspect(A)).

Note that {1} is maximal ideal independent; this accounts for the restriction abovethat X is infinite.

A simple way of coming up with maximal ideal independent subsets is givenin the following proposition.

Proposition 13.12. If X is infinite, ideal independent, and generates a maximalideal I, then X is maximal ideal independent.

Proof. Let y ∈ A\X . If y ∈ I, then y ≤∑

F for some finite F ⊆ I, and so X∪{y}is ideal dependent. Suppose that −y ∈ I. Say −y ≤

∑F with F ∈ [X ]<ω. Choose

z ∈ X\F . Then z ≤ 1 = y +∑

F , and so X ∪ {y} is ideal dependent. �Proposition 13.13. sspect(A) ∪ sspect(B) ⊆ sspect(A×B).

Proof. Let X be maximal ideal-independent in A. Define Y = {(x, 1) : x ∈ X}.Clearly Y is ideal-independent in A×B. To show that it is maximal, suppose that(u, v) ∈ A×B. Then there are two possibilities.

Case 1. There is a finite subset F of X such that u ≤∑

F . We may assume thatF �= ∅. Then (u, v) ≤

∑x∈F (x, 1), as desired.

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Case 2. There exist an x ∈ X and a finite subset F of X\{x} such that x ≤u +

∑F . Again we may assume that F �= ∅. Then (x, 1) ≤ (u, v) +

∑y∈F (y, 1),

as desired.

Hence the proposition follows by symmetry. �

Problem 122. Is sspect(A×B) = sspect(A) ∪ sspect(B)?

Problem 123. Is smm(A×B) = smm(A) ∪ smm(B)?

Theorem 13.14. If K is a nonempty finite set of infinite cardinals, then

sspect

(∏λ∈K

Fr(λ)

)= K.

Proof. ⊇ holds by Proposition 13.13. Suppose that κ ∈ sspect(∏

λ∈K Fr(λ))\K.

Let L = {λ ∈ K : λ < κ} and M = K\L. Assume that L �= ∅; some obviouschanges should be made in the following argument if L = ∅. Let X be a max-imal independent subset of

∏λ∈K Fr(λ) of size κ. For each λ ∈ M let uλ be a

free generator of Fr(λ) not in the subalgebra generated by {xλ : x ∈ X}. Now|∏

λ∈L Fr(λ)| < κ, so there is a q ∈∏

λ∈L Fr(λ) such that X ′ def= {x ∈ X : x � L =

q} has size greater than max(L). Let f = q ∪ 〈uλ : λ ∈ M〉. So f ∈∏

λ∈K Fr(λ)and f /∈ X (since clearly M �= ∅). Hence X ∪ {f} is ideal-dependent. This givestwo possibilities.

Case 1. There is a finite F ⊆ X such that f ≤∑

F . It follows that (∑

F )λ = 1for all λ ∈M . Choose g ∈ X ′\F . Then g ≤

∑F , contradiction.

Case 2. There exist a finite F ⊆ X and a g ∈ X\F such that g ≤ f +∑

F . Forany λ ∈M we have gλ · −uλ · −(

∑F )λ = 0, and hence gλ · −(

∑F )λ = 0. Choose

h ∈ X ′\(F ∪ {g}). Then g ≤ h+∑

F , contradiction. �

For the next result, we define reg to be the class of all regular cardinals.

Theorem 13.15. Suppose that K is an infinite set of infinite cardinals such that|K| ≤ min(K). Then there is a BA A such that K ⊆ sspect(A) and sspect(A) ∩reg ⊆ K.

Proof. Let μ = min(K), let λ map μ onto K, and let A =∏w

α<μ Fr(λα). We claimthat A is as desired.

The first inclusion in the proposition holds by Proposition 13.13. Now supposethat κ ∈ (sspect(A) ∩ reg)\K. Let X be maximal ideal-independent of size κ. LetL = {α < μ : κ < λα}, and let M = μ\L. For each α ∈ L let uα be a free generatorof Fr(λα) not in 〈{xα : x ∈ X}〉.

(1) M �= ∅.

For, suppose that M = ∅. Then κ < λα for each α < μ, and so κ < min(K) = μ.

(2) Some x ∈ X has type II.

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For, suppose not. Now⋃

x∈X supp(x) has size less than min(K) = μ, so we canchoose α < μ not in this union. Let y take the value uα at α and 0 elsewhere.Clearly y /∈ X and X ∪ {y} is still ideal-independent, contradiction. So (2) holds.

We take x as in (2). Now let yα = uα for all i ∈ supp(x), and yα = 0otherwise. Then y /∈ X , so X ∪ {y} is ideal-dependent.Case 1. y ≤

∑F for some finite F ⊆ X . We may assume that x ∈ F . Now for

α ∈ supp(x) we have uα ≤ (∑

F )α, and hence (∑

F )α = 1. Since x ∈ F , it followsthat

∑F = 1, contradiction.

Case 2. There exist a finite F ⊆ X and a g ∈ X\F such that g ≤ y +∑

F . Itfollows easily that g ≤

∑F , contradiction.

This proves (1).

In particular, κ > μ. Since κ is regular, it follows that there is a G ∈ [μ]<ω

such that X ′ def= {x ∈ X : supp(x) = G} has size κ. Now |

∏α∈G∩M Fr(λα)| < κ,

so there is a q ∈∏

α∈G∩M Fr(λα) such that Ydef= {x ∈ X ′ : x � (G ∩M) = q} has

size κ. Note also that G ∩ L �= ∅, as otherwise G = G ∩M and hence |X ′| < κ,contradiction. Let Y ′ = {y ∈ Y : y has type I} and Y ′′ = Y \Y ′. Now define

yα =

⎧⎨⎩uα if i ∈ G ∩ L,

qα if i ∈ G ∩M ,

0 otherwise.

Since G ∩ L �= ∅, we have y /∈ X . So X ∪ {y} is ideal-dependent. This gives twocases.

Case 1. There is a finite F ⊆ X such that y ≤∑

F . Then (∑

F ) � (G ∩ L) = 1.If |Y ′| = κ, choose g ∈ Y ′ such that g /∈ F . Then g ≤

∑F , contradiction. If

|Y ′′| = κ, choose distinct g, h ∈ Y ′′\F . Then g ≤∑

F + h, contradiction.

Case 2. There exist a finite F ⊆ X and a g ∈ X\F such that g ≤ y +∑

F . Theng � (G ∩ L) ≤ (

∑F ) � (G ∩ L) and also g � (μ\G) ≤ (

∑F ) � (μ\G). Choose

h ∈ Y \(F ∪ {g}). Then g ≤ h+∑

F , contradiction. �

Corollary 13.16. If K is an infinite set of regular cardinals and |K| ≤ min(K),then there is a BA A such that sspect(A) ∩ reg = K. �

Problem 124. Is the assumption |K| ≤ min(K) in Proposition 13.16 necessary?

Problem 125. How can Corollary 13.16 be extended to singular cardinals in K?

Proposition 13.17. r(A) ≤ smm(A) for any BA A.

Proof. Suppose that X is maximal ideal-independent. Let

Y = X ∪ {−∑

F : F ∈ [X ]<ω} ∪ {b · −∑

F : b /∈ F, F ∪ {b} ∈ [X ]<ω}.

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Clearly the members of Y are nonzero. We claim that Y is weakly dense in A.For, suppose that a ∈ A\X . Then X ∪ {a} is no longer ideal independent, so wehave two cases.

Case 1. a ≤∑

F for some F ∈ [X ]<ω. Then −∑

F ≤ −a, as desired.Case 2. There exist a finite subset F of X and a b ∈ X\F such that b ≤

∑F + a.

Then b · −∑

F ≤ a, as desired. �

Lemma 13.18. Suppose that Fr(ω1) is a subalgebra of A such that Idef= 〈{xα : ξ <

ω1〉idA is a maximal ideal of A, where 〈xα : α < ω1〉 is a system of free generators

of Fr(ω1). Also suppose that Xdef= {xα : α < ω1} is maximal ideal independent in

A. Suppose that Y is an infinite partition of unity in A, with |Y | ≤ ω1.

Then A has an extension B such that X is still maximal ideal independentin B, 〈{xα : ξ < ω1〉idB is a maximal ideal of B, and Y is not a partition of unityin B.

Proof. The main part of the proof is in establishing the following claim.

Claim. There is a b ∈ X such that b �≤∑

F for all F ∈ [Y ]<ω.

We suppose that the claim does not hold. Thus

(1) For every b ∈ X there is a finite Fb ⊆ Y such that b ≤∑

Fb.

Then

(2) y ∈ I for all y ∈ Y .

For, suppose that y ∈ Y and −y ∈ I. Thus there is a finite G ⊆ X such that−y ≤

∑G. Then 1 = y +

∑G = y +

∑b∈G Fb, contradiction. So (2) holds.

Thus for every y ∈ I we can choose a finite Gy ⊆ X such that y ≤ Gy.

Now if |Y | < ω1, choose xα not in the support of any element of⋃

y∈Y Gy.Now xα ≤

∑Fxα ≤

∑y∈Fxα

Gy, contradiction. Thus |Y | = ω1.

Let Γ ∈ [ω1]ω1 be such that 〈Fxα : α ∈ Γ〉 is a Δ-system, say with kernel H .

Then if α and β are distinct elements of Γ we have

(3) xα · xβ ≤(∑

Fxα

)·(∑

Fxβ

)≤∑y∈H

Gy .

Choose distinct α, β ∈ Γ so that xα, xβ /∈⋃

y∈GGy . Then (3) gives a contradiction.

This proves the claim.

Choose b ∈ X in accordance with the claim. Let A(x) be a free extension ofA, and let J be the ideal of A(x) generated by

{y · x : y ∈ Y } ∪ {x · −b}.

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Clearly A ∩ J = {0}. If x ∈ J , then we can write

x ≤ y1 · x+ · · ·+ ym · x+ x · −b.

Mapping x to 1 yields b ≤ y1 + · · ·+ ym, contradicting the choice of b.

Thus A(x)/J is as desired. �

Theorem 13.19. There is a BA A such that smm(A) = ω1 and a(A) = ω2.

Proof. This is obtained by an obvious iteration from Lemma 13.18. �

For A = Fr(κ) with κ an uncountable cardinal, we have a(A) = lengthmm(A) =ω < κ = smm(A). (This is easy to see.)

Let κ be uncountable, and let A be atomless an κ+-saturated (in the model-theoretic sense). Let B = κAw, the weak κ-power of A. Then the following is easyto check:

r(B) = i(B) = ω; smm(B) ≤ κ; a(B) = p(B) = κ; tow(B) ≥ κ+.

Proposition 13.20 (C. Bruns). p(A) ≤ smm(A) for any infinite BA A.

Proof. Let X be maximal ideal independent of size smm(A). Then∑

X = 1, sinceif a ∈ A+ and a · x = 0 for all x ∈ X it is clear that X ∪ {a} is still idealindependent. Also, if F ∈ [X ]<ω we have

∑F < 1; for if

∑F = 1 and x ∈ X\F

then x ≤∑

F , contradiction. Hence the inequality of the proposition follows. �

Problem 126. Is there an atomless BA A such that smm(A) < i(A)?

Now we show that it is consistent to have smm(P(ω)/fin) less than 2ω. The ar-gument is a modification of exercises (A12), (A13) in chapter VIII of Kunen [80];the essential argument is given in the following lemma.

Lemma 13.21. Let M be a c.t.m. of ZFC. Suppose that κ is an infinite cardinaland 〈ai : i < κ〉 is a system of infinite subsets of ω such that 〈[ai] : i < κ〉 is idealindependent, where [x] denotes the equivalence class of x modulo the ideal fin ofP(ω). Then there is a generic extension M [G] of M using a ccc partial order suchthat in M [G] there is a d ⊆ ω with the following two properties:

(i) 〈[ai] : i < κ〉�〈[ω\d]〉 is ideal independent.

(ii) If x ∈ (P(ω) ∩M)\({ai : i < κ} ∪ {ω\d}), then 〈[ai] : i < κ〉�〈[ω\d], [x]〉 isnot ideal independent.

Proof. Let I be the ideal of P(ω)/fin generated by {[ai] · [aj ] : i < j < κ}, andlet A be the quotient algebra (P(ω)/fin)/I. Let f be the natural homomorphismof P(ω) onto A. Note that f(ai) �= 0 for all i < κ, by ideal independence. LetB be the subalgebra of A generated by {f(ai) : i < κ}. Thus B is an atomicBA with {f(ai) : i < κ} its set of atoms. By Sikorski’s extension theorem, leth : P(ω)→ B extend f , where B is the completion of B.

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Let P = {(b, y) : b ∈ ker(h) and y ∈ [ω]<ω}. We define (b, y) ≤ (b′, y′) iffb ⊇ b′, y ⊇ y′, and y ∩ b′ ⊆ y′. Clearly this gives a ccc partial order of P . Let Gbe any P -generic filter over M , and let d =

⋃(b,y)∈G y.

(1) If R is a finite subset of κ and i ∈ κ\R, then ai ∩⋂

j∈R(ω\aj) ∩ d is infinite.

In fact, let R and i be as in the hypothesis of (1). For any natural number n let

En ={(b, y) ∈ P : ∃m > n

[m ∈ ai ∩

⋂j∈R

(ω\aj) ∩ y]}

.

Clearly it suffices to show that each such set En is dense in P . Suppose that(b, y) ∈ P . Then (ai ∩

⋂j∈R(ω\aj))\b is infinite. For, if it is a finite set c, then

ai ⊆⋃

j∈Raj ∪ b ∪ c,

and upon applying h we would get h(ai) ≤∑

j∈R h(aj), which is clearly impossible.Thus the indicated set is infinite. We can hence choosem in it with m > n. Clearly(b, y ∪ {m}) ∈ En and (b, y ∪ {m}) ≤ (b, y), proving (1).

(2) If R is a finite subset of κ, then ω\(d ∪

⋃i∈R ai

)is infinite.

In fact, let R be a finite subset of κ. For any natural number n let

Fn ={(b, y) ∈ P : ∃m > n

[m ∈ b\

(y ∪

⋃i∈R

ai

)]}.

We claim that Fn is dense in P . For, suppose that (b, y) ∈ P . Then the setω\(y ∪

⋃i∈R ai

)is infinite. For, if it is a finite set c, then we get

ω = c ∪ y ∪⋃

i∈Rai,

and applying h we get 1 =∑

i∈R h(ai), which is clearly impossible. Choose (b, y) ∈Fn ∩ G, and then choose m > n such that m ∈ b\

(y ∪

⋃i∈R ai

). We claim that

m /∈ d; by the arbitrariness of n, this will prove (2). Suppose that m ∈ d. Choose(c, z) ∈ G with m ∈ z. Then choose (d, w) ∈ G with (d, w) ≤ (b, y), (c, z). Thusm ∈ w since m ∈ z. Also, m ∈ b\y. This contradicts (d, w) ≤ (b, y). Hence (2)holds.

(3) 〈[ai] : i < κ〉�〈[ω\d]〉 is ideal independent.

For, suppose not. There are two possibilities.

Case 1. There are a finite subset R of κ and an i ∈ κ\R such that [ai] ≤ [ω\d] +∑j∈R[aj ]. This contradicts (1).

Case 2. There is a finite subset R of κ such that [ω\d] ≤∑

i∈R[ai]. This contra-dicts (2).

Thus (3) holds.

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(4) If b ∈ ker(h), then b ∩ d is finite.

In fact, clearly {(c, y) ∈ P : b ⊆ c} is dense in P , so we can choose (c, y) ∈ Gsuch that b ⊆ c. Then b ∩ d ⊆ y (as desired). For, suppose that m ∈ b∩ d. Choose(e, z) ∈ G such that m ∈ z. Then choose (r, w) ∈ G such that (r, w) ≤ (e, z), (c, y).Then m ∈ w ∩ c ⊆ y.

(5) If x ∈ (P(ω) ∩M)\({ai : i < κ} ∪ {ω\d}), then 〈[ai] : i < κ〉�〈[ω\d], [x]〉 isnot ideal independent.

To prove this, we consider two cases. First, if x ∈ ker(h), then [x] ≤ [ω\d] by(4), as desired. Second, if x /∈ ker(h), choose i < κ such that h(ai) ≤ h(x). Soai\x ∈ ker(h), and so by (4) we get [ai] ≤ [x] + [ω\d], as desired. �Theorem 13.22. It is consistent with 2ω > ω1 that smm(P(ω)/fin) = ω1.

Proof. Start with a c.t.m. M of ZFC+2ω > ω1. Iterate the construction of Lemma13.21 ω1 times, obtaining a generic filter G overM . Then M [G] is as desired, usingLemma 5.14 of Chapter VIII of Kunen [80]. �

Problem 127. What is the exact place of smm(P(ω)/fin) among the other contin-uum cardinals?

Turning to the relationships of spread to our other functions, we first list out thethings already proved: cH+(A) = s(A) by Theorem 3.30; Depthh+(A) = s(A) in

Theorem 4.26; t(A) ≤ s(A) in Theorem 5.18; and |A| ≤ 2s(A) in Theorem 13.10.Now we prove the important fact that π(A) ≤ s(A) · (t(A))+ for any infinite BAA, following Todorcevic [90a]. The result he proves is somewhat stronger, and tostate it we need two definitions. First, dd(A) is the least cardinality of a collectionof discrete subsets of Ult(A) whose union is dense in Ult(A). Second, f′(A) isthe smallest cardinal such that A does not have a free sequence of length f ′(A).Thus if t(A) is attained in the free sequence sense, then f′(A) = (t(A))+, whilet(A) = f′(A) otherwise.

Theorem 13.23. dd(A) ≤ f′(A), and π(A) ≤ s(A) · f′(A) for any infinite BA A.

Proof. Choose 〈aα : α < β〉 in accordance with Theorem 12.15. Let E = {aα : α <β}. Now we define Dγ ⊆ Ult(A) and Sγ ⊆ E for γ < f′(A) by induction. Supposethat they have been defined for all γ < δ. Let

Sδ = {x ∈ E : S(x) ∩Dγ = ∅ for all γ < δ}.

Then we let Dδ be a maximal subset of⋃

x∈SδS(x) having at most one element in

common with each S(x) for x ∈ Sδ. This finishes the construction. Note that Dδ

is discrete: if F ∈ Dδ, choose x ∈ Sδ such that F ∈ S(x). Then Dδ ∩ S(x) = {F}by the defining property of Dδ.

For each F ∈ Dδ choose g(F ) ∈ Sδ such that F ∈ S(g(F )). Then g is a

one-one function, and its range is S′γ

def= {x ∈ Sδ : Dδ ∩ S(x) �= ∅}. Now

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(1) S(x) ∩⋃

δ<f′(A) Dδ �= ∅ for all x ∈ E.

For, suppose that (1) fails for a certain x ∈ E. Then

(2) S(x) ⊆⋃

y∈S′δS(y) for all δ < f′(A).

For, suppose that (2) fails for a certain δ < f′(A). Choose F ∈ S(x)\⋃

y∈S′δS(y).

Now ifG ∈ Dδ∩S(y) with y ∈ Sδ, then y ∈ S′δ and so F /∈ S(y). Also,Dδ∩S(x) = ∅

by (1) failing. So Dδ ∪ {F} has at most one element in common with each S(y)for y ∈ Sδ, and F /∈ Dδ, contradicting the maximality of Dδ. Thus (2) holds.

Now if γ < δ < f′(A), then S′γ∩S′

δ = ∅, since if z ∈ S′γ∩S′

δ, then S(z)∩Dγ = ∅because z ∈ S′

δ ⊆ Sδ, but S(z) ∩Dγ �= ∅ by the definition of S′γ , contradiction. It

follows now that for any F ∈ S(x) we have F ∈ S(y) for a collection of f′(A) y’s,and this contradicts the condition of Theorem 12.15. So we have proved (1).

By (1) we have dd(A) ≤ f′(A), since E is dense in A. Moreover, E ⊆⋃δ<f′(A) S

′δ, since by (1), if x ∈ E then there is a δ < f′(A) such that S(x)∩Dδ �= ∅,

and hence x ∈ S′δ. Hence

(3)⋃

δ<f′(A) S′δ is dense in A.

Now |Dδ| = |S′δ| for all δ < f′(A). Hence

π(A) ≤∣∣∣∣⋃δ<f′(A)

S′δ

∣∣∣∣ ≤ s(A) · f′(A),

as desired. �

Note that t(A) can be much smaller than s(A), for example in the finite-cofinitealgebra on an infinite cardinal κ. Also note that, obviously, c(A) ≤ s(A); andthe difference is big in, e.g., free algebras. We have s(A) > Length(A) for A afree algebra; s(A) < Length(A) for A the interval algebra on the reals. Also,s(A) > π(A) for A = P(κ). The interval algebra of a Suslin line provides anexample of a BA A with s(A) = ω and d(A) > ω. In fact, clearly s(A) = c(A) forA an interval algebra, by the retractiveness of interval algebras.

An example with s(A) = ω < d(A) cannot be given in ZFC; this follows fromthe following rather deep results. Juhasz [71] showed that under the assumptionof MA+¬CH, for every compact Hausdorff space X , if s(X) = ω then hL(X) = ω.Todorcevic [83] showed that it is consistent with MA+¬CH that for every regularspace X , if s(X) = ω then hL(X) = ω. Hence it is consistent that for every BAA, if s(A) = ω then hL(A) = ω = hd(A).

Recall also Theorem 8.11, which says that Irr(A) ≤ s(A⊕A).

Bounded versions of spread can be defined as follows. For m a positive integer,a subset X of A is called m-ideal-independent if for all distinct x0, . . . , xm ∈ Xwe have x0 �≤ x1 + · · · + xm. Then we let sm(A) = sup{|X | : X ⊆ A and X ism-ideal-independent}. For these functions see Roslanowski, Shelah [98].

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14 Character

First note that we can define χ(A) as a sup; namely, for any ultrafilter F on A letχ(F ) = min{|X | : X is a set of generators of F} – then χ(A) = sup{χ(F ) : F isan ultrafilter on A}.

Character can increase in going from an algebra to a subalgebra. To constructan example of this sort, first notice that if A is the finite-cofinite algebra on aninfinite cardinal κ, then χ(A) = κ, as is easily seen. The algebra that we wantis the Alexsandroff duplicate of the free algebra A on κ free generators, where κis any infinite cardinal. By Theorem 14.6 below, χ(Dup(A)) = χ(A) = κ. Thefinite-cofinite algebra on Ult(A) is isomorphic to a subalgebra of Dup(A), and bythe previous remark it has character 2κ.

An atomless example of this sort can be obtained as follows. Take A andDup(A) as above, and let M be the set of atoms of Dup(A). Then there is anisomorphism from Dup(A) onto an algebra C of subsets of M , containing all

singletons {x} with x ∈M . For each x ∈M let Ax = Fr(ω). Then Ddef=∏C

x∈M Ax

has character κ by the above and Proposition 14.4 below. Now∏w

x∈M Ax is anatomless subalgebra of D, and its character is 2κ.

Below we show that χ(A ⊕ B) = max(χ(A), χ(B)). Hence if A ≤free B, thenχ(A) ≤ χ(B), and the difference can be arbitrarily large. Thus there are BAs A,Bwith χ(B) arbitrarily greater than χ(A), and with A ≤u B, A ≤proj B, A ≤rc B,A ≤σ B, and A ≤reg B.

The two examples given in detail above show that it is possible to haveC ≤π D with χ(C) > χ(D); so this applies to ≤reg also. On the other hand, wehave χ(A) < χ(B) for A = Fr(ω) and B = A; see the discussion of u below.

It is clear by topological duality that χ(A×B) = sup(χ(A), χ(B)). Then byPropositions 2.29 and 14.1 it follows that χ(B) ≤ χ(A) whenever A ≤s B. On theother hand, if F is an ultrafilter on B, with B = A(x), and xδ ∈ F with δ ∈ 2,and if X generates F ∩ A with |X | = χ(F ∩ A), then X ∪ {xδ} generates F ; itfollows that χ(A) ≤ χ(B). So χ(A) = χ(B) if A ≤s B. Hence also χ(A) = χ(B) ifA ≤m B.

It is clear from Propositions 2.52 and 14.18 that there exist A,B with A ≤mg

B and χ(B) arbitrarily larger than χ(A).

, ,OI 10.1007/978-3- - - _ ,


014


Basel 215

421

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422 Chapter 14. Character

The facts mentioned answer many natural questions concerning the specialsubalgebra notions; but other questions remain.

Problem 128. Completely describe the possibilities for character with respect to thevarious subalgebra notions.

Proposition 14.1. If A is a homomorphic image of B, then χ(A) ≤ χ(B).

Proof. Let f be a homomorphism from B onto A; if F ∈ Ult(A), then f−1[F ] ∈Ult(B), and if we choose X ⊆ f−1[F ] with |X | ≤ χ(B) such that X generatesf−1[F ], then f [X ] generates F . �

For a weak product we have χ(∏w

i∈I Ai) = max(|I|, supi∈Iχ(Ai)). To show this,it suffices to show that χ(F ) = |I| for the “new”ultrafilter F . This ultrafilter isdefined as follows. For each subset M of I, let xM be the element of

∏i∈I Ai such

that xM (i) = 1 if i ∈ M and xM (i) = 0 for i /∈ M . Then F is the set of ally ∈

∏wi∈I Ai such that xM ≤ y for some cofinite subset M of I. So, it is clear that

χ(F ) ≤ |I|. If X is a set of generators for F with |X | < |I|, we may assume thatX is closed under ·, and then there is a y ∈ X such that y ⊆ xM for infinitelymany cofinite subsets M of I; this is clearly impossible.

As usual, weak products enable us to discuss the attainment problem. Anyinfinite BA has a non-principal ultrafilter, and hence if A has character ω, then itis attained. Next, if κ is a singular cardinal, then we can construct a BA A withχ(A) = κ not attained. Namely, let 〈μξ : ξ < cf(κ)〉 be an increasing sequence ofinfinite cardinals with sup κ. For each ξ < cf(κ) let Aξ be the free BA on μξ freegenerators; thus χ(Aξ) = μξ. By the above remarks on weak products,

∏wξ<cfκ Aξ

has character κ not attained.

For regular limit cardinals, there is an old theorem of Parovicenko [67] whichcharacterizes attainment; this solves Problem 37 of Monk [90]. We give this the-orem here; it requires a definition. If X is a topological space and x ∈ X , thecharacter of x is min{|O| : O is a neighborhood base of x}. Clearly for any BA Aand any F ∈ Ult(A), the character of F as a point in Ult(A) coincides with χ(F )as defined above.

Theorem 14.2. Let κ be a limit cardinal. Then the following conditions are equiv-alent:

(i) κ is weakly compact.

(ii) For every compact Hausdorff space X of size at least κ, X has a point ofcharacter at least κ.

(iii) For every BA A, if χ(A) = κ, then A has an ultrafilter with character κ.

Proof. (i)⇒(ii): Assume (i), and suppose that X is a compact Hausdorff spacewith |X | ≥ κ such that X has no point of character ≥ κ.

(1) We may assume that |X | = κ.

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Chapter 14. Character 423

For, let Y ∈ [X ]κ. We claim that |Y | = κ. Since the character of a point of Yis clearly still < κ, (1) follows from the claim. By Theorem 9.49, it suffices toshow that if y ∈ Y , then there is a Z ∈ [Y ]<κ such that y ∈ Z, since thenY =

⋃Z∈[Y ]<κ Z, and by κ being strongly inaccessible |Y | = κ follows. Let U be

an open neighborhood base for y of size < κ. For every U ∈ U choose zU ∈ U ∩Y .

Clearly Zdef= {zU : u ∈ U } is as desired.

So we now assume that |X | = κ. Let X = {xα : α < κ}. For each α < κ, letUα be an open neighborhood base for xα of size < κ. Set Fα = {F : X\F ∈ Uα}.So Fα is a collection of closed sets,

⋃Fα = X\{xα}, and |Fα| < κ. Let T be

the collection of all functions f such that there is an α < κ such that dmn f = α,∀β < α(fβ ∈ Fβ), and

⋂β<α fβ �= 0. Thus T is a tree under ⊆.

(2) ∀α < δ∃f ∈ T (dmn f = α).

For,

0 �= X\{xβ : β < α} =⋂β<α

⋃Fα

=⋃

f ∈ ∏

β<αFβ

⋂β<α

fβ ,

so there is an f ∈∏

β<α Fβ such that⋂

β<α fβ �= 0. Thus f is as desired in (2).

(3) Every level of T has size < κ.

This is true since κ is strongly inaccessible, so that∣∣∣∏β<α Fβ

∣∣∣ < κ for everyα < κ.

Now by the weak compactness of κ, let f with domain κ be a branch throughT . By compactness,

⋂α<κ fα �= 0. But if y ∈

⋂α<κ fα, then y �= xα for all α < κ,

contradiction.

(ii)⇒(iii): obvious.

(iii)⇒(i): Assume that κ is not weakly compact; we want to find a BA A withcharacter κ not attained. By the comments on attainment above, we may assumethat κ is regular. Let L be a linear order of size κ such that neither κ nor κ∗ embedsin L. By replacing points of L by ordinals less than κ we may assume that eachordinal less than κ embeds in L. More precisely, write L = {aα : α < κ} with norepetitions. let M = {(β, aα) : β ≤ α, α < κ}, ordered anti-lexicographically. Weshow that M has no increasing chain of type κ. For, suppose that 〈(βξ, xξ) : ξ < κ〉is such a chain. Since κ is regular, the set {xξ : ξ < κ} has κ elements, and hencedetermines a chain in L of type κ, contradiction. Similarly, M has no decreasingchain of type κ. Let A = Intalg(M). Then by the description of character forinterval algebras below, A has the desired properties. �

To treat arbitrary direct products, note that obviously t(A) ≤ χ(A); henceInd(A) ≤ χ(A), and so clearly χ(

∏i∈I Ai) ≥ max(2|I|, supi∈Iχ(Ai)). Shelah and

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Peterson independently observed that strict inequality is possible. This solvesProblem 38 in Monk [90]. The same example used for spread works here: let κbe the first limit cardinal > 2ω, let A be the finite-cofinite algebra on κ, and con-sider ωA. Recall that character does not increase when going to a homomorphicimage.

We now discuss ultraproducts, giving some results of Douglas Peterson. Char-acter is a sup-min function, and so Theorems 6.5–6.9 apply. Then a proof simi-lar to that of Theorem 4.18 shows that if GCH holds then χ

(∏i∈I Ai/F

)≥∣∣∏

i∈I χAi/F∣∣ for F regular, and Donder’s theorem says that under V = L the

regularity assumption can be removed. The following result of Shelah, Spinas [00],part of Corollary 2.7, is relevant here, solving Problem 48 of Monk [96]:

There is a model in which there exist cardinals κ, μ, a system 〈Bi : i < κ〉 of BAs,and an ultrafilter D on κ such that∣∣∣∏

i<κχ(Bi)/D

∣∣∣ = μ++ and χ(∏

i<κBi/D

)≤ μ+.

On the other hand, it is easy to give an example in which > holds. Let κ be anyinfinite cardinal such that κω = κ. As we saw above, the Alexsandroff duplicate Aof a free BA on κ generators has character κ and cellularity 2κ. By Theorem 11.8

this implies that χ(A∗i) = κ for all i ∈ ω\1 while χ(∏

i∈ω\1 A∗i/F

)= 2κ for any

nonprincipal ultrafilter F on ω\1. (See below for the character of free products.)

The assumption κω = κ implies that∣∣∣∏i∈ω\1 χ(A

∗i)/F∣∣∣ = κ.

Proposition 14.3. If 〈Ai : i ∈ I〉 is a system of BAs each with at least four elements,then χ(⊕i∈IAi) = max(|I|, supi∈Iχ(Ai)).

Proof. For ≥, first let j ∈ I and let F ∈ Ult(Aj). Let G be any ultrafilter on⊕i∈IAi which includes F . Suppose that X ⊆ G generates G. Without loss ofgenerality, each member of X is a product of elements from distinct Ai’s. Thenit is clear that X ∩ Aj ⊆ F and X ∩ Aj generates F . So χ(F ) ≤ |X |. It followsthat χ(F ) ≤ χ(G) ≤ χ(⊕i∈IAi). Hence χ(Aj) ≤ χ(⊕i∈IAi). It is clear thatχH ≥ |I| for any ultrafilter H on ⊕i∈IAi. Altogether, this proves ≥. For ≤, forany ultrafilter G on ⊕i∈IAi, and for each i ∈ I let Xi ⊆ G ∩ Ai generate G ∩ Ai,with |Xi| = χ(G∩Ai). Clearly the set of all finite products of elements of

⋃i∈I Xi

generates G, as desired. �

Problem 129. Describe the behaviour of character under unions.

Proposition 14.4. Assume the framework for moderate products, where B and eachAi are infinite BAs. Then χ(

∏Bi∈I Ai) = max(χ(B), supi∈I χ(Ai)).

Proof. It suffices to look at the characterization of ultrafilters on∏B

i∈I Ai given in

Proposition 1.6(x) and show that the character of an ultrafilter U on∏B

i∈I Ai isthe same as the character of the associated ultrafilter on B or on Ai.

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Case 1. There exist an i ∈ I and an ultrafilter V on Ai such that U = {h(b, F, a) :(b, F, a) is normal, and i ∈ b or (i /∈ b and i ∈ F and ai ∈ V )}. Let X ⊆ V generateV , with |X | = χ(V ). Wlog X is closed under ∩. We claim that

{h({i}, ∅, ∅)} ∪ {h(∅, {i}, {(i, x)}) : x ∈ X}

is a subset of U which generates it. Clearly it is a subset of U . Now suppose that(b, F, a) is normal and h(b, F, a) ∈ U . If i ∈ b, then h({i}, ∅, ∅) ⊆ h(b, F, a), asdesired. Suppose that i /∈ b, i ∈ F , and ai ∈ V . Choose x ∈ X such that x ⊆ ai.Then {h(∅, {i}, {(i, x)}) ⊆ h(b, F, a), as desired. This proves our claim. It followsthat χ(U) ≤ χ(V ).

Now suppose that Y ⊆ U generates U and |Y | = χ(U). Wlog Y is closedunder ∩. We claim that

Zdef= {x : h(∅, {i}, {(i, x)}) ∈ Y }

is a subset of V which generates it. Clearly it is a subset of V . Now suppose thaty ∈ V . Then h(∅, {i}, {(i, y)}) ∈ U , so we can choose h(b, F, a) ∈ Y such thath(b, F, a) ⊆ h(∅, {i}, {(i, y)}). It follows that b = ∅, F = {i}, and ai ⊆ y. Thusai ∈ Z, as desired. This proves our claim. Hence χ(U) = χ(V ).

Case 2. There is a nonprincipal ultrafilter W on B such that U = {h(b, F, a) :(b, F, a) is normal and b ∈ W}. Let X ⊆W generate W , with |X | = χ(W ). WlogX is closed under ∩. Then we claim that

{h(b, ∅, ∅) : b ∈ X}

generates U . For, let (b, F, a) be normal with h(b, F, a) ∈ U . Choose v ∈ X suchthat v ≤ b. Then h(v, ∅, ∅) ⊆ h(b, F, a), as desired. This shows that χ(U) ≤ χ(W ).

Now let Y ⊆ U generate U , with |Y | = χ(U). Wlog Y is closed under ∩. We

claim that Zdef= {b : there is a normal (b, F, a) with h(b, F, a) ∈ Y } generates W .

Clearly it is a subset of W . Now suppose that c ∈ W . Then h(c, ∅, ∅) ∈ U , so thereis a normal (b, F, a) such that h(b, F, a) ∈ Y and h(b, F, a) ⊆ h(c, ∅, ∅). Then b ∈ Zand b ⊆ c, as desired.

This proves that χ(U) = χ(W ). �Proposition 14.5. Let C be the one-point gluing of BAs A,B with respect to ultra-filters F,G on A,B respectively. Then χ(C) = max(χ(A), χ(B)).

Proof. Let H be an ultrafilter on A×B.

(1) χ(H ∩C) ≤ max(χ(H), χ(F ), χ(G)).

In fact, wlog (1, 0) ∈ H . Let X ⊆ H generate H with |X | = χ(H). Let Z ⊆ Ggenerate G with |Z| = χ(G). Let

W = {(a, w) : a ∈ F,w ∈ Z, ∃b[(a, b) ∈ X ]} ∪ {(a, 0) : a /∈ F and ∃b[(a, b) ∈ X ]}.

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Suppose that (u, v) ∈ H ∩ C. Choose (a, b) ∈ X such that (a, b) ≤ (u, v).

Case 1. a ∈ F . Then u ∈ F and v ∈ G. Choose w ∈ Z such that w ≤ v. Then(a, w) ∈ W and (a, w) ≤ (u, v). Obviously (a, w) ∈ C. Since (a, b) ∈ H and(1, 0) ∈ H , we have (a, 0) ∈ H and hence (a, w) ∈ H .

Case 2. a /∈ F . Clearly (a, 0) ∈ H ∩ C and (a, 0) ≤ (u, v).

From (1) it follows that χ(C) ≤ χ(A×B).

Still with H an ultrafilter on A × B and (1, 0) ∈ H , choose X ⊆ H ∩ Cgenerating H ∩ C, with |X | = χ(H ∩ C). Let

W = X ∪ {(a, 0) : a ∈ F and ∃c[(a, c) ∈ X ]}.

Suppose that (a, b) ∈ H .

Case 1. a ∈ F and b ∈ G. Then there exists (c, d) ∈ X such that (c, d) ≤ (a, b).

Case 2. a ∈ F and b /∈ G. Then (a, 0) ∈W and (a, 0) ≤ (a, b).

Case 3. a /∈ F and b ∈ G. Then (a, 0) ∈W and (a, 0) ≤ (a, b).

Case 4. a /∈ F and b /∈ G. Then there exists (c, d) ∈ X such that (c, d) ≤ (a, b).

Clearly W ⊆ H .

So we have χ(H) ≤ χ(H ∩ C). Hence χ(A×B) ≤ χ(C). �Proposition 14.6. χ(A) = χ(Dup(A)) for any infinite BA A.

Proof. Let κ = χ(A). Take any nonprincipal ultrafilter F on Dup(A). By Propo-sition 1.19 there is an ultrafilter G on A such that F = {(a,X) : a ∈ G andS(A)�X is finite}. Let M generate G, M closed under ·, |M | = χ(G). We claimthat {(a,S(a)\{G} : a ∈M} generates F . For, suppose that (a, Y ) is any elementof F ; thus S(a)\Y is finite. For each H ∈ S(a)\(Y ∪{G}) choose aH ∈ F\H . Thenlet b = a ·

∏H∈S(a)\(Y ∪{G}) aH . Thus b ∈ F . Choose c ∈M such that c ≤ b. Then

(c,S(c)\{G}) ≤ (a, Y ), as desired.

Conversely, suppose that G is a nonprincipal ultrafilter on A. Let F ={(a,X) ∈ Dup(A) : a ∈ G}. So F is a nonprincipal ultrafilter on Dup(A). Let Mbe a subset of F of size at most πχ(Dup(A)) which is dense in F . For each a ∈ Gchoose (ba, Xa) ∈M such that (ba, Xa) ≤ (a,S(a)). We claim that {ba : a ∈ G} isdense in G. Suppose that a ∈ G. Then ba ≤ a, as desired. �Proposition 14.7. χ(A) ≤ χ(Exp(A)) for any infinite BA A.

Proof. Let G be an ultrafilter on A. Then the set {V(S(a)) : a ∈ G} has fip byLemma 1.22(iii). Let F be the filter on Exp(A) generated by this set. Then F is anultrafilter. To prove this, note that if a /∈ G then −V(S(a)) ∈ F . In fact, −a ∈ G,so S(−a) ∈ F . By Lemma 1.22(v) we then get −V(S(a)) ∈ F . Now by Proposition1.21 it follows that F is an ultrafilter.

Now suppose that X ⊆ G is dense in G, with X closed under · and |X | =χ(G). Then clearly {V(S(a)) : a ∈ G} is dense in F . So χ(F ) ≤ χ(G).

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Conversely, suppose that Y ⊆ F is dense in F with |Y | = χ(F ). We mayassume that each member of Y has the form V(S(a)). Then {a : V(S(a)) ∈ Y } isdense in G. So χ(G) ≤ χ(F ). �

Note that the example at the end of Chapter 1 gives an infinite BA A such thatχ(A) < χ(Exp(A)).

Lemma 14.8. χ(A) ≤ t(Exp (A)); in fact, every closed set in Ult(A) has a neigh-borhood basis with at most t(Exp (A)) elements.

Proof. Let F be a closed subset of Ult(A). Then

F ∈ {U : U clopen, F ⊆ U}.

For, suppose that F ∈ V (U0, . . . , Um−1) with each Ui clopen. Then U0 ∪ · · · ∪Um−1 ∈ V (U0, . . . , Um−1), as desired.

It follows that there is a subset O of {U : U clopen, F ⊆ U} such that|O| ≤ t(Exp (A) and F ∈ O. Then O is the desired neighborhood base for F . For,suppose F ⊆ W with W clopen. Then F ∈ V (W ), so there is a U ∈ O such thatU ∈ V (W ). So U ⊆W , as desired. �

Problem 130. Describe χ(Exp(A)).

We turn to the derived functions for character. By a remark above, we haveχH+(A) = χ(A) for any infinite BA A. Koszmider [99] has shown that it is consis-tent to have CardH−(A) = ω2 = 2ω while χH−(A) = ω1. This solves Problem 39in Monk [90].

We also do not know the status of χS+(A); we observed above that it canhappen that χS+(A) > χ(A). Clearly χS−(A) = ω for any infinite BA A. Thetopological version of character is this: for any space X and any x ∈ X , χ(x,X)is the minimum of the cardinalities of neighborhood bases for x in X , and χX =sup{χ(x,X) : x ∈ X}. Clearly then χh+(A) = χ(A), and χh−(A) = 1 for anyinfinite BA A, since A has an infinite discrete subspace. The function χinf is ofsome interest; recall from the introduction that χinf(A) = inf{χ(F ) : F ∈ UltA}for any infinite BA A. This function has value 1 if A has an atom. Hence it isnatural to modify it slightly; and in view of the special case P(ω)/fin, we useu(A) instead of χinf(A). Thus by definition,

u(A) = min{χ(F ) : F is a nonprincipal ultrafilter on A}.

There is a classical result of concerning u:

Theorem 14.9. 2u(A) ≤ |UltA| for any atomless BA A.

Proof. For brevity set κ = u(A). It clearly suffices to construct a function fmapping <κ2 into A such that

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(1) For each s ∈ <κ2, the set {f(s � α) : α ≤ dom s} has the finite intersectionproperty;

(2) f(s�0) · f(s�1) = 0 for each s ∈<κ 2.

Suppose s ∈ <κ2 and f(s � α) has been defined for all α ∈ dom s. By the inductionhypothesis, {f(s � α) : α ∈ dom s} has the finite intersection property; since thisset has < κ elements, it does not generate an ultrafilter, and hence there is a a ∈ Asuch that both a and −a fail to be in the filter generated by it. Hence if we setf(s�0) = a and f(s�1) = −a we extend our function f so that (1) and (2) willhold. This completes the proof. �

Kevin Selker has obtained the following results concerning u and moderate prod-ucts:

(1) Let 〈Ai : i ∈ I〉 be a system of infinite BAs, and B an infinite BA, satisfying

the hypotheses for the formation of a moderate product, and let C =∏B

i∈I Ai.Then uspect(C) = uspect(B) ∪

⋃i∈I uspect(Ai).

(2) If K is an infinite set of infinite cardinals, and either |K| < sup(K) or |K| =sup(K) is singular, then there is an atomless BA A such that uspect(A) = K.

It is easy to see that p ≤ u, and πχinf ≤ u is obvious. There is a BA A with u < i;see McKenzie, Monk[04]; and there is a BA A with u < smm; see Monk [08].

If A is complete, then ω < u(A). In fact, suppose that F is a nonprincipalultrafilter on A and X is a countable set which generates F . Let X = {xi : i < ω}.Without loss of generality, 1 = x0 > x1 > · · · . Then

(1)∏

X = 0.

In fact, suppose that∏

X �= 0. If∏

X ∈ F , then there is an i ∈ ω such thatxi ≤

∏F < xi+1, contradiction. If −

∏X ∈ F , then there is an i ∈ ω such that

xi ≤= −∏

X , and then∏

X = xi ·∏

X = 0, contradiction. So (1) holds.

Let ai = xi · −xi+1 for all i ∈ ω. Now take any nonzero b ∈ A. Let i bemaximum such that b ≤ xi. Then b ·ai = b ·xi ·−xi+1 = b ·−xi+1 �= 0. This showsthat

∑i∈ω ai = 1. If

∑i∈ω a2i ∈ F , choose j ∈ ω such that xj ≤

∑i∈ω a2i. We may

assume that j = 2k+1 for some k. Then a2k+1 = x2k+1 ·−x2k+2 ≤∑

i∈ω a2i ∈ F ,contradiction. A similar contradiction is reached if

∑i∈ω a2i+1 ∈ F .

There is an atomic BA A such that f(A) < u(A). Let A = P(ω). Thus by theabove, u(A) > ω. We claim that A = ω. For each n ∈ ω let an = ω\(n+ 1). Thusa0 �= ω and 〈an : n ∈ ω〉 is strictly decreasing; hence by Proposition 12.16 it isa free sequence. To show that it is maximal we apply Proposition 12.17. Supposethat x ⊆ ω. If x is finite, choose n ∈ ω with x ⊆ n+1. Thus an ∩x = 0, as desiredin Proposition 12.18(i). Suppose that x is infinite. Take any m ∈ x with m > 0.Then am−1 ∩ (ω\am) ∩ (ω\x) = 0, as desired in Proposition 12.17(ii).

Kevin Selker has shown under CH that there is an atomless BA A such thatf(A) < u(A).

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Problem 131. Can one prove in ZFC that there is an atomless BA A such thatf(A) < u(A)?

In Kunen [80], Exercise (A10) in Chapter VIII, a model is constructed withu(P(ω)/fin) < 2ω; that model also has f(P(ω)/fin) = u(P(ω)/fin).

We also have smm(A) = ω for A = P(ω). In fact, the set of all singletons is clearlymaximal ideal independent. Kevin Selker has shown under CH that there is anatomless BA A such that smm(A) < u(A). (Theorem 2.10 in Monk [08], assertingthat there is a BA satisfying this inequality, has a faulty proof.)

Problem 132. Can one prove in ZFC that there is an atomless BA A such thatsmm(A) < u(A)?

For the cardinal function alt, see Chapter 9.

Theorem 14.10. Let B be a homomorphic image of A, and G a nonprincipal ul-trafilter on B. Then alt(A) ≤ χ(G).

Proof. Let f be a homomorphism from A onto B, and let 〈bα : α < χ(G)〉 be anenumeration of a set of generators of G. For each α < χ(G) choose aα ∈ A suchthat f(aα) = bα. Now clearly for each α < χ(G) the set {aξ : ξ < α} ∪ {x ∈ A :f(x) = 1} has fip, so we can let Fα be the filter generated by this set. Clearly

(1) f−1[G] is an ultrafilter on A.

(2) Fα ⊆ f−1[G] for each α < χ(G).

In fact, if M is a finite subset of α, x ∈ A, f(x) = 1, and x ·∏

ξ∈M aξ ≤ c, then∏ξ∈M bξ ≤ f(c), hence f(c) ∈ G and c ∈ f−1[G]. So (2) holds.

Clearly

(3) α < β implies that Fα ⊆ Fβ .

(4) If α < χ(G), then Fα �= f−1[G].

In fact, choose c ∈ G such that c is not in the filter generated by {bξ : ξ < α}, andchoose d ∈ A such that f(d) = c. Then d ∈ f−1[G] but clearly d /∈ Fα. Next,

(5)⋃

α<χ(G) Fα = f−1[G].

In fact, take any c ∈ f−1[G], and choose a finite M ⊆ χ(G) such that∏

ξ∈M bξ ≤f(c). Then f(−

∏ξ∈M aξ + c) = 1, so that −

∏ξ∈M aξ + c ∈ Fα, where M ⊆ α <

χ(G). Now∏

ξ∈M aξ · (−∏

ξ∈M aξ + c) ≤ c, so c ∈ Fα, proving (5).

From (1)–(5) we see that a subsequence of 〈Fα : α < χ(G)〉 is strictly in-creasing with union f−1[G]. �Corollary 14.11. alt(A) ≤ χH−A ≤ cardH−(A) for any infinite BA A. �Corollary 14.12. alt(A) = ω iff p-alt(A) = ω iff CardH−(A) = ω iff χH−(A) = ω,for any infinite BA A.

Proof. See Proposition 9.4. �

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We turn to the relationships of character with our previously treated functions.Obviously t(A) ≤ χ(A) for any infinite BA A; the difference can be big – forexample for the finite-cofinite algebra on an infinite cardinal κ. Now consider thepossibility that χ(A) > s(A). By the comment at the end of the last Chapter,plus the fact that χ(A) ≤ hL(A) (easy, and proved in Chapter 15), it is consistentthat s(A) = ω implies χ(A) = ω. The Kunen line, constructed in Chapter 8, hasuncountable character but countable spread; it was constructed using CH. To showthat it has uncountable character, assume the notation in the construction.

(∗) If U ∈ τω1 is open and {ξ < ω1 : xξ ∈ U} is uncountable, then U is notcompact in τω1 .

To see this, note that for each ξ ∈ U we have ξ ∈ U∩P(ξ) ∈ τω1 . Thus {U∩P(ξ) :ξ ∈ U} is an open cover of U , and it clearly has no finite subcover, proving (∗).

Now suppose that the new point y of Y has a countable base {{y} ∪ Um :m ∈ ω}; we may assume that these are clopen. Hence Y \({y} ∪ Um) = X\Um isopen in Y and hence in X ; and it is also compact. It follows from (∗) that X\Um

is countable. Choose α ∈⋂

m∈ω Um. Then Y \{α} is an open neighborhood of ynot containing any of the basis sets {y} ∪ Um, contradiction.

Under ♦ there is a BA A such that s(A) < χ(A); see Chapter 17. The followingproblem is open, however.

Problem 133. Can one construct in ZFC a BA A such that s(A) < χ(A)?

This was Problem 49 in Monk [96]. It is equivalent to the problem whether onecan construct in ZFC a BA A such that s(A) < hL(A); see the end of Chapter 15.

An example of a BA A with c(A) > χ(A) is provided by the Alexsandroffduplicate of the free algebra on κ free generators, as described at the beginningof this chapter. The interval algebra on R gives an example of an algebra A withLength(A) > χ(A).

Now we turn to Arhangelskiı’s theorem that |Ult(A)| ≤ 2χ(A) for any infinite BAA. We need some lemmas.

Lemma 14.13. If Y ⊆ UltA and F ⊆⋃Y has the finite intersection property, then

there is an ultrafilter G such that F ⊆ G ⊆⋃Y .

Proof. Let G be maximal among the filters H such that F ⊆ H ⊆⋃Y . Suppose

that G is not an ultrafilter; say a,−a /∈ G. Then 〈G ∪ {a}〉fi �⊆⋃Y . Say b ∈ G

and b · a /∈⋃Y . Similarly obtain c ∈ G such that c · −a /∈

⋃Y . Choose H ∈ Y

such that b · c ∈ H . Then b · c · a /∈ H and b · c · −a /∈ H , contradiction. �

Note that for any subset Y of Ult(A), the closure of Y is {F ∈ Ult(A) : F ⊆⋃Y }.

Lemma 14.14. If Z ⊆ Ult(A) is closed, then Ult(A)\Z is the union of at most

max{ω, |Z|, supG∈Z

χ(G)}

clopen sets.

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Proof. For every G ∈ Z let {aGα : α < χ(G)} be a set of generators of G, closedunder multiplication. Let

B = {aG0α0

+ · · ·+ aGn−1αn−1

: n ∈ ω, G0, . . . , Gn−1 ∈ Z, αi < χ(Gi) for all i < n},

and let C = {y : −y ∈ B ∩⋂Z}. We claim that

Ult(A)\Z =⋃y∈C

S(y),

which gives the desired result. ⊇ is clear. Now suppose that F ∈ Ult(A)\Z. Forevery G ∈ Z choose bG ∈ F\G; say aGα(G) ≤ −bG. Then

(∗) There exist an integer n ∈ ω and elements G0, . . . , Gn−1 ∈ Z with the property

that aG0

α(G0)+ · · ·+ a

Gn−1

α(Gn−1)∈ H for all H ∈ Z.

Otherwise, Ldef= {−aG0

α(G0)· . . . · −aGn−1

α(Gn−1): n ∈ ω, G0, . . . , Gn−1 ∈ Z} has the

finite intersection property and is contained in⋃Z. Hence by Lemma 14.13, there

is an ultrafilter K such that L ⊆ K ⊆⋃Z. Hence K ∈ Z and −aKα(K) ∈ K,

contradiction.

We choose n ∈ ω and G0, . . . , Gn−1 ∈ Z as in (∗). Let y be the element

−aG0

α(G0)· . . . · −aGn−1

α(Gn−1). Then y ∈ C and F ∈ S(y), as desired. �

Lemma 14.15. If Y ⊆ Ult(A) and |Y | ≤ χ(A), then |Y | ≤ 2χ(A).

Proof. For every ultrafilter G on A let {aGα : α < χ(A)} be a set of generators ofG, and set f(G) = {{F ∈ Y : aGα ∈ F} : α < χ(A)}. Thus f(G) ∈ [P(Y )]≤χ(A).Hence it is enough to show that f � Y is one-one. Suppose that G and H aredistinct ultrafilters on A such that G,H ∈ Y . Say aGα ∈ G\H , and choose aHβ ≤−aGα . Suppose that f(G) = f(H); then there is a γ < χ(A) such that {F ∈ Y :aGγ ∈ F} = {F ∈ Y : aHβ ∈ F}. Then aGα ·aGγ ∈ G; say then aGα ·aGγ ∈ F ∈ Y . Then

aHβ ∈ F , aGα ∈ F , and −aGα ∈ F , contradiction. �

Now we are ready for the proof of Arhangelskiı’s theorem.

Theorem 14.16. |Ult(A)| ≤ 2χ(A) for any infinite BA A.

Proof. Suppose that 2χ(A) < |Ult(A)|. Fix an ultrafilter F on A. For each f ∈⋃α<(χ(A))+

α(2χ(A)) we define a closed set Xf ⊆ Ult(A) and a Gf ∈ Ult(A). Let

X0 = Ult(A) and G0 = F . For dmn(f) limit let Xf =⋂

α<dmn f Xf�α, and ifXf �= 0 choose Gf ∈ Xf , and otherwise let Gf = F . Now suppose that dmn(f)is a successor ordinal α + 1. Let g = f � α, and set Yg = {Gg�β : β ≤ α}.Thus |Yg| ≤ χ(A), so by Lemma 14.15, |Y g| ≤ 2χ(A), and so by Lemma 14.14we can let 〈agβ : β < 2χ(A)〉 be such that Ult(A)\Y g =

⋃β<2χ(A) S(agβ) and set

Xf = Xg ∩ S(agf(α)

). Again let Gf ∈ Xf if Xf �= 0, and Gf = F otherwise. This

finishes the construction.

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Now choose

H ∈ Ult(A)\⋃⎧⎨⎩{Gf�β : β ≤ dmn(f)} : f ∈

⋃α<(χ(A))+

α(2χ(A))

⎫⎬⎭ .

Now we define f mapping (χ(A))+ into 2χ(A) by induction. Suppose that f(β)has been defined for all β < α. Now H /∈ {Gf�β : β ≤ α}, so there is a γ < 2χ(A)

such that H ∈ X(f�α)∪{(α,γ)}; set f(α) = γ. Thus H ∈ Xf�α for all α < (χ(A))+.We claim that 〈Gf�(β+1) : β < (χ(A))+〉 is a free sequence, which contradictst(A) ≤ χ(A). Let α < (χ(A))+ and suppose that

K ∈ {Gf�(β+1) : β < α} ∩ {Gf�(β+1) : α ≤ β < (χ(A))+}.

Then K ∈ {Gf�β : β ≤ α}, so K ∈ Ult(A)\Xf�(α+1), and this set is open, sothere is a β ≥ α such that Gf�(β+1) ∈ Ult(A)\Xf�(α+1) ⊆ Ult(A)\Xf�(β+1),contradiction. �

Problem 134. Characterize χHs(A).

Problem 135. Characterize χSs(A).

Problem 136. Characterize χSr(A).

Problem 137. Characterize χHr(A).

We describe character for interval algebras; this corrects the description given inMonk [96].

Proposition 14.17. Let L be a linear order with first element 0, let F be an ul-

trafilter on Adef= Intalg(L), and let T = {a ∈ L : [0, a) ∈ F} be the end segment

determined by F . Note that 0 /∈ T . Then:

(i) If T = [b,∞) and b has an immediate predecessor c, then F is the principalultrafilter determined by {c}.

(ii) If T = [b,∞) and b does not have an immediate predecessor, then the char-acter of F is the cofinality of [0, b).

(iii) If T = (b,∞) �= ∅ and b does not have an immediate successor, then thecharacter of F is the coinitiality of [b,∞).

(iv) If T is empty and L has a greatest element b, then F is the principal ultrafiltergenerated by {b}

(v) If T is empty and L does not have a greatest element, then the character ofF is the cofinality of L.

(vi) If T is nonempty and there is no glb for T in L, then the character of F isthe maximum of the left and right characters of the gap (L\T, T ). �

Theorem 14.18. p(A) = u(A) for every interval algebra A.

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Proof. As remarked above, p(A) ≤ u(A) for any atomless BA A. Hence it sufficesto show that u(A) ≤ p(A) for any atomless interval algebra. So, let L be a denselinear order with first element 0, and let A = Intalg(L). Suppose that X ⊆ A,∑

X = 1,∑

F �= 1 for every finite subset F of X , and |X | = p(A). Wlog eachmember of X has the form [u, v) with u < v ≤ ∞. Clearly {−x : x ∈ X} has fip,so this set is included in an ultrafilter U . It suffices now to show that the characterof U is ≤ |X |. Let

M = {w ∈ L : ∃v([v, w) ∈ X and [w,∞) ∈ U)};N = {v ∈ L : ∃w([v, w) ∈ X and [0, v) ∈ U}.

Let T = {a ∈ L : [0, a) ∈ U} be the end segment determined by U . We mayassume that U is nonprincipal, and hence (ii), (iii), (v), or (vi) of Proposition14.17 holds.

Case 1. (ii) holds; so T = [b,∞), where b does not have an immediate predecessor.Thus the character of U is the cofinality of [0, b). It suffices now to show that M iscofinal in [0, b). Suppose not; take c < b such that M ∩ (c, b) = ∅, and also choosed with c < d < b. Choose [x, y) ∈ X such that [c, d)∩ [x, y) �= ∅. Then max(c, x) <min(d, y), so x < b and hence [x,∞) ∈ U . Hence −[x, y)∩ [x,∞) = [y,∞) ∈ U . Soy ∈M and c < y < b, contradiction.

Case 2. (iii) holds; so T = (b,∞), where b does not have an immediate successor.The character of U is the coinitiality of (b,∞). Then N is coinitial in (b,∞). For,suppose not. Take b < d with N ∩ (b, d) = ∅, and take c with b < c < d. Choose[x, y) ∈ X such that [c, d) ∩ [x, y) �= ∅. Then max(c, x) < min(d, y), so b < c < y,hence [0, y) ∈ U . So −[x, y) ∩ [0, y) = [0, x) ∈ U , and so b < x < d and x ∈ N ,contradiction.

Case 3. (v) holds. So T is empty and L does not have a greatest element. Hencethe character of U is the cofinality of L. We claim that M is cofinal in L. Supposenot. Then there is a c ∈ L such that M ∩ [c,∞) = ∅. Take d with c < d. Choose[x, y) ∈ X such that [x, y) ∩ [c, d) �= ∅. Then y ∈M ∩ (c,∞), contradiction.

Case 4. (vi) holds, so that T is nonempty and there is no glb for T in L; we alsoassume that the left character of the gap (L\T, T ) is≥ the right character. Then weclaim thatM is cofinal in L\T . Suppose not; say c ∈ L\T andM∩[c,∞)∩(L\T )) =∅. Choose d > c with d ∈ L\T . Let [x, y) ∈ X be such that [c, d)∩ [x, y) �= ∅. Thenx < d, so [x,∞) ∈ U ; hence [y,∞) = −[x, y)∩ [x,∞) ∈ U . So y ∈ L\T , c < y, andy ∈M , contradiction.

Case 5. (vi) holds, so that T is nonempty and there is no glb for T in L; we alsoassume that the right character of the gap (L\T, T ) is ≥ the left character. Thenwe claim that N is coinitial in T . Suppose not. Say d ∈ L and N ∩ [0, d) ∩ L = ∅.Choose c ∈ L such that c < d. Then choose [x, y) ∈ X such that [x, y)∩ [c, d) �= ∅.Then c < y, so y ∈ L and [0, y) ∈ U . Hence [0, x) = −[x, y) ∩ [0, y) ∈ U . Thenx ∈ d and x ∈ N , contradiction. �

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For tree algebras we have the following theorem (Brenner [82]):

Theorem 14.19. Let T be a tree, and set A = Treealg(T ). Then χ(A) = sup{|{x : xis an immediate successor of C}|, cf(C) : C an initial chain of T }.

Proof. Let κ = sup{|{x : x is an immediate successor of C}|, cf(C) : C an initialchain of T }. Let F be an ultrafilter of A and let C be the associated initial chain.We shall show that F has a set of generators of size at most κ; this will prove ≤.

If C has a maximal element x, then {(T ↑ x)\⋃

y∈F (T ↑ y) : F a finite set ofimmediate successors of x} generates F , as desired.

Suppose that C has no maximal element. Let 〈xα : α < cf(C)〉 be an increas-ing cofinal sequence in C. Then the set

{(T ↑ xα)\⋃y∈F

(T ↑ y) : α < cf(C), F a finite set of immediate successors of C}

generates F , as desired.

Conversely, Let C be an initial chain of T and F the associated ultrafilter.Suppose X generates F and |X | < max(|{x : x is an immediate successor ofC}|, cf(C)). Without loss of generality each element x ∈ X has the form (T ↑tx)\

⋃y∈Fx

(T ↑ y), where Fx is a finite set of successors of tx. If C has a maximalelement z, then |X | < {u : u is an immediate successor of z}, so there is animmediate successor u of z such that z /∈

⋃x∈X Fx. Then (T ↑ z)\(T ↑ u) ∈ F , but

no element of X is ≤ it, contradiction. Suppose that C has no maximal element.If |X | < cf(C), choose z ∈ C such that tx < z for all x ∈ X ; then T ↑ z is in Fbut has no element of X less than it. Finally, if cfC ≤ |X |, then |X | < {u : u isan immediate successor of C}, and we get a contradiction as above. �

χ(A) for pseudo-tree algebras has been characterized by Jennifer Brown.

Problem 138. Characterize χ(A) for A a complete BA.

Concerning superatomic algebras we have the following result (due to Monk):

Theorem 14.20. χ(A) = |A| for A superatomic.

Proof. Let λ be a regular cardinal ≤ |A|; we want to show that χ(A) ≥ λ. LetR be a complete system of representatives of atoms (of all levels) of A. We mayassume that the top atoms of R form a finite partition of unity. Recall the notionof rank of an element of A: this is the least α such that a ∈ Iα+1. An elementa ∈ A is big if |{x ∈ R : x ≤∗ a}| ≥ λ. (u ≤∗ v means that if β is the rank ofu then u/Iβ ≤ v/Iβ). Note that at least one of the top atoms of R is big. Let αbe minimum such that there is a big a ∈ R of rank α, and fix such an a, and theultrafilter F associated with it. Note that if x is any element of rank less than α,then x is small. In fact, otherwise suppose that x is of smallest rank β < α suchthat x is big. Then x/Iβ = c1/Iβ + · · · + cm/Iβ for certain c1, . . . , cm ∈ R such

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that each ci/Iβ is an atom. Thus x · −c1 · . . . · −cm ∈ Iβ , and hence it is small. Ify ∈ R and y ≤∗ x, say y has rank γ. Then

y/Iγ ≤ x/Iγ = (x · c1)/Iγ + · · ·+ (x · cm)/Iγ + (x · −c1 · . . . · −cm)/Iγ

≤ c1/Iγ + · · ·+ cm/Iγ + (x · −c1 · . . . · −cm)/Iγ ,

and so y/Iγ is ≤ one of these last summands. This means that one of c1, . . . , cm, x ·−c1 · . . . · −cm is big, contradiction.

We claim that χ(F ) ≥ λ (as desired). In fact, suppose that X ⊆ F generatesF , where |X | < λ. If b ∈ R, b ≤∗ a, and b has rank less than α, then −b ∈ F , andhence there is an xb ∈ X such that xb ≤ −b. Since λ is regular there is an S ⊆ Rwith |S| ≥ λ and an x ∈ X such that each member of S has rank less than α andx ≤ −b for each b ∈ S. Thus b ≤ −x, and b ≤∗ a · −x. So, a · −x is big.

Case 1. a/Iα ≤ x/Iα. Then a · −x ∈ Iα, so a · −x is small, contradiction.

Case 2. a/Iα · x/Iα = 0. Then a · x ∈ Iα, and hence −a+ −x ∈ F and −x ∈ F ,contradiction. �

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15 Hereditary Lindelof Degree

Recall from just before Proposition 13.4 the definition of a right-separated se-quence in a topological space. Let α be an ordinal. A sequence 〈aξ : ξ < α〉 ofelements of a BA A is right separated provided that the following two conditionshold:

(i) If F ∈ [α]<ω, then∏

ξ∈F −aξ �= 0.

(ii) If F ∪ {η} ∈ [α]<ω with ∀ξ ∈ F [ξ < η], then aη ·∏

ξ∈F −aξ �= 0.

Note that if α is a limit ordinal, then (i) follows from (ii).

Theorem 15.1. For any infinite BA A, hL(A) is equal to each of the followingcardinals:

sup{κ:there is an ideal not generated by less than κ elements};sup{κ:there is a strictly increasing sequence of ideals of length κ};sup{κ:there is a strictly increasing sequence of filters of length κ};sup{κ:there is a strictly increasing sequence of open sets of length κ};sup{κ:there is a strictly decreasing sequence of closed sets of length κ};sup{κ:there is a right-separated sequence in Ult(A) of length κ};sup{κ:there is a right-separated sequence in A of length κ;

min{κ:every open cover of a subspace of Ult(A) has a subcover of size ≤ κ}.

Proof. Nine cardinals are mentioned; let them be denoted by κ0, . . . , κ8 in theirorder of mention (starting with hL itself). First we take care of easy relations:

κ2 = κ3 since, if I is an ideal then Ifdef= {a ∈ A : −a ∈ I} is a filter, and I ⊂ J iff

If ⊂ Jf ; similarly, going from filters to ideals. So κ2 = κ3 follows. Next, κ2 ≤ κ4.For, if I is an ideal, let Iu =

⋃a∈I S(a). Then Iu is open, and I ⊂ J implies

Iu ⊂ Ju. (If a ∈ J\I, then S(a) ⊆ Ju, of course, but S(a) �⊆ Iu, since otherwisecompactness of S(a) would easily yield a ∈ I.) This shows κ2 ≤ κ4. It is clear thatκ4 = κ5, by taking complements. κ4 ≤ κ6: If 〈Uα : α < κ〉 is a strictly increasingsequence of open sets, for every α < κ choose xα ∈ Uα+1\Uα. Clearly 〈xα : α < κ〉is right-separated. κ6 = κ7: First suppose that 〈Fα : α < κ〉 is right-separated inUlt(A). For all α < κ choose aα ∈ A such that Fα ∈ S(aα)∩{Fβ : β < κ} ⊆ {Fβ :β ≤ α}. We claim that 〈aα : α < κ〉 is right-separated in A. To see this, supposethat Γ is a finite subset of κ, α < κ, and β < α for all β ∈ Γ. Thus aα ∈ Fα.

, ,OI 10.1007/978-3- - - _ ,


014


Basel 216

437

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438 Chapter 15. Hereditary Lindelof Degree

If β ∈ Γ, then aβ /∈ Fα by the choice of aβ , and hence −aβ ∈ Fα. Therefore theelement aα ·

∏β∈Γ−aβ is in Fα, and hence it must be non-zero, as desired. Second,

suppose that 〈aα : α < κ〉 is right-separated in A. Then for each α < κ the set{aα}∪{−aβ : β < α} has the finite intersection property, and hence is included inan ultrafilter Fα. It is easy to check that 〈Fα : α < κ〉 is right-separated in Ult(A),as desired. κ8 ≤ κ0: for any subspace X of Ult(A), any cover of X has a subcoverof power ≤ L(X) ≤ κ0, so κ8 ≤ κ0.

It remains only to prove that κ0 ≤ κ1, κ1 ≤ κ2, and κ7 ≤ κ8. For the firstone, suppose that X ⊆ Ult(A) and O is an open cover of X with no subcover ofpower λ; we construct an ideal not generated by λ or fewer elements. Let

I = 〈{a ∈ A : ∃U ∈ O(S(a) ∩X ⊆ U)}〉Id.

Suppose that I is generated by J , where |J | ≤ λ. For every a ∈ J there is a finitesubset Pa of O such that S(a)∩X ⊆

⋃Pa. Let O′ =

⋃a∈J Pa. We claim that O′

covers X , which is the desired contradiction. Indeed, let x ∈ X . Say x ∈ U ∈ O.Say x ∈ S(a) ∩X ⊆ U . Choose a finite subset F of J such that a ≤

∑F . Then

x ∈ S(a) ∩X ⊆⋃

b∈F

⋃Pb, as desired.

Next, κ1 ≤ κ2: suppose that I is an ideal not generated by fewer than λelements. Then it is easy to construct a sequence 〈aα : α < λ〉 of elements of Isuch that aα /∈ 〈{aβ : β < α}〉Id for all α < λ. Thus 〈〈{aβ : β < α}〉Id : α < λ〉 isa strictly increasing sequence of ideals, as desired.

For κ7 ≤ κ8, suppose that λ is a regular cardinal ≤ κ7 and 〈xα : α < λ〉is right separated. Thus for each α < λ we can choose an open set Uα such thatUα ∩{xξ : ξ < λ} = {xξ : ξ < α+1}. Then {Uα : α < λ} is a cover of {xξ : ξ < λ}with no subcover of size < λ. Hence λ < κ8, and this shows that κ7 ≤ κ8. �

In Theorem 15.1, eight of the nine equivalents involve sups, and thus give rise toattainment problems. The proof of the theorem shows the following: attainment isthe same for κ2 and κ3, for κ4 and κ5, and for κ6 and κ7; moreover, attainment inthe sense κ2 implies attainment in the sense κ4, attainment in the sense κ4 impliesattainment in the sense κ6, and attainment in the sense κ1 implies attainmentin the sense κ2. It is also easy to see that attainment in the sense κ4 impliesattainment in the sense κ2. In fact, if 〈Uα : α < κ3〉 is an increasing sequence ofopen sets, for each α < κ3 let Iα = {a : Sa ⊆ Uα}. Clearly Iα is an ideal. To showproperness, pick F ∈ Uα+1\Uα. Say F ∈ Sa ⊆ Uα+1. Thus a ∈ Iα+1\Iα. Andattainment in the sense κ6 implies attainment in the sense κ4. In fact, supposethat 〈Fα : α < κ〉 is right separated. For each α < κ choose aα ∈ Fα such thatSaα ∩ {Fβ : β > α} = 0, and let Uα =

⋃β<α Saβ . Note that Fα ∈ Uα+1\Uα, as

desired. Also note that if hLA is regular, then attainment in the right-separatedsense implies attainment in the defined sense.

Thus we have seen that there are only three versions of the definition of hLthat might lead to different attainment properties: hL as defined, in the ideal-generated sense, and in the right-separated sense, where we know only that at-tainment in the ideal-generated sense implies attainment in the right-separated

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Chapter 15. Hereditary Lindelof Degree 439

sense. In Roslanowski Shelah [01a] further results are obtained. By Hypothesis 1.1and Theorem 1.4 of this paper we have:

Suppose that 〈χi : i < cf(λ)〉 is a strictly increasing sequence of infinite cardinalssuch that (2cf(λ))+ < χ0 and λ = supi<cf(λ) χi. Then for any BA A such thathL(A) = λ, all (three) versions of hL are attained.

On the other hand, in 3.7 they show that it is consistent to have a BA with hLattained in the right-separated sense but not in the ideal-generation sense.

These results leave the following problem open; this is a version of Problem50 on Monk [96].

Problem 139. Completely describe the relations between these attainment possibil-ities for hL.

There are two important positive results concerning attainment.

Theorem 15.2. If hL(A) is singular strong limit, then it is attained in the ideal-generated sense.

Proof. Since hL(A) ≤ |A| ≤ |Ult(A)|, we can apply Theorem 13.2 to get a discretesubset X of Ult(A) of size hL(A). Let I = {a ∈ A : ∃F ∈ [X ]<ω[S(a) ∩X = F ]}.Clearly I is an ideal in A; we claim that it is not generated by fewer than hL(A)elements. In fact, suppose that Y ∈ [I]<hl(A) generates I. For each y ∈ Y chooseFy ∈]X ]<ω such that S(y)∩X = Fy. Take any x ∈ X\

⋃y∈Y Fy, and then choose

a ∈ A such that S(a) ∩X = {x}. Thus a ∈ I. Let Z be a finite subset of Y suchthat a ≤

∑Z. Then

x ∈ X ∩⋃z∈Z

S(z) =⋃z∈Z

(X ∩ S(z)) =⋃z∈Z

Fz ,

contradiction. �

The second attainment result is given in Proposition 13.4: hL(A) is attained forhL(A) singular of countable cofinality.

We turn to algebraic operations.

If A is a subalgebra or homomorphic image of B, then hL(A) ≤ hL(B). Forspecial subalgebras, note that one can have A ≤free B with hL(B) arbitrarilygreater than hL(A); hence the same applies to ≤u, ≤proj, ≤rc, ≤σ, and ≤reg. IfA ≤π B, then |B| ≤ 2|A|; so hL cannot jump arbitrarily. But for A = intalg(R) wehave hL(A) = ω (see the remarks on interval algebras at the end of this chapter),hL(A) = 2ω by the Balcar, Franek theorem since obviously Ind(B) ≤ hL(B) forany BA B, and A ≤π A. For A ≤s B we clearly have hL(A) = hL(B); one can useProposition 2.29 to see this. Clearly one can have A ≤mg B with hL(A) arbitrarilysmaller than hL(B).

Proposition 15.3. hL(A×B) = max(hL(A), hL(B)).

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Proof. We use the ideal-generation equivalent of the definition of hL. If I is anideal in A not generated by less than κ elements, then {(a, b) ∈ A×B : a ∈ I} isan ideal in A × B not generated by less than κ elements. Hence by symmetry ≥holds. Conversely, let I be an ideal in A×B not generated by less than κ elements.Define

J = {a ∈ A : (a, 0) ∈ I};K = {b ∈ B : (0, b) ∈ I}.

Then J and K are ideals in A,B respectively. If J is generated by X with |X | < κand K is generated by Y with |Y | < κ, let Z = {(a, 0 :)a ∈ X} ∪ {(0, b) : b ∈ Y }.Then Z generates I, since if (a, b) ∈ I, then (a, 0), (0, b) ∈ I, hence a ∈ J andb ∈ I; choosing U ∈ [X ]<ω and V ∈ [Y ]<ω such that a ≤

∑U and b ≤

∑V , let

W = {(a, 0) : a ∈ U} ∪ {(0, b) : b ∈ V }; then W ⊆ Z and (a, b) ≤∑

W . From this≤ follows. �

Proposition 15.4. For I infinite, hL(∏w

i∈I Ai

)= max (|I|, supi∈I hL(Ai)).


i∈I Ai. By Proposition 15.3 we have hL(Aj) ≤ hL(B)for each j ∈ I. Now let K = {a ∈ B : {i ∈ I : ai �= 0} is finite}; we claim that Kis not generated by less than |I| elements. For, suppose that X ∈ [K]<ω. Choosei ∈ I\

⋃a∈X{j ∈ I : aj �= 0}. Then the element of B which is 1 at i and 0 elsewhere

is not in the ideal generated by X . This proves ≥ in the proposition.

For ≤, let κ be the right side of our equation, and let K be an ideal on B;we want to show that K is generated by a set with at most κ elements. For eachi ∈ I let Li = {a ∈ Ai : ∃b ∈ K[bi = a]}. Then Li is an ideal in Ai, so it can begenerated by a set Xi with at most κ elements. Let Y =

⋃i∈I Li. Thus |Y | ≤ κ.

Now we have two cases.

Case 1. Every element of K is of type I. Then we claim that Y generates K. For,suppose that b ∈ K. For each i ∈ I there is a finite Yi ⊆ Xi such that bi ≤

∑Yi.

Let J = {i ∈ I : bi �= 0}. Then b ≤∑

i∈J Yi, as desired.

Case 2. K has an element c of type II. Let Jdef= {i ∈ I : ci �= 1}. Then for any

element b ∈ K, b ≤ c+∑

i∈I\J Yi with Yi as in Case 1, as desired. �

Note that Ind(A) ≤ hL(A), using the equivalent concerning algebraic right-sepa-rated sequences. Hence it is clear that hL

(∏i∈I Ai

)≥ max(2|I|, supi∈IhL(Ai)).

Strict inequality is possible, as was noticed by Shelah and Peterson independently,solving Problem 44 of Monk [90]. Again the example used for spread applies here.

Concerning ultraproducts, note that hL is an order-independence function, andhence Theorem 12.6 holds. By the proof of Theorem 12.7 it follows that underGCH we have hL

(∏i∈I Ai/F

)≥∣∣∏

i∈I hL(Ai)/F∣∣ for F regular, and Donder’s

theorem says that under V = L the regularity assumption can be removed. Theexample of Laver for depth works for hL also: it is consistent to have a situation

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where hL(∏

i∈I Ai/F)>∣∣∏

i∈I hL(Ai)/F∣∣ for F regular; see also Roslanowski.

Shelah [98] for another consistent example.

Also note the following results. In Shelah [99], part of 15.13, the following isproved:

If D is a uniform ultrafilter on κ then there is a class of cardinals χ with χκ = χsuch that there are BAs Bi for i < κ such that:

(i) hL(Bi) ≤ χ, hence |∏

i<κ hL(Bi)/D| ≤ χ;

(ii) hL(∏

i<κ Bi/D) = χ+.

This solves Problem 51 of Monk [96]. In Shelah, Spinas [00], part of Corollary 2.4says the following:

There is a model in which there exist cardinals κ, μ, a system 〈Bi : i < κ〉 of BAs,and an ultrafilter D on κ such that∣∣∣∏

i<κhL(Bi)/D

∣∣∣ = μ++ and hL(∏

i<κBi/D

)≤ μ+.


Next come free products:

Theorem 15.5. If 〈Ai : i ∈ I〉 is a system of BAs each with at least 4 elements,then

max(|I|, supi∈I

hL(Ai)) ≤ hL (⊕i∈IAi) ≤ max(|I|, 2supi∈I s(Ai)

).

Proof. The first inequality is easy; use the right-separated equivalent to see thathL(Aj) ≤ hL (⊕i∈IAi) for each j ∈ I, and use the fact that Ind (⊕i∈IAi) ≤hL (⊕i∈IAi) to see that |I| ≤ hL (⊕i∈IAi).

For the second,

hL (⊕i∈IAi) ≤ |⊕i∈IAi| = |I| · supi∈I|Ai|

≤ |I| · supi∈I

2s(Ai)

≤ max(|I|, 2supi∈I s(Ai)

). �

The inequalities in Theorem 15.5 are sharp, in the sense that all possibilities canoccur. Thus both are equalities if I = ω1 and each Ai is a four-element algebra.The first is an equality and the second not for I = ω1 and each Ai the free BA onω1 free generators. The first is a strict inequality and the second an equality forA⊕ A, where A = IntalgR. Finally, both inequalities are strict for A ⊕A, whereA is the tree algebra on a Suslin tree in which each element has infinitely manyimmediate successors, and ¬CH holds. It is clear that ¬CH is needed to get anexample where both inequalities are strict.

Proposition 15.6. hL(∏B

i∈I Ai

)= max(hL(B), |I|, supi∈I hL(Ai)).

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Proof. Let C =∏B

i∈I Ai. Then hL(B) ≤ hL(C) since B is isomorphic to a subalge-bra of C. Also, hL(Ai) ≤ hL(C) by Proposition 15.3. By Proposition 15.4 we havehL(∏w

i∈I Ai

)≥ |I|; since

∏wi∈I Ai is a subalgebra of C, it follows that |I| ≤ hL(C).

So we have proved ≥.Now let κ be the indicated maximum. We show that an arbitrary ideal K

on C can be generated by ≤ κ elements; this will prove ≤. Let L = {b ∈ B :h(b, 0, 0) ∈ K}. Then L is an ideal on B, and hence can be generated by a set Xof size ≤ κ. For each i ∈ I let

Mi = {a ∈ Ai : h(∅, {i}, {(i, a)}) ∈ K}.

Then Mi is an ideal of Ai. Say Mi is generated by a set Yi of size at most κ.Now let

Ndef= {h(b, 0, 0) : b ∈ X} ∪ {h(0, {i}, {(i, a) : a ∈ Yi})}.

Clearly |N | ≤ κ. We claim that it generates K. Let h(b, F, a) be any element ofK. Then also h(b, 0, 0) ∈ K, and so there is a finite P ⊆ X such that b ≤

∑P .

Also, for each i ∈ F we have h(0, {i}, {(i, ai)}) ∈ K, hence ai ∈ Mi, so there is afinite Qi ⊆ Yi such that ai ≤

∑Qi. Then

h(b, F, a) ≤∑

c∈Ph(c, 0, 0) +

∑i∈F

∑c∈Qi

H(0, {i}, {(i, c)}),

as desired. �Proposition 15.7. If C is the subalgebra of A×B obtained by one-point gluing withrespect to ultrafilters F,G on A,B respectively, then hL(C) = max(hL(A), hL(B)).

Proof. Clearly hL(C) ≤ max(hL(A), hL(B)), by Proposition 15.3. Now if I is anideal on A not generated by less than κ elements, then {(a, b) ∈ C : a ∈ I} is anideal on C not generated by less than κ elements. By symmetry, the equation inthe proposition follows. �

Since clearly c(A) ≤ hL(A) for any BA A, we must have hL(B) = |B| if B is theAlexsandrov duplicate of an algebra A.

By Proposition 2.6 we have hL(A) ≤ hL(Exp(A)). It is open whether ≤ canbe replaced by = here:

Problem 140. Is there a BA A such that hL(A) < hL(Exp(A))?

Theorem 15.8. hL(A) ≤ s(Exp (A)).

Proof. Let 〈Fα : α < κ〉 be a strictly decreasing sequence of closed subsets of

Ult(A). We claim that Ddef= {Fα+1 : α < κ} is a discrete set of points of Exp (A).

For, let α < κ. Choose x ∈ Fα\Fα+1 and y ∈ Fα+1\Fα+2. Then there is a clopensubset S of Ult(A) such that y ∈ S, Fα+2 ∩ S = ∅, and x /∈ S. And thereis a clopen U such that Fα+1 ⊆ U and x /∈ U . Now V (U, S) ∩ D = {Fα+1}.For, obviously Fα+1 ∈ V (U, S). Suppose that α < β and Fβ+1 ∈ V (U, S). ThenFβ+1 ⊆ Fα+2 and Fβ+1 ∩ S �= ∅, contradicting Fα+2 ∩ S = ∅. Suppose that β < α

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and Fβ+1 ∈ V (U, S). Then Fα ⊆ Fβ+1, so x ∈ Fβ+1. But x /∈ U and x /∈ S,contradiction. �

Concerning derived functions of hL, we mention these obvious facts:

hL(A) = hLH+(A) = hLS+(A) = hLh+(A) = dhLS+(A);

and hLS−(A) = hLh−(A) = ω. The following theorem is a corollary of 14.12, usingthe fact given below that χ(A) ≤ hL(A) for any infinite BA.

Theorem 15.9. If hL(A) = ω, then CardH−(A) = ω. �Now we consider the derived function hLmm. Here we have several choices as to howto define this function, depending on which equivalent definition from Theorem15.1 we take. We consider three possibilities: right-separated in the algebraic andin the topological senses, and increasing sequences of ideals.

A right-separated sequence 〈aξ : ξ < α〉 of elements of A is maximal iffthere does not exist a b ∈ A such that 〈aξ : ξ < α〉�〈b〉 is right-separated. right-separated sequences exist by Zorn’s lemma.

Proposition 15.10. Let 〈aξ : ξ < α〉 be a right-separated sequence in a BA A. Thenit is maximal iff {aξ : ξ < α} generates a maximal ideal.

Proof. Let I be the ideal generated by {aξ : ξ < α}. Clearly I is a proper ideal.

⇒: take any b ∈ A. Then 〈aξ : ξ < α〉�〈b〉 is not right-separated, and thisgives two possibilities.

Case 1. There is a finite F ⊆ α such that −b ·∏

ξ∈F −aξ = 0. Then −b ∈ I.

Case 2. There is a finite F ⊆ α such that b ·∏

ξ∈F −aξ = 0. Then b ∈ I.

⇐: Given b ∈ A, if b ∈ I, then b ·∏

ξ∈F −aξ = 0 for some finite F ⊆ α. If−b ∈ I, then −b ·

∏ξ∈F −aξ = 0 for some finite F ⊆ α. �

Proposition 15.11. u(A) is equal to min{|α| : there is a maximal right-separatedsequence of length α}, for any BA A.

Proof. Let λ = min{|α| : there is a maximal right-separated sequence of lengthα}. From Proposition 15.10 it follows that u(A) ≤ λ. Now let κ = u(A), andlet X be a set of size κ which generates a nonprincipal maximal ideal I. WriteX = {xα : α < κ}. We now define a sequence 〈ya : α < κ〉 by recursion. If yβ ∈ Xhas been defined for all β < α, then {yβ : β < α} does not generate I, so we canlet yα be xγ with γ minimum such that xγ /∈ 〈{yβ : β < α〉id. This finishes thedefinition. Clearly 〈{yα : α < κ}〉id = I. We claim that 〈{yα : α < κ}〉 is rightseparated. To see this, first suppose that F ∈ [κ]<ω. Then

∑α∈F yα �= 1 since I is

nonprincipal. Second, suppose additionally that F < β. Then yβ �≤∑

α∈F yα, asdesired. Thus we have produced a maximal right-separated sequence of length κ.So λ ≤ u(A). �

By Proposition 15.11, the function hLmm defined in terms of right-separated se-quences of elements of a BA is not a new function.

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Looking at the equivalent in Theorem 15.1 involving a sequence of ideals,from the definition of the function alt in Chapter 9 we see that alt is the hLmm

function in this case.

We turn to a variant of minimum-maximum functions associated with hL,involving a right-separated sequence of ultrafilters. We modify this notion so thatit can be applied to sequences of non-limit length. We call a sequence 〈Fξ : ξ ≤ α〉of ultrafilters right-separated iff the sequence does not have any repetitions, andfor every ξ < α the set {Fη : η ≤ ξ} is open in {Fη : η ≤ α}. Note that we insistthat every such sequence have length a successor ordinal.

Proposition 15.12. Suppose that α is a limit ordinal, and 〈Fξ : ξ < α〉 is a sequenceof distinct ultrafilters on A. Then the following conditions are equivalent:

(i) For every ξ < α, the set {Fη : η ≤ ξ} is open in {Fη : η < α}.(ii) {Fξ : ξ < α} �= Ult(A), and for every G /∈ {Fξ : ξ < α}, the sequence

〈Fξ : ξ < α〉�〈G〉 is right-separated.

(iii) For some G /∈ {Fη : η < α}, the sequence 〈Fξ : ξ < α〉�〈G〉 is right-separated.

Proof. (i)⇒(ii): First we show that {Fξ : ξ < α} �= Ult(A). For each ξ < α wehave Fξ ∈ {Fη : η ≤ ξ}, which is open in {Fη : η < α}. Hence there is an aξ ∈ Fξ

such that

(∗) S(aξ) ∩ {Fη : η < α} ⊆ {Fη : η ≤ ξ}.

We claim that {−aξ : ξ < α} has f.i.p. For, suppose that ξ(0) < · · · < ξ(m) < αand −aξ(0) · . . . · −aξ(m) = 0. Choose ξ < α with ξ(m) < ξ; this is possible since αis a limit ordinal. Then aξ(0) + · · ·+ aξ(m) = 1 ∈ Fξ, so aξ(i) ∈ Fξ for some i. Thiscontradicts (∗). Clearly any ultrafilter containing {−aξ : ξ < α} is different fromeach Fξ.

Next, suppose that G /∈ {Fη : η < α}. Take any ξ < α. Let U be an open setin Ult(A) such that U ∩ {Fη : η < α} = {Fη : η ≤ ξ}. Then (U\{G}) ∩ ({Fη : η <α} ∪ {G}) = {Fη : η ≤ ξ}.

Finally, clearly {Fξ : ξ < α} is open in {Fξ : ξ < α} ∪ {G}. This checks that〈Fξ : ξ < α〉�〈G〉 is right-separated.

(ii)⇒(iii): Trivial.

(iii)⇒(i): Assume (iii), and suppose that ξ < α. Let U be an open set suchthat U ∩ ({Fη : η < α} ∪ {G}) = {Fη : η ≤ ξ}. Then U ∩ {Fη : η < α} = {Fη : η ≤ξ}, as desired. �

Proposition 15.13. Let 〈Fξ : ξ < α〉 be a sequence of distinct ultrafilters on A, andlet G be an ultrafilter on A. Then the following conditions are equivalent:

(i) 〈Fξ : ξ < α〉�〈G〉 is a maximal right-separated sequence.

(ii) G is the only ultrafilter on A such that 〈Fξ : ξ < α〉�〈G〉 is a right-separatedsequence.

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Proof. (i)⇒(ii): Assume (i), and suppose that H �= G is an ultrafilter such that〈Fξ : ξ < α〉�〈H〉 is right-separated. We claim then that 〈Fξ : ξ < α〉�〈G,H〉 isright-separated. To prove this, first take any ξ < α. Then there are open sets U, Vin Ult(A) such that

U ∩ ({Fη : η < α} ∪ {G}) = {Fη : η ≤ ξ} and

V ∩ ({Fη : η < α} ∪ {H}) = {Fη : η ≤ ξ}.

Then U ∩ V ∩ ({Fη : η < α} ∪ {G,H}) = {Fη : η ≤ ξ}, as desired. We also needto show that {Fη : η < α} ∪ {G} is open in {Fη : η < α} ∪ {G,H}; but this isobvious.

(ii)⇒(i): Assume (ii), and suppose that 〈Fξ : ξ < α〉�〈G,H〉 is right-separated. In particular, G �= H . So we will get a contradiction by showing that〈Fξ : ξ < α〉�〈H〉 is right-separated. Take any ξ < α. Let U be open such thatU ∩ ({Fη : η < α} ∪ {G,H}) = {Fη : η ≤ ξ}. Then U ∩ ({Fη : η < α} ∪ {H}) ={Fη : η ≤ ξ}, as desired. �Proposition 15.14. Let 〈Fξ : ξ ≤ α〉 be a maximal right-separated sequence ofultrafilters. For each ξ < α let aξ be such that Fξ ∈ S(aξ) ∩ {Fη : η ≤ α} ⊆ {Fη :η ≤ ξ}. Then 〈aξ : ξ < α〉 is a maximal right-separated sequence of elements of A.

Proof. Since −aξ ∈ Fα for all ξ < α, condition (i) in the definition of right-separated holds. For condition (ii), suppose that ξ < α and Γ is a finite set suchthat Γ < ξ. Then aξ ·

∏η∈Γ−aη ∈ Fξ, so (ii) holds. For maximality, suppose

that {aξ : ξ < α} does not generate a maximal ideal. Then {−aξ : ξ < α} doesnot filter-generate an ultrafilter, but it is a subset of Fα, so there is an ultrafilterG �= Fα such that {−aξ : ξ < α} ⊆ G. Clearly 〈Fξ : x < α〉�〈G〉 is right-separated,contradiction. �Proposition 15.15. Intalg(κ) has a maximal right-separated sequence of ultrafilters,of length κ+ 1.

Proof. For each ξ < κ, the set

{[η, κ) : η < ξ} ∪ {[0, η) : ξ ≤ η}

clearly has f.i.p., and we let Fξ be an ultrafilter containing this set. Furthermore,the set

{[η, κ) : η < κ}has f.i.p., and we let Fκ be an ultrafilter containing this set.

Clearly all of these ultrafilters are distinct. To show that the sequence 〈Fξ :ξ ≤ κ〉 is right-separated, suppose that η < κ. Then clearly S([0, η)) ∩ {Fξ : ξ ≤κ} = {Fξ : ξ ≤ η}, as desired.

Finally, we claim that {Fξ : ξ ≤ κ} = Ult(Intalg(κ)); hence 〈Fξ : ξ ≤ κ〉 ismaximal. For, take any ultrafilter G. If [η, κ) ∈ G for all η < κ, then G = Fκ.Otherwise, take the least ξ such that [ξ, κ) /∈ G. Then G = Fξ. �

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Problem 141. Does every BA have a maximal right-separated sequence of ultrafil-ters?

Problem 142. Describe hLHs(A) in cardinal number terms.

Problem 143. Describe hLSr(A) in cardinal number terms.

Problem 144. Describe hLHr(A) in cardinal number terms.

On the relationship of hL with the previously defined functions: obviously s(A) ≤hL(A) for any infinite BA A. Next, χ(A) ≤ hL(A). In fact, suppose that F isany ultrafilter on A; we want to find a subset X of F which generates F andhas at most hL(A) elements. The set {S(a) : −a ∈ F} covers Ult(A)\{F}. Hencethere is a subset X of F such that {S(a) : −a ∈ X} also covers Ult(A)\{F},and |X | ≤ hL(A). We claim that X generates F . For suppose that a ∈ F . ThenX ∪ {−a} does not have the finite intersection property; otherwise, there wouldexist an ultrafilter G containing this set – then G �= F , so b ∈ G for some bsuch that −b ∈ X , contradiction. But X ∪ {−a} not having the finite intersectionproperty means that a is in the filter generated by X , as desired.

The following theorem is due to Todorcevic [90]:

Theorem 15.16. |A| ≤ Irr(A) · (hL(A))+ for any infinite BA A.

Proof. Let θ = Irr(A) · (hL(A))+ and κ = (hL(A))+. Assume that |A| > θ, inorder to work for a contradiction. Wlog |A| = θ+. Write A as a strictly increasingsequence 〈Aξ : ξ < θ+〉 of subalgebras of size ≤ θ. Let S0 = {δ < θ+ : cf(δ) = κ}.So S0 is stationary in θ+. For each δ ∈ S0 choose aδ ∈ Aδ+1\Aδ. Now fix δ ∈ S0.Define

Iδ = {b ∈ Aδ : b · aδ = 0}; Jδ = {b ∈ Aδ : b · −aδ = 0}.Note that Iδ and Jδ are ideals in Aδ. Let I ′δ be the ideal of A generated by Iδ.Thus I ′δ = {a ∈ A : a ≤ b for some b ∈ Iδ}. Now I ′δ has a generating set of size≤ hL(A); so Iδ itself has such a generating set. Since κ is a regular cardinal >hL(A), there is an f(δ) < δ such that a generating set for Iδ is a subset of Af(δ).So by Fodor’s theorem there is a ξ0 < θ+ and a stationary subset S1 of S0 suchthat f(δ) = ξ0 for all δ ∈ S1. Similarly, we can get a stationary subset S2 of S1

and a ξ1 < θ+ such that every ideal Jδ for δ ∈ S2 has a generating set in Aξ1 .Let ξ2 be the maximum of ξ0, ξ1. We now claim that 〈aδ : δ ∈ S2〉 is irredundant,which, of course, is a contradiction. To prove this claim we first show

(∗) Suppose ξ2 < δ < δ0 < · · · < δn are elements of S2 and ε ∈ n+12. Supposethat c ∈ Aδ and

c ·∏i≤n

aεiδi ≤ aδ.

Then there exist b0, . . . , bn ∈ Aξ2 such that

c ·∏i≤n

aε(i)δi

≤ c ·∏i≤n

bi ≤ aδ.

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We prove (∗) by induction on n; the following argument will work when n = 0 andalso for the inductive step. Assume the hypothesis of (∗). Let d =

(c ·∏

i<n aεiδi

)·

−aδ. Then d ∈ Aδn and d ≤ −aε(n)δn.

Case 1. ε(n) = 1. Then d ∈ Iδn . Hence there is an x ∈ Aξ0 ∩ Iδn such that d ≤ x.

So d ≤ x ≤ −aε(n)δn.

Case 2. ε(n) = 0. Then d ∈ Jδn , so there is an x ∈ Aξ1 ∩ Jδn such that d ≤ x. So

again d ≤ x ≤ −aε(n)δn.

So, in either case we get an x ∈ Aξ2 such that d ≤ x ≤ −aε(n)δn. It follows that

c ·∏i≤n

aε(i)δi

≤ c · −x ·∏i<n

aε(i)δi

≤ aδ,

so we have started the induction if n = 0, and continued the induction otherwise.

Now suppose that 〈aδ : ξ2 < δ ∈ S2〉 is redundant. So we can find δ < δ0 <· · · < δn with ξ2 < δ such that aδ is generated by Aδ ∪ {aδ0 , . . . , aδn}. Thereforeaδ is a finite sum of elements of the form

c ·∏i≤n

aεiδi ,

where c ∈ Aδ. By (∗), every such intersection can be replaced by one of the form

c ·∏i≤n

bi

for some b0, . . . , bn ∈ Aξ2 . It follows that aδ ∈ Aδ, contradiction. �

The BA of the Kunen line constructed in Chapter 8 (assuming CH) has size andcharacter ω1 (see Chapter 14), hence hereditary Lindelof degree ω1, and countablespread.

If one can construct in ZFC a BA A such that s(A) < hL(A), then one canalso construct in ZFC a BA B such that s(B) < χ(B). For, let I be an ideal of Asuch that I is not generated by fewer than (s(A))+ elements, and let B = I ∪−I.

An example where χ(A) < hL(A) is provided by the Alexsandroff duplicateof a free algebra; see Chapter 14. An example with hL(A) < d(A) is provided bythe interval algebra on a complete Suslin line, using the argument of Lemma 3.43.On the other hand, we have the following result of Juhasz [75]:

Theorem 15.17. (MA + ¬CH) If hL(A) = ω, then hd(A) = ω.

Proof. Suppose that hL(A) = ω but hd(A) > ω. Note that A satisfies ccc, sincec(A) ≤ hL(A). By Theorem 6.15, there is a homomorphism f from A onto a BAB such that π(B) > ω. We define 〈bξ : ξ < ω1〉 by recursion, as follows. If bξhas been defined for all ξ < η, then |〈{bξ : ξ < η}〉| < π(B), and hence there is

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an element bη ∈ B+ such that no nonzero element of {∏

ξ∈F bξ : F ∈ [η]<ω} isbelow bη. Clearly this constructs a sequence of distinct elements. For each ξ < ω1

let aξ be such that f(aξ) = bξ. Now by Lemma 3.5 there is an S ∈ [ω1]ω1 such

that {−aξ : ξ ∈ S} has fip. Clearly then a strictly increasing enumeration of{−aξ : ξ ∈ S} gives a right-separate sequence, contradiction. �

These observations leave the following question open; this is Problem 53 inMonk [96]:

Problem 145. Is there an example in ZFC of a BA A such that hL(A) < d(A)?

This problem is equivalent to the problem of constructing in ZFC a BA A suchthat hL(A) < hd(A); see the end of Chapter 16.

Bounded versions of hL can be defined as follows. For m a positive integer, asequence 〈xα : α < κ〉 of elements of A is said to be m-right-separated providedthat if Γ ∈ [κ]m, α < κ, and β < α for all β ∈ Γ, then aα ·

∏β∈Γ−aβ �= 0. Then

we define

hLm(A) = sup{κ : there is an m-right-separated sequence in A}.

For this notion see Roslanowski Shelah [98].

For an interval algebra A we have hL(A) = c(A). In fact, suppose that A is theinterval algebra on L and I is an ideal of A. Define a ≡ b iff a, b ∈ L and eithera = b or else if, say, a < b, then [a, b) ∈ I. Then ≡ is a convex equivalence relationon L. For each ≡-class k having more than one element, let 〈akα : α < λk〉 be astrictly decreasing coinitial sequence in k (with λk = 1 if k has a first element),and let 〈bkα : α < μk〉 be a strictly increasing cofinal sequence in k (with μk = 1if k has a greatest element), and with ak0 < bk0 . Note that there are at most c(A)≡-classes with more than one element, and always λk, μk < (c(A))+. Hence

{[akα, bkβ) : k an ≡-class with more than one element, α < λk, β < μk}

is a collection of at most c(A) elements which generates I; so hL(A) ≤ c(A) byTheorem 15.1.

For any tree algebra B on an infinite pseudo-tree T we also have c(B) =hL(B). For, Treealg(T ) embeds in an interval algebra A, and we may assumethat Treealg(T ) is dense in A (extend the identity from Treealg(T ) onto itself toa homomorphism from A into the completion of Treealg(T ), and then take theimage of A). Hence

hL(A) ≥ hL(Treealg(T )) ≥ c(Treealg()T ) = c(A) = hL(A).

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16 Hereditary Density

We begin again with some equivalent definitions, which are similar to the case ofhereditary Lindelof degree. Recall the definition of left-separated sequence fromChapter 6, before Theorem 6.12.

Theorem 16.1. For any infinite BA A, hd(A) is equal to each of these cardinals:

sup{κ:there is a strictly decreasing sequence of ideals of length κ};sup{κ:there is a strictly decreasing sequence of filters of length κ};sup{κ:there is a strictly decreasing sequence of open sets of length κ};sup{κ:there is a strictly increasing sequence of closed sets of length κ};sup{κ:there is a left-separated sequence of length κ};min{κ:every subspace S of Ult(A) has a dense subset of power ≤ κ};sup{π(B) : B is a homomorphic image of A};sup{d(B) : B is a homomorphic image of A}.

(Note that left-separated can be taken in the topological or algebraic sense.)

Proof. This time there are nine cardinals, named κ0, . . . , κ8 in their order of men-tion, starting with hd itself. The following relationships are easy, following thepattern of the proof of Theorem 15.1: κ1 = κ2; κ1 ≤ κ3; κ3 = κ4; κ3 ≤ κ5; andκ0 = κ6. Furthermore, κ8 = κ0 by Theorem 5.20, κ0 = κ5 by Theorem 6.12, andκ0 = κ7 by Theorem 6.15. Hence only two inequalities remain.

κ6 ≤ κ2: Suppose that X is a subspace of Ult(A), and d(X) = κ; we constructa strictly decreasing sequence of filters of type κ. By induction let

Fα ∈ X\{Fβ : β < α}

for each α < κ. Then set Cα =⋂

β≤α Fβ . Thus 〈Cα : α < κ〉 is a decreasingsequence of filters. It is strictly decreasing, since if α < κ we can choose a ∈ Fα+1

such that S(a) ∩ {Fβ : β ≤ α} = 0, so that −a ∈ Cα\Cα+1.

κ5 ≤ κ6: Suppose 〈xα : α < κ〉 is left separated, where κ is regular. Clearlythen {xα : α < κ} has no dense subset of power < κ. �

The equivalents in Theorem 16.1 give rise to eight possible attainment problems,on the face of it. However, proofs of previous results set some limits:

, ,OI 10.1007/978-3- - - _ ,


014


Basel 217

449

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450 Chapter 16. Hereditary Density

Proof of Theorem 5.20:

Attainment in the κ8 sense implies attainment in the κ0 sense;


Attainment in the κ0 sense implies attainment in the κ5 sense;

For hd(A) regular, attainment in the κ5 sense implies attainment in the κ0

sense;


attainment in the κ7 sense implies attainment in the κ5 sense;

attainment in the κ0 sense implies attainment in the κ7 sense;


attainment for κ1, κ2, κ3, κ4 are equivalent;

attainment in the κ0 sense implies attainment in the κ2 sense.

Now we note two other implications.

κ1 attained implies κ5 attained. Suppose that 〈Iα : α < κ1〉 is a strictlydecreasing sequence of ideals. For each α < κ1 choose aα ∈ Iα\Iα+1. Then 〈aα :α < κ1〉 is left-separated.

κ5 attained implies κ1 attained. Similarly.

In Roslanowski, Shelah [01a] there are two results about attainment for hd.

(1) Theorem 1.5 says:

If 2cf(λ) < λ, then in any BA with hd equal to λ, all equivalent versions areattained.

(2) In Section 4, it is shown that it is consistent to have a singular cardinal λ anda BA A with hd(A) = λ, with attainment in the left-separated sense but not inthe sense κ7.

Thus the following problem is open; this is Problem 54 in Monk [96].

Problem 146. Completely describe the relationships between the attainment prob-lems for hd.

Corollary 16.2. hd is attained in the sense of left-separation for strong limit sin-gular cardinals.

Proof. Suppose that λ is strong limit singular and hd(A) = λ. By Theorem 13.2,Ult(A) has a discrete subset of size λ. By a remark before Theorem 13.2, A hasan atomic homomorphic image B with λ atoms; hence π(B) = λ, and the remarkson attainment above give the desired result. �

Also recall from Proposition 13.4 that hd is attained in the left-separated sensefor singular cardinals of confinality ω.

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Chapter 16. Hereditary Density 451

If A is a subalgebra or homomorphic image of B, then hd(A) ≤ hd(B). Wenow consider special subalgebras. One can have A ≤free B with hL(B) arbitrarilylarger than hL(A). Hence the same applies to ≤proj, ≤u, ≤rc, ≤reg, and ≤σ. Thesituation for ≤π is similar to that for hL. Also, the arguments for hL and ≤s and≤mg work for hd too.

Proposition 16.3. hd(A×B) = max(hd(A), hd(B)).

Proof. Suppose that 〈Iα : α < κ〉 is a strictly decreasing sequence of ideals in A.For each α < κ let Jα = {(a, 0) : a ∈ Iα}. Then 〈Jα : α < κ〉 is a strictly decreasingsequence of ideals in A×B. A similar argument holds for B, so ≥ holds.

Now let κ = max(hd(A), hd(B)), and suppose that 〈Kα : α < κ+〉 is astrictly decreasing sequence of ideals in A × B; we want to get a contradiction.Let Lα = {a ∈ A : (a, 0) ∈ Kα} and Mα = {b ∈ B : (0, b) ∈ Kα}, for each α < κ+.Then Lα and Mα are ideals in A and B respectively. Moreover, if α < β, thenLβ ⊆ Lα and Mβ ⊆ Mα, with Lβ �= Lα or Mβ �= Mα. Since hd(A) ≤ κ, the set

{Lα : α < κ+} has size at most κ, and so there is a Γ ∈ [κ+]κ+

such that Lα = Lβ

for all α, β ∈ Γ. Then 〈Mα : α ∈ Γ〉 is one-one, contradicting hd(B) ≤ κ. �Proposition 16.4. hd

(∏wi∈I Ai

)= max(|I|, supi∈I hd(Ai)).

Proof. It is convenient to assume that I is a cardinal κ, and by Proposition 16.3we may assume that κ is infinite. For brevity let B =

∏wα<κ Aα. By Proposition

16.3 we have hd(Aβ) ≤ hd(B) for all β < κ. Now for each α < κ let Jα consistof all x ∈

∏wα∈κ Aα : x is of type 1, with support contained in κ\α}. Clearly this

gives a strictly decreasing sequence of ideals. Thus we have proved ≥.Now let λ = max(|I|, supi∈I hd(Ai)). Suppose that 〈Kα : α < λ+〉 is a strictly

decreasing sequence of ideals in B; we want to get a contradiction.

Case 1. There is an α < λ+ such that some a ∈ Kα has type 2. Then with M thesupport of a, let Lβ = {a � M : a ∈ Kβ} for each β ∈ [α, λ+). Then

(1) 〈Lβ : β ∈ [α, λ+)〉 is a strictly decreasing sequence of ideals in∏

γ∈M Aγ .

In fact, suppose that α ≤ β < γ < λ+. Take any b ∈ Kβ\Kγ . Then b � M ∈ Lβ .Suppose that b � M ∈ Lγ . Say b � M = c � M with c ∈ Kγ . Then also a+ c ∈ Kγ ,and b ≤ a+ c, so b ∈ Kγ , contradiction. Thus b � M /∈ Lγ . This proves (1).

But (1) contradicts Proposition 16.3.

Case 2. For each α < λ+, every element of Kα has type 1. For each α < λ+ andβ < κ let Nβ

α = {a ∈ Aβ : χβa ∈ Kα}, where we define χβ

a ∈ B by

χβa(γ) =

{a if γ = β,

0 otherwise.

Clearly 〈Nβα : α < λ+〉 is a decreasing sequence of ideals in Aβ . Hence there is

a γα < λ+ such that Nβδ = Nγα for all δ ∈ [γα, λ

+). Let ε < λ+ be such that

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γα < ε for all α < κ. Choose a ∈ Kε\Kε+1. Let a have support P . For each β ∈ P

we have aβ ∈ Nβε = Nβ

ε+1, and so χβaβ∈ Kε+1. Now a =

∑β∈P χβ

aβ∈ Kε+1, so

a ∈ Kε+1, contradiction. �

Obviously s(A) ≤ hd(A), and hence Ind(A) ≤ hd(A). It follows that for arbi-trary products we have, as usual, hd

(∏i∈I Ai

)≥ max(2|I|, supi∈Ihd(Ai)). Shelah

and Peterson independently noticed that strict inequality is possible; this answersProblem 49 of Monk [90]. The example for spread can be used here.

The situation for ultraproducts is similar to that for hL. Problems 55 and 56 ofMonk [96] were solved by Shelah [99] and Spinas [00] as follows:

Shelah [99], part of 15.13: Suppose that D is a uniform ultrafilter on κ. Thenthere is a class of cardinals χ such that χκ = χ and there are BAs Bi fori < κ such that hd(Bi) ≤ χ for each i < κ, hence

∏i<κ hd(Bi)/D ≤ χ, while

hd(∏

i<κ Bi/D) = χ+.

Shelah, Spinas [00], part of 2.4:

It is consistent that there exist cardinals κ, μ, BAs Bi for i < κ, and an ultrafilterD on κ such that

∏i<κ hd(Bi)/D = μ++ while hd(

∏i<κ Bi/D) = μ+.

For free products, the analog of Theorem 15.5 holds, with essentially the sameproof:

Theorem 16.5. If 〈Ai : i ∈ I〉 is a system of non-trivial BAs, for brevity letλ = supi∈Ihd(Ai); then

max(|I|, supi∈I

hd(Ai)) ≤ hd(⊕i∈IAi) ≤ max(|I|, 2supi∈I s(Ai)

). �

The inequalities in Theorem 16.5 are sharp. This is seen as for Theorem 15.5,except for both <; for this case one can take S such that Q ⊆ S ⊆ R, |S| = ℵ1,let A = Intalg(S), and consider A⊕A, assuming ¬CH.

Another important fact about free products is given in the following theorem.

Theorem 16.6. For infinite BAs A and B we have s(A⊕B) ≥ min(hL(A), hd(B)).

Proof. Let κ = min(hL(A), hd(B)), and let λ+ ≤ κ. Let 〈aα : α < λ+〉 be right-separated in A, and let 〈bα : α < λ+〉 be left-separated in B. We claim that〈aα · bα : α < λ+〉 is ideal independent. For, suppose that Γ ∈ [λ+]<ω and α ∈λ+\Γ. Let Δ = {β ∈ Γ : β < α}. then

aα · bα ·∏β∈Γ

−(aβ · bβ) ≥

⎛⎝aα ·

∏β∈Δ

−aβ

⎞⎠ ·

⎛⎝bα ·

∏β∈Γ\Δ

−bβ

⎞⎠ �= 0,

as desired. �

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Proposition 16.7. In the setup for moderate products,

hd

(∏B

i∈IAi

)= max(|I|, hd(B), sup

i∈Ihd(Ai)).

Proof. For brevity let C =∏B

i∈I Ai.

Clearly ≥ holds; see the proof of Proposition 15.6. Now let κ be the right sideof the above equation, and suppose that 〈Kα : α < κ+〉 is a strictly decreasingsequence of ideals of C. Let Lα = {b ∈ B : h(b, ∅, ∅) ∈ Kα}. Clearly Lα is an idealin B, and Lβ ⊆ Lα whenever α < β < κ+. Hence there is an ordinal α such thatLβ = Lα for all β ∈ (α, κ+). For each i ∈ I and α < κ+ let Miα = {a ∈ Ai :h(∅, {i}, {(i, a)}) ∈ Kα}. Then Miα is an ideal in Ai, and Miβ ⊆ Miα wheneverα < β < κ+. Hence there is an ordinal βi < κ+ such that Miγ = Miβi for allγ ∈(βi, κ

+. Let γ = sup(α, supi∈I βi). Then γ < κ, and Kδ = Kγ for all δ ∈ (γ, κ+)(contradiction). In fact, if h(b, F, a) ∈ Kδ, then h(b, ∅, ∅) ∈ Kδ, hence b ∈ Lδ,hence b ∈ Lγ , hence h(b, ∅, ∅) ∈ Kγ . Also, if i ∈ F , then h(∅, {i}, {(i, ai)}) ∈ Kδ,hence ai ∈Miδ, hence ai ∈Miγ , hence h(∅, {i}, {(i, ai)}) ∈ Kγ . So

h(b, F, a) = h(b, ∅, ∅) ∪⋃i∈F

h(∅, {i}, {(i, ai)}) ∈ Kγ . �

Proposition 16.8. Let C be the one-point gluing of A and B with respect to ultra-filters F on A and G on B. Then max(hd(A), hd(B)) ≤ hd(C).

Proof. Suppose that 〈Iα : α < κ〉 is a strictly decreasing sequence of ideals in A.Define Jα = {(a, b) ∈ C : a ∈ Iα}. Then 〈Jα : α < κ〉 is a decreasing sequenceof ideals in C. If a ∈ Iα\Iα+1, then (a, 1) ∈ Jα\Jα+1 if a ∈ F , and (a, 0) ∈Jα\Jα+1 if a /∈ F . Thus 〈Jα : α < κ〉 is strictly decreasing. A similar argumentworks for B. �

Problem 147. Do there exist BAs A,B and ultrafilters F,G on A,B respec-tively such that if C is the associated one-point gluing, then max(hd(A), hd(B)) <hd(C)?

Similarly to hL, it is clear that hd(A) = |A| for A the Alexsandroff duplicate ofa BA.

Theorem 16.9. hd(A) ≤ t(Exp (A)).

Proof. We use the fact that hd(A) = sup{d(B) : B a homomorphic image of A},given in Theorem 16.1. Let B be any homomorphic image of A. Then by Propo-sition 2.7, Exp (B) is a homomorpic image of Exp (A), so d(B) ≤ t(Exp (B)) ≤t(Exp (A)), using Lemma 12.14. �Lemma 16.10. For any infinite BA A and any X ⊆ A the following are equivalent:

(i) X generates A.

(ii) {S(x) : x ∈ X} separates points in Ult(A).

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Proof. (i)⇒(ii): Suppose that F,G ∈ Ult(A), F �= G. If ∀x ∈ X(x ∈ F iff x ∈ G),then ∀x ∈ 〈X〉(x ∈ F iff x ∈ G), by an easy argument.

(ii)⇒(i): Suppose that a ∈ A\〈X〉. Let A = {xε00 ·. . .·x

εm−1

m−1 : each xi ∈ X andxε00 · . . . · x

εm−1

m−1 ≤ a}. Then⋃

b∈A S(b) ⊂ S(a) by compactness, since a /∈ 〈X〉, sochoose F ∈ S(a)\

⋃b∈A S(b). Now (F ∩ 〈X〉) ∪ {−a} has fip, and so is contained

in an ultrafilter G. But then F and G are distinct ultrafilters which cannot beseparated by {S(x) : x ∈ X}. �

Concerning derived functions, we have the following obvious facts:

hd(A) = hdH+(A) = hdS+(A) = hdh+(A) = dhdS+(A);

and hdS−(A) = hdh−(A) = ω.

Now we consider min-max versions of hd. We extend the definition of left separatedsequences, algebraic or topological, to sequences indexed by any infinite ordinal;see Chapter 6, discussion around Theorem 6.12. We take the definition of hdspectand hdmm in the algebraic sense. Thus

hdspect(A) = {|α| : α ≥ ω and there is a maximal left-separated sequence

of elements of A of length α};hdmm(A) = min(hdspect(A)).

Proposition 16.11. A sequence 〈aξ : ξ < α〉 of elements of A is a maximal left-separated sequence iff it is a left-separated sequence and{

aξ ·∏

η∈F−aη : ξ ∈ α, F ∈ [α]<ω, and ξ < F

}is dense in A. �Proposition 16.12. hdmm(A) = π(A) for any infinite BA A.

Proof. This is immediate from Proposition 16.11 and Lemma 6.14. �

There are BAs A with |hdspect(A)| > 1. For example, hdspect(P(ω)) has at leastthe two members ω, 2ω, since π(P(ω)) = ω and P(ω) has an independent sub-set of size 2ω. The existence of a BA A with |hdspect(A)| > 2 is necessarily aconsistency problem, since |A| ≤ 2π(A).

Problem 148. Is it consistent to have an infinite BA A such that |hdspect(A)| > 2?

If we generalize the notion of a left-separated sequence of ultrafilters to arbitraryordinal lengths, we see that such a sequence is maximal iff it contains all ultrafilters.Thus the min-max function is trivial for hd in this sense.

We consider one more variant of hd and its min-max function. Clearly astrictly decreasing sequence of ideals in a BA A is maximal iff its intersection isthe trivial ideal {0}. Let hdid

mm(A) be the least size of the index set of a maximalstrictly decreasing sequence of ideals in A.

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Proposition 16.13. hdidmm(A) ≤ cf(d(A)).

Proof. By Theorem 5.1 we may assume that A is a subalgebra of P(κ), whereκ = d(A). For each α < κ let Iα = {a ∈ A : a ⊆ [α, κ)}. Clearly Iα is an ideal inA, and Iβ ⊆ Iα if α < β. Suppose that Iα = {0}. Then ∀a ∈ A+[a ∩ α �= 0]. Thenthe mapping a !→ (a ∩ α) is an isomorphism from A into P(α), contradicting thechoice of κ. �

Problem 149.What is the exact place of hdidmm among the other cardinal functions?

Now we want to go into a result of Fedorchuk [75], which provides an examplefor several of the questions in Monk [90]: assuming ♦, there is a BA A withhd(A) = ω and CardH−A = ω1. This is a weakened form of his main theorem. Wegive a construction of such a BA due to Kunen [75] (quite different from that ofFedorchuk but done upon looking at that article and noticing some problems withthe construction).

Special ♦-sequences. This material is from Kunen [75] (except for the name specialand the proof of the lemma). If f ∈ ω1(ω1ω1), let f��α = 〈fξ � α : ξ < α〉. A special♦-sequence is a sequence 〈fα : α < ω1〉 such that each fα ∈ α(αω1) and for allf ∈ ω1(ω1ω1) the set {α < ω1 : f��α = fα} is stationary.

Lemma 16.14. ♦ implies that there is a special ♦-sequence.

Proof. Let H be a one-one function from ω1 onto ω1×ω1. If x ∈ ω1×ω1, we writex = (x0, x1). For f ∈ ω1(ω1ω1) we define f ∈ ω1ω1 by f(α) = f(H(α))0((H(α))1).Define G : ω1ω1 → ω12 by

Gh(β) =

{1 if h

((H(β))0

)= (H(β))1,

0 otherwise.

Define F : ω12→ ω1ω1 by

(F (k))(β) =

{γ if k(H−1(β, γ)) = 1 and k(H−1(β, δ)) = 0 for all δ �= γ,

0 if there is no such γ.

Let χ : P(ω1)→ ω12 be the natural bijection. Let C0 = {α < ω1 : H [α] = α×α}.So, C0 is club. In fact, suppose α < ω1 is limit and C0 ∩ α is unbounded in α.If β < α, choose γ ∈ C0 ∩ α such that β < γ. Then H(β) ∈ γ × γ ⊆ α × α.On the other hand, if β, γ < α, choose δ ∈ C0 ∩ α such that β, γ < δ. Then(β, γ) ∈ δ× δ, so there is an ε ∈ δ such that H(ε) = (β, γ). Since d < α, this showsthat (β, γ) ∈ H [α]. Thus C0 is closed. To show that C0 is unbounded, let β < ω1.Let γ0 = β. Choose γ1 > γ0 such that H [γ0] ⊆ γ1 × γ1, and choose γ2 > γ1 suchthat H−1[γ1 × γ1] ⊆ γ2. Continuing this way, the supremum of all γi is in C0.

Let 〈Aα : α < ω1〉 be a ♦-sequence. For α ∈ C0 let fα ∈ α(αω1) be definedby

fαβ (γ) = (F (χAα))(H

−1(β, γ)).

Let fα ∈ α(αω1) be arbitrary if α /∈ C0.

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Now suppose that f ∈ ω1(ω1ω1). Let B = χ−1(Gf ). Now

C1def= {α : ∀ξ < α(fξ � α ∈ αα)}

is club. In fact, to show that C1 is closed, suppose that α < ω1 is a limit ordinaland C1 ∩ α is unbounded in α. Take any ξ < α. For any β ∈ C1 with ξ < β < αwe have fξ � β ∈ ββ, and C1 ∩α is unbounded in α, so fξ � α ∈ αα. Thus α ∈ C1.To show that C1 is unbounded in ω1, let β < ω1. Choose γ1 such that for allξ < β we have fξ � β ∈ βγ1. Then choose γ2 such that for all ξ < γ1 we havefξ � γ1 ∈ γ1γ2. Continuing this way, we let δ be the supremum of all γi. We claimthat δ ∈ C1. For, suppose that ξ < δ. Then for all i ∈ ω such that ξ < γi wehave fξ � γi ∈ γiγi+1. Hence fξ � δ ∈ δδ. In fact, let ε < δ. Say ε < ϕi. Thenfξ(ε) ∈ γi+1 ⊆ δ, as desired.

Hence C0 ∩ C1 ∩ {α < ω1 : α ∩B = Aα} is stationary. We claim that if α isin this set, then f��α = fα; this will finish the proof. Suppose ξ < α. We want toshow that fξ � α = fα

ξ . Let γ < α. Then

f(H−1(ξ, γ)) = fξ(γ);

Gf (H−1(H−1(ξ, γ), fξ(γ))) = 1;

H−1(H−1(ξ, γ), fξ(γ)) ∈ α ∩B;

H−1(H−1(ξ, γ), fξ(γ)) ∈ Aα;

χAα(H−1(H−1(ξ, γ), fξ(γ))) = 1.

It is easily checked that if δ �= fξ(γ) then χAα(H−1(H−1(ξ, γ), δ)) = 0. In fact, we

need to show that H−1(H−1(ξ, γ), δ) /∈ Aα; equivalently, thatH−1(H−1(ξ, γ), δ) /∈

B, i.e., χB(H−1(H−1(ξ, γ), δ)) = 0, which means that Gf (H

−1(H−1(ξ, γ), δ)) = 0,

i.e., f(H−1(ξ, γ) �= δ, i.e., fξ(γ) �= δ, as desired. Therefore (FχAα)H−1(ξ, γ) =

fξ(γ). It follows that fαξ (γ) = fξ(γ). �

Kunen’s construction. We assume ♦; so CH is available also. Fix a special ♦-sequence 〈fα : α < ω1〉. For any space Y , a point y ∈ Y is a strong limit point ofH ⊆ P(Y ) if for all neighborhoods V of y there is an H ∈ H such that y /∈ Hand H ⊆ V . For α ≤ β ≤ ω1 define πβ

α : β2 → α2 by πβα(g) = g � α; thus πβ

α iscontinuous. We claim

(1) there is an enumeration 〈qα : α < ω1〉 of⋃

σ<ω1

σ2 such that every element isrepeated ω1 times and qα ∈ σα2 with σα ≤ α.

To prove (1), for each σ < ω1 let σ2 = {hσξ : ξ < ω1} with each element repeatedω1 times. Let H enumerate ω1 × ω1 under its natural order ((α, β) < (γ, δ) iff[max(α, β) < max(γ, δ) or (max(α, β) = max(γ, δ) and α < γ) or (max(α, β) =max(γ, δ) and α = γ and β < δ)]). As is well known, H has domain ω1. Byinduction on α one can show that H(α)0, H(α)1 ≤ α for all α < ω1. In fact,

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clearly H(0) = (0, 0), so it holds for α = 0, Suppose that it holds for α. Letβ = H(α)0 and γ = H(α)1. Then H(α+1) is determined as in the following table.

Case: β + 1 < γ β + 1 = γ γ < β γ = β

H(α+ 1): (β + 1, γ) (γ, 0) (β, γ + 1) (0, γ + 1)

ThusH(α+1)0, H(α+1)1 ≤ α+1. Now suppose that α is limit andH(β)0, H(β)1 ≤β for all β < α. For each β < α let K(β) = sup(H(β)0, H(β)1).

Case 1. For all β < α there is a γ ∈ (β, α) such that K(β) < K(γ). ThenH(α) = (0, supβ<αK(β)), and the inductive step works.

Case 2. There is a β < α such that K(β) = K(γ) for all γ ∈ (β, α).

Subcase 2.1. H(β)0 < H(β)1. Thus H(β)1 = K(β).

Subsubcase 2.1.1. δdef= supβ<γ<αH(γ)0 <K(γ). Then H(α) = (δ,K(β)), as

desired.

Subsubcase 2.1.2. δ = K(γ). Then H(α) = (0, δ + 1), as desired.

Subcase 2.2. H(β)0 > H(β)1. So H(β)0 = K(β). Let θ = supβ<γ<αH(γ)1.

Subsubcase 2.2.1. θ < K(β). Then H(α) = (K(β), θ), as desired.

Subsubcase 2.2.2. θ = K(β). Then H(α) = (0,K(β) + 1), as desired.

This finishes our inductive proof. Now let qα = hH(α)0H(α)1 for all α < ω1. Clearly(1) holds.

Our space will be a closed subspace of ω12. By induction on α ≤ ω1 we willdefine Xα ⊆ α2 and pα ∈ Xα so that:

(2) Xα is closed and nonempty.

(3) If α ≤ β then πβα[Xβ] = Xα.

(4) If qα ∈ Xσα , then pα � σα = qα.

(5) pα�0, pα

�1 ∈ Xα+1.

(6) If α ≤ β, {fαξ : ξ < α} ⊆ Xα, and h ∈ Xα is an accumulation point of

{fαξ : ξ < α}, then every point k in Xβ ∩ (πβ

α)−1[{h}] is a strong limit point of

{Xβ ∩ (πβα)

−1[{fαξ }] : ξ < α}.

Before actually making this construction we check that (2)–(6) yield the desired

properties of Xdef= Xω1 .

X is hereditarily separable: If not, let f = 〈fξ : ξ < ω1〉 be a left-separatedsequence in X . Let

C = {α < ω1 : for all clopen N ⊆ Xα

(N ∩ {fξ � α : ξ < α} = 0 iff (πω1α )−1[N ] ∩ {fξ : ξ < ω1} = 0) and

(|N ∩ {fξ � α : ξ < α}| = 1 iff |(πω1α )−1[N ] ∩ {fξ : ξ < ω1}| = 1)}.

Then C is club. To prove this, first note, obviously:

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(7) N ∩ {fβ � α : β < α} �= 0 implies that (πω1α )−1[N ] ∩ {fβ : β < ω1} �= 0, if

α < ω1 and N is a clopen subset of Xα.

(8) |N ∩ {fβ � α : β < α}| ≥ 2 implies that |(πω1α )−1[N ] ∩ {fβ : β < ω1}| ≥ 2, if

α < ω1 and N is a clopen subset of Xα.

Now to prove that C is closed, suppose that α < ω1 is a limit ordinal and α ∩ Cis unbounded in α. Suppose that N is clopen in Xα and (πω1

α )−1[N ] ∩ {fξ : ξ <

ω1} �= 0. Say ξ < ω1 and fξ � α ∈ N . Write N = Uαg

def= {h ∈ Xα : g ⊆ h}, where

g ∈ F 2 for some finite F ⊆ α. Say F ⊆ β < α, β ∈ C. Then fξ � β ∈ Uβg , i.e.,

(πω1

β )−1[Uβg ]∩{fη : η < ω1} �= 0 so, since β ∈ C, we get Uβ

g ∩{fη � β : η < β} �= 0.

Hence choose η < β with fη � β ∈ Uβg . Hence fη � α ∈ Uα

g = N , and N ∩ {fη � α :η < α} �= 0, as desired. For the other part of C, suppose that N is clopen in Xα

and (πω1α )−1[N ] ∩ {fξ : ξ < ω1} ≥ 2. Say ξ, η < ω1, ξ �= η, and fξ � α, fη � α ∈ N .

Write N = Uαg

def= {h ∈ Xα : g ⊆ h}, where g ∈ F 2 for some finite F ⊆ α. Say

F ⊆ β < α, β ∈ C. Then fξ � β, fη � β ∈ Uβg , i.e., |(πω1

β )−1[Uβg ]∩{fη : η < ω1}| ≥ 2

so, since β ∈ C, we get |Uβg ∩ {fη � β : η < β}| ≥ 2. Hence choose distinct

ρ, σ < β with fρ � β, fσ � β ∈ Uβg . Hence fρ � α, fσ � α ∈ Uα

g = N , and|N ∩ {fη � α : η < α}| ≥ 2, as desired. This proves that C is closed.

To prove that C is unbounded, suppose that α0 < ω1. Choose α1 such thatα0 < α1 and for all clopen N ⊆ Xα0 we have

(πω1α0)−1[N ] ∩ {fξ : ξ < ω1} �= 0⇒ ∃ξ < α1(fξ � α0 ∈ N) and

|(πω1α0)−1[N ] ∩ {fξ : ξ < ω1}| ≥ 2⇒ |N ∩ {fξ � α0 : ξ < α1}| ≥ 2.

This is possible since there are only countably many clopen sets N ⊆ Xα0 . Con-tinuing in this fashion with α2, α3, . . ., we see that αω = supn<ω αn is the desiredmember of C.

By the definition of special ♦-sequences, fix α ∈ C such that f��α = fα.

(9) fη � α ∈ {fαβ : β < α} for all η < ω1.

For, if fη � α ∈ N with N clopen, then (πω1α )−1[N ] ∩ {fν : ν < ω1} �= 0 so, since

α ∈ C, N ∩ {fβ � α : β < α} �= 0, as desired.

(10) For all η < ω1, if fη � α is an isolated point of {fβ � α : β < α}, then η < α.

For, letN be clopen such thatN∩{fβ � α : β < α} = {fη � α}. Say fη � α = fγ � αwith γ < α. Now since α ∈ C we get |(πω1

α )−1[N ] ∩ {fβ : β < ω1}| = 1. Since fηand fγ are both in (πω1

α )−1[N ] ∩ {fβ : β < ω1}, it follows that η = γ, as desired.

From (9) and (10) it follows that fα � α is an accumulation point of {fβ � α :β < α} = {fα

β : β < α}. Hence by (6) fα is a strong limit point of {Xω1 ∩(πω1

α )−1[{fαξ }] : ξ < α}. So fα is a limit point of {fξ : ξ < α}, which contradicts

the left-separatedness.

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Now we show that χH−A ≥ ω1, where A = ClopXω1 . By Corollary 14.9 it suf-fices to show that Xω1 has no one-one convergent sequences. So, assume thatlimn→∞ gn = h with all gn distinct and different from h. Choose f ∈ ω1(ω12) sothat {fα : α < ω1} = {gn : n ∈ ω}. Choose α so that f��α = fα, all of thefunctions gn � α, h � α are distinct, and {fβ : β < α} = {gn : n ∈ ω}. Then h � αis an accumulation point of {fα

ξ : ξ < α}. Say h � α = qβ with α ≤ β. Then pβ�0

and pβ�1 are both in Xβ+1 and extend h � α. This gives by (6) two distinct points

h, l which are strong limit points of Xω1 ∩{(πω1α )−1[{fα

ξ }] : ξ < α}. Thus both arelimit points of {gn : n ∈ ω}, contradiction.Next we do the construction to yield (2)–(6). As soon as a spaceXα is constructed,fix pα ∈ Xα such that (4) holds. Let X0 be the one-point space. For δ limit, letXδ = {g ∈ δ2 : ∀α < δ(g � α ∈ Xα)}. It is straightforward to check (2)–(6) then.Now we do the crucial step from Xδ to Xδ+1. We now define a nested clopen basis〈Kn : n ∈ ω〉 of pδ. First let 〈K ′

n : n ∈ ω〉 be any such basis, with K ′0 = Xδ. If

there is no α ≤ δ such that {fαξ : ξ < α} ⊆ Xα and pδ � α is an accumulation point

of {fαξ : ξ < α}, let Kn = K ′

n for all n ∈ ω. Otherwise, let {αn : n ∈ ω} enumerateall α ≤ δ such that {fα

ξ : ξ < α} ⊆ Xα and pδ � α is an accumulation point of{fα

ξ : ξ < α}, each one enumerated infinitely many times by both even and oddintegers. Now define Kn by induction as follows. K0 = Xδ. If Kn has been defined,by (6) for β = δ, pδ is a strong limit point of {Xδ ∩ (πδ

αn)−1[{fαn

ξ }] : ξ < αn}, sothere is a ξ < αn such that pδ /∈ Xδ ∩ (πδ

αn)−1[{fαn

ξ }] ⊆ Kn. We let pδ ∈ Kn+1 ⊆K ′

n ∩ (Kn\(πδαn

)−1[{fαn

ξ }]), Kn+1 clopen. Thus the following condition holds:

(11) {Kn : n ∈ ω} is a nested clopen base for pδ, and if α ≤ δ is such that{fα

ξ : ξ < α} ⊆ Xα and pδ � α is an accumulation point of {fαξ : ξ < α},

then there are infinitely many even and infinitely many odd n such that ∃ξ(pδ /∈Xδ ∩ (πδ

α)−1[{fα

ξ }] ⊆ Kn\Kn+1).

Finally, we set

Xδ+1 ={g ∈ δ+12 : g � δ = pδ}∪⋃n even

{g ∈ δ+12 : g � δ ∈ Kn\Kn+1, gδ = 0}∪⋃

n odd

{g ∈ δ+12 : g � δ ∈ Kn\Kn+1, gδ = 1}.

It remains only to check (2)–(6) for δ + 1. All except (6) are easy, and (6) isobvious if α = δ + 1. So assume that α ≤ δ, h ∈ Xα is an accumulation point of{fα

ξ : ξ < α}, k ∈ Xδ+1 ∩ (πδ+1α )−1[{h}]. To show that k is a strong limit point of

{Xδ+1 ∩ (πδ+1α )−1[{fα

ξ }] : ξ < α}, let k ∈ U δ+1g with g ∈ F 2, F ⊆ δ + 1, F finite.

Without loss of generality, δ ∈ F .

Case 1. k � δ �= pδ. Say k � δ ∈ Kn\Kn+1 with n even. Thus k(δ) = 0 = g(δ).Thus k � δ ∈ U δ

g�δ ∩ (Kn\Kn+1), so by (6) for δ choose ξ < α such that

k � δ /∈ (πδα)

−1[{fαξ }] ⊆ U δ

g�δ ∩ (Kn\Kn+1).

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It follows that k /∈ (πδ+1α )−1[{fα

ξ }] ⊆ U δ+1g , since if l ∈ (πδ+1

α )−1[{fαξ }] then

l � δ ∈ Kn\Kn+1 and n is even, so l(δ) = 0 = g(δ).

Case 2. k � δ = pδ. Say g(δ) = 0. Choose m even such that Km ⊆ U δg�δ

and pδ /∈ Xδ ∩ (πδα)

−1[{fαξ }] ⊆ Km\Km+1 for some ξ < α. Then k /∈ Xδ+1 ∩

(πδ+1α )−1[{fα

ξ }] ⊆ U δ+1g , as desired.

This completes Kunen’s construction.

Another generalization of Fedorchuk’s example is found in Koszmider [99]; seeChapter 3, (20) in the discussion of cHr.

On the relationships of hd with the other functions, note also that by Theorem 16.1we have π(A) ≤ hd(A). π(A) is strictly less than hd(A) in P(κ), for example. Andwe have s(A) < hd(A) forA the interval algebra on a Suslin line, and hd(A) < χ(A)for a Kunen line (Chapter 8).

There is a model of ZFC with a BA A such that s(A), d(A) < hd(A) (aremark of I. Juhasz (email message in February, 1995). Namely, take a model withMA(σ-centered) + ∃ a 0-dimensional Suslin line S, let K be a compactification ofω such that K\ω = S, and let A = clop(K). (See Weiss, van Mill [84].)

From the result that π(A) ≤ s(A) · (t(A))+ it follows that hd(A) ≤ s(A) ·(t(A))+. It is also true that hd(A) ≤ Irr(A). In fact, we have hd(A) = πH+A, andfor any homomorphic image B of A we have π(B) ≤ Irr(B) ≤ Irr(A).

If one can construct in ZFC a BA A such that hL(A) < hd(A), then onecan also construct in ZFC a BA B such that hL(B) < d(B) (see problem 145). Infact, A has a homomorphic image such that hL(A) < d(B), and hL(B) ≤ hL(A).Similarly for hL < π.

The following problems are open; these are Problems 57 and 58 in Monk [96].

Problem 150. Can one construct in ZFC a BA A such that s(A) < hd(A)?

Problem 151. Can one construct in ZFC a BA A such that hd(A) < χ(A)?

Bounded versions of hd can be defined as follows. For m a positive integer, asequence 〈xα : α < κ〉 of elements of A is said to be m-left-separated provided thatif Γ ∈ [κ]m, α < κ, and α < β for all β ∈ Γ, then aα ·

∏β∈Γ−aβ �= 0. Then we

define

hdm(A) = sup{κ : there is an m-left-separated sequence in A}.

For this notion, see Roslanowski, Shelah [98].

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17 Incomparability

We begin with one important equivalent definition:

Theorem 17.1. For any infinite BA A we have Inc(A) = sup{|T | : T is a treeincluded in A}.

(Note that when we say that T is a tree included in A, we mean merely that T isa subset of A which is a tree under the induced ordering; there is no assumptionthat incomparable elements (in T ) are disjoint (in the dual of A).)

Proof. Since any incomparable set is a tree having only roots, the inequality ≤is clear. To show equality, suppose that κ is regular and A has no incomparableset of size κ; we show that A has no tree of size κ. Suppose T is a tree of sizeκ. By Theorem 4.25 of Part I of the BA handbook, A has a dense subset D ofsize < κ. Now each level of T is an incomparable set, and hence has fewer thanκ elements. Hence T has at least κ levels. Let T ′ be a subset of T of power κconsisting exclusively of elements of successor levels. For each d ∈ D let

Md = {t ∈ T ′ : if s is the immediate predecessor of t, then d ≤ t · −s}.

Thus T ′ =⋃

d∈D Md, so there is a d ∈ D such that |Md| = κ. But then Md

is incomparable, contradiction: if y, z ∈ Md and y < z, then y ≤ u where u isthe immediate predecessor of z, and d ≤ z · −u, hence d · y = 0, contradictingd ≤ y. �

Note that if Inc(A) is attained, then it is obviously attained in the tree sense. Theconverse also holds, as Todorcevic pointed out in a letter to the author severalyears before Monk [90] appeared; this solves Problem 52 in Monk [90]. We givethis result here, following the proof in an email message from Shelah of December1990.

Theorem 17.2. If A is an infinite BA and there is a tree T ⊆ A with |T | = Inc(A),then A has an incomparable subset of power Inc(A).

Proof. By the proof of Theorem 17.1 we may assume that λdef= Inc(A) is singular.

Let 〈κα : α < cf(λ)〉 be an increasing sequence of cardinals with supremum λ

, ,OI 10.1007/978-3- - - _ ,


014


Basel 218

461

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462 Chapter 17. Incomparability

and with cf(λ) < κ0. Without loss of generality, T has no level of size λ. Now weconsider two cases.

Case 1. For every α < cf(λ) there is a β such that T has at least κα elements oflevel β. For any ordinal β let levβ(T ) be the set of elements of T of level β. Byan easy construction we obtain a strictly increasing sequence 〈βα : α < cf(λ)〉 ofordinals such that |levβα+1T | > max(κα, |levβα(T )|) for all α < cf(λ). For everyα < cf(λ) let Sα be a subset of levβα+1T of power (max(κα, |levβα |))+ such thatall elements of Sα have the same predecessors at level βα. Note that if α < cfλthen

Rαdef= {t ∈ Sα : t ≤ s for some s ∈ Sγ with α < γ < cf(λ)}

has power at most cf(λ). Now the set⋃

α<cfλ(Sα\Rα) is incomparable of size λ,as desired.

Case 2. Case 1 fails to hold. Then clearly T must have at least λ levels. HenceDepth(A) = λ and c(A) = λ, so by the Erdos–Tarski theorem A has a disjointsubset of power λ; it is also an incomparable subset. �

Concerning attainment of Inc, several things are known. Milner and Pouzet [86]proved a general result, of which a special case is that if Inc(A) = λ with cf(λ) = ω,then Inc(A) is attained. has shown that if 2ω is weakly inaccessible, then there isa BA of size 2ω with incomparability 2ω not attained. The statement in Monk [90]about attainment due to Shelah has been withdrawn by him. Instead, in Shelah[92] he proves that Inc(A) is always attained for singular cardinals. We give thisinteresting proof here.

Theorem 17.3. Inc is attained for singular cardinals.

Proof. Suppose that Inc(B) = λ, with λ singular. Now we choose 〈λi : i < cf(λ)〉and 〈Ai : i < cf(λ)〉 so that the following conditions hold:

(1) 〈λi : i < cf(λ)〉 is a strictly increasing sequence of regular cardinals all greaterthan cf(λ) and with supremum λ.

(2) 〈Ai : i < cf(λ)〉 is a system of incomparable sets in B with |Ai| = λ+i for all

i < cf(λ).

It is clear that this can be done. In addition, if possible we choose these things sothat the following condition holds:

(3) If i < j < cf(λ), x ∈ Ai, and y ∈ Aj , then y �≤ x.

Now let A =⋃

i<cf(λ) Ai. The following notation will also be useful. Let C ⊆ B.

For any x ∈ B, C ↑ x = {y ∈ C : y ≥ x}, C ↓ x = {y ∈ C : y ≤ x}, and for anycardinal μ, C ↑μ= {x ∈ C : |C ↑ x| < μ} and C ↓μ= {x ∈ C : |C ↓ x| < μ}.Case 1. There is a μ < λ such that |A ↑μ | = λ. For each x ∈ (A ↑μ) letf(x) = (A ↑μ) ↑ x. Thus f : (A ↑μ) → P(A ↑μ) and |f(x)| < μ < λ for allx ∈ (A ↑μ). Hence by Hajnal’s free set Theorem (see the Handbook, Part III, p.

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Chapter 17. Incomparability 463

1231) we get E ⊆ (A ↑μ) of size λ such that x /∈ f(y) for all distinct x, y ∈ E. SoE is incomparable, as desired.

Case 2. There is a μ < λ such that |A ↓μ | = λ. Similarly.

Case 3. For every i < cf(λ) there is an xi ∈ A such that λi < |A ↑ xi| < λ. Wenow define a function μ : cf(λ) → cf(λ) by induction. Having defined μj for allj < i, choose μi < cfλ so that λμi >

∑j<i(|A ↑ xμj | + λμj ). Now let A′

i = Aμi ,λ′i = λμi , x

′i = xμi for all i < cf(λ). Then (1)–(2) hold for A′

i and λ′i, (3) holds for

the A′i’s if it held for the Ai’s, and

(4) If j λ′

i, cf λ < λi, and A =⋃j<cf λ Aj , so there is an α(i) < cf λ such that

∣∣∣∣∣∣Aα(i) ∩ (A ↑ x′i)\

⋃j λ′i.

Clearly α(i) ≥ μi. We define β : cf(λ) → cf(λ) by induction. If β(j) has beendefined for all j < i, choose β(i) < cf(λ) such that supj<i max{α(β(j)), μβ(j)} <μβ(i). Now choose

A∗i ⊆ Aα(β(i)) ∩ (A ↑ x′

β(i))\⋃j<i

(A ↑ x′β(j))

of size λ′β(i). Note that α(β(j)) < μβ(i) ≤ α(β(i)) for j < i < cf λ. Let λ∗

i = λ′β(i).

Then (1)–(2) hold for A∗i and λ∗

i , (3) holds for the A∗i ’s if it held for the Ai’s (since

A∗i ⊆ Aα(β(i))), and

(5) if i < j < cf(λ), x ∈ A∗i , and y ∈ A∗

j , then x �≤ y.

For, otherwise x′β(i) ≤ x ≤ y /∈ (A ↑ x′

β(i)), contradiction. Now let A�i = {x :

−x ∈ A∗i }. Then (1)–(3) hold for A�

i and λ∗i . Therefore by the initial choices, (3)

itself holds. So, (3) and (5) hold for A∗i and λ∗

i . It follows that⋃

i<cf(λ) A∗i is an

incomparable set of size λ, as desired.

Case 4. For every i < cf(λ) there is an x ∈ A such that λi < |A ↓ x| < λ. Similarto Case 3.

Case 5. None of the previous cases. By ¬Case 3, there is an i(∗) < cf(λ) such thatfor all x ∈ A we have ¬(λi(∗) < |A ↑ x| < λ). By ¬Case 2, |A ↓λ+

i(∗)| < λ, and by

¬Case 1, |A ↑λ+i(∗)| < λ. Choose x∗ ∈ A\((A ↓λ+

i(∗)) ∪ (A ↑λ+

i(∗))). Thus |A ↑ x∗| ≥

λ+i(∗), so by the choice of i(∗) we have |A ↑ x∗| = λ. Also, |A ↓ x∗| ≥ λ+

i(∗) > cf(λ),

so there is a j(∗) < cf(λ) such that |(A ↓ x∗) ∩ Aj(∗)| ≥ cf(λ). Choose distinctyi ∈ (A ↓ x∗) ∩ Aj(∗) for i < cf(λ).

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For each i < cf(λ) let A′i = Ai ∩ (A ↑ x∗). Thus A′

i is an incomparable setand ∣∣∣∣∣∣

⋃i<cf(λ)

A′i

∣∣∣∣∣∣ =∣∣∣∣∣∣(A ↑ x∗) ∩

⋃i<cf(λ)

Ai

∣∣∣∣∣∣ = |(A ↑ x∗) ∩A| = λ.

Finally, {yi + x · −x∗ : i < cf(λ), x �= x∗, x ∈ A′i} is an incomparable set of size

λ, as desired. �

These results leave the following problem open.

Problem 152. Is it true that for every regular limit cardinal κ there is a BA withInc(A) = κ not attained?

Now we turn to algebraic operations, as usual. If A is a subalgebra or homomorphicimage of B, then Inc(A) ≤ Inc(B). The difference here can clearly be large. Forour special subalgebras, the difference can be large when A ≤free B, and hencealso when A ≤proj B, A ≤u B, A ≤rc B, A ≤reg B, and A ≤σ B. Clearly this isalso true for A ≤π B and for A ≤mg B. But we have the following problem.

Problem 153. Completely describe the behaviour of Inc for special subalgebras.

If A is a subalgebra of B, then, easily, Inc(A × B) ≥ |A|; in fact, {(a,−a) :a ∈ A} is an incomparable set in A × B. Hence if A is cardinality-homogeneousand has no incomparable set of size |A|, then A is rigid (this follows from someelementary facts concerning automorphisms; see the article in the BA handbookabout automorphisms). Thus the incomparability of a product can jump from thatin a factor – for example, if A is such that Inc(A) < |A|, we have Inc(A×A) = |A|.Finally, Inc(A⊕B) = max(|A|, |B|) if |A|, |B| ≥ 4, since A⊕C ∼= A×A if |C| = 4.

Ultraproducts: Inc is an ultra-sup function, so Theorems 3.20–3.22 hold, The-orem 3.22 saying that Inc

(∏i∈I Ai/F

)≥∣∣∏

i∈I IncAi/F∣∣ for F regular, and Don-

der’s theorem says that under V = L the regularity assumption can be removed.In Shelah [97a] it is shown that > is consistent, but we do not know whether thiscan be done in ZFC; this is problem 59 in Monk [96],

Problem 154.Do there exist in ZFC a system 〈Ai : i ∈ I〉 of infinite BAs, I infinite,and a regular ultrafilter F on I such that Inc

(∏i∈I Ai/F

)>∣∣∏

i∈I IncAi/F∣∣?

In the other direction, we have the following result of Shelah and Spinas [00], partof 1.7:

It is consistent that there exist cardinals κ, μ, an ultrafilter D on κ, and a sys-tem 〈Bi : i < κ〉 of interval algebras, such that

∣∣∏i<κ Inc(Bi)/D

∣∣ = μ++ whileInc(

∏i<κ Bi/D) ≤ μ+.

This answers problem 60 of Monk [96].

Concerning derived functions of incomparability, we mention only a result of She-lah (email message of December 1990), solving Problem 53 in Monk [90]:

[email protected]

Theorem 17.4. If Inc(A) = ω, then CardH−(A) = ω.

Proof. Suppose that Inc(A) = ω < CardH−(A). Without loss of generality, assumethat A is a subalgebra of P(ω) containing all of the finite subsets of ω. Hencethere is an a ∈ A such that both a and ω\a are infinite. Let F be a nonprincipalultrafilter on A � a. Then we can construct 〈aα : α < χ(F )〉, each aα ⊆ a, such thatfor each α < χ(F ), the element aα is not in the filter generated by {aβ : β < α};in particular, β < α ⇒ aβ �≤ aα. Similarly we get a nonprincipal ultrafilter G onA � −a and a sequence 〈bα : α < χ(G)〉 of subelements of −a such that β < α⇒ bβ �≤ bα. Say χ(F ) ≤ χ(G). Then 〈aα + −bα · −a : α < χ(F )〉 is a system ofincomparable elements. By 14.10, alt(A) ≤ χ(F ), so a(A) = ω. Hence 14.12 givesa contradiction. �

Next we consider “small” versions of Inc. Thus

Incspec(A) = {|X | : X ⊆ A is maximal incomparable};Incmm(A) = min(Incspec(A)).

Charles Scherer has studied these notions and proved the following:

(1) For any infinite cardinal κ, Incmm(finco(κ)) = ω.

(2) Incmm(P(κ)) = ω.

(3) Incmm(A) = ω if A has Fr(ω) as a dense subalgebra.

The following is a slight generalization of (2);

Proposition 17.5. [ω, κ] ⊆ Incspect(P(κ)). Moreover, if κ < 2μ ≤ 2κ, then 2μ ∈Incspect(P(κ)).

Proof. Suppose that ω ≤ ν ≤ κ. Let M ⊆ κ with |M | = ν. Then {{α} : α ∈M}∪{κ\M} is maximal incomparable of size ν. Now suppose that κ < 2μ ≤ 2κ. Thenwe may assume that μ ≤ κ. Let N be a subset of κ of size μ, and let A be maximalincomparable in P(N) of size 2μ. (For example, extend an independent subset ofP(N) of size 2μ to a maximal incomparable set.) Then {{α} : α ∈ κ\N} ∪ A ismaximal incomparable of size 2μ. �

Problem 155. If κ is an infinite cardinal and κ < μ < 2κ, μ not a power of 2, isthere a maximal incomparable set in P(κ) of size μ?

Proposition 17.6. Incspect(Fr(κ)) = {κ}. �Proposition 17.7. Suppose that X is incomparable in A. Then X is maximal in-comparable iff ∀a ∈ A\X∃x ∈ X(a ≤ x or x ≤ a). �Proposition 17.8. Suppose that X is maximal incomparable in A. Then

(i)∑

X = 1.

(ii)∏

X = 0.

(iii) {a ∈ A : −a ∈ X} is maximal incomparable.

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Proof. (i): if∑

X �= 1, then there is an a ∈ A such that a · x = 0 for all x ∈ X ,and a �= 0. Then X ∪ {a} is still incomparable, contradiction.

(iii): clear; and (ii) follows from (i) and (iii). �

The equivalent version of Inc given in Theorem 17.1 gives rise to a another “small”function. If A is a BA and T1, T2 are trees contained in A (i.e., trees in the inducedordering from A), then we say that T2 extends T1 iff T1 ⊆ T2 and for every x ∈ T2

and y ∈ T1, if x ≤ y then x ∈ T1; so T2 is an end-extension of T1, but there may benew roots and elements above them. We can apply Zorn’s lemma to get maximaltrees in A under this partial ordering. We define

Inctreespect(A) = {|T | : T is an infinite maximal tree

contained in A and 0 ∈ T, 1 /∈ T };Inctreemm(A) = min(Inctreespect(A)).

We require that 1 /∈ T in order to make the notion nontrivial. Since we take endextensions, if we allowed 1 as an element we could take a well-ordered chain of anylength that is present in the algebra and adjoin 1 to it to obtain a maximal tree.

Proposition 17.9. {κ, 2κ} ⊆ Inctreespect(P(κ)) for any infinite cardinal κ.

Proof. {∅} ∪ {{α} : α < κ} is a maximal tree. If X ⊆ P(κ) is independent with|X | = 2κ, then we can extend X to a maximal tree. �

Problem 156. Is Inctreespect(P(κ)) = {κ, 2κ}?

Proposition 17.10. Inctreespect(Fr(κ)) = {κ} for any infinite cardinal κ.

Proof. Suppose that T is a tree in Fr(κ) and |T | < κ; we want to show that T isnot maximal. Let 〈xα : α < κ〉 be a system of free generators for Fr(κ). Choose αsuch that xα is not in the support of any element of T . Clearly T ∪ {xα} is a treeextending T . �

Proposition 17.11. Let T be a tree in A. Then the following are equivalent:

(i) T is maximal.

(ii) For every x ∈ A\T one of the following conditions holds:

(a) There are incomparable u, v ∈ T such that u+ v ≤ x.

(b) There is a u ∈ T such that x < u.

Proof. (i)⇒(ii): Assume (i), and suppose that x ∈ A\T . Then either x = 1, hence(a) holds, T ∪ {x} is not a tree, or it is a tree but is not an end extension of T .If T ∪ {x} is not a tree, then there exist u, v ∈ T such that either u and v areincomparable and both are less than x, giving (a), or v and x are incomparableand both are less than u, and this gives (b). If T ∪ {x} is a tree but not an endextension of T , this gives (b).

(ii)→(i): obvious. �

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Some more remarks:

1. For A = Fr(κ) with κ > ω we have: Inctreemm(A) = κ, a(A) = ω, Lengthmm(A)=ω, and s(A) = ω.

2. For A = P(ω) we have: Inctreemm(A) = ω and i(A) > ω.

We finish our discussion of small incomparability with some facts about P(ω)/fin.

Theorem 17.12. If T is a nonempty countable collection of infinite, co-infinitesubsets of ω, then there is an infinite subset X of ω such that X\Y and Y \X areinfinite for all Y ∈ T .

Proof. Let T = {Yi : i < ω} possibly with repetitions (necessary if T is finite).We define increasing finite subsets F0, F1, . . . and G0, G1, . . . of ω by recursion.Suppose that Fj and Gj have been defined for all j < i. Let Hi =

⋃j<i Fj and

Ki =⋃

j<i Gj . Now for all j < i choose

mij ∈ ω\(Hi ∪Ki ∪ Yj ∪ {nik : k < j}) andnij ∈ Yj\(Hi ∪Ki ∪ {mik : k ≤ i}).

Let Fi = Hi ∪ {mij : j < i} and Gi = Ki ∪ {nij : j < i}.

(1) Fi ∩Gi = ∅ for all i.

We prove this by induction on i. Assume that it is true for all j < i. ThenHi ∩Ki = ∅. We have mij /∈ Ki for all j < i, so Fi ∩Ki = ∅. Also, nij /∈ Hi for allj < i, so Gi ∩Hi = ∅. And nij /∈ {mik : k < i}, so Fi ∩Gi = ∅.

Now let X =⋃

i<ω Fi and Z =⋃

i<ω Gi.

(2) X\Yj is infinite for all j < ω.

For, if i > j, then mij ∈ X\Yj, and if i > k > j, then mkj ∈ Fk ⊆ Hi, and somij �= mkj . So (2) holds.

(3) Yj\X is infinite for all j < ω.

For, if i > j, then nij ∈ Z ∩ Yj , and if i > k > j, then nkj ∈ Gk ⊆ Ki, and sonij �= nkj . Hence Yj ∩ Z is infinite. By (1), (3) follows. �

Corollary 17.13. ω < Incmm(P(ω)/fin), Inctreemm(P(ω)/fin). �Theorem 17.14. (MA) If T is a collection of infinite, co-infinite subsets of ω and|T | < 2ω, then there is an infinite subset X of ω such that X\Y and Y \X areinfinite for all Y ∈ T .

Proof. Again we apply MA to the partially ordered set Fn(ω, 2). For each a ∈ Tand i ∈ ω let

Dai = {f ∈ Fn(ω, 2) : ∃j > i[j ∈ dmn(f) ∩ a and f(j) = 0]}.

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Then each set Dai is dense. For, suppose that g ∈ Fn(ω, 2). Choose j > i withj ∈ a and j /∈ dmn(g), and let f = g ∪ {(j, 0)}. Clearly g ⊆ f ∈ Dai.

For each a ∈ T and i ∈ ω let

Eai = {f ∈ Fn(ω, 2) : ∃j > i[j ∈ dmn(f)\a and f(j) = 1]}.

Then each set Eai is dense, with proof as above.

Now if G is a generic filter intersecting all of these dense sets, then Xdef=

{j ∈ ω : j ∈ dmn(⋃G) and (

⋃G)(j) = 1} is the set desired in the theorem. �

Corollary 17.15. (MA) Incmm(P(ω)/fin) = Inctreemm(P(ω)/fin) = 2ω. �Problem 157. Is it consistent that Incmm(P(ω)/fin) < 2ω or Inctreemm(P(ω)/fin) <2ω?

Problem 158. Is it consistent that Incmm(P(ω)/fin) �= Inctreemm(P(ω)/fin)?

Now we turn to connections with our other functions. From the Handbook The-orem 4.25 it follows that π(A) ≤ Inc(A) for any infinite BA A; hence hd(A) ≤Inc(A), by an easy argument. An example in which they are different is the intervalalgebra A on the reals. In fact, hd(A) = ω by Theorem 16.1, and an incomparableset of size 2ω is provided by

{[0, r) ∪ [1 + r, 2) : r ∈ (0, 1)}.

Much effort has been put into constructing BAs A in which Inc(A) < |A|. Anexample with |A| a successor cardinal is found in Shelah [97a]. An example ofsuch an algebra is an algebra of Bonnet, Shelah [85]; their algebra is an intervalalgebra, and has power cf(2ω), so that if cf(2ω) is not limit one gets such an algebra.In any case, every incomparable set in their algebra has size less than cf(2ω). Wegive this construction here.

Throughout the next results, let λ = 2ω and κ = cf(λ).

A subset A of nS is good iff each member of A is one-one.

If (L,≤) is a linear order, n ≥ 2, and ε ∈ n2, then we define x ≤ε y iffx, y ∈ nL and for all k < n, ε(k) = 1 → xk ≤ yk, while ε(k) = 0 → yk ≤ xk.Suppose that |L| = λ. We say that L is hyper-rigid iff ∀n ≥ 2∀ε ∈ n2[every goodantichain of (nL,≤ε) has size less than λ].

We say that a subset P of R is κ-dense iff every nonempty open interval inR contains κ elements of P .

Lemma 17.16. If P is a κ-dense hyper-rigid subset of R, then Inc(Intalg(P )) < κ.

Proof. Suppose that A is an antichain in Intalg(P ) of size κ. For each a ∈ A write

a = [xa0 , x

a1) ∪ · · · ∪ [xa

2m(a), xa2m(a)+1)

with −∞ ≤ xa0 < xa

1 < · · · < xa2m(a) < xa

2m(a)+1 ≤ ∞. Then there exist a subset

A′ of A of size κ and an integer n such that m(a) = n for all a ∈ A′. Let B be

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the set of all sequences 〈(ri, r′i) : i ≤ 2n〉 of pairs of rational numbers such thatr0 < r′0 < · · · < r2n < r′2n. Then

A′ =⋃{{a ∈ A′ : xa

0 < r0 < r′0 < xa1 < · · · < xa

2n < r2n < r′2n < xa2n+1}

: 〈(ri, r′i) : i ≤ 2n〉 ∈ B}.

It follows that there exist a A′′ ∈ [A]κ and a 〈(ri, r′i) : i ≤ 2n〉 ∈ B such that forall a ∈ A′′,

xa0 < r0 < r′0 < xa

1 < · · · < xa2n < r2n < r′2n < xa

2n+1.

Now take any k < 2n + 2. We define a ≡ a′ iff a, a′ ∈ A′′ and xak = xa′

k . Thisgives an equivalence relation, and hence there is a A′′′ ∈ [A′′]κ such that one ofthe following conditions holds:

(1) xak = xa′

k for all a, a′ ∈ A′′′.

(2) xak �= xa′

k for all distinct a, a′ ∈ A′′′.

Repeating this argument finitely many times, we arrive at a set A0 ∈ [A′′′]κ suchthat one of the following conditions hold for all k ≤ 2n+ 1:

(3) xak = xa′

k for all a, a′ ∈ A0.

(4) xak �= xa′

k for all distinct a, a′ ∈ A0.

Let R = {k ≤ 2n + 1 : ∀a, a′ ∈ A0[a �= a′ → xak �= xa′

k ]}. Clearly R �= ∅. WriteR = {k0, . . . , kp−1} with k0 < · · · < kp−1. Define ε ∈ p2 by setting

ε(i) =

{1 if ki is odd,

0 if ki is even.

Now define ca ∈ pP for any a ∈ Intalg(P ) by setting

ca(i) = xaki.

Then we claim

(5) For any a, a′ ∈ A0, a ≤ a′ iff ca ≤ε ca′.

In fact, suppose that a, a′ ∈ A0. Then

a ≤ a′ iff ∀i ≤ n[[i even → xa′i ≤ xa

i ]

and [i odd → xai ≤ xa′

i ]]

iff ca ≤ε ca′.

Since A0 is an antichain, it follows that {ca : a ∈ A0} is a good antichain of(pP,≤ε). Since it is of size κ, this is a contradiction. �

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Let P ⊆ R, n ≥ 2, and ε ∈ nP .

A subset A of nP is separate iff the following two conditions hold:

(1) There is a sequence 〈ri : i < 2n〉 of rational numbers such that for all a ∈ A,

r0 < a0 < r1 < r2 < a1 < · · · < an−2 < r2n−3 < r2n−2 < an−1 < r2n−1.

(2) If a, b ∈ A are distinct and k < n, then ak �= bk.

Proposition 17.17. Suppose that A is separate, with notation as above. For eachk < n let Ik = (r2k, r2k+1). Then the Ik’s are pairwise disjoint, and ak ∈ Ik forall k < n. �

Suppose that A is separate, with notation as above, and suppose that 0 ≤ l < n.Then we define

a[l] = 〈a0, a1, . . . , al−1, al+1, . . . , an〉 for any a ∈ A;

A[l] = {a[l] : a ∈ A};Al = {al : a ∈ A};

ψl(a) = a[l];

πl(a[l]) = al.

Note here that ψl is one-one and πl(a[l]) = πl(ψ−1l (a[l])).

Now we define a ≤lε b iff a, b ∈ n−1R and the following two conditions hold:

(1) If k < l and ε(k) = 1, then ak ≤ bk;

If k < l and ε(k) = 0, then bk ≤ ak.

(2) If l ≤ k < n− 1 and ε(k + 1) = 1, then ak ≤ bk;

If l ≤ k < n− 1 and ε(k + 1) = 0, then bk ≤ ak.

Also we define u ≤1 v iff u, v ∈ R and u ≤ v, and u ≤0 v iff u, v ∈ R and v ≤ u.

Proposition 17.18. For a, b ∈ A we have

a ≤ε b iff a[l] ≤lε b[l] and al ≤ε(l) bl.

Proof.

a[l] ≤lε b[l] and al ≤ε(l) bl iff

∀k < l[[ε(k) = 1→ ak ≤ bk] and [ε(k) = 0→ bk ≤ ak]]

and ∀k[l ≤ k < n− 1 and ε(k + 1) = 1→ (a[l])k ≤ (b[l])k]

and ∀k[l ≤ k < n− 1 and ε(k + 1) = 0→ (b[l])k ≤ (a[l])k]

and al ≤ε(l) bl

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iff ∀k < l[[ε(k) = 1→ ak ≤ bk] and [ε(k) = 0→ bk ≤ ak]]

and ∀k[l ≤ k < n− 1 and ε(k + 1) = 1→ ak+1 ≤ bk+1]

and ∀k[l ≤ k < n− 1 and ε(k + 1) = 0→ bk+1 ≤ ak+1]

and al ≤ε(l) bl

iff a ≤ε b. �

Proposition 17.19. Let A be a separate subset of nR. Then the following are equiv-alent:

(i) A is an antichain of (nR,≤ε).

(ii) There is an l < n such that πl is a decreasing function from (A[l],≤lε) to

(Al,≤ε(l)).

(iii) For every l < n, πl is a decreasing function from (A[l],≤lε) to (Al,≤ε(l)).

Proof. (i)⇒(iii): Assume (i), and suppose that l < n and a[l] ≤lε b[l]. Then a �≤ε b,

and so by Proposition 17.18, al �≤ε(l) bl. Hence bl <ε(l) al.

(iii)⇒(ii): obvious.

(ii)⇒(i): Assume that (i) is false. Say a �= b and a ≤ε b. Then by Proposition17.18, al ≤ε(l) bl. �

A subset A of nR is a nice antichain iff the following conditions hold:

(1) A is good and separate.

(2) A is an antichain of size κ.

(3) For every l < n, the set A[l] does not have an antichain of size κ in (n−1R,≤lε).

Proposition 17.20. Let P be a κ-dense subset of R. Then the following conditionsare equivalent:

(i) P is hyper-rigid.

(ii) For each n ≥ 2 and ε ∈ n2 there is no nice antichain in (nP,≤ε).

Proof. Obviously (i) implies (ii). Now assume that P is not hyper-rigid. Let n ≥ 2be smallest such that for some ε ∈ n2 the set (nP,≤ε) has a good antichain A ofsize κ.

Since A is good, every member of A is one-one, and so for every a ∈ Athere is a permutation σa of n such that aσa(0) < · · · < aσa(n−1). Hence thereexist an A′ ∈ [A]κ and a permutation τ of n such that ∀a ∈ A′[σa = τ ]. letA′′ = {a ◦ τ : a ∈ A′.

Now for each a ∈ A′′ choose a strictly increasing sequence 〈rai : i < 2n〉of rational numbers such that ra2k < ak < ra2k+1 for all k < n. Then there exista single strictly increasing sequence 〈si : i < 2n〉 of rational numbers and anA′′ ∈ [A]]κ such that s2k < ak < s2k+1 for all k < n.

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Now we claim

(1) S0def= {a0 : a ∈ A′′} has size κ.

In fact, suppose not. Then there exist A′′′ ∈ [A′′]κ and t such that a0 = t forall a ∈ A′′′. By Proposition 17.20, (A′′′,≤ε) is isomorphic to (A′′′[0],≤0

ε). ClearlyA′′′[0] is a good antichain in n−1P , contradicting the minimality of n.

By (1), there is a subset A′′′ of A′′ of size κ such that for any two distincta, b ∈ A′′′ we have a0 �= b0.

Now we apply this argument to 1, . . . , n−1; we end up with a set A0 ∈ [A′′]κ

such that for all distinct a, b ∈ A0 and all k < n, ak �= bk. Thus A0 is nice. �

Theorem 17.21. (Corominas, Shelah) Suppose that λ is an infinite cardinal, and(E,≤) is a poset of size λ. Suppose that D is a subset of E of size < cf(λ) suchthat the following conditions hold:

(i) If x < y in E, then there is a d ∈ D such that x ≤ d ≤ y.

(ii) For any x ∈ E there are d1, d2 ∈ D such that d1 ≤ x ≤ d2.

Then every subset F of E of size λ either contains an antichain of size λ or achain isomorphic to Q.

Proof. Assume the hypothesis, with F a subset of E of size λ. Let G be a subset ofF of size λ, and let G∗ = G ∪D. For each x ∈ G∗ we define εx = (ε−x , ε

+x ) ∈ 2× 2

as follows.

Let G−x = {t ∈ G : t ≤ x} and G+

x = {t ∈ G : x ≤ t}. Then we set

ε−x =

{0 if |G−

x | < λ,

1 if |G−x | = λ;

ε+x =

{0 if |G+

x | < λ,

1 if |G+x | = λ;

G−∗ = {x ∈ G∗ : ∃d ∈ D[x ≤ d and ε−d = 0]};

G+∗ = {x ∈ G∗ : ∃d ∈ D[d ≤ x and ε+d = 0]};

Since |D| < cf(λ) and |G−d | < λ for each d ∈ D, it follows that |G−∗ | < λ. Similarly,

|G+∗ | < λ. Hence N(G)def= G\(G−∗ ∪G+∗ ) has size λ.

(1) ε−x (G) = ε−x (N(G)).

In fact,

{t ∈ G : t ≤ x} = {t ∈ G−∗ : t ≤ x} ∪ {t ∈ G+

∗ : t ≤ x} ∪ {t ∈ N(G) : t ≤ x}.

Now |G−∗ | < λ and |G+∗ | < λ; so |{t ∈ G : t ≤ x}| = λ iff |{t ∈ N(G) : t ≤ x}| = λ.So (1) holds.

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Similarly,

(2) ε+x (G) = ε+x (N(G)).

Case 1. For every subset G of F of size λ there is an x ∈ G such that εx = (1, 1).

Choose a(1/2) ∈ F such that εa(1/2) = (1, 1). Then the sets F1/4def= {x ∈ F :

x ≤ a1/2} and F3/4def= {x ∈ F : a1/2 ≤ x} have size λ. Choose a(1/4) ∈ F1/4 and

a(3/4) ∈ F3/4 so that εa(1/4) = εa(3/4) = (1, 1). Then the following sets have size λ:

F1/8def= {x ∈ F1/4 : x ≤ a(1/4)},

F3/8def= {x ∈ F1/4 : a(1/4) ≤ x} = {x ∈ F1/4 : a(1/4) ≤ x ≤ a(1/2)},

F5/8def= {x ∈ F3/4 : x ≤ a(3/4)} = {x ∈ F3/4 : a(1/2) ≤ x ≤ a(3/4)},

F7/8def= {x ∈ F3/4 : a(3/4) ≤ x}.

Now suppose that k ≥ 3 and we have defined a(2s−1)/2k−1 for all s such that

1 ≤ s ≤ 2k−2 so that the following sets have size λ:

F1/2k ⊆ {x ∈ F : x ≤ a(1/2k−1)};F(2s−1)/2k ⊆ {x ∈ F : a((s− 1)/2k−1) ≤ x ≤ a(s/2k−1)} for 1 < 2s− 1 < 2k − 1

F(2k−1)/2k ⊆ {x ∈ F : a((2k−1 − 1)/2k−1) ≤ x}.

Note that these conditions hold for k = 3. We now choose a((2s − 1)/2k) for alls such that 1 ≤ s ≤ 2k−1 so that εa((2s−1)/2k)(F(2s−1)/2k ) = (1, 1). It follows thatthe following sets have size λ:

F1/2k+1 = {x ∈ F1/2k : x ≤ a(1/2k)};F3/2k+1 = {x ∈ F1/2k : a(1/2k) ≤ x};

F(2s−1)/2k+1 = {x ∈ F(2t−1)/2k : x ≤ a((2t− 1)/2k)}if s = 2t− 1 and 5 ≤ 2s− 1 < 2k+1 − 2;

F(2s−1)/2k+1 = {x ∈ F(2t−1)/2k : a((2t− 1)/2k) ≤ x}if s = 2t and 5 ≤ 2s− 1 < 2k+1 − 2;

F(2k+1−2)/2k+1 = {x ∈ F(2k−1)/2k : a((2k − 1)/2k) ≤ x};F(2k+1−1)/2k+1 = {x ∈ F(2k−1)/2k : x ≤ a((2k − 1)/2k)}.

It follows that

F1/2k+1 ⊆ {x ∈ F : x ≤ a(1/2k)};F(2s−1)/2k+1 ⊆ {x ∈ F : a((s− 1)/2k) ≤ x ≤ a(s/2k)} for 1<2s−1<2k+1−1

F(2k+1−1)/2k+1 ⊆ {x ∈ F : a((2k − 1)/2k) ≤ x}.

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This finishes the inductive construction, and it shows that Q is embedded in Fvia the mapping s/2k !→ a(s/2k) for s < 2k.

Case 2. There is a subset G of F of size λ such that εx �= (1, 1) for every x ∈ G.

(3) ∀x, y ∈ N(G)[x < y → εx = (0, 1) and εy = (1, 0)].

In fact, choose d ∈ D such that x ≤ d ≤ y. Now x /∈ G−∗ , so ε−d = 1, i.e., |G−d | = λ.

Now G−d = {t ∈ G : t ≤ d}, and {t ∈ G : t ≤ d} ⊆ {t ∈ G : t ≤ y}. Hence

|{t ∈ G : t ≤ y}| = λ, so ε−y = 1. Similarly, y /∈ G+∗ . so ε+d = 1, i.e., |G+

d | = λ.

Now G+d = {t ∈ G : d ≤ t}, and t ∈ G : d ≤ t} ⊆ {t ∈ G : x ≤ t}. Hence

|{t ∈ G : x ≤ t}| = λ, so ε+x = 1. By the case we are in, εx �= (1, 1) �= εy, soεx = (0, 1) and εy = (1, 0), as desired in (3).

(4) There do not exist a, b, c ∈ N(G) such that a < b < c.

In fact, otherwise by (3) we get εb = (1, 0) = (0, 1), contradiction.

Now it follows that N(G) has an antichain of size λ. In fact, let X be amaximal antichain in N(G). We may assume that |X | < λ. Now for each u ∈N(G)\X there is an x ∈ X such that u and x are comparable. Let Y = {u ∈N(G)\X : ∃x ∈ X [x ≤ u]} and Z = {u ∈ N(G)\X : ∃x ∈ X [u ≤ x]}. ThenN(G)\X = Y ∪ Z, so |Y | = λ or |Z| = λ. Now Y is an antichain. For, supposethat u, v ∈ Y and u < v. Choose x ∈ X such that x ≤ u. Then x < u < v,contradicting (4). Similarly Z is an antichain. �

Let p ≥ 2, B ⊆ pR, and let f be an increasing function from (B,≤) into (R,≤).Let b ∈ pR. We define

f(b−) = sup{f(t) : t ∈ B and t ≤ b},f(b+) = inf{f(t) : t ∈ B and b ≤ t}.

Note that f(b−) = −∞ if there is no t ∈ B such that t ≤ b, and f(b+) = ∞ ifthere is no t ∈ B such that b ≤ t.

Now let D be a countable subset of pR, and let f be an increasing functionfrom (D,≤) into (R,≤). Then an element b ∈ pR is a good point for f iff eitherb ∈ D or f(b−) = f(b+). Let D∗ = G(D) be the set of good points for f . Wedefine a function D(f) = f∗ as follows. If b ∈ D, then f∗(b) = f(b). If b ∈ D∗\D,then f∗(b) = f(b−) (= f(b+)).

Proposition 17.22. f∗ is an increasing function.

Proof. Suppose that b, c ∈ D∗ and b < c. If b, c ∈ D, then f∗(b) = f(b) ≤f(c) = f∗(c). If b ∈ D and c ∈ D∗\D, then f∗(b) = f(b) ≤ f(c−) = f∗(c). Ifb ∈ D∗\D, then f∗(b) = f(b+) ≤ f(c) = f∗(c). Finally, if b, c ∈ D∗\D, thenf∗(b) = f(b−) ≤ f(c−) = f∗(c). �

The function f∗ is called the entire extension of f .

The same definitions work for a decreasing function f .

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Proposition 17.23. Let n ≥ 2 and ε ∈ 2n. Let A be a nice antichain in (nR,≤ε).Let l < n. Let Dl be a countable topologically dense subset of A[l]. Let πl be thedecreasing function from (A[l],≤l

ε) onto (Al,≤ε(l)) according to Proposition 17.19,and let ϕl be its restriction to Dl. Let D

∗l be the domain of the entire extension

ϕ∗l of ϕl.

Then A[l]\D∗l has size less than cf(2ω).

Proof. Let κ = cf(2ω). Wlog ε(k) = 1 for all k. Let S = A[l]\D∗l ; so we want to

show that |S| < κ. Let c ∈ A[l]. Then c = a[l] for a unique a ∈ A, and πl(c) = al.Now suppose that c ∈ S. In particular, c /∈ Dl. Now we have

ϕl(c−) = inf{ϕl(v) : v ∈ Dl and v < c};

ϕl(c+) = sup{ϕl(v) : v ∈ Dl and c < v};

πl(c−) = inf{πl(v) : v ∈ A[l] and v < c};

πl(c+) = sup{πl(v) : v ∈ A[l] and c < v}.

Then

(1) ϕl(c+) < ϕl(c

−).

(2) ϕl(c−) ≥ πl(c

−) ≥ πl(c+) ≥ ϕl(c

+).

Now let S− = {c ∈ S : ϕl(c−) > πl(c

−)}, S0 = {c ∈ S : πl(c−) > πl(c

+)},S+ = {c ∈ S : πl(c

+) > ϕl(c+). Clearly S = S− ∪ S0 ∪ S+. So it suffices to show

that each of S−, S0, S+ has size less than cf(λ).

S0 has size less than κ. Suppose not. For each c ∈ S0 let r(c) be a rationalsuch that πl(c

−) > r(c) > πl(c+). Then there exist a rational s and a subset S′

of S0 of size κ such that r(c) = s for all c ∈ S′. Then S′ does not have a chain oflength 3 or more. For, suppose that c < d < e in S′. Then

s > πl(c+) ≥ πl(d) ≥ πl(e

−) > s,

contradiction. It follows that S′ has an antichain of size κ, contradiction.

S− has size less than κ. In fact, take any c ∈ S−. Thus ϕl(c−) > πl(c

−),so there is a dc ∈ A[l] with dc < c and ϕl(c

−) > πl(dc) ≥ πl(c−). Let U(dc, c) =∏

k<n−1(dk, ck). Then U(dc, c) is a nonempty subset of n−1R and U(dc, c)∩Dl = ∅.In fact, suppose that w ∈ U(dc, c) ∩Dl. So dc < w < c, hence

πl(w) ≤ πl(dc) < ϕl(c−) ≤ ϕl(w) = πl(w),

contradiction. It follows that also U(dc, c) ∩A[l] = ∅. Now for each c ∈ S− chooser(c) ∈ n−1Q such that dc < r(c) < c. Suppose that |S−| ≥ κ. Then there exists ∈ n−1Q and T ⊆ S− with |T | = κ such that r(c) = s for all c ∈ T . Then T is anantichain. For, suppose that c, c′ ∈ T with c < c′. Then dc′ < r(c′) = r(c) < c < c′,hence c ∈ (dc′ , c

′) ∩ A[l], contradiction.

S+ has size less than κ: this is similar to S−. �

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Suppose that p ≥ 1, η ∈ p2, θ ∈ 2, and D is a countable subset of pR. Let ≤1

be the usual order on R and ≤0 its converse. Let M(p, η, θ,D) be the set of alldecreasing functions from (D,≤η) into (R,≤θ). Let M

∗(p, η, θ,D) be the set of allentire extensions of members of M(p, η, θ,D). Clearly M(p, η, θ,D) is of size 2ω,and hence so is M∗(p, η, τ,D). Let M∗(p, η, θ) =

⋃{M∗(p, η, θ,D) : D a countable

subset of pR}, M∗(p) =⋃{M∗(p, η, θ) : η ∈ p2, θ ∈ 2}, and M∗ =

⋃{M∗(p) : p ≥

1}. Clearly all of these sets have size 2ω. For each f ∈M∗ let n(f) be the uniquepositive integer such that f ∈M∗(n(f)).

Lemma 17.24. There is a κ-dense hyperrigid subset of R.

Proof. Let M∗ =⋃

α<κ M∗α with each M∗

α of size less than 2ω, and 〈M∗α : α < κ〉

increasing. Let 〈Iα : α < κ〉 enumerate all intervals (r, r′) with r, r′ rational, eachinterval repeated κ times.

Now we define 〈xα : α < κ〉 by recursion. Let x0 be any element of R. Supposethat xα has been constructed for all α < β. Let Pβ = {xα : α < β}. Let Tβ bethe set of all f(a) for f ∈ M∗

β and a ∈ n(f)Pβ . Clearly |Tβ | < 2ω and so also|Pβ ∪ Tβ| < 2ω. Choose xβ ∈ Iβ\(Pβ ∪ Tβ).

Let P = {xα : α < κ}. Then P is κ-dense, since each interval (r, r′) isrepeated κ many times. To show that P is hyperrigid, suppose not. Then byProposition 17.22, there exist n ≥ 2, ε ∈ n2, and a nice antichain A for (nP,≤ε).Let a ∈ A. Then a is one-one. For each i < n choose α(i) < κ such that ai = xα(i).Let q(a) be the k such that α(k) is the largest element of {α(0), . . . , α(n − 1)}.Let A′ be a subset of A of size κ and l < n such that q(a) = l for all a ∈ A′. ByProposition 17.21 let πl be the decreasing function from (A′[l],≤l

ε) to (Al,≤ε(l)).

Let θ = ε(l) and let ≤lε be εη, where η(k) = ε(k) for k < l and η(k) = ε(k+1) for

l ≤ k < n−1. Let f = ϕ∗l ∈M∗. Say f ∈M∗

α. Let D = dmn(f). Then A′[l]\D hassize less than κ, so A′[l]∩D has size κ. Choose a so that a[l] ∈ (A′[l] ∩D)\Pn−1

α .Say ai = xγ(i) for all i < n. Then if i < n and i �= l we have ϕ(i) < γ(l). Letβ = γ(l). Since a /∈ P−1

α , it follows that α < γ(k) for some k < n. So α < β.Let b = a[l]. Then f ∈ M∗

β and b ∈ Pn−1β , so xβ = f(b) ∈ Tβ, contradicting the

definition of xβ . �

Theorem 17.25. There is a subset P of R of size cf(2ω) such that intalg(P ) doesnot have an incomparable subset of size cf(2ω). �

Another example of a BA with incomparability less than cardinality is Rubin’salgebra [83] (constructed assuming ♦). Baumgartner [80] showed that it is con-sistent to have MA, 2ω = ω2, and every uncountable BA has an uncountableincomparable subset.

We give a construction, using ♦, of a BA A with IncA < |A|; the example is dueto Baumgartner, Komjath [81], and settles another question which is of interest.It depends on some lemmas. For all the lemmas let A be a denumerable atomlesssubalgebra of P(ω), and let I be a maximal ideal in A. We consider a partial

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ordering P = {(a, b) : a ∈ I, b ∈ A\I, a ⊆ b}, ordered by: (a, b) ( (c, d) iff a ⊇ cand b ⊆ d.

Lemma 17.26. The following sets are dense in P:

(i) For each m ⊆ I, the set D1(m)def= {(a, b) ∈ P : either ∀c ∈ m(c �⊆ b) or

∃c ∈ m(c ⊆ a)}.(ii) For each c ∈ A, the set D2(c)

def= {(a, b) ∈ P : ¬(a ⊆ c ⊆ b}).

(iii) For each c ∈ I, the set D3(c)def= {(a, b) ∈ P : c ⊆ a ∪ (ω\b)}.

(iv) The set D4def= {(a, b) ∈ P : a �= 0}.

Proof. Suppose (a, b) ∈ P . (i) If ∀c ∈ m(c �⊆ b), then (a, b) ∈ D1(m), as desired.Otherwise there is a c ∈ m such that c ⊆ b, and then (a∪c, b) ∈ P , (a∪c, b) ( (a, b),and (a ∪ c, b) ∈ D1(m), as desired.

(ii) If a ⊆ c ⊆ b, then there are two cases:

Case 1. c ∈ I. Thus c ⊂ b. Choose disjoint non-empty d, e such that b\c = d ∪ e.Since 1 /∈ I, one of d, e is in I; say d ∈ I. Then (a ∪ d, b) ∈ P , (a ∪ d, b) ( (a, b),and (a ∪ d, b) ∈ D2(c).

Case 2. c /∈ I. We can similarly find d ⊂ (c\a) such that d /∈ I; then (a, a ∪ d) isthe desired element showing that D2(c) is dense.

(iii) The element (a ∪ (c ∩ b), b) shows that D3(c) is dense.

(iv) If a �= 0, we are through. Otherwise choose non-empty disjoint c, d suchthat c ∪ d = b. One of c, d, say c, is in I; then (c, b) is as desired. �

Lemma 17.27. Suppose that D is dense in P. Then so are the following sets:

(i) S(D)def= {(a, b) ∈ P : (ω\b, ω\a) ∈ D}.

(ii) For e, f ∈ I, T (D, e, f)def= {(a, b) ∈ P : ((a\e) ∪ f, (b\e) ∪ f) ∈ D}.

Proof. Suppose (a, b) ∈ P . (i) We can choose (c, d) ( (ω\b, ω\a) so that (c, d) ∈ D.Thus ω\b ⊆ c and d ⊆ ω\a, so ω\c ⊆ b and a ⊆ ω\d, which shows that (ω\d, ω\c)is a desired element.

(ii) By Lemma 17.26(iii), choose (a′, b′) ( (a, b) such that e∪f ⊆ a′ ∪ (ω\b′).Let e′ = a′ ∩ e and f ′ = a′ ∩ f . By density of D, choose (x, y) ∈ D such that(x, y) ( ((a′\e) ∪ f, (b′\e) ∪ f). Now let

a′′ = (x\(e ∪ f)) ∪ e′ ∪ f ′ and b′′ = (y\(e ∪ f)) ∪ e′ ∪ f ′.

It is easy to check that (a′′, b′′) ∈ T (D, e, f) and (a′′, b′′) ( (a, b). �

Now suppose that M is a countable collection of subsets of I; then we let D(M)be the smallest collection of dense sets in P such that

(1) every set D1(m), D2(c), D3(e), D4 is in D(M) for m ∈M , c ∈ A, e ∈ I.

(2) if D ∈ D(M) and e, f ∈ I, then S(D),T (D, e, f) ∈ D9M).

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A subset x ⊆ ω is M -generic if for all D ∈ D(M) there is an (a, b) ∈ D such thata ⊆ x ⊆ b. Note that because D2(c) ∈ D(M) for all c ∈ A we have x /∈ A in sucha case.

Lemma 17.28. For every M as above, there is a subset x ⊆ ω which is M-generic.

Proof. Let 〈D0, D1, . . .〉 enumerate all members of D(M). Now we define (a0, b0),(a1, b1), . . . by induction: a0 = 0 and b0 = ω. Having defined (ai, bi), choose(ai+1, bi+1) so that (ai+1, bi+1) ( (ai, bi) and (ai+1, bi+1) ∈ Di. Let x =

⋃i<ω ai.

Clearly x is as desired. �Lemma 17.29. Let x be M-generic and set B = 〈A ∪ {x}〉. Then every element ofB\A is M-generic. B is atomless, and B ⊃ A. Moreover, for any a ∈ I we havex ∩ a ∈ I and a\x ∈ I.

Proof. First we prove the final statement. In fact, choose (c, d) ∈ D3(a) so thatc ⊆ x ⊆ d. Thus a ⊆ c ∪ (ω\d). It follows that a ∩ x ⊆ c ∩ a ⊆ a ∩ x, as desired.And a\x ⊆ a\d ⊆ a\x, as desired.

Now we claim:

(1) Every element of B\A has one of the two forms (x\e)∪ f or ((ω\x)\e)∪ f forsome e, f ∈ I.

In fact, take any element y of B\A; we can write it in the form y = (e∩x)∪ (f\x),where e, f ∈ A. By the above we cannot have e, f ∈ I. Now −y = ((ω\e) ∩ x) ∪((ω\f)\x), so by the above we also cannot have −e,−f ∈ I. So we have two cases:

Case 1. e ∈ I and f /∈ I. Then y = ((ω\x)\(ω\f))∪ (e∩ x), which is in one of thedesired forms.

Case 2. e /∈ I and f ∈ I. Then y = (x\(ω\e))∪ (f\x), which again is in one of thedesired forms. So (1) holds.

Next

(2) If e, f ∈ I, then (x\e) ∪ f is M -generic.

For, given D ∈ D(M), we also have T (D, e, f) ∈ D(M), and hence there is an(a, b) ∈ T (D, e, f) such that a ⊆ x ⊆ b. Then (a\e) ∪ f ⊆ (x\e) ∪ f ⊆ (b\e) ∪ f ,and ((a\e) ∪ f, (b\e) ∪ f) ∈ D, as desired.

Finally, since D(M) is closed under the operation S, it follows easily thatω\x is M -generic. From (1) and (2) the first conclusion of the lemma now follows.

That B is atomless follows from what has already been shown, plus the factthat D4 ∈ D(M). And B is a proper extension of A since D2(c) ∈ D(M) for everyc ∈ A. �Lemma 17.30. Under the hypotheses of Lemma 17.29,

(i) for all a ∈ I and all b ∈ B, if b ⊆ a then b ∈ A.

(ii) I ′ def= 〈I ∪ {x}〉Id is a maximal ideal in B.

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Proof. (i): By Lemma 17.29, if b /∈ A then b is M -generic, so by the last commentof Lemma 17.29, a ∩ b ∈ A; so, of course, b �⊆ a.

(ii): First suppose that I ′ is not proper; write ω = x ∪ a with a ∈ I. Thusω\a ⊆ x so, since a\x ∈ A by Lemma 17.29, its complement is also in A, andx = (ω\a) ∪ x ∈ A, contradiction. So, I ′ is a proper ideal.

Since u ∈ I ′ or −u ∈ I ′ for every u ∈ A ∪ {x}, and A ∪ {x} generates B, itfollows that I ′ is maximal. �

Example 17.31. (The Baumgartner–Komjath algebra.) We construct a BA A suchthat Inc(A) = ω = Length(A), while χ(A) = ω1. ♦ is assumed.

Let 〈Sα : α ∈ ω1〉 be a ♦-sequence, and let 〈aα : α ∈ ω1〉 be a one-oneenumeration of P(ω). For each β < ω1 let mβ = {aα : α ∈ Sβ}.

We define sequences 〈Aα : α < ω1〉, 〈Iα : α < ω1〉, 〈Mα : α < ω1〉 byinduction, as follows. Let A0 be a denumerable atomless subalgebra of P(ω),and let I0 be a maximal ideal in A0. If we have defined a denumerable atomlesssubalgebra Aα of P(ω) and a maximal ideal Iα of Aα, we let Mα = {mβ : β ≤ α,and mβ ⊆ Iα}. Let xα be Mα-generic (with respect to Aα and Iα), and let Aα+1 =〈Aα ∪ {xα}〉 and Iα+1 = 〈Iα ∪ {xα}〉Id. For α a limit ordinal let Aα =

⋃β<α Aβ

and Iα =⋃

β<α Iβ . Finally, let A =⋃

α<ω1Aα and I =

⋃α<ω1

Iα.

From the above lemmas it is clear that each Aα is atomless, and hence A isatomless. Furthermore, I is a maximal ideal of A, and |A � a| = ω for every a ∈ I.Moreover, |A| = ω1. The filter dual to I has character ω1, so χ(A) = ω1. In fact, letF = {a ∈ A : −a ∈ I}. Thus F is an ultrafilter on A. Assume that χ(F ) = ω; sayN is a countable generating set for F . Say N ⊆ Aα, α < ω1. Choose b ∈ Aα+1\A.Wlog b ∈ F . Then there exist a0, . . . , am−1 ∈ N such that a0 · . . . · am−1 ≤ b. So−b ≤ −a0 + · · ·+−am−1 ∈ Aα ∩ I, so by Lemma 17.32, −b ∈ Aα, contradiction.

Suppose that m is an uncountable incomparable set. Now trivially a ⊆ biff ω\b ⊆ ω\a, so we may assume that m ⊆ I. And wlog m ∩ A0 = 0. LetS = {α : aα ∈ m}, and let Z be the set of all α satisfying the following twoconditions:

(1) {aβ : β ∈ S ∩ α} = m ∩ Aα.

(2) For all b ∈ Aα\Iα, if there is a c ∈ m such that c ⊆ b, then there is a β ∈ S ∩αsuch that aβ ⊆ b.

Clearly Z is club. Hence by the ♦ property, choose α ∈ Z such that Sα = S ∩ α.

(3) mα ⊆ Aα ∩m.

For, let x ∈ mα. Say x = aβ with β ∈ Sα = S∩α. Since α ∈ Z, we get x ∈ m∩Aα.

For each c ∈ A\A0 let ρ(c) be the least β such that c ∈ Aβ+1\Aβ. Now pickc ∈ m\mα. Thus c /∈ A0. Write ρ(c) = β. Now β ≥ α : if β < α, then c ∈ Aα ∩m,so, since α ∈ Z, we have c = aγ for some γ ∈ S∩α = Sα, so c ∈ mα, contradiction.

Since c is Mβ-generic (with respect to Aβ and Iβ) and mα ∈ Mβ, thereis a (a, b) ∈ D1mα such that a ⊆ c ⊆ b. By the definition of D1mα and since

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c is not comparable with any element of mα, we must have ∀c′ ∈ mα(c′ �⊆ b).

Choose b′ with b′ ∈ A0 or (b′ /∈ A0 and ρb′ minimum) such that c ⊆ b′ /∈ Iα and∀c′ ∈ mα(c

′ �⊆ b′).

Now b′ /∈ A0 and ρ(b′) ≥ α: suppose not. Then b′ ∈ Aα\Iα. Now for any γ, ifγ ∈ S ∩ α then γ ∈ Sα, aγ ∈ mα, and aγ �⊆ b′. This contradicts (2) for α.

Say ρ(b′) = γ ≥ α. Now b′ is Mγ-generic (with respect to Aγ and Iγ) andmα ∈ Mγ , so there is a (a′′, b′′) ∈ D1mα such that a′′ ⊆ b′ ⊆ b′′; note thata′′, b′′ ∈ Aγ . For any c′ ∈ mα we have c′ �⊆ a′′; hence by the definition of D1mα

we have ∀c′ ∈ mα(c′ �⊆ b′′). Since c ⊆ b′′ /∈ Iα and b′′ ∈ A0 or (b′′ /∈ A0 and

ρb′′ < γ), this contradicts the minimality of ρb′. Thus we have shown that A hasno uncountable incomparable set.

If C is an uncountable chain in A, we may assume that C ⊆ I. We define〈cα : α < ω1〉. Suppose cβ ∈ C has been constructed for all β < α. Say {cβ :β < α} ⊆ Aγ . Then {c : c ∈ C and c ≤ cβ for some β < α} ⊆ Aγ by Lemma17.30(i). So, we can choose cα ∈ C such that cβ < cα for all β < α. The sequenceso constructed shows that DepthA = ω1; hence IncA = ω1, contradiction. �

Problem 159. Can one construct in ZFC a BA A such that Inc(A) < χ(A)?

This is Problem 61 in Monk [96]. This problem is equivalent to constructing in ZFCa BA A such that Inc(A) < hL(A); see the argument at the end of Chapter 15.

We should mention that Shelah [80], and independently van Wesep, showedthat it is consistent to have 2ω arbitrarily large and to have a BA of size 2ω whoselength and incomparability are countable.

We conclude this chapter with some remarks about incomparability in subalgebrasof interval algebras. By Theorem 15.22 of Part I of the BA handbook, if κ isuncountable and regular, and B is a subalgebra of an interval algebra and |B| = κ,then B has a chain or incomparable subset of size κ. M. Bekkali has shown thatit is consistent that this no longer holds for singular cardinals.

An important combinatorial equivalent for incomparability in interval alge-bras has been established by Shelah. Let μ be an infinite cardinal and let L bea linear order. We say that L is μ-entangled if for every n ∈ ω, every system〈tiζ : i < n, ζ < μ〉 of pairwise distinct elements of L with t0ζ < t1ζ < · · · < tn−1

ζ ,

and every w ⊆ n there exist ζ < ξ < μ such that ∀i < n(i ∈ w iff tiζ < tiξ).Shelah [90] showed that the following conditions are equivalent, for μ regular anduncountable:

(1) L is μ-entangled;

(2) If 〈aα : α < μ〉 is a sequence of elements of Intalg(L) then there exist α <β < μ such that aα ≤ aβ ;

(3) There is no incomparable subset of Intalg(L) with μ elements.

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18 Hereditary Cofinality

Theorem 18.1. For any infinite BA A, h-cofA is equal to each of:

sup{|T | : T ⊆ A, T well-founded};sup{κ : there is an a ∈ κA such that for all α, β < κ, if α < β then aα �≥ aβ}.

Proof. Call these three cardinals κ0, κ1. κ2 respectively. Suppose that κ1 < κ0.Let X be a subset of A having no cofinal subset of power ≤ κ1. We constructelements 〈xα : α < κ+

1 〉 by induction: if xα has been defined for all α < β, withβ < κ+

1 , then {xα : α < β} is not cofinal in X , so there is an xβ ∈ X such thatxβ �≤ xα for all α < β. This finishes the construction. Now {xα : α < κ+

1 } is notwell founded, so there exist α0, α1, . . . , < κ+

1 such that xα0 > xα1 > · · · . Choosei < j such that αi < αj. Then xαj < xαi is a contradiction.

Suppose κ0 < κ1. Let T be a well-founded subset of A of power κ+0 . If T

has κ+0 incomparable elements, this is a contradiction. So T has ≥ κ+

0 levels. LetT ′ consist of all elements of T of level < κ+

0 . Let X ⊆ T ′ be a cofinal subset ofT ′ of cardinality ≤ κ0. Then choose a ∈ T ′ of level greater than the levels of allmembers of X ; clearly this is impossible.

Thus we have shown that κ0 = κ1. Next, we show that κ2 ≤ κ1. Supposethat κ and a are as in the definition of κ2; we show that {aα : α < κ} is wellfounded. Suppose not: say aα0 > aα1 > · · · . Then there exist m < n such thatαm < αn. Since aαm > aαn , this contradicts the defining property of a.

Finally, suppose that κ2 < κ1; we shall get a contradiction. Let T be wellfounded, with |T | > κ2. Write T = {aα : α < κ}, with κ > κ2 and a one-one.Now because κ > κ2, it follows that for each Γ ∈ [κ]κ there are α, β ∈ Γ suchthat α < β and aα > aβ . Let Δ = {{α, β} : α < β < κ and not(aα > aβ)}. Thenfrom the partition relation κ → (κ, ω)2 we obtain α0 < α1 < · · · in κ such thataα0 > aα1 > · · · , contradicting T well founded. �

Concerning ultraproducts, h-cof is a sup-min function, so Theorems 6.5–6.9 hold.Moreover, h-cof

(∏i∈I Ai/F

)≥∣∣∏

i∈I h-cofAi/F∣∣ under GCH for F regular, by

a proof similar to that of Theorem 4.18, and Donder’s theorem says that underV = L the regularity assumption can be dropped. The inequality > is consistentlypossible by Roslanowski, Shelah [98]. < is consistently possible by Shelah [03]; seethe comment at the end of that paper. This solves problem 62 of Monk [96].

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014


Basel 219

481

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482 Chapter 18. Hereditary Cofinality

The equivalences in Theorem 18.1 give rise to two “small” versions of h-cof.

First, a general h-cof sequence in a BA A is a one-one sequence 〈aξ : ξ < α〉of elements of A such that {aξ : ξ < α} is well founded under the order < of theBoolean algebra, and for all ξ, η < α, if ξ < η then rk(aξ) ≤ rk(aη).

Proposition 18.2. Suppose that 〈aξ : ξ < α〉 is a general h-cof sequence and aξ �= 1for all ξ < α. Then 〈aξ : ξ < α〉�〈1〉 is a maximal general h-cof sequence. �

Thus this notion of maximal general h-cof sequence is rather trivial. So we definean h-cof sequence to be a general h-cof sequence with all entries different from 1.Zorn’s lemma can be applied to see that maximal h-cof sequences exist. Now wedefine

h-cofspect(A) = {|α| : A has a maximal h-cof sequence of length α},h-cofmm(A) = min(h-cofspect(A)).

Now consider Adef= Finco(κ) with κ an infinite cardinal. Then the following is an

h-cof sequence: ai = {1, . . . , i + 1} for each i < ω, and aω = κ\{0}. For atomlessalgebras we have:

Proposition 18.3. If A is atomless and 〈aξ : ξ < α〉 is a maximal h-cof sequence,then α is a limit ordinal.

Proof. If α = β + 1, let aα be such that aβ < aα < 1. �

Proposition 18.4. If 〈aξ : ξ < α〉 is an h-cof sequence, then∑

ξ<α aξ = 1.

Proof. Suppose that∑

ξ<α aξ �= 1. Let b ∈ A+ be disjoint from each aξ. Then〈aξ : ξ < α〉�〈−b〉 is an h-cof-sequence. �

Proposition 18.5. h-cofmm(A) ≤ tow(A) for any infinite BA A. �

Problem 160. Is h-cofmm = p?

The second version of a small h-cof function is as follows. A sequence 〈aξ : ξ < α〉is an h-cof-sequence of the second kind provided that aξ �= 1 for all ξ < α, and forall ξ < η < α we have aξ �≥ aη. Note that the condition aξ �= 1 is implied by thesecond condition except if α = ξ+1. So for limit α this condition can be omitted.Now we set

h-cof2spect(A) = {|α| : there is a maximal h-cof-sequence of

the second kind 〈aξ : ξ < α〉};h-cof2mm(A) = min(h-cof2spect(A)).

Proposition 18.6. An h-cof-sequence of the second kind 〈aξ : ξ < α〉 is maximal iff{−aξ : ξ < α} is dense in A.

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Chapter 18. Hereditary Cofinality 483

Proof. ⇒: Suppose that {−aξ : ξ < α} is not dense. Choose 0 �= b ∈ A suchthat −aξ �≤ b for all ξ < α. Then −b �= 1 and −b �≤ aξ for all ξ < α, and so〈aξ : ξ < α〉�〈−b〉 is an h-cof-sequence and it follows that 〈aξ : ξ < α〉 is notmaximal.

⇐: Suppose that 〈aξ : ξ < α〉 is not maximal. Say that 〈aξ : ξ < α〉�〈b〉is still an h-cof-sequence. Choose ξ < α such that −aξ ≤ −b. Then b ≤ aξ,contradiction. �

Corollary 18.7. π(A) = h-cof2mm(A).

Proof. ≤ holds by Proposition 18.6. For the other direction, let 〈aξ : ξ < π(A)〉enumerate a system of elements of A+ which is dense in A. Now we define a system〈bξ : ξ < π(A)〉 of elements of A+. Let b0 = a0. Suppose that bη has been definedfor each η < ξ so that

(∗) ∀ρ < ξ∃η < ξ[bη ≤ aρ].

Now {bη : η < ξ} is not dense in A, so there is a ρ < π(A) such that bη �≤ aρ forall η < ξ. Let bξ be aρ, ρ minimum among such ordinals. By (∗), ξ ≤ ρ, so (∗)holds for ξ+1. Thus bη �≤ bξ whenever η < ξ. So −bξ �≤ −bη whenever η < ξ. So itsuffices to show that {bξ : ξ < π(A)} is dense in A. In fact, we claim that for everyρ < π(A) there is an η < π(A) such that bη ≤ aρ. This follows from (∗). �

Thus {κ, 2κ} ⊆ h-cof2spect(P(κ)) ⊆ [κ, 2κ]card.

Problem 161. What are the possibilities for h-cof2spect(A)?

Concerning relationships to our other functions, the main facts that Inc(A) ≤h-cof(A) ≤ |A| and hL(A) ≤ h-cof(A). To see that hL(A) ≤ h-cof(A), supposethat 〈xα : α < κ〉 is a right-separated sequence of elements of A. Then it is alsowell founded. For, suppose that xα0 > xα1 > · · · . Choose i < j with αi < αj . Thenxαi > xαj , so xαj ·−xαi , contradiction. It is obvious that Inc(A) ≤ h-cof(A) ≤ |A|.An example in which hLA < h-cofA is provided by the interval algebra A on thereals. In fact, in Lemma 3.45 we showed that hL(A) = ω, and clearly Inc(A) = 2ω,and hence h-cof(A) = 2ω. Since χ(A) ≤ hL(A) ≤ h-cof(A), the Baumgartner,Komjath algebra of Chapter 17 provides an example where Inc(A) < h-cof(A).

Problem 162. Can one construct in ZFC a BA A with the property that Inc(A) <h-cof(A)?

This is problem 63 in Monk [96].

For interval algebras the equality h-cof(A) = Inc(A) holds. This was provedby Shelah [90]. In fact, let A = Intalg(I), and suppose that Inc(A) < h-cof(A).Let μ = (Inc(A))+, and by Theorem 18.1 let T be a well-founded subset of A ofpower μ. Now each level of T is an incomparable set, so there are at least μ levels.Let 〈aα : α < μ〉 be a sequence of elements of T such that aα has level α for each

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α < μ. Then by the result at then end of Chapter 17 there exist α < β < μ suchthat −aα ≤ −aβ. So aβ < aα, which is impossible.

To complete the picture, it remains to provide an example in which h-cof isless than cardinality. We are going to describe a construction due to Rubin [83].It requires ♦, but it will be used later too. It is relevant to many of our functionsand problems.

Example 18.8. Rubin’s construction is not direct, but goes by way of more generalconsiderations. Let A be a BA. A configuration for A is, for some n ∈ ω, an (n+3)-tuple 〈a, c1, c2, b1, . . . , bn〉 such that a, b1, . . . , bn are pairwise disjoint, each bi �= 0,c1 ⊆ a+

∑ni=1 bi, a+ c1 ≤ c2, and (c2− c1) · bi �= 0 for all i = 1, . . . , n. (See Figure

18.9.)

Now we call a subset P of A nowhere dense for configurations in A, for brevitynwdc in A, if for every n ∈ ω\1 and all disjoint a, b1, . . . , bn with each bi �= 0, thereexist c1, c2 such that 〈a, c1, c2, b1, . . . , bn〉 is a configuration and P ∩ (c1, c2) = 0.Rubin’s theorem that we are aiming for says that, assuming ♦, there is an atomlessBA A of power ω1 such that every set which is nwdc in A is countable. Beforeproceeding to the proof of this theorem, let us check that for such an algebra wehave h-cof(A) = ω. Suppose that P is an uncountable subset of A. Thus P is notnwdc, so we get n ∈ ω\1 and disjoint a, b1, . . . , bn with each bi �= 0 such that

(1) For all c1, c2, if 〈a, c1, c2, b1, . . . , bn〉 is a configuration then P ∩ (c1, c2) �= 0.

Let c0 = a + b1 + · · · + bn. Choose d0 such that a ≤ d0 ≤ c0 and d0 · bi �= 0 �=−d0 · bi for all i; this is possible since A is atomless. Thus 〈a, d0, c0, b1, . . . , bn〉 is aconfiguration, hence choose c1 ∈ P with d0 < c1 < c0. Then c1 · bi ≥ d0 · bi �= 0 forall i, and a ≤ c1. Choose d1 so that a ≤ d1 ≤ c1 and d1 · bi �= 0 �= c1 ·−d1 · bi for alli. then 〈a, d1, c1, b1, . . . , bn〉 is a configuration, so choose c2 ∈ P with d1 < c2 < c1.Continuing in this fashion, we get elements c1, c2, . . . of P such that c1 > c2 > · · · ,which means that P is not well founded. This shows that h-cofA = ω.

To do the actual construction leading to Rubin’s theorem, we need another defi-nition and two lemmas. Let A be a BA and assume that P ⊆ B ⊆ A. We say thatP is B-nowhere dense for configurations in A, for brevity P is B-nwdc in A, iffor every n ∈ ω\1 and all disjoint a, b1, . . . , bn ∈ A with each bi �= 0, there existc1, c2 ∈ B such that 〈a, c1, c2, b1, . . . , bn〉 is a configuration and P ∩ (c1, c2) = 0.Thus to say that P is nwdc in A is the same as saying that P is A-nwdc in A.

An important tool in the construction is the general notion of the free ex-tension A(x) of a BA A obtained by adjoining an element x (and other elementsnecessary when it is adjoined); this is the free product of A with a BA with fourelements 0, x,−x, 1. We need only this fact about this procedure:

Lemma 18.9. Let A be a BA and A(x) the free extension of A by an element x.Suppose that 〈ai : i ∈ I〉 is a system of disjoint elements of A, 〈bi : i ∈ I〉 is anothersystem of elements of A, and bi ≤ ai for all i ∈ I. Let I = 〈{(ai · x)�bi : i ∈ I〉Id,

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Figure 18.9

and let k be the natural homomorphism from A(x) onto A(x)/I. Then k � A isone-one.

Proof. Suppose kc = 0, with c ∈ A. Then there exist i(0), . . . , i(m) ∈ I such that

c ≤ (ai(0) · x)�bi(0) + · · ·+ (ai(m) · x)�bi(m);

letting f be a homomorphism of A(x) into A such that f is the identity on A andfx = bi(0) + · · ·+ bi(m), we infer that c = 0, as desired. �

Note that the effect of the ideal I in Lemma 18.10 is to subject x to the conditionthat x · ai = bi for all i ∈ ω. Now we prove the main lemma:

Lemma 18.10. Let A be a denumerable atomless BA and for each i < ω we havePi ⊆ Bi ⊆ A with Pi is Bi-nwdc for A. Then there is a countable proper extensionA′ of A such that A is dense in A′ and Pi is Bi-nwdc for A′ for all i < ω.

Proof. Let A(x) be a free extension of A by an element x; we shall obtain thedesired algebra A′ by the procedure of Lemma 18.10; thus we will let A′ = A(x)/I,with I specified implicitly by defining aj ’s and bj’s. Let 〈sn : n < ω〉 be anenumeration of the following set:

{〈0, a〉 : a ∈ A} ∪ {〈1, a, b, c〉 : a, b, c are disjoint elements of A}∪{〈2, a, b, c, b1, . . . , bk, i〉 : a, b, c are disjoint elements of A, k ∈ ω\1

b1, . . . , bk are disjoint non-zero elements of A, and i < ω}.

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As we shall see, 〈sn : n < ω〉 is a list of things to be done in coming up with theideal I. We will take care of the objects si by induction on i. Suppose that we havealready taken care of si for i < n, having constructed aj and bj for this purpose,j ∈ J , so that J is a finite set, bj ≤ aj for all j ∈ J , the aj ’s are pairwise disjoint,and

∑j∈J aj < 1. Let u =

∑j∈J aj and v =

∑j∈J bj . We want to take care of sn

so that these conditions (called the “list conditions”) will still be satisfied. Notethat under I, x · u will be equivalent to v, and −x · u will be equivalent to u · −v.Now we consider three cases, depending upon the value of the first term of sn.

Case 1. The first term of sn is 0; say sn = 〈0, a〉, where a ∈ A. We want to addnew elements ak and bk to our lists in order to insure that [x] �= [a] in A(x)/I,where in general [z] denotes the equivalence class of z ∈ A(x) with respect to I.Thus the fact that this case is taken care of for all sn of this type in our list willinsure merely that A(x)/I is a proper extension of A. If a+u �= 1, choose e so that0 < e < −(a+u), and set ak = bk = e. Then in the end we will have (e ·x)�e ∈ I,hence 0 < [e] ≤ [x], and [e] · [a] = 0, so [x] �= [a]. Clearly the list conditions stillhold. Now suppose that a + u = 1. Thus −u ≤ a, and −u �= 0. Choose e with0 < e < −u. Let ak = e and bk = 0. Then in the end we will have e · x ∈ I, hence[e] · [x] = 0, and 0 < [e] ≤ [a], so [a] �= [x]. And again the list conditions hold.

Case 2. The first term of sn is 1; say sn = 〈1, a, b, c〉, where a, b, c are disjoint

elements of A. We consider the element tdef= a + b · x + c · −x; we want to fix

things so that if [t] is non-zero then there will be some element w ∈ A such that0 < [w] ≤ [t]. This will insure that A will be dense in A(x)/I. Now

t = a+ b · x · u+ b · x · −u+ c · −x · u+ c · −x · −u,

and under I this is equivalent to

a+ b · v + u · −v · c+ b · −u · x+ c · −u · −x.

Let a′ = a + b · v + u · −v · c, b′ = b · −u, c′ = c · −u; thus a′, b′, c′ are disjoint.If a′ �= 0, we don’t need to add anything to our lists. Suppose that b′ �= 0. Thenchoose e with 0 < e < b′, and add ak, bk to our lists, where ak = bk = e; thisassures that [e] ≤ [x], hence 0 < [e] ≤ [t]; clearly the list conditions hold. If c′ �= 0a similar procedure works. Finally, if a′ = b′ = c′ = 0, then [t] = 0, and again wedo not need to add anything.

Case 3. The first term of sn is 2; say sn = 〈2, a, b, c, b1, . . . , bk, i〉, where a, b, c aredisjoint elements of A, k ∈ ω\1, b1, . . . , bk are disjoint non-zero elements of A,and i < ω. Let t be as in case 2. Case 3 is the crucial case, and here we will doone of three things: (1) make t equivalent to an element of A; (2) make sure that[t] · [bj] �= 0 for some j = 1, . . . , k; (3) find c1, c2 ∈ Bi with Pi ∩ (c1, c2) = 0 so that〈[t], [c1], [c2], [b1], . . . , [bk]〉 is a configuration. Thus this step will assure in the endthat Pi is Bi-nwdc for A′. In fact, assume that the construction is completed. Toshow that Pi is Bi-nwdc for A

′, suppose that k ∈ ω, a′, b′1, . . . , b′k ∈ A′ are disjoint

with each b′i �= 0. Since A is dense in A′, choose bi ∈ A with 0 < bi ≤ b′i for each i.

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Write a′ = a + b · [x] + c · [−x] with a, b, c pairwise disjoint elements of A. Say〈2, a, b, c, b1, . . . , bk〉 = sn. If (1) was done, the desired conclusion follows since Pi

is Bi-nwdc for A. Since a′ is disjoint from each bi, (2) could not have been done.If (3) was done, the desired conclusion is clear.

Let a′, b′, c′ be as in Case 2. If∑k

j=1 bi · a′ �= 0, then (2) will automaticallyhold, and we do not need to add anything to our lists. If there is a j, 1 ≤ j ≤ k,such that bj · b′ �= 0, let e be such that 0 < e < bj · b′, and adjoin al, bl to ourlists, where al = bl = e; then we will have [e] ≤ [x], and [bj] · [t] �= 0, which meansthat (2) holds – and the list conditions are ok. Similarly if bj · c′ �= 0 for some j.If b′ + c′ = 0, then [t] = [a′], i.e., (1) holds. Thus we are left with the essentialsituation: b′+c′ �= 0, and (a′+b′+c′) ·bj = 0 for all j = 1, . . . , k. First of all we usethe fact that Pi is Bi-nwdc for A, applied to a′, b′+ c′, b1, . . . , bk, to get c1, c2 ∈ Bi

such that Pi ∩ (c1, c2) = 0 and 〈a′, c1, c2, b′ + c′, b1, . . . , bk〉 is a configuration. Thistime we add elements al, am, bl, bm to our lists, where al = c1 · (b′+ c′), bl = c1 · b′,am = (b′ + c′) · −c2, and bm = c′ · −c2. Clearly al · am = 0 and both elementsare disjoint from previous aj ’s. Obviously bl ≤ al and bm ≤ am. Next, since〈a′, c1, c2, b′ + c′, b1, . . . , b+ k〉 is a configuration, c2 · −c1 · (b′ + c′) �= 0, and sincethis element is disjoint from all previous aj ’s as well as from al and am it followsthat u+al+am < 1. Thus the list conditions hold. It remains only to show that inthe end 〈[t], [c1], [c2], [b1], . . . , [bk]〉 is a configuration. The only things not obvious

are that [c1] ≤ [t +∑k

j=1 bj] and [t] ≤ [c2]. Since 〈a′, c1, c2, b′ + c′, b1, . . . , bk〉 is aconfiguration, we have c1 ≤ a′ + b′ + c′ + b1 + · · ·+ bk. Hence to show that [c1] ≤[t+

∑kj=1 bj], it suffices to prove that [c1 · (b′ + c′)] ≤ [t], which is done as follows.

First note that our added elements al, am, bl, bm assure that [x·c1 ·(b′+c′)] = [c1 ·b′]and [x · (b′ + c′) · −c2] = [c′ · −c2], hence [c1 · b′] ≤ [x] and [c′ · −c2] ≤ [x]. Now,

[t] ≥ [b′ · x+ c′ · −x]≥ [b′ · c1 · x+ c′ · −x · c1]= [c1 · b′ + c′ · c1 · −x]= [c1 · b′ + (c′ · c1) · −(c′ · c1 · x)]≥ [c1 · b′ + (c′ · c1) · −((b′ + c′) · c1 · x)]= [c1 · b′ + (c′ · c1) · −(c1 · b′)]= [c1 · b′ + c1 · c′]= [c1 · (b′ + c′)].

To show that [t] ≤ [c2], it suffices to show that [t · (b′+ c′)] ≤ [c2], and that is donelike this:

[t · (b′ + c′)] = [t · (b′ + c′) · c2 + t · (b′ + c′) · −c2] ≤ [c2 + t · (b′ + c′) · −c2]= [c2 + b′ · x · (b′ + c′) · −c2 + c′ · −x · (b′ + c′) · −c2]= [c2 + b′ · c′ · −c2 + c′ · −c2 · −x] = [c2].

This completes the construction and the proof. �

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Example 18.8 (Conclusion). Recall that we are trying to construct, using ♦, anatomless BA A of power ω1 such that every nwdc subset of A is countable. We shalldefine by induction an increasing sequence 〈Aα : α < ω1, α limit〉 of countableBAs, and a sequence 〈Pα : α < ω1, α limit〉, such that: the universe of Aα is α; Aω

is atomless and is dense in Aα for all limit α < ω1; Pα ⊆ Aα for all limit α < ω1,and Pα is Aα-nwdc for Aβ whenever α, β are limit ordinals < ω1 with α ≤ β.

Let 〈Sα : α < ω1〉 be a ♦-sequence. Let Aω be a denumerable atomless BA.If λ is a limit of limit ordinals, λ < ω1, let Aλ =

⋃α<λ Aα. If Sλ is nwdc for Aλ,

let Pλ = Sλ, and let Pλ = 0 otherwise. Now suppose that α is a limit ordinal < ω1,and Aβ and Pβ have been defined for all limit ordinals β ≤ α. By Lemma 18.13let Aα+ω be a BA with universe α + ω such that Aα is dense in Aα+ω and Pβ isAβ-nwdc for Aα+ω for all limit β ≤ α. And again choose Pα+ω = Sα+ω if Sα+ω isnwdc for Aα+ω, and let it be 0 otherwise. This completes the inductive definition.Let A =

⋃{Aα : α limit, α < ω1}.

Clearly A is atomless and of power ω1. Now suppose, in order to get a con-tradiction, that P is an uncountable nwdc subset of A. Let

F = {α : α < ω1, α limit, and (Aα, P ∩ α) (ee (A,P )}.

Here (ee means “elementary substructure”. Clearly F is club in ω1. Now by the

♦-property, the set Sdef= {α < ω1 : α ∩ P = Sα} is stationary, so we can choose

α ∈ F ∩S. Clearly nwdc can be expressed by a set of first-order formulas; so P ∩αis nwdc in Aα. Since P ∩ α = Sα, the construction then says that Pα = Sα. SinceP is uncountable, choose a ∈ P\Pα, and then choose c1, c2 ∈ Aα so that 〈a, c1, c2〉is a configuration (this means just so that c1 ≤ a ≤ c2) and Pα ∩ (c1, c2) = 0;this is possible, since if a ∈ Aβ with α ≤ β, then Pα is Aα-nwdc for Aβ by theconstruction. But then we have

(Aα, Pα) |= ∀x[P (x)→ x /∈ (c1, c2)];

(A,P ) |= P (a) ∧ x ∈ (c1, c2).

This contradicts the fact that (Aα, Pα) (ee (A,P ). �

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19 Number of Ultrafilters

This cardinal function is rather easy to describe, at least if we do not try togo into the detail that we did for cellularity, for example. If A is a subalgebraor homomorphic image of B, then |Ult(A)| ≤ |Ult(B)|. For weak products wehave |

∏wi∈I Ai| = max(ω, supi∈I |Ult(Ai)|). The situation for full products is more

complicated: ∣∣∣∣∣Ult(∏

i∈I

Ai

)∣∣∣∣∣ ≤ 22κ

,

where κ =∑

i∈I d(Ai). This follows from the following two facts:

∏i∈I

Ai �∏i∈I

P(d(Ai)) ∼= P

( •⋃i∈I

d(Ai)

),

where “�” means “is isomorphically embeddable in”, and “•⋃” means “disjoint

union”. Next, clearly |Ult(⊕i∈IAi)| =∏

i∈I |UltAi|. We give some observationsdue to Douglas Peterson concerning ultraproducts and the number of ultrafilters.Clearly Ult

(∏i∈I Ai/F

)≥∣∣∏

i∈I UltAi/F∣∣. And if ess.supFi∈I |Ai| ≤ |I| and F is

regular, then |Ult(∏

i∈I Ai/F)| = 22

|I|. This follows from one of the results stated

for independence, for example.

Concerning relationships to our other functions, we mention only that |A| ≤|Ult(A)|; and 2Ind(A) ≤ |Ult(A)| if Ind(A) is attained. This last assumption isneeded. For example, if κ is an uncountable strong limit cardinal and Aα is thefree BA of size |α + ω| for each α < κ, then

∏wα<κAα has independence κ and

only κ ultrafilters. (These remarks are due to L. Heindorf, and correct a mistakein Monk [90].)

About |Ult(A)| for A in special classes of BAs: first recall from Theorem 17.10of Part I of the BA handbook that |Ult(A)| = |A| for A superatomic. If A is notsuperatomic, then |Ult(A)| ≥ 2ω, since A has a denumerable atomless subalgebraB, and obviously |Ult(B)| = 2ω.

, ,OI 10.1007/978-3- - - _ ,


014


Basel 220

489

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492 Chapter 20. Number of Automorphisms

then fa(a, 0) = (0, a) while fb(a, 0) = (a · −b, a · b) �= (0, a). This proves that|A| ≤ |Aut(A× A)|. The second fact is that the group Aut(A) ×Aut(B) embedsisomorphically into Aut(A × B); an isomorphism F is defined like this, for anyf ∈ Aut(A), g ∈ Aut(B), a ∈ A, b ∈ B: (F (f, g))(a, b) = (fa, gb). Putting thesetwo elementary facts together, we have |A|, |Aut(A)| both ≤ |Aut(A×A)|. Shelahin an email message of December 1990 showed that actually equality holds (thissolves Problem 56 of Monk [90]):

Theorem 20.1. If A is an infinite BA, then |Aut(A×A)| = max(|A|, |Aut(A)|).

Proof. First note that A′ def= (A ×A) � (1, 0) and A′′ def

= (A× A) � (0, 1) are bothisomorphic to A. Now for any b ∈ A×A let Gb = {g ∈ Aut(A×A) : g(1, 0) = b}.Then

(∗) For any b ∈ A×A, |Gb| ≤ |Aut(A)|2.For, take any b ∈ A × A and fix f ∈ Gb (if Gb �= 0). Note that for any g ∈ Gb,f−1g(1, 0) = (1, 0); so (f−1 ◦ g) � A′ ∈ Aut(A)′, and similarly (f−1 ◦ g) � A′′ ∈Aut(A)′′. Now the map

g !→ ((f−1 ◦ g) � A′, (f−1 ◦ g) � A′′)

is clearly one-one, so (∗) follows.By (∗),

|Aut(A×A)| =∑

b∈A×A|Gb| ≤ |A×A| · |AutA|2,

and the theorem follows. �

For weak products, we have supi∈I |Aut(Ai)| ≤ |Aut(∏w

i∈I Ai

)| by the above

remarks; a similar statement holds for full products – in fact, the full direct productof groups

∏i∈I Aut(A)i is isomorphically embeddable in Aut

(∏i∈I Ai

).

The situation for free products is much like that for products. By Proposition11.11 of the BA handbook, Part I, every automorphism of A extends to one ofA⊕ B; so |Aut(A ⊕ B)| ≥ max(|Aut(A)|, |Aut(B)|). And |A| ≤ |Aut(A ⊕ A)|. Infact, choose a ∈ A with 0 < a < 1. Then |(A ⊕ A) � (a × −a)| = |A|, (A ⊕ A) �(a×−a) ∼= (A⊕A) � (−a× a), and

A⊕A ∼= [(A⊕A) � (a×−a)]× [(A⊕A) � (−a× a)]× [(A⊕A) � c]

for some c, so our statement follows from the above considerations on products.Actually, Shelah showed that there is an infinite BA A such that |A| and |Aut(A)|are both smaller than |Aut(A ⊕ A)| (in an email message of December 1990; S.Koppelberg supplied some details and simplifications in January 1992). This solvesProblem 57 in Monk [90]. Namely, we start with an uncountable cardinal κ anda system 〈Bα : α < κ〉 of rigid BAs of size κ such that Bα � b �∼= Bβ � c ifα < β < κ and b ∈ B+

α , c ∈ B+β ; for the existence of such a system see Shelah

[83]. Let A =∏w

α<κ Bα. Then A is also rigid, as is easy to check. We claim that

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Chapter 20. Number of Automorphisms 493

A ⊕ A has 2κ automorphisms. To see this we use duality. Recall that Ult(A) ishomeomorphic to the one-point compactification of

⋃α<κ Ult(Bα). For each Γ ⊆ κ

we define fΓ : Ult(A) ×Ult(A)→ Ult(A)×Ult(A) by

fΓ(F,G) =

{(G,F ) if F,G ∈ UltBα for some α ∈ Γ,

(F,G) otherwise.

Obviously fγ is one-one and onto, and it is easy to check that it is continuous.Since fΓ �= fΔ for Γ �= Δ, this exhibits 2κ autohomeomorphisms, as desired.

Concerning the relationship between ultraproducts and automorphisms, it iseasy to check that the inequality Aut

(∏i∈I Ai/F

)≥∣∣∏

i∈I Aut(A)i/F∣∣ holds. If

CH holds and A is a rigid BA of power ℵ1, then ωA/F has at least ℵ1 automor-phisms; this is true because ωA/F is ω1-saturated and of power ℵ1.As mentioned at the beginning of this section, |Aut(A)| is not strongly related toour previous cardinal functions. An example with the property that |Aut(A)| <Depth(A) is provided by embedding the interval algebra on κ into a rigid BA A.A similar procedure can be applied for independence and π-character, and thesethree examples show similar things for all of our preceding functions. And recallfrom the chapter on incomparability that if A is cardinality-homogeneous and hasno incomparable subset of size |A|, then A is rigid.

Concerning automorphisms of special kinds of BAs, first note that |Aut(A)| =2κ for A the interval algebra on κ. In fact, every automorphism of A is inducedby a permutation of κ; so we just need to describe 2κ permutations of κ that giverise to automorphisms of A. For each α < κ we can consider the transposition(ω · α+ 1, ω · α+ 2). For each ε ∈ κ2 let fε be the permutation of κ which, on theinterval [ω · α, ω · α+ ω), is this transposition if εα = 1, and is the identity thereotherwise. It is easy to see that the function on A induced by fε maps into A, andhence is an automorphism, as desired.

If A is infinite and superatomic, then |Aut(A)| ≥ 2ω. In fact, we may assumethat A is a subalgebra of some power-set algebra P(κ), and {α} ∈ A for allα < κ. Let a be a representative of an atom of A at level 1. Suppose that f isa permutation of a such that f2 is the identity. Extend f to all of κ by lettingfα = α if α ∈ κ\a. Now we claim that x ∈ A implies that f [x] ∈ A; this will showthat f induces an automorphism of A, hence proving the theorem.

Case 1. a/I1 ≤ x/I1, where I1 is the ideal of A generated by its atoms. Then a\xis finite. Hence

f [x] = f [x ∩ a] ∪ f [x\a] = (f [a]\f [a\x]) ∪ f [x\a]= (a\f [a\x]) ∪ (x\a) ∈ A,

since f [a\x] is finite.Case 2. a/I1 · x/I1 = 0. Thus a ∩ x is finite. Hence

f [x] = f [a ∩ x] ∪ f [x\a] = f [a ∩ x] ∪ (x\a) ∈ A,

since f [a ∩ x] is finite.

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496 Chapter 21. Number of Endomorphisms

Corollary 21.6. Let λ be strong limit, let L consist of all members of λ2 which arenot eventually 1, and let A be the interval algebra on L (which is lexicographicallyordered). Then |A| = |End(A)| = 2λ.

Proof. Let D consist of all members f ∈ λ2 such that there is an α with fα = 0and fβ = 1 for all β > α. Then D is dense in L and Theorem 21.2 applies. �

Corollary 21.6 was pointed out by Shelah (answering Problems 58 and 59 in Monk[90].) We mention one more result connecting |A| and |End(A)|:Theorem 21.7. If A is infinite, then |End(A)| ≥ 2ω.

Proof. If A has an atomless subalgebra, then |End(A)| ≥ |Ult(A)| ≥ 2ω. So sup-pose that A is superatomic. Then there is a homomorphism f from A onto B, thefinite-cofinite algebra on ω: if [a] is an atom of A/〈At(A)〉Id, then f can be takento be the composition of the natural onto mappings

A→ A � a→ C → B,

where C is a finite-cofinite algebra and B is the finite-cofinite algebra on ω. Thereis an isomorphism g of B into A. If X is any subset of ω with ω\X infinite, thenB/〈{i} : i ∈ X〉Id is isomorphic to B, and so there is an endomorphism kX of Bwith kernel 〈{i} : i ∈ X〉Id. Clearly the endomorphisms g ◦kX ◦ f of A are distinctfor distinct X ’s. �Corollary 21.8. (ω1 < 2ω). There is no BA A with the property that |A| =|End(A)| = ω1. �

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22 Number of Ideals

The main relationships with our earlier functions are: |Ult(A)| ≤ |Id(A)| and2s(A) ≤ |Id(A)|; both of these facts are obvious. Also recall the deep Theorem 10.10from Part I of the BA handbook: if A is an infinite BA, then |Id(A)|ω = |Id(A)|.This result is due to Shelah [86b]; there he also proves that if κ is a strong limitcardinal of size at most |A|, then |Id(A)|<κ = |Id(A)|. Note that |Ult(A)| < |Id(A)|for A the finite-cofinite algebra on an infinite cardinal κ.

Next, we show that |Id(A)| = 2ω for the interval algebra A on the reals; thusA has the property that |A| = |Ult(A)| = |Id(A)|. For each ideal I on A, let ≡I bedefined as follows: a ≡I b iff a = b or else if, say a < b, then [a, b) ∈ I. Thus ≡I isa convex equivalence relation on R. Now define the function f by setting, for anyideal I,

f(I) = {(r, s, ε) : there is an equivalence class a under ≡I such that |a| > 1

and a has left endpoint r, right endpoint s, and

ε = 0, 1, 2, 3 according as a is [r, s], [r, s), (r, s], or (r, s)}.

Clearly f is a one-one function; since f(I) ∈ (R×R×4)≤ω, it follows that |Id(A)| =2ω, as desired.

A rigid BA A shows that |Aut(A)| < |Id(A)| is possible. Koppelberg, Shelah[95] show that if μ is a strong limit cardinal satisfying cf(μ) = ω and 2μ = μ+, then

there is a Boolean algebra B such that |B| = |End(B)| = μ+ and |Id(B)| = 2μ+

.This answers Problem 60 of Monk [90]. Also, in an email message of December1990 Shelah showed that under suitable set-theoretic hypotheses there is a BA Asuch that |Id(A)| < |AutA|, answering problem 61 from Monk [90]. This result iseasy to see from known facts. Namely, let T be a Suslin tree in which each elementhas infinitely many immediate successors, and with more than ω1 automorphisms.

Assume CH. Then Adef= Treealg(T ) has more than ω1 automorphisms. By the

characterization of the cellularity of tree algebras given in Chapter 3, c(A) = ω,and since hL(A) = c(A) (see the end of Chapter 15), by the equivalents at thebeginning of Chapter 15 every ideal in A is countably generated, and hence A hasonly ω1 ideals. The following problem is open.

Problem 163. Can one construct in ZFC a BA A such that |Id(A)| < |Aut(A)|?

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23 Number of Subalgebras

First we note the following simple result:

Proposition 23.1. If B is a homomorphic image of A, then |Sub(B)| ≤ |Sub(A)|.

Proof. Let f be a homomorphism from A onto B. With each subalgebra C of Bassociate the subalgebra f−1[C] of A. �

It is also obvious that if B is a subalgebra of A, then |Sub(B)| ≤ |Sub(A)|.Now we give some results from Shelah [92]; the main fact is that |End(A)| ≤

|Sub(A)|. This answers Problem 63 from Monk [90]. Let Psub(A) be the collectionof all subsets of A closed under +, ·, and − (as a binary operation – namely,a− b = a · −b). The part |Id(A)| ≤ |Sub(A)| in the next theorem is due to JamesLoats, and can be proved more easily.

Theorem 23.2. |Psub(A)|= |Sub(A)| for any infinite BA A. In particular, |Id(A)|≤|Sub(A)|.

Proof. First, it is clear that |A| ≤ |Sub(A)|, since a !→ {0, 1, a,−a} (a ∈ A) is a2-to-1 mapping from A into SubA. Hence for the theorem it suffices to show thatthe set of infinite members of Psub(A) has cardinality at most |Sub(A)|. To thisend, choose for every infinite X ∈ Psub(A) an element aX of X such that thereare elements u, v ∈ X with 0 < u < aX < v < 1; then let Y [X ] be the subalgebragenerated by X � aX and let Z[X ] be the subalgebra generated by X � −aX . Notethat Y [X ] consists of all elements of the form y+−z with y ∈ X � aX , and eitherz ∈ X � aX or z = 1, and similarly for Z[X ]. Now we claim that for X �= X ′ wehave Y [X ] �= Y [X ′] or Z[X ] �= Z[X ′], from which the desired conclusion clearlyfollows. Suppose that this claim fails; say X\X ′ �= 0. Let a = aX and b = aX′ . Weclaim next that −a ≤ b or a ≤ −b. For, take any x ∈ X\X ′. Then we can write

x · a = y +−z, y ∈ X ′ � b, and z ∈ X ′ � b or z = 1,

x · −a = u+−v, u ∈ X ′ � −b, and v ∈ X ′ � −b or v = 1.

Since x /∈ X ′, we have z �= 1 or v �= 1; this gives in the first case −x + −a =z · −y ≤ b, hence −a ≤ b, and a ≤ −b in the second case, proving our latest claim.

Say without loss of generality that −a ≤ b. Choose b′ ∈ X ′ such that b < b′ <1. So 0 < −b′ < −b. Thus −b′ ∈ X � −b, so −b′ = s+ −t, where s ∈ X � −a, and

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500 Chapter 23. Number of Subalgebras

t ∈ X � a or t = 1. If t = 1, then −b′ = s ≤ −a ≤ b and −b′ ≤ −b, contradiction.If t �= 1, then b′ = t · −s ≤ −a ≤ b, so −b ≤ −b′, contradiction. �

Lemma 23.3. |Aut(A)| ≤ |Sub(A)| for A infinite.

Proof. The idea is to associate with each automorphism of A a sequence of 8members of Psub(A), in a one-one fashion; clearly this will prove the theorem. Letf be any automorphism. Define

Jf = {x ∈ A : f(y) = y for all y ≤ x};If = {x ∈ A : x · f(x) = 0}.

Clearly we have:

(1) If x ∈ If then f(x) ∈ If .

(2) Jf ∪ If is dense in A.

To prove (2), let a ∈ A+, and suppose that a /∈ Jf . Then there is a b ≤ a suchthat b �= f(b). If b �≤ f(b), then 0 �= b · −f(b) ≤ a and b · −f(b) ∈ If . If f(b) �≤ b,then 0 �= b · −f−1(b) ≤ a and b · −f−1(b) ∈ If .

Next, let X be a maximal subset of If such that x ·f(y) = 0 for all x, y ∈ X .

Let If1 = 〈X〉Id and If2 = 〈{f(x) : x ∈ X〉Id}. Then let

If0 = {y ∈ If : f(y) ∈ If1 and y · x = 0 for all x ∈ If1 }.

Let If3 = 〈{f(x) : x ∈ If2 }〉Id. Thus

(3) Ifi is an ideal ⊆ If for i < 4.

(4) Ifi ∩ Ifi+1 = {0} for i < 3.

(5) If0 ∪ If1 ∪ If2 is dense in If .

To prove (5), suppose that x ∈ (If )+ but there is no non-zero member of theindicated union which is below it. Since x ∈ If , we have x · f(x) = 0. Now since

there is no nonzero element of If1 below x, we have x /∈ X , and this gives twocases.

Case 1. There is a z ∈ X such that x · f(z) �= 0. But x · f(z) ∈ If2 , contradiction.

Case 2. There is a z ∈ X such that z ·f(x) �= 0. Choose w such that f(w) = z ·f(x).Since w ≤ x, we have w ∈ (If )+, and since f(w) ≤ z we have f(w) ∈ If1 . For any

t ∈ If1 we have t · x = 0 (otherwise 0 �= t · x ≤ x and t · x ∈ If1 ), hence t · w = 0.

Thus w ∈ If0 , contradiction, proving (5).

For i < 3 we now define a member Cfi of PsubA: Cf

i = {x+ f(x) : x ∈ Ifi }.Clearly each Cf

i is closed under +. For any x, y ∈ Ifi we have x · f(y) = 0, and

from this it follows that Cfi is closed under · and −.

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Chapter 23. Number of Subalgebras 501

We have now defined our sequence

〈Jf , If0 , If1 , I

f2 , I

f3 , C

f0 , C

f1 , C

f2 〉

of 8 members of Psub(A). It remains just to show that if two automorphisms f andg give rise to the same sequence 〈J, I0, I1, I2, I3, C0, C1, C2〉, then f = g. By (2) and(5) it is enough to show that they agree on J∪I0∪I1∪I2. Suppose that y ∈ I0. Thusy+ f(y) ∈ C0, so there is a z ∈ I0 such that y+ f(y) = z + g(z). Now 0 = y · g(z)since g(z) ∈ I1, so y ≤ z. Similarly z ≤ y, so y = z. So y + f(y) = y + g(y). Buty ∈ If , so y · f(y) = 0. Similarly y · g(y) = 0, so f(y) = g(y). The cases of I1 andI2 are similar. �Theorem 23.4. |End(A)| ≤ |Sub(A)| for any infinite BA A.

Proof. Let μ = |Sub(A)|, and suppose that μ < |End(A). With each f ∈ End(A)associate the pair (Kernel(f),Range(f)). The number of such pairs is at most|Id(A)|× |Sub(A)|, which by 23.2 is μ. Thus there is a set E of μ+ endomorphismswith the same kernel I and range R. For each f ∈ E we define a mapping gf fromA/I onto R by gf(x/I) = f(x); clearly gf is well defined and is an isomorphismfrom A/I onto R. Fix h ∈ E. Then {gf ◦ g−1

h : f ∈ E} is a set of μ+ differentautomorphisms of R. Thus by 23.3,

μ < |Aut(R)| ≤ |Sub(R)| ≤ |SubA|,

contradiction. �

Another result in Shelah [92] is that |Aut(A)|ω ≤ |Sub(A)|; we shall not give theproof.

For any BA A, let Pend(A) = {f : f is a homomorphism from a subalge-bra of A onto another subalgebra of A}. Such homomorphisms are called partialendomorphisms of A.

Theorem 23.5. |Pend(A)| = |Sub(A)| for any infinite BA A.

Proof. ≥ is clear, since with each subalgebra B one can associate the identitymapping on B, a partial endomorphism of A. For ≤ we proceed as in the proofof Theorem 23.4: this time we associate with each partial endomorphism a tripleconsisting of its domain, kernel, and range; otherwise the details are similar. �Theorem 23.6. If A×B is infinite, then |Sub(A×B)| = max(|Sub(A)|, |Sub(B)|).

Proof. We prove the equivalent statement that if A is infinite and a ∈ A, then|Sub(A)| = max(|Sub(A � a)|, |Sub(A � −a)|). The inequality ≥ is obvious. Letμ = max(|Sub(A � a)|, |Sub(A � −a)|), and suppose that μ < |Sub(A)|. Now weassociate with each subalgebra B of A five objects:

CB0 = {x · a : x ∈ B}, a subalgebra of A � a;

CB1 = {x · −a : x ∈ B}, a subalgebra of A � −a;

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IB0 = {x ∈ B : x ≤ a}, an ideal of CB0 ;

IB1 = {x ∈ B : x ≤ −a}, an ideal of CB1 ;

and gB, the isomorphism from CB0 /IB0 onto CB

1 /IB1 such that gB((x · a)/IB0 ) =(x · −a)/IB1 for all x ∈ B. Now B can be reconstructed from these five objects:

B = 〈IB0 ∪ IB1 ∪ {x+ y : x ∈ CB0 , y ∈ CB

1 , and gB(x/IB0 ) = y/IB1 }〉

In fact, the direction ⊆ is clear. For ⊇ it suffices to show that if x ∈ CB0 , y ∈ CB

1 ,and gB(x/I

B0 ) = y/IB1 then x + y ∈ B. Say x = x′ · a and y = y′ · −a, with

x′, y′ ∈ B. Now (x′ · −a)/IB1 = (y′ · −a)/IB1 , so (x′ · −a)�(y′ · −a) ∈ B. Now

(x + y)�(x′ · −a)�(y′ · −a) = (x′ · a)�(y′ · −a)�(x′ · −a)�(y′ · −a)= (x′ · a)�(x′ · −a) = x′ ∈ B,

so x+ y ∈ B.

Now there is a setX of μ+ subalgebras of A such that CBi = CD

i and IBi = IDifor all B,D ∈ X and all i < 2. Fix B ∈ X . Now {g−1

D ◦ gB : D ∈ X} is a set ofautomorphisms of CB

0 /IB0 , and by 23.1 and 23.3,

|Aut(CB0 /I0)| ≤ |Sub(CB

0 /I0)| ≤ |SubCB0 | ≤ μ,

so there are distinct D,E ∈ X for which gD = gE . This contradicts the noted factabout reconstruction. �

Note that 2Irr(A) ≤ |Sub(A)| if Irr(A) is attained. An example A for which|Id(A)| < |Sub(A)| is provided by the interval algebra A on the reals. We notedin the last chapter that |Id(A)| = 2ω. Since Irr(A) = 2ω attained, we have|SubA| = 22

ω

.

The above theorems imply that |Sub(A)| is our biggest cardinal function. Itssize is, of course, always at most 2|A|. For most algebras, this value is actuallyattained. It is also quite interesting to construct a BA A in which |Sub(A)| is assmall as possible. The only algebra we know of where this is the case is Rubin’salgebra A from Chapter 18. We now go through the proof that |Sub(A)| = ω1;thus |A| = |Ult(A)| = |Id(A)| = |Sub(A)|. We call an element a ∈ A countableprovided that A � a is countable.

Lemma 23.7. A has only countably many countable elements.

Proof. Let P be the set of all countable elements of A, and suppose that P isuncountable. Then it is easy to construct 〈aα : α < ω1〉 ∈ ω1P such that if

α < β < ω1 then aβ �≤ aα. Since h-cof(A) = ω, the set P ′ def= {aα : α < ω1} is

not well founded; say aα(0) > aα(1) > · · · . Choose i, j ∈ ω such that i < j andα(i) < α(j). Then aα(i) > aα(j) contradicts the choice of the aβ’s. �

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Chapter 23. Number of Subalgebras 503

Note that the collection of countable elements of A forms an ideal, which we denoteby C(A). To proceed further, we have to go back to the main property of Rubin’salgebra. For this purpose we introduce the following notation. A preconfigurationin a BA A is a sequence 〈a, b1, . . . , bn〉 of pairwise disjoint elements of A witheach bi �= 0, n > 0. Given such a preconfiguration, a subset P of A is dense at〈a, b1, . . . , bn〉 provided that for all c1, c2, if 〈a, c1, c2, b1, . . . , bn〉 is a configurationthen P ∩ (c1, c2) �= 0. Thus the main property of Rubin’s BA A is that if P is anuncountable subset of A then there is a preconfiguration 〈a, b1, . . . , bn〉 of A suchthat P is dense at 〈a, b1, . . . , bn〉. For both of the next two lemmas we advise thereader to draw a diagram along the lines of the one in Chapter 18 to see what isgoing on.

Lemma 23.8. Assume that P ⊆ A, P is uncountable, 〈a, b1, . . . , bn〉 is a precon-figuration of A, P is dense at 〈a, b1, . . . , bn〉, a ≤ b ≤ a +

∑ni=1 bi, b · bi �= 0 for

i = 1, . . . , n, and b · −a /∈ C(A). Then P ∩ [a, b] is uncountable.

Proof. Suppose that P ∩ [a, b] is countable, and let Q be the closure of P ∩ [a, b]under +; so Q is countable also. Say b · bi /∈ C(A). Pick any c ≤ b · bi, c �= 0,

such that c′ def= a + c +

∑j �=i bj /∈ Q. Then pick c1, c2 so that a + ci < c′ and

〈a, ci, c′, b1, . . . , bn〉 is a configuration for i = 1, 2, and c1 + c2 = c′. Then pickd1, d2 ∈ P so that ci < di < c′ for i = 1, 2. But d1, d2 ∈ [a, b] and d1 + d2 = c′, soc′ ∈ Q, contradiction. �

Lemma 23.9. Every subalgebra of A is the union of countably many closed inter-vals.

Proof. Suppose that B is a subalgebra of A which is not the union of countablymany closed intervals. Let 〈[xα, yα] : α < ω1〉 enumerate all of the closed intervalscontained in B. Now B contains ω1 elements pairwise inequivalent with respectto C(A); hence it is easy to construct a sequence 〈zα : α < ω1〉 ∈ ω1B with thefollowing two properties:

(1) zα /∈⋃

β<α[xβ , yβ ] for each α < ω1;

(2) zα�zβ /∈ C(A) for distinct α, β < ω1.

Let D = {zα : α < ω1}. Since D is somewhere dense, let 〈a, b1, . . . , bn〉 be apreconfiguration of A such that D is dense at 〈a, b1, . . . , bn〉. Choose any b suchthat a ≤ b ≤ a +

∑ni=1 bi and b · bi �= 0 �= bi · −b for all i = 1, . . . , n. We show

that [a, b] ⊆ B. Take any d ∈ [a, b]. Choose e1, e2 with the following properties:d = e1 · e2; a ≤ ei ≤ a +

∑nj=1 bj ; ei · bj �= 0 for i = 1, 2, j = 1, . . . , n. Thus

〈a, d, ei, b1, . . . , bn〉 is a configuration, so we can choose fi ∈ D∩ (d, ei) for i = 1, 2.Then f1 · f2 = d, and so d ∈ B (since D ⊆ B). Thus, indeed, [a, b] ⊆ B. By aneasy argument, [a, b] ∩D has at least two elements. Since distinct elements of Dare inequivalent mod C(A), it follows that b · −a /∈ C(A). Hence by Lemma 23.8,D ∩ [a, b] is uncountable. But this clearly contradicts the construction of D. �

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With these lemmas available we can now prove that |Sub(A)| = ω1. We claim thateach subalgebra B of A is generated by an ideal along with a countable set. Infact, write B =

⋃i<ω[ai, bi]. Let I be the ideal generated by {bi · −ai : i < ω}.

Then clearly B is generated by I ∪ {bi : i < ω}, as required. Now every idealis countably generated. This follows from the fact that hL(A) ≤ h-cof(A) = ω,proved in Chapter 18, and one of the equivalents of hL given in Chapter 15. Thisbeing the case, it follows that there are exactly ω1 ideals in A, since ωω

1 = ω1 byvirtue of CH (which follows from ♦, which we are assuming). Now |SubA| = ω1 isclear, again using CH.

In Cummings, Shelah [95] it is shown that it is consistent (relative to a largecardinal assumption) that every infinite BA A has 2|A| subalgebras. This answersProblem 62 in Monk [90]. We give part of this consistency proof.

Lemma 23.10. If κ is a strong limit cardinal ≤ |A|, then A has an irredundantsubset of size κ.

Proof. Let B be a subalgebra of A of size κ. Then B has an irredundant subset ofsize κ; this subset is irredundant in A too. �

Let (∗) be the conjunction of the following statements:

(1) For every infinite cardinal κ, 2κ is weakly inaccessible.

(2) For all infinite cardinals λ, κ, if κ ≤ λ < 2κ, then 2λ = 2κ.

(3) For every infinite cardinal κ and every BA A of size 2κ, A has an irredundantsubset of size 2κ.

Cummings and Shelah show that there is a model in which (∗) holds, assumingGCH + six supercompact cardinals.

Theorem 23.11. Assuming GCH + six supercompact cardinals, there is a model inwhich every infinite BA A has 2|A| subalgebras.

Proof. We assume (∗). Let

C = {μ : μ is strong limit or ∃θ ≥ ω[μ = 2θ]}.

Clearly C is a closed unbounded class of cardinals. Let A be any infinite BA, andlet ν ∈ C be minimum such that |A| < ν. Note that ω ∈ C. Let μ = sup{ν ∈C : ν < |A|}. Thus μ ∈ C and μ ≤ |A|. Now 2μ ∈ C, so |A| < 2μ. Let B be asubalgebra of A of size μ.

Case 1. μ = 2θ for some θ. Then by (∗)(3), B has an irredundant subset of size μ.

Case 2. μ is strong limit. Then by Lemma 23.10, B has an irredundant subset ofsize μ.

Now μ ≤ |A| < 2μ, so by (∗)(2), 2μ = 2|A|, and the desired conclusion follows. �

It is possible to have |Aut(A)| < |Sub(A)|: take a rigid BA. One can even have|End(A)| < |Sub(A)|, for example in the interval algebra on the reals.

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24 Other Cardinal Functions

There are many other cardinal functions besides the 21 that we have discussed inthe preceding chapters. In this chapter we give a list of some natural ones; someof these have been explicitly mentioned earlier. We also mention some facts andproblems about them, without trying to be exhaustive.

Functions mentioned in the previous text

1. a, aspect, 121

2. cSr, 137

3. cHr, 141

4. dDepthS−, 1785. tow, towspect, 178

6. spl, 188

7. h, 188

8. p and pspect, 188,

9. DepthSr, 199

10. DepthHr, 203

11. dn, 227

12. dSr, 232

13. dHr, 232

14. πS+, 247

15. r, 247

16. rm, 248

17. rmn, 249

18. πχinf , 261

19. hwd, 264

20. πSr, 276

21. πHr, 276

22. LengthH+, 284

23. LengthH−, 289

24. Lengthh−, 28925. Lengthh+, 289

26. dLengthS−, 28927. Lengthspect and Lengthmm, 290

28. LengthSr, 290

29. LengthHr, 290

30. Irrmm, 309

31. Irrmn, 310

32. CardH−, 31133. cf, 311

34. alt, 311

35. p-alt, 311

36. Cardh+, 316

37. IndH−, 34638. Indh+, 346

39. Indh−, 34640. dIndS−, 34741. i, isp, 347

42. Freecal, 355

43. precal, 361

44. Indn, 372

45. πχS+, 382

46. dπχS−, 382

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47. πχSr, 384

48. πχHr, 384

49. tH−, 39750. dtS−, 39751. f, fsp, 398

52. tSr, 400

53. tHr, 400

54. tmn, 401

55. utmn, 401

56. utm, 401

57. sH−, 41358. dsS−, 41359. smm, sspect, 413

60. dd, 419

61. sm, 420

62. χH−, 42763. χS+, 427

64. u, 427

65. χSs, 432

66. χHs, 432

67. χSr, 432

68. χHr, 432

69. hLmm, 443

70. hLHs, 446

71. hLSr, 446

72. hLHr, 446

73. hLm, 448

74. hdmm, hdspect, 454

75. hdid, 454

76. hdm, 460

77. Incmm, Incspect, 465

78. Inctreemm, Inctreespect, 466

79. h-cofmm, h-cofspect, 482

80. h-cof2mm, h-cof2spect, 482

Some additional natural functions

81. The tree algebra number. For any BA A, the tree algebra number of A, de-noted by ta(A), is the supremum of cardinalities of subalgebras of A isomor-phic to tree algebras. This number is clearly greater or equal to cellularity.It is dominated by d, since if A is isomorphic to a tree algebra and A is asubalgebra of B, then |A| = d(A) ≤ d(B). Note that an infinite free algebraalways has tree algebra number ℵ0. So ta(B) can be much smaller than d(B).

82. The pseudo-tree algebra number. For any BA A, the pseudo-tree algebranumber of A, denoted by pta(A), is the supremum of cardinalities of subalge-bras of A isomorphic to pseudo-tree algebras. Clearly Length(A) ≤ pta(A),ta(A) ≤ pta(A), and pta(A) ≤ Irr(A). Any infinite free algebra has pseudo-tree number ℵ0. In a large free algebra we thus have pta(A) = ℵ0 while d(A)is large.

83. The semigroup algebra number. For any BA A, the semigroup algebra numberof A, denoted by sa(A), is the supremum of cardinalities of subalgebras of Awhich are semigroup algebras. We have Ind(A) ≤ sa(A) and pta(A) ≤ sa(A).It is not clear whether sa(A) ≤ Irr(A).

84. The tail algebra number. For any BA A, the tail algebra number of A, denotedby tla(A), is the supremum of cardinalities of subalgebras of A isomorphic totail algebras. Clearly sa(A) ≤ tla(A). It may be that tla(A) = |A| for everyinfinite BA A.

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Chapter 24. Other Cardinal Functions 507

85. Disjunctiveness. The disjunctiveness of A, dj(A), is the supremum of cardi-nalities of disjunctive subsets of A. Clearly tla(A) ≤ dj(A) and s(A) ≤ dj(A).It may be that dj(A) = |A| for every infinite BA A.

86. Minimality. The minimality of A, m(A), is the supremum of cardinalities ofminimally generated subalgebras of A. Clearly pta(A) ≤ m(A). For everyinfinite free BA A we have m(A) = ℵ0.

87. Initial chain algebra number. This number, denoted by ic(A), is the supre-mum of cardinalities of subalgebras of A isomorphic to the initial chain al-gebra on some tree. If A is free, then ic(A) = ω.

88. Initial chain algebra number for pseudo trees. Similarly, for pseudo-trees;denoted by icp(A) Thus ic(A) ≤ icp(A). If A is free, then icp(A) = ω.

89. Superatomic number. This is the supremum of cardinalities of superatomicsubalgebras of a BA; denoted by spa(A). If A is free, then spa(A) = ω. Wehave ic(A) ≤ spa(A).

90. The order-ideal number. This is the supremum of the cardinality of a systemof ideals ordered by inclusion; we denote it by oi(A) Clearly hL(A) ≤ oi(A)and hd(A) ≤ oi(A). It is possible to have |A| < oi(A); this is true for A =Intalg(Q), for example. It is not clear whether oi(A) < |A| is possible.

Dimensions of Boolean algebras

Heindorf [91] introduced an interesting notion of dimension of Boolean algebras,which gives rise to several cardinal functions. Let A be a non-empty class ofnontrivial BAs. Since every BA is embeddable in a product of two-element BAs,every BA is embeddable in a product of members of A . Hence the followingdefinition makes sense: the A -dimension of a BA A is the smallest cardinal numberκ such that A can be embedded in a product of κ members of A (not necessarilydistinct members). This A -dimension is denoted by A -dimA. It is natural toconsider this notion for natural classes A . Various one-element classes A arenatural, of course. At the opposite extreme we can consider the notion for ournatural proper classes – the class of all free algebras, of all superatomic algebras,etc. Heindorf investigated the three cases (1) all free algebras, (2) all superatomicalgebras, (3) all interval algebras. Some cases are trivial; for example, with A theclass of all complete BAs, the A -dimension of any BA is 1, since any BA canbe embedded in a complete BA. We list some dimension functions which may beinteresting.

91. int-dim, where A is the class of all interval algebras.

92. sa-dim, where A is the class of all superatomic algebras.

93. free-dim, where A is the class of all free algebras.

94. tree-dim, where A is the class of all tree algebras.

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508 Chapter 24. Other Cardinal Functions

95. ptree-dim, where A is the class of all pseudo-tree algebras.

96. sg-dim, where A is the class of all semigroup algebras.

97. mg-dim, where A is the class of all minimally generated algebras.

98. ic-dim, where A is the class of all initial chain algebras.

99. dj-dim, where A is the class of all disjunctively generated algebras.

100. finco-dim, where A is the class of all finite-cofinite algebras on some infiniteset.

101. {Finco(ω)}-dim.

102. {Intalg(R)}-dim.

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25 Diagrams

In this chapter we give several diagrams for the relationships between the main 21functions that we have considered. Thus this chapter summarizes the main text.But it also turns out that we have some new things to say upon considering theserelationships thoroughly.

For each of the diagrams, we need to do the following:

(1) For each edge, indicate where the relation is proved, and give an examplewhere the functions involved are different. Also, if the difference is indicatedas “small”, indicate where that is stated in the text, while if the differenceis “large”, indicate an example. Recall that a difference is “small” if there issome limitation on the difference. It is “large” if for every infinite cardinal κ,there is an example where the difference is at least κ.

(2) Show that there are not any relations except those indicated in the diagrams.It suffices to do this just for crucial places in the diagram. For example, if inthe diagram for the general case we give an algebra A in which LengthA <πχA, this will also be an example for LengthA < h-cofA and DepthA < πχA.As will be seen, we have not been completely successful in either of these twotasks; there are several open problems left.

The main diagram, edges and “large” and “small” indications

See the diagram on top of the next page.

25.1. Depth ≤ Length. This relation is obvious from the definitions. The differenceis small by the Erdos–Rado theorem.

25.2. Depth ≤ c. Again, this is obvious from the definitions. The difference islarge in the finite-cofinite algebra on an infinite cardinal κ.

25.3. Depth ≤ t. This is proved in Chapter 4; see 4.26. The difference is big in afree algebra.

25.4. πχ ≤ t. See Chapter 11, Theorem 11.18. The difference can be large in aninterval algebra. Namely, given an uncountable cardinal κ, one can take a completelinear order L in which the elements have character (ω, λ) or (λ, ω) for each λ ≤ κ,

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510 Chapter 25. Diagrams

The general case

l = difference can be large; s =“small” difference

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Chapter 25. Diagrams 511

with the cofinality and coinitiality of L being ω; see Hausdorff [1908]. See the endsof Chapters 11 and 12.

25.5. πχ ≤ π. Obvious from the definitions. The difference is large in a finite-cofinite algebra; see Chapter 11.

25.6. c ≤ s. This is an easy consequence of Theorem 3.30. The difference is largein free algebras.

25.7. c ≤ d. See Corollary 5.2. The difference is large in free algebras.

25.8. Length ≤ Irr. Obvious from the definitions. The difference is large in freealgebras.

25.9. Ind ≤ t. This is clear by the free sequence characterization of tightness. Thedifference can be large in some interval algebras, since Depth ≤ t.

25.10. d ≤ π. Obvious from the topological versions of these functions. Thedifference is small. See the end of Chapter 6 for an example where they differ.

25.11. t ≤ χ. Obvious from the definitions. The difference is big in a finite-cofinitealgebra on a large cardinal κ.

25.12. t ≤ s. See Theorem 5.18. The difference is big in a finite-cofinite algebraon a cardinal κ.

25.13. π ≤ hd. See Theorem 6.15. The difference is small, since π(A) ≤ hd(A) ≤|A| ≤ 2π(A). The functions differ in P(κ), for example.

25.14. hd ≤ Irr. See the end of Chapter 16. They differ in the interval algebra onthe reals.

25.15. χ ≤ hL. See after Problem 144. The difference is small, since hL(A) ≤|A| ≤ |UltA| ≤ 2χ(A). They differ on the Alexsandroff duplicate of a free algebra;see Proposition 14.7.

25.16. s ≤ hL. Obvious from the definitions. The difference is small, since |A| ≤2s(A) for any BA A by 13.10. They differ in a Kunen line (constructed under CH;see page 378). Whether there is an example in ZFC is an open question

Problem 164. Can one construct in ZFC a BA A such that s(A) < hL(A)?

25.17. s ≤ hd. Obvious from the definitions. The difference is small (see above).They differ on the interval algebra of a Suslin line. Whether there is an examplein ZFC is open (Problem 153).

25.18. Irr ≤ Card. Obvious from the definitions. The difference is small; fromTheorem 4.23 of Part I of the Boolean algebra handbook it follows that |A| ≤2Irr(A). A compact Kunen line (constructed under CH) gives a BA in which theyare different (see Chapter 8). It is open to give an example in ZFC. (Problem 91).

25.19. hL ≤ h-cof. See Chapter 18. The difference is small, since χ(A) ≤ hL(A),

and so hL(A) ≤ h-cof(A) ≤ |A| ≤ 2χ(A)≤2hL(A), using Chapter 14. They differ onthe interval algebra on the reals; see Chapter 18.

25.20. hd ≤ Inc. This is an easy consequence of Theorem 4.25 of the BA handbook,part I. The difference is small, since s ≤ hd, Inc ≤ Card, and |A| ≤ 2s(A) for anyBA. They differ on Intalg(R).

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25.21. Inc ≤ h-cof. Obvious from Theorem 18.1. The difference is small since bythe above |A| ≤ 2Inc(A). They differ on the Baumgartner, Komjath algebra (seeChapter 18); this was constructed using ♦, and it remains a problem to get anexample with weaker assumptions. (Problem 162)

25.22. h-cof ≤ Card. Obvious from the definitions. The difference is small (seeabove). Since Inc(A) = h-cof(A) for an interval algebra (see Chapter 18), we getan example with h-cof(A) < |A|.25.23. Card ≤ |Ult|. This is well known; see the Handbook Part I, Theorem 5.31.The difference is, of course, small. They differ in an infinite free algebra.

25.24. |Ult| ≤ |End|. This is obvious. The difference is small. They differ for thefinite-cofinite algebra on ω; see Theorem 21.7.

25.25. |Aut| ≤ |End|. Also obvious. The difference is large, as shown by a rigidBA.

25.26. |Ult| ≤ |Id|. Again obvious. The difference is small. They differ on thefinite-cofinite algebra on an infinite cardinal.

25.27. |Id| ≤ |Sub|. See Chapter 23. The difference is small. They differ for theinterval algebra on the reals.

25.28. |End| ≤ |Sub|. See Chapter 23. The difference is small. They differ for theinterval algebra on the reals.

The main diagram: no other relationships

Keep in mind that we only treat “crucial” relations; other possibilities are supposedto follow from these.

25.29. Length < πχ: an uncountable free algebra: see Chapter 11.

25.30. Length < c: the finite-cofinite algebra on an uncountable cardinal.

25.31. Length < Ind: an uncountable free algebra.

25.32. πχ < Depth: see the example in Chapter 11.

25.33. d < πχ: some free algebras.

25.34. Ind < Depth: the interval algebra on an uncountable cardinal.

25.35. Ind < πχ: true in the interval algebra on an uncountable cardinal; seeChapter 11.

25.36. π < Ind: P(κ). The difference is small.

25.37. χ < c: the Alexsandroff duplicate of an infinite free algebra; Chapter 14.

25.38. hL < d: The interval algebra of a complete Suslin line. It is not known ifthis is possible in ZFC. (Problem 145)

25.39. hL < Inc: The interval algebra on the reals; see Chapter 17.

25.40. Inc < χ: The Baumgartner-Komjath algebra constructed under ♦; seeChapter 17. It is not known if this is possible under weaker hypotheses. See Prob-lem 159.

25.41. Inc < Length: Constructed by Shelah in ZFC using entangled linear orders.

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25.42. Irr < χ: The compact Kunen line, constructed using CH. No example isknown in ZFC; this is problem 65 in Monk [96].

Problem 165. Can one construct in ZFC a BA A such that IrrA < χA?

This problem is equivalent to the problem of constructing in ZFC a BA A suchthat Irr(A) < hL(A); see the argument at the end of Chapter 15.

25.43. h-cof < Length: Constructed by Shelah in ZFC using entangled linearorders; see Shelah [91].

25.44. |Aut| < Depth: embed a large interval algebra in a rigid algebra.

25.45. |Aut| < πχ: a rigid complete BA of large cellularity gives an example.

25.46. |Aut| < Ind: embed a large free algebra in a rigid algebra.

25.47. |Id| < |Aut|: see Chapter 22; possible under some set-theoretic assumptions.No example is known in ZFC (Problem 163)

25.48. |Ult| < |Aut|: the finite-cofinite algebra on ω; see Chapter 20.

25.49. |End| < |Id|: See Chapter 22.

The interval algebra diagram: the edges, indicated equalities,and the “large” and “small” indications

(See the next page for the diagram.)

25.50. Ind = ω: this is one of the main results about interval algebras; see Part Iof the BA handbook.

25.51. ω ≤ πχ, difference possibly large. See the description of πχ for intervalalgebras in Chapter 11.

25.52. Depth = t = χ: see Chapter 14.

25.53. πχ ≤ Depth, difference possibly large. See Chapter 11.

25.54. c=s=hL: see Chapter 15.

25.55. Depth ≤ c, with the difference small. The difference is small since |A| ≤2Depth(A) for an interval algebra A; this implies smallness for the next few that weconsider also. For an example where they differ, see Chapter 4.

25.56. d = π = hd: obvious from the retractiveness of interval algebras.

25.57. c ≤ d. They differ in the interval algebra of a Suslin line; see Chapter 5. Infact, for any infinite cardinal κ the following two conditions are equivalent:

(1) there is an interval algebra A such that κ = cA < dA;

(2) there is a κ+-Suslin line (or tree).

Thus the problem of getting an example in ZFC of a BA A such that c(A) < d(A)is equivalent to the set-theoretical question of proving in ZFC that there is forsome infinite κ a κ+-Suslin tree.

25.58. d ≤ Inc. They differ in the interval algebra on the reals.

25.59. Inc = h-cof. See Chapter 18.

25.60. h-cof ≤ Card. Shelah [91] constructed an example where they differ in ZFC.

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Interval algebras


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25.61. Card ≤ |Ult|. They differ for the interval algebra on the rationals.

25.62. |Ult| ≤ |Id|. They differ on the interval algebra on κ.

25.63. |Aut| ≤ |End|, the difference large: take an infinite rigid interval algebra.

25.64. |Id| ≤ |End|: follows from retractiveness. A Suslin line with more than ω1

automorphisms gives an example where they are different, assuming CH. And ifan example of an interval algebra A such that |Id(A)| < |End(A)| can be givenin ZFC, then assuming GCH + (there is no uncountable inaccessible) one canshow that for some infinite κ there is a κ+-Suslin tree. For, suppose that A isan interval algebra such that |Id(A)| < |End(A)|, GCH holds, and there are nouncountable inaccessibles. Thus |A| = |Ult(A)| = |Id(A)| and |End(A)| = |A|+. Ifc(A) = |A|, then c(A) is attained (since there are no uncountable inaccessibles),and hence |A| < |Id(A)|, contradiction. Thus c(A) < |A|, and |A| = |c(A)|+.If d(A) < |A|, then |End(A)| ≤ |Ult(A)|d(A) = |A|d(A) = |A|, contradiction. Soc(A) < d(A), and A must be the interval algebra on a (c(A))+-Suslin tree. So, theproblem of existence of such interval algebras A in ZFC is stronger than the aboveset-theoretical question; this is problem 69 in Monk [96]:

Problem 166. Can one construct in ZFC an interval algebra A such that |Id(A)| <|End(A)|?

25.65. |End| ≤ |Sub|. They differ on the interval algebra on the reals.

25.66. |Aut| < Ind: an infinite rigid interval algebra.

25.67. |Id| < |Aut|: as in the case of Suslin trees (see Chapter 22), one canconstruct a Suslin line with more than ω1 automorphisms. Again, the questionof existence of such algebras in ZFC is a strong set-theoretical hypothesis; this isproblem 70 in Monk [96]:

Problem 167. Can one construct in ZFC an interval algebra A such that |IdA| <|AutA|?

The tree algebra diagram: the indicated equalities and inequalities,and the “large” and “small” indications

See the diagram on the next page.

Let T be a tree, and let A = Treealg(T ).

25.68. IndA = ω, since A can be embedded in an interval algebra.

25.69. The difference between Ind and πχ can be arbitrarily large by the descrip-tion of πχ for tree algebras.

25.70. πχ ≤ t in general.

25.71. Depth = t, since tree algebras are retractive; see Chapter 4.

25.72. Length = Depth by the Brenner, Monk theorem (Handbook, p. 269).

25.73. The difference between πχ and Depth can be arbitrarily large; Chapter 11.

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Tree algebras


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25.74. χA = sup{|set of immed. succ. of C|, cf(C) : C an initial chain}; seeChapter 14. Finco κ is an example where χ is high and depth low.

25.75. In Chapter 3 we showed that

c(A) = max{|{t ∈ T : t has finitely many immed. succ.}|, Inc(T )}.

From this it is easy to see that χ ≤ c. In fact, suppose that C is an initial chainof T whose order type is an infinite regular cardinal. Obviously |set of immediatesucc. of C| ≤ c. If |{x ∈ C : x has finitely many immediate successors}| = |C|,clearly |C| ≤ c. If this set has power less than |C|, one can choose an element tothe side of x for |C| many elements x ∈ C, and this gives an incomparable subsetof T of size |C|, and so again |C| ≤ c. So this shows that χ ≤ c. An Aronszajnnon-Suslin tree with every element having infinitely many immediate successors isan example in which χ < c. The difference has to be small by the general diagram.

25.76. s=c by Chapter 3 plus the fact that tree algebras are retractive.

25.77. To see that s=hL, take any infinite tree T . Now Treealg(T ) embeds inan interval algebra A, and we may assume that Treealg(T ) is dense in A (ex-tend the identity from Treealg(T ) onto itself to a homomorphism from A into thecompletion of Treealg(T ), and then take the image of A). Hence

hL(A) ≥ hL(Treealg(T )) ≥ c(Treealg(T )) = c(A) = hL(A).

25.78. Suppose that c(A) < |A|. Then clearly T is a tree such that Inc(T ) < |T |and |T | has height |T | but no chains of length |T |. Moreover, since c(A) < |A|, thecardinal |A| must be a successor. Thus T is a generalized Suslin tree on a successorcardinal.

25.79. d=hd, since tree algebras are retractive, d(B) ≤ d(A) for B a subalgebraof A, and by Theorem 16.1.

25.80. Recall from Chapter 5 that d(A) = |A| for A a tree algebra; hence by thegeneral diagram,

|A| = h-cof(A) = Inc(A) = hd(A) = π(A) = d(A) = Irr(A).

25.81. For T a chain with order type a regular cardinal we have |A| = |Ult(A)|,since A is superatomic. For T the full binary tree of height ω we have |A| = ω and|Ult(A)| = 2ω.

25.82. For A = Finco(κ) we have |Ult(A)| = κ and |Id(A)| = 2κ = |Aut(A)|.25.83. There are rigid tree algebras.

25.84. |Id(A)| ≤ |End(A)| by retractiveness. See Chapter 22 for an example wherethey differ (consistently). If c(A) = |A| and cellularity is attained, then |Id(A)| =2|A|, so that such an example is impossible. On the other hand, if c(A) < |A|, thensee 25.78. So no such example is possible in ZFC alone.

25.85. 2|A| = |Sub(A)| since {T ↑ t : t ∈ T } is an irredundant set.

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25.86. |End(A)| ≤ |Sub(A)| in general; no example of a tree algebra is knownwhere they differ. By 25.85, the problem reduces to the following question; this isproblem 71 of Monk [96].

Problem 168. Is there a tree algebra A such that |End(A)| < 2|A|?

The tree algebra diagram: no other relationships

25.87. |AutA| < others: there are rigid tree algebras.

25.88. See Chapter 22 for an example of a tree algebra where |Id(A)| < |Aut(A)|(consistently). It is open to construct such an example in ZFC (Problem 166).

25.89. |Ult(A)| < |Aut(A)| for the interval algebra on an infinite cardinal.

The complete BA diagram: the indicated equalities andinequalities and the “large” and “small” indications

In fact, the “small” indications are clear.

25.90. c = Depth: obvious.

25.91. c < Length: P(ω).

Complete BAs

l = difference can be small; s =“small” difference

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25.92. c < d: completions of free algebras.

25.93. c < πχ: see 25.94.

25.94. d < π: completion of the free algebra on ω1 free generators.

25.95. πχ < π: under GCH these are equal, by an argument of Bozeman. It is notknown if this is true in ZFC. See Problem 74.

25.96. Length < Card: a large ccc algebra.

25.97. π < Card: P(κ).25.98. card = Ind = t = s = χ = hL = Irr = s = hd = Inc = h-cof: theseequalities all follow from |Ind| = Card (attained), which is a consequence of theBalcar, Franek theorem.

25.99. |Ult(A)| = |Id(A)| = |Sub(A)| = |End(A)| = 2|A|: again true since anyinfinite complete BA A has an independent subset of size |A|.25.100. |Aut| < |Ult|: a rigid complete algebra.

The complete BA diagram: no other relations

25.101. π < Length: P(ω).

25.102. Length < πχ: the completion of a free algebra.

25.103. Length < d: the completion of a free algebra.

25.104. Under GCH we have d ≤ πχ, by the result of Bozeman and Chapter 11;it is not known whether this holds in ZFC.

25.105. |Aut| < c: embed P(κ) in a rigid BA.

25.106. Card < |Aut|: the completion of the free BA of size 2ω.

Diagram for superatomic BAs: the indicated relations,and the “large” and “small” indications

See below.

25.107. Ind = ω since every superatomic BA has countable independence.

25.108. Ind < Depth: Any interval algebra on an uncountable cardinal providesan example; the difference can be arbitrarily large.

25.109. Depth = Length by Rosenstein [82] Corollary 5.29, p. 88.

25.110. Ind < πχ: the interval algebra on a cardinal provides an example withthe difference arbitrarily large.

25.111. Depth < t. There is an even stronger example, with c < t. Let 〈aα : α <ω1〉 be a system of subsets of ω such that for α < β < ω1 we have aα\aβ finiteand aβ\aα infinite. For the existence of such a sequence see Blass [10], 6.4. LetA be the subalgebra of P(ω) generated by the singletons together with the aα’s.Clearly A is as desired. On the other hand, in Dow, Monk [94] it is shown that ifκ→ (κ)<ω

2 , then any superatomic BA with tightness κ+ also has depth at least κ.This shows that the difference between tightness and depth cannot be arbitrarilylarge.

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Superatomic BAs

l = difference can be small; s =“small” difference

25.112. πχ < t: see Chapter 11; the difference can be large.

25.113. Obviously c = d = π = number of atoms.

25.114. Depth < c: they differ in a finite-cofinite algebra, where the difference canbe large.

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25.115. πχ < c: see Chapter 11; the difference can be large.

25.116. hd = s: see the characterizations of hd and s.

25.117. t < s, and the difference can be arbitrarily large: a finite-cofinite algebra.

25.118. c < s: Take a family A of 2ω almost disjoint subsets of ω, and considerthe BA generated by A ∪ {{i} : i ∈ ω}.25.119. s < Inc. Under ♦, Shelah constructed a thin-tall BA A with countablespread. Then A×A has countable spread too, while its incomparability is ω1. Theexample of Bonnet, Rubin [11], also constructed using ♦, can also be used for thispurpose. We do not know whether there is an example in ZFC; this is Problem 72in Monk [96]:

Problem 169. Is there an example in ZFC of a superatomic BA A such that sA <IncA?

25.120. Roslanowski, Shelah [00] showed that it is consistent to have a superatomicBA in which s is less than Irr. We do not know whether this can be done in ZFC;this is a version of Problem 73 of Monk [96]:

Problem 170. Can one construct in ZFC a superatomic BA A such that s(A) <Irr(A)?

25.121. card = h-cof = hL = χ: χ = Card by Chapter 14, and the other equalitiesfollow.

25.122. In Bonnet, Rubin [11] a superatomic algebra of power ω1 is constructedusing ♦ in which Inc is countable.

The following is Problem 74 in Monk [96]:

Problem 171. Can one construct in ZFC a superatomic algebra A with the propertythat Inc(A) < |A|?

25.123. In Roslanowski, Shelah [00] a superatomic BA is forced in which Irr(A) <|A|.

Problem 75 in Monk [96] is:

Problem 172. Can one construct in ZFC a superatomic algebra A with the propertythat Irr(A) < |A|?

25.124. |A| = |UltA|: see the Handbook.

25.125. |A| < |End(A)| in a finite-cofinite algebra.

25.126. M. Rubin constructed under ♦ a BA A such that |Aut(A)| < |A| (un-published, December 1992) in particular, |Aut(A)| < |End(A)|. In Shelah [01] analgebra A with |Aut(A)| < |End(A)| was constructed in ZFC. This solves Problem76 of Monk [96].

25.127. c < |Aut|. The BA of finite and cofinite subsets of κ gives an examplewhere they differ.

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25.128. |End(A)| ≤ |Id(A)| for A superatomic. For, let κ be the number of atomsof A. Then |Ult(A)| = |A| ≤ 2κ ≤ |Id(A)|, and hence by Chapter 21, |End(A)| ≤|Ult(A)|d(A) ≤ 2κ ≤ |Id(A)|. In the example of 25.118 we have |End(A)| = 2ω and|Id(A|) = 22

ω

.

25.129. Roslanowski, Shelah [00] showed that it is consistent to have a superatomicBA A where |Id(A)| < |Sub(A)|. We do not know whether this can be done inZFC; this is a version of Problem 77 of Monk [96]:

Problem 173. In ZFC, can one have |Id(A)| < |Sub(A)| in a superatomic BA?

Superatomic BAs, no additional relationships

25.130. πχ < Depth: see Chapter 11.

25.131. Depth < πχ: Dow, Monk [94] constructed an example.

25.132. t < c: the finite-cofinite algebra on κ.

25.133. Roslanowski, Shelah [00] showed that it is consistent to have a superatomicBA A with the property that Inc(A) < Irr(A). It is open whether this can be donein ZFC; this is a version of Problem 78 of Monk [96]:

Problem 174. Can one construct in ZFC a superatomic BA A such that Inc(A) <Irr(A)?

25.134. Also Roslanowski, Shelah [00] showed that it is consistent to have asuperatomic BA A with the property that Irr(A) < Inc(A). It is open whetherthis can be done in ZFC; this is a version of Problem 79 of Monk [96]:

Problem 175. Can one construct in ZFC a superatomic BA A such that Irr(A) <Inc(A)?

25.135. Card < |Aut|: a finite-cofinite algebra.

25.136. |Aut| small relative to “lower” functions. Recall 25.127. Shelah [01] con-structs a superatomic BA A such that |Aut(A)| < |A|. This solves problem 80 ofMonk [96].

Roslanowski, Shelah [00] showed that it is consistent to have a superatomic BAA such that |Aut(A)| < t(A). It is open to do this in ZFC; this is a version ofProblem 81 of Monk [96]:

Problem 176. In ZFC, is there a superatomic BA A such that |Aut(A)| < t(A)?

25.137. The relationship between Card and |Aut| is not completely clear, butwe indicate some other facts; see 25.136. Recall from Chapter 20 that an infinitesuperatomic BA has at least 2ω automorphisms. The initial chain algebra A on≤ω2 is such that |A| = |Aut(A)| = 2ω.

Now assume that 2ω = ω2, 2ω1 = ω3, and 2ω2 = ω4. Let T be the tree ≤ωω1,

and let A = Init(T ), the initial chain algebra on T . Note that |A| = ω2. For eachpermutation ϕ of ω1 there is an automorphism ϕ′ of A such that ϕ′(T ↓ t) =

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T ↓ (ϕ ◦ t) for every t ∈ T . If ϕ �= ψ, then ϕ′ �= ψ′. Thus this gives 2ω1 = ω3

automorphisms; since A has only ω1 atoms, it follows that |AutA| = ω3. Note that2|A| = ω4. Thus |A| < |Aut(A)| < 2|A|.

Initial chain algebras over a pseudo-tree

For this diagram, see Baur [00]. Examples in interval algebras exist for ω < πχ,πχ < t, t < c, χ < s, s < hd, hd < Inc, Depth < Length, c < d. An examplewith d < hd is given by the full binary tree <ω2; this also works for t < χand Length < card. The Bonnet, Shelah interval algebra gives an example withInc < card, assuming that cf(2ω) is a successor cardinal.

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Problem 177. Is there an example in ZFC with Inc <card for initial chain algebras?

A rigid interval algebra shows that one can have |Aut(A)| < |End(A)| for initialchain algebras.

Initial chain algebras over a tree

See Baur [00]. t < c for init(T ), T having only roots, uncountably many of them.Ind < πχ for init(ω1).

Small functions, atomless BAs

See the diagram on top of the next page.

25.138. r ≤ i: see Proposition 10.21.

25.139. r ≤ πχinf : see Theorem 6.28.

25.140. p ≤ f: Theorem 12.21.

25.141. p ≤ u: this is easy to check.

25.142. p ≤ smm: Proposition 13.20.

25.143. p ≤ a: Proposition 4.47(iii).

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Small functions, atomless Boolean algebras

25.144. p ≤ tow: Proposition 4.47(iii).

25.145. πχinf ≤ f: Proposition 12.20.

25.146. πχinf ≤ u: obvious.

25.147. πχinf ≤ smm: We will apply Theorem 6.28. Let X be maximal ideal inde-pendent with |X | = smm(A). Suppose that 2 ≤ m < ω. We claim that{

y ·∏

x∈F−x : y ∈ X,F ∈ [X\{y}]<ω

}

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is m-dense. For, suppose that 〈bi : i < m〉 is a weak partition of A. For any i < mwe have two possibilities:

(1) There is a finite F ∈ [X ]<ω such that bi ≤∑

F .

(2) There exist x ∈ X and a finite F ⊆ X\{x} such that x ≤ bi +∑

F .

If (2) holds for some i, clearly the desired conclusion follows. If (1) holds for all i,ideal independence of X is contradicted.

25.148. πχinf ≤ π: obvious.

25.149. a ≤ c: obvious.

25.150. tow ≤ c: obvious.

25.151. tow ≤ h: Proposition 4.62.

25.152. h ≤ spl: Proposition 4.63.

25.153. π ≤ Irrmm: Handbook, Proposition 4.23.

Concerning counterexamples, we have the following results and problems.

25.154. It is a problem to find a BA A such that i(A) < πχ(A). (Problem 101).

25.155. An example with i(A) < p(A) is given in Monk [01a].

25.156. For an example with f(A) < i(A) see example 12.22.

25.157. Under CH there is a BA A with f(A) < u(A). It is a problem to do this inZFC. (Problem 131)

25.158. A BA A with f(A) < smm(A) is described in Monk [11].

25.159. Concerning f and a we have the following problem:

Problem 178. Is there an atomless BA A such that f(A) < a(A)?

25.160. A BA A with f(A) < tow(A) is given in Monk [11].

25.161. For a BA A such that u(A) < i(A) see McKenzie, Monk [04].

25.162. It is a problem to get a BA with u < f:

Problem 179. Is there an atomless BA A such that u(A) < f(A)?

25.163. An example with u < smm is given in Monk [08].

25.164. An example with u < a is given in Monk [01a].

25.165. An example with u < tow is given in Monk [01a].

25.166. It is a problem to get a BA with smm < i. (Problem 126)

25.167. It is a problem to get a BA with smm < f:

Problem 180. Is there an atomless BA A such that smm(A) < f(A)?

25.168. Under CH there is a BA A such that smm(A) < u(A), but it is a problemto get one in ZFC. (Problem 134)

25.169. A BA A such that smm(A) < a(A) is given in Theorem 13.19.

25.170. See Chapter 13 for a BA A such that smm(A) < tow(A).

25.171. The exact place of Irrmm in the diagram is not known:

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Problem 181. Are there examples with Irrmm less than the other functions?

25.172. An example with spl(A) < r(A) is given in Monk [01a].

25.173. An example with spl(A) < a(A) is given in McKenzie, Monk [04].

Small functions, atomless interval algebras

25.174. i = r = ω in interval algebras, since an interval algebra does not have anuncountable independent subset.

25.175. πχinf(A) = ω for A an interval algebra, since there is a point or gap withleft or right character ω in any linear order; see the description of πχinf for intervalalgebras at the end of Chapter 11.

25.176. For p = f = u for interval algebras, see Monk [12].

25.177. p ≤ smm: true in general.

25.178. f ≤ tow for interval algebras: see Corollary 12.24.

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25.179. For smm ≤ a for interval algebras, see Monk [12].

25.180. For spl = π = d(L) see Monk [01a].

25.181. For an example of an interval algebra with ω < p see Monk [01a].

25.182. For an example of an interval algebra with a < tow see Monk [02].

25.183. It is a problem to find an interval algebra with h < smm:

Problem 182. Is there an atomless interval algebra A such that h(A) < smm(A)?

25.184. For an example of an atomless interval algebra A such that h(A) < spl(A),see Monk [01a].

25.185. It is a problem whether there is an atomless interval algebra A with π(A) <Irrmm(A):

Small functions, atomless complete BAs

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Problem 183. Is there is an atomless interval algebra A with π(A) < Irrmm(A)?

Small functions, atomless complete BAs

25.186. Clearly p = tow = a = ω in any complete BA.

25.187. It is shown in Monk [01a] that h ≤ r for complete BAs.

25.188. See Theorem 6.46 for πχinf = r for complete BAs.

25.189. In general r ≤ i, f, u, smm, π, and π ≤ Irrmm.

Small functions, superatomic Boolean algebras

25.190. There are several questions concerning whether the diagram is exactly asindicated. We mention only examples that are known, described in Monk [01a]:ω < h, i < u, u < spl, i < spl, spl < r, i, u.

Here spl′(A) = min{|X | : ∀a ∈ A[|A � a| ≥ ω → ∃x ∈ X [x · a �= 0 �= −x · a]]}.25.191. f = ω is shown in Monk [11].

25.192. u = p is shown in Monk [01a].

25.193. There are examples in Monk [12] with smm < spl, tow < smm, and smm <tow. Other possible improvements of the diagram are unknown.

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26 Examples

We determine our cardinal functions on the following examples, as much as pos-sible; see also the following table:

1. The finite-cofinite algebra on κ.

2. The free algebra on κ free generators.

3. The interval algebra on the reals.

4. P(κ).

5. The interval algebra on κ.

6. P(ω)/fin.

7. The Alexsandroff duplicate of a free algebra.

8. The completion of a free algebra.

9. The countable-cocountable algebra on ω1.

10. A compact Kunen line.

11. The Baumgartner-Komjath algebra.

12. The Rubin algebra.

We do not have to consider all of our 21 functions for each of them, since usuallythe determination of some key functions says what the rest are; see the diagrams.

1. The finite-cofinite algebra on κ

1. c(A) = κ.

2. t(A) = ω. See the beginning of Chapter 12.

3. |Ult(A)| = κ.

4. |Aut(A)| = 2κ.

2. The free BA on κ free generators

1. Length(A) = c(A) = ω. See Handbook, Part I, Corollaries 9.17 and 9.18.

2. d(A) = the smallest cardinal λ such that κ ≤ 2λ; see Corollary 5.6.

3. πχ(A) = κ: see Corollary 11.12.

4. Ind(A) = κ.

5. |Ult(A)| = 2κ.

6. |Aut(A)| = 2κ.

, ,OI 10.1007/978-3- - - _ ,


014


Basel 227

531

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532 Chapter 26. Examples

3. The interval algebra on the reals

1. π(A) = ω.

2. Inc(A) = 2ω. For example, {[r, r + 1) : r ∈ R} is incomparable.

3. |End(A)| = 2ω; Corollary 21.3.

4. |Aut(A)| = 2ω; this is clear by the above, since it is easy to exhibit 2ω

automorphisms.

4. P(κ)

1. c(A) = κ.

2. π(A) = κ.

3. πχ(A) = κ. An easy argument gives this.

4. Length(A) = Ded(κ). See Chapter 7.

5. |Ult(A)| = 22κ

.

6. |Aut(A)| = 2κ.

5. The interval algebra on κ

1. πχ(A) = κ. See the end of Chapter 11. πχ(A) is attained if κ is regular,otherwise not.

2. |Ult(A)| = κ. See Theorem 17.10 of Part I of the BA handbook.

3. |Aut(A)| = 2κ. See the end of Chapter 20.

6. P(ω)/fin

1. Depth(A) ≥ ω1. It is consistent that it is ω1 and ¬CH holds. Under MA, itis 2ω. See the end of Chapter 4.

2. Length(A) = 2ω.

3. c(A) = 2ω.

4. πχ(A) ≥ cf2ω. Con(2ω = ℵω1 + πχ(A) = ω1); see van Mill [84], p. 558.

5. Ind(A) = 2ω.

6. |Ult(A)| = 22ω

.

7. |Aut(A)| can consistently be 2ω or 22ω

; see van Mill [84], p. 537.

7. The Alexsandroff duplicate of a free BA

We use notation as in Chapter 14. Thus B is a free BA of size κ and Dup(B) isits Alexsandroff duplicate.

1. c(Dup(B)) = 2κ.

2. χDup(B) = κ. See Chapter 14.

3. Length(Dup(B)) = ω. In fact, suppose that Y ⊆ Dup(B), Y a chain, |Y | =ω1. Define (a,X) ≡ (b, Z) iff (a,X), (b, Z) ∈ Y and a = b. Then since B hasno uncountable chains, there are only countably many ≡-classes. So there isa class, say K, which has ω1 elements; say that a is the first member of each

ordered pair in K. Then Mdef= {X : (a,X) ∈ K} is of size ω1, is a chain

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Chapter 26. Examples 533

under inclusion, and if X,Z ∈ M with X ⊆ Z, then Z\X is finite. Clearlythis is impossible.

4. Ind(Dup(B)) = κ.

5. πχ(Dup(B)) = ω. (This corrects a mistake in Monk [90].) For, let H be anon-principal ultrafilter on UltDup(B). By the description of ultrafilters inProposition 1.19, there is an ultrafilter G on B such that

H = {(a,X) : a ∈ G, X ⊆ Ult(B), S(a)�X is finite}.

Let 〈xα : α < κ〉 be the system of free generators of B (without repetitions).Choose ε ∈ κ2 such that xεα

α ∈ G for all α < κ. For each n < ω let Fn bean ultrafilter on B such that xεα

α ∈ Fn for all α �= n and x1−εnn ∈ Fn. We

claim that {(0, {Fn}) : n < ω} is dense in H . For, let y ∈ H . Without loss ofgenerality y has the form

(xεα1α1

· . . . · xεαmαm

, X), S(xεα1α1

· . . . · xεαmαm

)�X finite.

Choose n < ω so that n �= α1, . . . , αm and Fn /∈ S(xεα1α1

· . . . · xεαmαm

)�X).Since Fn ∈ S(xεα1

α1· . . . · xεαm

αm), it follows that Fn ∈ X , as desired.

6. |Ult(Dup(B))| = 2κ. See the description of ultrafilters in Chapter 14.

7. |Id(Dup(B))| = 22κ

.

8. In an email message of January 1992, Sabine Koppelberg shows thatAut(Dup(A)) has exactly 2κ elements, answering Problem 64 in Monk [90].She also showed that |End(Dup(A))| = 22

κ

. We give her proofs here.

(a) For any BA A, let

Dup′(A) = 〈S[A] ∪ {{F} : F ∈ Ult(A)〉PUlt(A).

For A atomless, Dup(A) is isomorphic to Dup′(A); an isomorphism isgiven by f(a,X) = X for all (a,X) ∈ Dup(A), as is easily checked.

(b) Any automorphism of A induces an automorphism of Dup(A). Namely,if f is an automorphism of A, define f+(a,X) = (f(a), {f [F ] : F ∈ X});it is easy to check that f+ is an automorphism of Dup(A). Clearlyf+ �= g+ for distinct f, g.

(c) In Dup′(A), {F} =⋂

a∈F S(a). Hence if f and g are automorphisms ofDup′(A) and f � S[A] = g � S[A], then f = g.

(d) From (a)–(c) it follows that |Aut(Dup(A))| = 2κ for A free on κ freegenerators.

(e) Suppose that A is atomless. Let f be a homomorphism from S[A] intoDup′(A) and g a homomorphism from Finco(Ult(A)) into Dup′(A).Then f ∪ g extends to an endomorphism of Dup′(A) iff the followingcondition holds:

(∗) If a ∈ A, M is a finite subset of Ult(A), and M ⊆ S(a), theng(M) ⊆ f(S(a)).

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In fact, ⇒ is obvious. For ⇐, to apply Sikorski’s criterion suppose thatS(a) ∩ N = 0, where a ∈ A and N ∈ Finco(Ult(A)). If N is cofinite,then S(a) is finite, hence a = 0, so f(S(a)) ∩ g(N) = 0. If N is finite,then f(S(a)) ∩ g(N) = 0 by (∗).

(f) Suppose that π is a one-one mapping of Ult(A) into Ult(A). Then wecan define an endomorphism π+ of Finco(Ult(A)) by setting π+(F ) ={π(F ) : F ∈ F} for F a finite subset of Ult(A), and π+(F ) =Ult(A)\π+(Ult(A)\F ) for F a cofinite subset of Ult(A). Then for fa homomorphism from S[A] into Dup′(A) the criterion (∗) for f ∪ π+

to extend to an endomorphism of Dup′(A) becomes

(∗∗) If a ∈ A and F ∈ S(a), then π(F ) ∈ f(S(a)).

(g) Now suppose that A is free on κ free generators. We show that Dup(A)has 22

κ

endomorphisms. Now A is isomorphic to A ⊕ A; let h be anisomorphism of A ⊕ A onto A. Let k be the embedding of A onto thefirst factor of A ⊕ A, and l the embedding onto the second factor. Setf = h◦k. Thus f is a one-one endomorphism of A, so the dual mappingf−1 from Ult(A) to Ult(A) is onto. Now

(�) For each G ∈ Ult(A), the set {F : f−1[F ] = G} has at least twoelements.

In fact, we claim that for any non-zero element b of A the set f [G] ∪{h(l(b))} has the finite intersection property. If not, there is an elementa ∈ G such that f(a) · h(l(b)) = 0; since f = h ◦ k, it follows thatk(a) · l(b) = 0, contradiction. This claim being true, if we take b ∈A\{0, 1}, we can get ultrafilters F, F ′ in A such that f [G]∪{h(l(b))} ⊆ Fand f [G] ∪ {h(l(−b))} ⊆ F ′. Then F �= F ′ and f−1[F ] = G = f−1[F ′],as desired in (�).

Now take a function π from Ult(A) into Ult(A) such that πG ∈ {F :f−1[F ] = G} for all G ∈ UltA. Let f ′[S(a)] = S(f(a)) for all a ∈ A.Then f ′ is a homomorphism from S[A] into Dup′(A). If G ∈ S(a),then π(G) ∈ {F : f−1[F ] = G}, so f−1[π(G)] = G, a ∈ f−1[π(G)],f(a) ∈ π(G), and π(G) ∈ S(f(a)) = f ′[S(a)]. This means that (∗∗)holds for π and f ′, and so f ′ ∪ π+ extends to an endomorphism ofDup′(A). Clearly π+ �= σ+ for distinct π, σ, so we have exhibited 22

κ

endomorphisms of Dup′(A), as desired.

8. The completion of a free algebra

Let B be a free algebra of size κ, A its completion.

1. c(A) = ω.

2. Length(A) = 2ω. In fact, ≥ is clear. Suppose that L is a chain of size (2ω)+.Using a well-ordering of L and the partition relation (2ω)+ → (ω1)

2ω, we get

an uncountable well-ordered chain in A, contradiction.

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3. d(A) is the least cardinal λ such that κ ≤ 2λ. For, this is true for B itself byChapter 5, and an application of Sikorski’s extension theorem shows that itis true of A.

4. πχ(A) = κ. For, ≤ is clear. Suppose that F is an ultrafilter on A and D is aπ-base for F . Without loss of generality D ⊆ B. Then D is dense in F ∩ B,so |D| ≥ κ.

5. π(A) = κ by the same argument.

6. |A| = κω.

7. |Ult(A)| = 2κω

.

8. |Aut(A)| = 2κ. In fact, any automorphism of A is uniquely determined by itsrestriction to B, and there are only 2κ mappings of B into A. On the otherhand, there are at least 2κ automorphisms of A.

9. The countable-cocountable algebra on ω1

1. Depth(A) = ω1, by an easy argument.

2. Length(A) = 2ω.

3. π(A) = ω1.

4. Ind(A) = 2ω.

5. πχ(A) = ω1: let F be the ultrafilter of cocountable sets. Suppose that D isdense in F , with |D| ≤ ω. Without loss of generality the members of D aresingletons. But then ω1\

⋃D ∈ F , contradiction.

6. |A| = 2ω.

7. |Ult(A)| = 22ω

.

8. |Aut(A)| = 2ω1 .

10. A compact Kunen line

Recall that this Boolean algebra was constructed using CH.

1. Irr(A) = ω. See Chapter 8.

2. χ(A) = ω1. This was proved in the discussion following Corollary 14.12.

3. |A| = |Ult(A)| = ω1. Clear from the construction.

4. |End(A)| = ω1 by 3 and Theorem 21.1.

5. Although we have not been able to determine Inc(A), the algebra A × Ahas incomparability ω1, and still has the other important properties of A:|A × A| = |Ult(A × A)| = ω1, χ(A × A) = ω1, and Irr(A × A) = ω. The set{(a,−a) : a ∈ A} is incomparable, showing that Inc(A×A) = ω1. Obviously|A×A| = |Ult(A×A)| = ω1 and χ(A×A) = ω1. To see that Irr(A×A) = ω,note from the Handbook volume 1, example 11.6, that (A×A)⊕ (A×A) ∼=(A⊕ A)4, and then apply Heindorf’s theorem in Chapter 8.

Problem 184. Determine Inc(A), |Aut(A)|, |Id(A)|, and |Sub(A)| for the compactKunen line of Chapter 8.

This is problem 96 in the Monk [96].

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Table of examples

Example Depth πχ c Length Ind d t π χ s Irr

Fincoκ ω ω κ ω ω κ ω κ κ κ κ

Frκ ω κ ω ω κ (1) κ κ κ κ κ

IntalgR ω ω ω 2ω ω ω ω ω ω ω 2ω

Pκ κ κ κ Dedκ 2κ κ 2κ κ 2κ 2κ 2κ

Intalgκ κ κ κ κ ω κ κ κ κ κ κ

Pω/fin (2) (3) 2ω 2ω 2ω 2ω 2ω 2ω 2ω 2ω 2ω

Dup ω ω 2κ ω κ 2κ κ 2κ κ 2κ 2κ

Fr(κ) ω κ ω 2ω κω (1) κω κ κω κω κω

Cblcoω1 ω1 ω1 ω1 2ω 2ω ω1 2ω ω1 2ω 2ω 2ω

CKL ω ω ω ω ω ω ω ω ω1 ω ω

BK ω ω ω ω ω ω ω ω ω1 ω ?

Rubin ω ω ω ω ω ω ω ω ω ω ω

Notes:Dup is the Alexsandroff duplicate of the free BA of size κ.CKL is the compact Kunen line constructed in chapter 8.BK is the Baumgartner, Komjath algebra constructed in chapter 17.Rubin is the algebra constructed in chapter 18.

(1) The least λ such that κ ≤ 2λ.(2) The depth is ≥ ω1; various possibilities are consistent.(3) ≥ cf2ω.

(Table continued on the next page)

11. The Baumgartner, Komjath algebra

Recall that this BA was constructed using ♦. See Chapter 17.

1. Inc(A) = ω.

2. Length(A) = ω.

3. χ(A) = ω1.

4. |Ult(A)| = ω1. To see this, first note that each ultrafilter on A is determinedby the membership of the elements xα or their complements. Hence thisequality follows from the following fact:

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(∗) If F and G are ultrafilters on A, xα ∈ F ∩G, and F ∩Aα+1 = G∩Aα+1,then F = G.

In fact, suppose that β ∈ (α, ω1). If xβ ∈ F , then xα ∩ xβ ∈ F ; but byconstruction xα ∩ xβ ∈ Aα+1, so xα ∩ xβ ∈ G and so xβ ∈ G. The sameargument works if ω\xβ ∈ F , so F ⊆ G and hence F = G.

5. |End(A)| = ω1 by 4 and Theorem 21.1.

6. Since A has a nonzero element a such that A � a is countable, A � a has ω1

automorphisms, and so the same is true of A itself.

Problem 185. Determine Irr(A), |Id(A)|, and |Sub(A)| for the Baumgartner, Kom-jath algebra.

This is problem 97 in Monk [96].

Example hL hd Inc h-cof Card |Ult| |Aut| |Id| |End| |Sub|

Fincoκ κ κ κ κ κ κ 2κ 2κ 2κ 2κ

Frκ κ κ κ κ κ 2κ 2κ 2κ 2κ 2κ

IntalgR ω ω 2ω 2ω 2ω 2ω 2ω 2ω 2ω 22ω

Pκ 2κ 2κ 2κ 2κ 2κ 22κ

2κ 22κ

22κ

22κ

Intalgκ κ κ κ κ κ κ 2κ 2κ 2κ 2κ

Pω/fin 2ω 2ω 2ω 2ω 2ω 22ω

(1) 22ω

22ω

22ω

Dup 2κ 2κ 2κ 2κ 2κ 2κ 2κ 22κ

22κ

22κ

Fr(κ) κω κω κω κω κω 2κω

2κ 2κω

2κω

2κω

Cblcoω1 2ω 2ω 2ω 2ω 2ω 22ω

2ω1 22ω

22ω

22ω

CKL ω1 ω ? ω1 ω1 ω1 ? ? ω1 ?

BK ω1 ω ω ω1 ω1 ω1 ω1 ? ω1 ?

Rubin ω ω ω ω ω1 ω1 ? ω1 ω1 ω1

Notes:Dup is the Alexsandroff duplicate of the free BA of size κ.CKL is the compact Kunen line constructed in chapter 8.BK is the Baumgartner, Komjath algebra constructed in chapter 17.Rubin is the algebra constructed in chapter 18.

(1) Consistently 2ω or 22ω

.

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12. The Rubin algebra

This algebra was also constructed using ♦.1. h-cof(A) = ω; see Chapter 18.

2. Irr(A) = ω; see Rubin [83].

3. |A| = ω1. This is clear from the construction in Chapter 18.

4. |Sub(A)| = ω1. See Chapter 23.

We do not know about |Aut(A)|, although the construction can be changed tomake A rigid; see Shelah [91] for more details.

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27 Problems

Problem 1. Is there in ZFC an infinite minimally generated BA with no countablyinfinite homomorphic image? Page 69

Problem 2. Is every disjunctively generated BA isomorphic to a tail algebra?Page 70

Problem 3. For which infinite cardinals κ, λ are there BAs A, B such that c′(A) =κ, c′(B) = λ, while c′(A⊕B) > max(κ, λ)? Page 84

Problem 4. Can one construct BAs A,B such that c′(B) < c′(A) < c′(A ⊕ B)with c′(B) regular limit? Page 106

Problem 5. Can one prove in ZFC that if κ is inaccessible but not weakly compact,then there a BA A such that c′(A) = κ while c′(A⊕A) > κ? Page 106

Problem 6. Describe the possibilities for chain conditions in ultraproducts withrespect to countably complete ultrafilters. Page 109

Problem 7. Does A ≤s B imply that a(B) ≤ a(A)? Page 123

Problem 8. Is it true that for all infinite BAs A,B one has

a(A⊕B) = min(a(A), a(B))? Page 125

Problem 9. Suppose that K is a set of infinite cardinals without greatest element,with sup(K) inaccessible. Is there a BA A such that aspect(A) = K? Page 127

Problem 10. Describe in terms not involving BAs the sets K,L such that thereexist BAs A,B with A ≤reg B, aspect(A) = K, and aspect(B) = L. Page 127

Problem 11. Describe aspect(⊕i∈IAi) in terms of the sets aspect(⊕i∈FAi) for F afinite subset of I. Page 127

Problem 12. Describe the behaviour of a(A) under ultraproducts. Page 127

Problem 13. Give a purely cardinal number characterization of cSr. Page 137

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540 Chapter 27. Problems


cSr(A) = {(ω, ω), (ω, ω1), (ω1, ω1), (ω1, ω2)}? Page 140


cSr(A) = {(ω, ω), (ω, ω1), (ω1, ω1), (ω1, ω2), (ω2, ω2)}? Page 141

Problem 16. Give a cardinal number characterization of cHr(A). Page 141

Problem 17. Is it consistent that cHr(A) = {(ω, ω1), (ω1, ω1), (ω1, ω2)} for someBA A? Page 147

Problem 18. Can one prove in ZFC that there is a BA A such that

cHr(A) = {(ω, ω), (ω, ω1), (ω1, ω1), (ω1, ω2)}? Page 148

Problem 19. Is it consistent that cHr(A) = {(ω, ω1), (ω, ω2), (ω1, ω1), (ω2, ω2)} forsome BA A? Page 149


cHr(A) = {(ω, ω), (ω, ω1), (ω1, ω1), (ω1, ω2), (ω2, ω2)}? Page 150

Problem 21. Can one prove in ZFC that there is a BA A of size ω2 such that(ω1, ω2) ∈ cHr(A) but (ω, ω2) /∈ cHr(A)? Page 150

Problem 22. Are the following conditions equivalent, for any linearly orderedset L?

(i) a(Intalg(L)) ≥ κ.

(ii) L is κ-saturated. Page 153

Problem 23. For each infinite cardinal κ, give a criterion, purely in terms of thelinear order L, for there to exist a partition of size κ in Intalg(L). Page 153

Problem 24. If K is a nonempty set of infinite cardinals, is there a tree algebraA such that aspect(A) = K? Page 153

Problem 25. For each infinite cardinal κ, give a criterion, purely in terms of thetree T , for there to exist a partition of size κ in Treealg(T ). Page 153

Problem 26. If K is a nonempty set of infinite cardinals, is there a pseudo-treealgebra A such that aspect(A) = K? Page 154

Problem 27. For each infinite cardinal κ, give a criterion, purely in terms of thepseudo-tree T , for there to exist a partition of size κ in Treealg(t). Page 154

Problem 28. Find necessary and sufficient conditions on a BA A for there to existextensions B,C of A such that Depth′(B ⊕A C) > max(Depth′(B),Depth′(C)).Page 164

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Chapter 27. Problems 541

Problem 29. Determine the possibilities for Depth(∏

i∈I Ai/F ) in terms of

〈Depth(Ai) : i ∈ I〉

with F countably complete. Page 164

Problem 30. Is an example with Depth(∏

i∈I Ai/F)>∣∣∏

i∈I Depth(Ai)/F∣∣ pos-

sible in ZFC? Page 166

Problem 31. Is it true that for all infinite cardinals κ ≤ λ there is a BA A of sizeλ with dDepthS−(A) = κ? Page 178

Problem 32. Let C be a nonempty set of infinite cardinals such that the followingcondition holds:

(∗) fIf κ ∈ C, λ < κ, and cf(λ) = cf(κ), then λ ∈ C.

Is there an atomless BA A such that towwspect(A) = C? Page 180

Problem 33. Characterize those sets K of limit ordinals for which there is anatomless BA A such that K is the collection of all order types of very weak towersof A. Page 180

Problem 34. Given nonempty sets K ⊆ L of cardinals, are there BAs A ≤reg Bsuch that towspect(A) = K and towspect(B) = L? There is a similar question forA ≤free B, A ≤proj B, A ≤rc B, A ≤π B. Page 180

Problem 35. Given nonempty sets K ⊆ L of cardinals with ω /∈ K, are there BAsA ≤σ B such that towspect(A) = K and towspect(B) = L? Page 181

Problem 36. Given nonempty sets K ⊆ L of cardinals, are there BAs A ≤m Bsuch that towspect(A) = K and towspect(B) = L? Page 181

Problem 37. Does A ≤s B imply that towspect(A) ⊆ towspect(B) or tow(B) ≤t(A)? Page 182

Problem 38. Are there BAs A,B such that A ≤m B and tow(B) < tow(A)?Page 182

Problem 39. Describe the behaviour of tow under ultraproducts. Page 186

Problem 40. Describe in cardinal number terms the pairs M,N of cardinals suchthat there is a BA A with towspect(A) = M and aspect(A) = N . Page 189

Problem 41. Given nonempty sets K ⊆ L of cardinals, are there BAs A ≤reg Bsuch that pspect(A) = K and pspect(B) = L? There is a similar question forA ≤free B, A ≤proj B, A ≤rc B, A ≤π B. Page 189

Problem 42. Given nonempty sets K ⊆ L of cardinals with ω /∈ K, are there BAsA ≤σ B such that pspect(A) = K and pspect(B) = L? Page 189

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Problem 43. Given nonempty sets K ⊆ L of cardinals with ω /∈ K, are there BAsA ≤mm B such that pspect(A) = K and pspect(B) = L? Page 189

Problem 44. Does A ≤s B imply that pspect(A) ⊆ pspect(B)? Page 190

Problem 45. Are there BAs A,B such that A ≤m B and p(B) < p(A)? Page 190

Problem 46. Is it true that for all infinite BAs A,B we have p(A ⊕ B) =min(p(A), p(B))? Page 191

Problem 47. Given cardinals κ, λ, μ with ω < κ ≤ λ, μ, is there a BA A such thatp(A) = κ, tow(A) = λ, and a(A) = μ? Page 192

Problem 48. Is

p(A) = min{|X | : X is a maximal ramification set in A}? Page 192

Problem 49. Is it consistent that p(P(ω)/fin) < tow(P(ω)/fin)? Page 192

Problem 50. Can one prove in ZFC that for each regular cardinal κ there is a BAA such that h(A) = κ? Page 195

Problem 51. What is the relationship, if any, between h(A⊕B) and h(A), h(B)?Page 198

Problem 52. Is spl(A⊕B) = min(spl(A), spl(B)) for atomless A,B? Page 198

Problem 53. What is the relationship between h of algebras and their ultraproduct?Page 198

Problem 54. What is the relationship between spl of algebras and their ultraprod-uct? Page 198

Problem 55. Suppose that κ is regular and κ ≤ λ. Is there an atomless BA A suchthat h(A) = κ and spl(A) = λ? Page 198

Problem 56. Can one construct in ZFC a BA A such that t(A) /∈ DepthHs(A)?Page 199

Problem 57. Can one show in ZFC that there is a cardinal κ such that there is aBA A of size (2κ)+ with Depth(A) = κ while (ω, (2κ)+) /∈ DepthSr(A)? Page 200

Problem 58. Characterize the relation DepthSr. Page 200

Problem 59. Can one prove in ZFC that for every infinite cardinal λ there is aninterval algebra A such that |A| = λ+ and every subalgebra of A of size λ+ hasdepth λ? Page 202


DepthSr(A) = {(ω, ω), (ω, ω1), (ω1, ω1), (ω1, ω2)}? Page 202

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Problem 61. Characterize the relation DepthHr. Page 207

Problem 62. Describe the topological density of amalgamated free products of BAs.Page 222

Problem 63. Is it true that [ω, hd(A)) ⊆ dHs(A) for every infinite BA A? Page 231

Problem 64. Completely describe dHs. Page 232

Problem 65. Characterize in cardinal number terms the sets dSr. Page 232

Problem 66. Is there a BA A such that dSr(A) = {(ω, ω), (ω1, ω1), (ω1, ω2)}?Page 232

Problem 67. Characterize in cardinal number terms the sets dHr. Page 232

Problem 68. Is it consistent to have a BA A such that dHr(A) = {(ω, ω), (ω1, ω2)}?Page 233

Problem 69. Describe d(A) for A a pseudo-tree algebra. Page 235

Problem 70. Are there BAs A,B such that A ≤σ B and π(A) > π(B)? Page 238

Problem 71. Are there BAs A,B such that A ≤s B and π(A) > π(B)? Page 238

Problem 72. Are there BAs A,B such that A ≤u B and π(A) > π(B)? Page 238

Problem 73. Characterize the density of amalgamated free products of BAs.Page 239

Problem 74. Can one prove in ZFC that πχ(B) = π(B) for any atomless completeBA? Page 275

Problem 75. Characterize πSr in cardinal number terms. Page 276

Problem 76. Characterize πHr in cardinal number terms. Page 277

Problem 77. Let ω < cf(κ) < κ, and let L be a dense linear ordering of size cf(κ)with no dense subset of size less than cf(κ), and with no family of cf(κ) pairwisedisjoint open intervals. Let A be the interval algebra on L, and suppose that B isa BA with Length(B) = κ. Does A⊕ B have a chain of size κ? Page 282

Problem 78. Investigate the length of amalgamated free products. Page 282

Problem 79. Is there a BA A such that Length(A) < Length(Exp(A))? Page 283

Problem 80. Is LengthH+(A)=t(A)·Length(A) for every infinite BAA? Page 285

Problem 81. Is always Lengthh−(A) = ω? Page 289

Problem 82. What are the possibilities for LengthHs(A)? Page 289

Problem 83. Can one prove in ZFC that if K is a nonempty set of infinite cardinalsthen there is a BA A such that Lengthspect(A) = K? Page 290

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Problem 84. Characterize in cardinal number terminology the sets LengthSr(A).Page 290

Problem 85. Characterize in cardinal number terminology the sets LengthHr(A).Page 290

Problem 86. Can one have Irr(A) < Irr(B) for A ≤s B or A ≤m B? Page 291

Problem 87. For 〈Ai : i ∈ I〉 a system of infinite BAs with I infinite, determine,if possible, Irr

(∏i∈I Ai

)in terms of 〈Irr(Ai) : i ∈ I〉. Page 296

Problem 88. For infinite BAs A,B is it true that max(Irr(A), Irr(B)) = Irr(A ⊕B)? Page 297

Problem 89. Can one construct in ZFC a BA A such that IrrA < |A|? Page 309

Problem 90. Is Irrmm(A) = π(A) for every infinite BA A? Page 310

Problem 91. Can one prove in ZFC that cf(A) ≤ ω1 for every infinite BA A?Page 315

Problem 92. Is alt(A) = p-alt(A) for every infinite BA A? Page 315

Problem 93. Is Theorem 9.51 best possible? In particular, is it consistent thatthere is a BA A such that CardHs(A) = {ω2, ω4}? Page 334

Problem 94. Do there exist infinite BAs A,B such that A ≤mg B and Ind(A) <Ind(B)? Page 335

Problem 95. Describe the independence of the of a system of BAs. In particular,does the analog of Theorem 10.6 hold? Page 345

Problem 96. Can one construct in ZFC a BA A with the property that IndH−A <CardH−A? Page 346

Problem 97. Is Indh+A = Cardh+A for every infinite BA A? Page 346

Problem 98. Is Indh−A = IndH−A for every infinite BA A? Page 347

Problem 99. Determine i(∏

i∈I Ai

)in terms of I and 〈i(Ai) : i ∈ I〉, where I is

infinite and each Ai is atomless. Page 355

Problem 100. Determine i(∏B

i∈I Ai

)in terms of B and 〈i(Ai) : i ∈ I〉, where

I is infinite, Finco(I) ≤ B ≤ P(I), each Ai is atomless (moderate products).Page 355

Problem 101. Is πχinf(A) ≤ i(A) for every atomless BA A? Page 355

Problem 102. Describe Freecal(∏

i∈I Ai

)in terms of 〈Freecal(Ai) : i ∈ I〉.

Page 359

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Problem 103. For K any nonempty set of regular cardinals, are the followingconditions equivalent?

(i) There is a BA A such that K = {κ : κ ∈ Freecal(A) and κ is regular}.(ii) The following conditions hold, where μ = minK and ν = supK:

(a) μ is uncountable.

(b) For all λ ∈ (μ, ν], if λ is regular and is not the successor of a singularcardinal, then λ ∈ K.

(c) For all λ ∈ (μ, ν], if λ = σ+ for some singular σ with μ ≤ cf σ, thenλ ∈ K. Page 371

Problem 104. Assume that ρ < ν < κ ≤ 2ρ < λ ≤ 2ν with κ and λ regular. Isthere a κ-cc BA A of power λ with no independent subset of power λ? Page 371

Problem 105. Can one prove the following in ZFC? For every m ∈ ω with m ≥ 2there is an interval algebra having a subset P of size ω1 such that for all Q ∈[P ]ω1 , Q has m pairwise comparable elements and also m independent elements.Page 372

Problem 106. Describe the relationship between πχ(A) and πχ(B) for the variousspecial notions of A a subalgebra of B. Page 374

Problem 107. What is the relationship between πχ(A) and πχ(B) for A the one-point gluing of B? Page 378

Problem 108. Characterize πχ(Dup(A)) in terms of A. Page 378

Problem 109. Give a purely cardinal number characterization of πχSr. Page 384

Problem 110. Give a purely cardinal number characterization of πχHr. Page 384

Problem 111. Does attainment of tightness in the πχH+ sense imply attainmentin the sense of the definition? Page 391

Problem 112. Is the following true? Let κ and λ be infinite cardinals, with λregular. Then the following conditions are equivalent:

(i) cf(κ) = λ.

(ii) There is a strictly increasing sequence 〈Aα : α < λ〉 of Boolean algebras eachhaving no ultrafilter with tightness κ such that

⋃α<λ Aα has an ultrafilter

with tightness κ. Page 396

Problem 113. Do there exist a system 〈Ai : i ∈ I〉 of BAs and a system 〈Fi : i ∈ I〉of ultrafilters, each Fi an ultrafilter on Ai, such that, with B the one-point gluingusing these inputs, t(B) < t

(∏i∈I Ai

)? Page 397

Problem 114. Is there a BA A such that t(A) < t(Dup(A))? Page 397

Problem 115. Describe the possibilities for tHs. Page 400

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Problem 116. Describe tSr(A) in cardinal number terms. Page 400

Problem 117. Describe tHr(A) in cardinal number terms. Page 400

Problem 118. Can one prove in ZFC that for every uncountable cardinal λ, t′(B) ≥λ+ implies that Depth′(B) ≥ λ? Page 404

Problem 119. Completely describe the behaviour of spread under one-point gluing.Page 413

Problem 120. Is there a BA A such that s(A) < s(Exp(A))? Page 413

Problem 121. Completely describe dsS− in terms of the other cardinal functions.Page 413

Problem 122. Is sspect(A×B) = sspect(A) ∪ sspect(B)? Page 414

Problem 123. Is smm(A×B) = smm(A) ∪ smm(B)? Page 414

Problem 124. Is the assumption |K| ≤ min(K) in Corollary 13.16 necessary?Page 415

Problem 125. How can Corollary 13.16 be extended to singular cardinals in K?Page 415

Problem 126. Is there an atomless BA A such that smm(A) < i(A)? Page 417

Problem 127. What is the exact place of smm(P(ω)/fin) among the other contin-uum cardinals? Page 419

Problem 128. Completely describe the possibilities for character with respect tothe various subalgebra notions. Page 422

Problem 129. Describe the behaviour of character under unions. Page 424

Problem 130. Describe χ(Exp(A)). Page 427

Problem 131. Can one prove in ZFC that there is an atomless BA A such thatf(A) < u(A)? Page 429

Problem 132. Can one prove in ZFC that there is an atomless BA A such thatsmm(A) < u(A)? Page 429

Problem 133. Can one construct in ZFC a BA A such that s(A) < χ(A)?Page 430

Problem 134. Characterize χHs(A). Page 432

Problem 135. Characterize χSs(A). Page 432

Problem 136. Characterize χSr(A). Page 432

Problem 137. Characterize χHr(A). Page 432

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Problem 138. Characterize χ(A) for A a complete BA. Page 434

Problem 139. Completely describe the relations between these attainment possi-bilities for hL. Page 439

Problem 140. Is there a BA A such that hL(A) < hL(Exp(A))? Page 442

Problem 141. Does every BA have a maximal right-separated sequence of ultrafil-ters? Page 446

Problem 142. Describe hLHs(A) in cardinal number terms. Page 446

Problem 143. Describe hLSr(A) in cardinal number terms. Page 446

Problem 144. Describe hLHr(A) in cardinal number terms. Page 446

Problem 145. Is there an example in ZFC of a BA A such that hL(A) < d(A)?Page 448

Problem 146. Completely describe the relationships between the attainment prob-lems for hd. Page 450

Problem 147. Do there exist BAs A,B and ultrafilters F,G on A,B respec-tively such that if C is the associated one-point gluing, then max(hd(A), hd(B)) <hd(C)? Page 453

Problem 148. Is it consistent to have an infinite BA A such that |hdspect(A)| > 2?Page 454

Problem 149. What is the exact place of hdidmm among the other cardinal func-tions? Page 455

Problem 150. Can one construct in ZFC a BA A such that s(A) < hd(A)?Page 460

Problem 151. Can one construct in ZFC a BA A such that hd(A) < χ(A)?Page 460

Problem 152. Is it true that for every regular limit cardinal κ there is a BA withInc(A) = κ not attained? Page 464

Problem 153. Completely describe the behaviour of Inc for special subalgebras.Page 464

Problem 154. Do there exist in ZFC a system 〈Ai : i ∈ I〉 of infinite BAs, I infi-nite, and a regular ultrafilter F on I such that Inc

(∏i∈I Ai/F

)>∣∣∏

i∈I IncAi/F∣∣?

Page 464

Problem 155. If κ is an infinite cardinal and κ < μ < 2κ, μ not a power of 2, isthere a maximal incomparable set in P(κ) of size μ? Page 465

Problem 156. Is Inctreespect(P(κ)) = {κ, 2κ}? Page 466

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Problem 157. Is it consistent that Incmm(P(ω)/fin) < 2ω or Inctreemm(P(ω)/fin) <2ω? Page 468

Problem 158. Is it consistent that Incmm(P(ω)/fin) �= Inctreemm(P(ω)/fin)?Page 468

Problem 159. Can one construct in ZFC a BA A such that Inc(A) < χ(A)?Page 480

Problem 160. Is h-cofmm = p? Page 482

Problem 161. What are the possibilities for h-cof2spect(A)? Page 483

Problem 162. Can one construct in ZFC a BA A with the property that Inc(A) <h-cof(A)? Page 483

Problem 163. Can one construct in ZFC a BA A such that |Id(A)| < |Aut(A)|?Page 497

Problem 164. Can one construct in ZFC a BA A such that s(A) < hL(A)?Page 511

Problem 165. Can one construct in ZFC a BA A such that IrrA < χA? Page 513

Problem 166. Can one construct in ZFC an interval algebra A such that |Id(A)| <|End(A)|? Page 515

Problem 167. Can one construct in ZFC an interval algebra A such that |IdA| <|AutA|? Page 515

Problem 168. Is there a tree algebra A such that |End(A)| < 2|A|? Page 518

Problem 169. Is there an example in ZFC of a superatomic BA A such thatsA < IncA? Page 521

Problem 170. Can one construct in ZFC a superatomic BA A such that s(A) <Irr(A)? Page 521

Problem 171. Can one construct in ZFC a superatomic algebra A with the propertythat Inc(A) < |A|? Page 521

Problem 172. Can one construct in ZFC a superatomic algebra A with the propertythat Irr(A) < |A|? Page 521

Problem 173. In ZFC, can one have |Id(A)| < |Sub(A)| in a superatomic BA?Page 522

Problem 174. Can one construct in ZFC a superatomic BA A such that Inc(A) <Irr(A)? Page 522

Problem 175. Can one construct in ZFC a superatomic BA A such that Irr(A) <Inc(A)? Page 522

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Problem 176. In ZFC, is there a superatomic BA A such that |Aut(A)| < t(A)?Page 522

Problem 177. Is there an example in ZFC with Inc <card for initial chain algebras?Page 524

Problem 178. Is there an atomless BA A such that f(A) < a(A)? Page 526

Problem 179. Is there an atomless BA A such that u(A) < f(A)? Page 526

Problem 180. Is there an atomless BA A such that smm(A) < f(A)? Page 526

Problem 181. Are there examples with Irrmm less than the other functions?Page 527

Problem 182. Is there an atomless interval algebra A such that h(A) < smm(A)?Page 528

Problem 183. Is there is an atomless interval algebra A with π(A) < Irrmm(A)?Page 529

Problem 184. Determine Inc(A), |Aut(A)|, |Id(A)|, and |Sub(A)| for the compactKunen line of Chapter 8. Page 535

Problem 185. Determine Irr(A), |Id(A)|, and |Sub(A)| for the Baumgartner, Kom-jath algebra. Page 537

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References

Abraham, U.; Magidor, M. [10] Cardinal arithmetic. Chapter 14 in Handbook ofSet Theory, 1149–1227, Eds. Foreman, Kanamori, xiv + 2177 pp.

Arhangelskiı, A. [78]. The structure and classification of topological spaces andcardinal invariants. Russian Mathematical Surveys 33, no. 6 (1978), 33–96.

Balcar, B.; Simon, P. [92] Reaping number and π-character of Boolean algebras.Discrete Math. 108, no. 1 (1992), 5–12.

Baumgartner, J. [76] Almost disjoint sets, the dense set problem and the partitioncalculus. Ann. Math. Logic 9 (1976), 401–439.

Baumgartner, J. [80] Chains and antichains in Pω. J. Symb. Logic 45 (1980),85–92.

Baumgartner, J.; Komjath, P. [81] Boolean algebras in which every chain andantichain is countable. Funda. Math. 111 (1981), 125–133.

Baumgartner, J.; Taylor, A.; Wagon, S. [82]. Structural properties of ideals. Dis-sert. Math. 197 (1982), 95 pp.

Baur, L. [00] Cardinal functions on initial chain algebras on pseudotrees. Order17 (2000), no. 1, 1–21.

Baur, L.; Heindorf, L. [97] Initial chain algebras on pseudotrees. Order 14, no. 1,21–38.

Bekkali, M. [91] Topics in set theory. Springer-Verlag Lecture Notes in Mathemat-ics 1476, 120 pp.

Bell, M.; Ginsburg, J.; Todorcevic, S. [82] Countable spread of expY and λY .Topol. Appl. 14 (1982), 1–12.

Blass, A. [10] Combinatorial characteristics of the continuum. In Handbook of SetTheory, Springer 2010, 395–489.

Bonnet, R.; Rubin, M. [04] On poset Boolean algebras of scattered posets with finitewidth. Arch. Math. Logic 43 (2004), no. 4, 467–476.

Bonnet, R.; Rubin, M. [11] A thin-tall Boolean algebra which is isomorphic to eachof its uncountable subalgebras. Topol. Appl. 158 (2011), 1503–1525.

, ,OI 10.1007/978-3- - -


014

J.D. Monk Cardinal Invariants on Boolean Algebras: Second Revised Edition142 0348 0730 2,

Basel 2

551

[email protected]

552 References

Bonnet, R.; Shelah, S. [85] Narrow Boolean algebras. Ann. Pure Appl. Logic 28(1985), 1–12.

Bozeman, K. [91] On the relationship between density and weak density in Booleanalgebras. Proc. Amer. Math. Soc. 112, no. 4 (1991), 1137–1141.

Brenner, G. [82]. Tree algebras. Ph. D. thesis, University of Colorado (1982).

Brown, J. [05] Cardinal functions on pseudo-tree algebras and a generalization ofhomogeneous weak density. Ph. D. thesis, Univ. of Colo.

Brown, J. [06] Cellularity of pseudo-tree algebras. Notre Dame J. Formal Logic, 47(2006), no. 3, 353–359.

Bruns, C. [13] A simultaneous generalization of independence and disjointness inBoolean algebras. Order 30, no. 1 (2013), 211–231.

Chang, C.C.; Keisler, H.J. [73]. Model Theory. North-Holland, 550 pp.

Chestnut, R. [12] Independent partitions in Boolean algebras. Ph. D. dissertation,Univ. of Colo., Boulder, 2012.

Comfort, W.W. [71]. A survey of cardinal invariants. Gen. Topol. Appl. 1 (1971),163–199.

Comfort, W.W.; Negrepontis, S. [74]. The theory of ultrafilters. Springer-Verlag,x+482 pp.

Comfort, W.W.; Negrepontis, S. [82]. Chain conditions in topology. CambridgeUniversity Press, xi + 300 pp.

Cramer, T. [74] Extensions of free Boolean algebras. J. London Math. Soc. (2) 8(1974), 226–230.

Cummings, J.; Shelah, S. [95] A model in which every infinite Boolean algebra hasmany subalgebras. J. Symb. Logic 60, no. 3 (1995), 992–1004.

Day, G.W. [67] Superatomic Boolean algebras. Pacific J. Math. 23, no. 3 (1967),479–489.

Devlin, K. [84] Constructibility. Springer-Verlag. xi + 425 pp.

Donder, H.-D. [88] Regularity of ultrafilters and the core model. Israel J. Math. 63,289–322.

van Douwen, E.K. [89] Cardinal functions on Boolean spaces. In Handbook ofBoolean algebras, North-Holland (1989), 417–467.

Dow, A.; Monk, J.D. [97]Depth, π-character, and tightness in superatomic Booleanalgebras.; Erratum to: “Depth, π-character, and tightness in superatomicBoolean algebras”. [Topology Appl. 75 (1997), no. 2, 183–199; Topology Appl.105 (2000), no. 1, 121.

Dow, A.; Steprans, J.; Watson, S. [96] Reaping numbers of Boolean algebras. Bull.London Math. Soc. 28, no. 6 (1996), 591–599.

Engelking, R. [89] General Topology. Heldermann Verlag, viii + 529 pp.

[email protected]

References 553

Erdos, P.; Hajnal, A.; Mate, A., Rado, R. [84]. Combinatorial set theory: partitionrelations for cardinals. Adademiai Kiado, 347 pp.

Fedorchuk, V. [75] On the cardinality of hereditarily separable compact Hausdorffspaces. (Russian) Dokl. Akad. Nauk SSSR 222, no. 2 (1975), 651–655. Englishtranslation: Sov. Math. Dokl. 16, 651–655.

Fedorchuk, V.; Todorcevic, S. [97] Cellularity of covariant functors. Topol. and itsappl. 76 (1997), 125–150.

Fleissner, W. [78] Some spaces related to topological inequalities proven by theErdos–Rado theorem. Proc. Amer. Math. Soc. 71 (1978), 2, 313–320.

Foreman, M.; Laver, R. [88] Some downwards transfer properties for ℵ2. Adv. inMath. 67, no. 2 (1988), 230–238.

Gardner, R.; Pfeffer, W. [84] Borel measures. In Handbook of set-theoretic topol-ogy (1984), 961–1044. (North-Holland)

Garti, S.; Shelah, S. [08] On Depth and Depth+ of Boolean algebras. Alg. Univ.58 (2008), 243–248.

Garti, S.; Shelah, S. [∞] Depth of Boolean algebras in the constructible universe.

Gratzer, G.; Lakser, H. [69] Chain conditions in the distributive free product oflattices. Trans. Amer. Math. Soc. 144 (1969), 301–312.

Gurevich, Y. [82] Crumbly spaces. Logic, Methodology, and Philosophy of Sci-ence VI (Hannover 1979), 179–191. Studies in Logic and the Foundations ofMathematics 104, North-Holland 1982.

Hausdorff,, F. [1908]Grundzuge einer Theorie der geordneten Mengen.Math. Ann.65 (1908), 435–505.

Heindorf, L. [85] Boolean algebras whose ideals are disjointly generated. Demons.Math. 18, no. 1, 43–64.

Heindorf, L. [86] Strongly retractive Boolean algebras. Fund. Math. 126, 253–259.

Heindorf, L. [89] Boolean semigroup rings and exponentials of compact zero-dim-ensional spaces. Fund. Math. 135, no. 1, 37–47.

Heindorf, L. [89a] A note on irredundant sets. Alg. Univ. 26 (1989), 216–221.

Heindorf, L. [91] Dimensions of Boolean algebras. Seminarberichte No. 112, Hum-boldt-Universitat zu Berlin, 48–56.

Heindorf, L. [92] Moderate families in Boolean algebras. Ann. Pure Appl. Logic57(3) (1992), 217–250.

Heindorf, L. [94] Graph spaces and ⊥-free Boolean algebras. Proc. Amer. Math.Soc. 121, no. 3, 657–665.

Heindorf, L. [97] On subalgebras of Boolean interval algebras. Proc. Amer. Math.Soc. 125, no. 8, 2265–2274.

[email protected]

554 References

Heindorf, L.; Shapiro, L. [94] Nearly projective Boolean algebras. Lecture Notesin Mathematics, Springer-Verlag, no. 1596, 202 pp.

Hodel, R. [84]. Cardinal functions I. In Handbook of set-theoretic topology, North-Holland (1984), 1–61.

Jech, T. [86] Multiple forcing. Cambridge University Press, viii + 136 pp.

Jech, T. [∞] Stationary sets. In Handbook of set-theory, to appear.

Juhasz, I. [75]. Cardinal functions in topology. Math. Centre Tracts 34, Amster-dam, 150 pp. (1975).

Juhasz, I. [80]. Cardinal functions in topology – ten years later. Math. CentreTracts 123, Amsterdam, 160 pp.

Juhasz, I. [84]. Cardinal functions II. In Handbook of set-theoretic topology,North-Holland, 63–109.

Juhasz, I. [92] The cardinality and weight spectrum of a compact space. Recentdevelopments in general topology and its applications, Math. Res. 67, Aka-demie-Verlag.

Juhasz, I. [93] On the weight spectrum of a compact space. Israel J. Math. 81, no.3 (1993), 369–379.

Juhasz, I.; Kunen, K.; Rudin, M.E. [76] Two more hereditarily separable non-Lindelof spaces. Canad. J. Math. 28, no. 5 (1976), 998–1005.

Juhasz, I.; Shelah, S. [98] On the cardinality and weight spectra of compact spaces.II. Funda. Math. 155 (1998), 91–94. Publ. 612 of Shelah.

Juhasz, I.; Szentmiklossy, Z. [92] Convergent free sequences in compact spaces.Proc. Amer. Math. Soc. 116, no. 4 (1992), 1153–1160.

Just, W.; Koszmider, P. [91] Remarks on cofinalities and homomorphism types ofBoolean algebras. Alg. Univ. 28, no. 1 (1991), 138–149.

Keisler, H.J.; Prikry, K. [74] A result concerning cardinalities of ultraproducts. J.Symb. Logic 39, no. 1 (1974), 43–48.

Koppelberg, S. [77] Boolean algebras as unions of chains of subalgebras, Alg. Univ.7 (1977), 195–203.

Koppelberg, S. [88] Counterexamples in minimally generated Boolean algebras.Acta Univ. Carolin. Math. Phys. 29, no. 2, 27–36.

Koppelberg, S. [89] General theory of Boolean algebras, Part I of Handbook ofBoolean algebras. North-Holland, 312 pp. (1989).

Koppelberg, S. [89a] Minimally generated Boolean algebras. Order, 5, 393–406.

Koppelberg, S. [89b] Projective Boolean algebras. Chapter 20 in Handbook ofBoolean algebras, Part III, 741–773.

Koppelberg, S. [93] A construction of Boolean algebras from first-order structures.Ann. Pure Appl. Logic 59, no. 3, 239–256.

[email protected]

References 555

Koppelberg, S.; Monk, J.D. [92] Pseudo-trees and Boolean algebras. Order 8, 359–374.

Koppelberg, S.; Shelah, S. [95] Densities of ultraproducts of Boolean algebra.Canad. J. Math. 47 (1995), 132–145. Publ. 415 of Shelah.

Koszmider, P. [90] The consistency of ¬CH+pa ≤ ω1. Alg. Univ. 27, no. 1 (1990),80–87.

Koszmider, P. [99] Forcing minimal extensions of Boolean algebras. Trans. Amer.Math. Soc. 351 (1999), no. 8, 3073–3117.

Kunen, K. [78] Saturated ideals. J. Symb. Logic 43 (1978), 65–76.

Kunen, K. [80] Set theory. North-Holland, xvi + 313 pp.

Kuratowski, K. [58] Topologie, vol. 1. Panstwowe Wydawnictwo Naukowe, xiii +494 pp.

Kurepa, G. [35] Ensembles ordonnes et ramifies. Publ. Math. Univ. Belgrade 4(1935), 1–138.

Kurepa, G. [57] Partitive sets and ordered chains. “Rad” de l’Acad. Yougoslave302 (1957), 197–235.

Laver, R. [84] Products of infinitely many perfect trees. J. London Math. Soc. 29(1984), 385–396.

Magidor, M.; Shelah, S. [98] Length of Boolean algebras and ultraproducts. Math.Japon. 48 (1998), 301–307. Publication 433 of Shelah.

Malyhin, V. [72] On tightness and Suslin number in expX and in a product ofspaces. Sov. Math. Dokl. 13, 496–499.

Martınez, J.C. [02] Attainment of tightness in Boolean spaces. Math. Logic Quar-terly 48 (2002), no. 4, 555–558.

McKenzie, R.; Monk, J.D. [82] Chains in Boolean algebras. Ann. Math. Logic 22,137–175.

McKenzie, R.; Monk, J.D. [04] On some small cardinals for Boolean algebras. J.Symbolic Logic 69 (2004), no. 3, 674–682.

van Mill, J. [84] An introduction to βω. In Handbook of set-theoretic topology,North-Holland, 503–567.

Milner, E; Pouzet, M. [86] On the width of ordered sets and Boolean algebras. Alg.Univ. 23 (1986), 242–253.

Mizokami, T. [79] Cardinal function on hyperspaces. Colloq. Math. 41 (1979),201–205.

Monk, J.D. [83] Independence in Boolean Algebras. Periodica Mathematica Hun-garica 14, 1983, 269–308.

Monk, J.D. [84] Cardinal functions on Boolean algebras. In Orders: Descriptionsand Roles, Annals of Discrete Mathematics 23 (1984), 9–37.

[email protected]

556 References

Monk, J.D. [89] Endomorphisms of Boolean algebras. In Handbook of Boolean al-gebras, North-Holland, 491–516.

Monk, J.D. [90] Cardinal functions on Boolean algebras. 152 pp. Birkhauser Verlag(1990).

Monk, J.D. [96] Cardinal invariants on Boolean algebras. 298 pp. Birkhauser Ver-lag (1996).

Monk, J.D. [96a] Minimum-sized infinite partitions of Boolean algebras. Math.Logic Quart. 42, no. 4, 537–550.

Monk, J.D. [01] The spectrum of partitions of a Boolean algebra. Arch. Math.Logic 40 (2001), no. 4, 243–254.

Monk, J.D. [01a] Continuum cardinals generalized to Boolean algebras. J. Symb.Logic 66, no. 4 (2001), 1928–1958.

Monk, J.D. [02] An atomless interval Boolean algebra A such that a(A) < t(A).Algebra Universalis 47 (2002), no. 4, 495–500.

Monk, J.D. [04] The spectrum of maximal independent subsets of a Boolean algebra.Ann. Pure Appl. Logic 126 (2004), no. 1–3, 335–348.

Monk, J.D. [07] Towers and maximal chains in Boolean algebras. Algebra Univer-salis 56 (2007), no. 3-4, 337–347.

Monk, J.D. [08]Maximal irredundance and maximal ideal independence in Booleanalgebras. J. Symb. Logic 73, no. 1 (2008), 261–275.

Monk, J.D. [10] Special subalgebras of Boolean algebras. Math. Log. Q. 56 (2010),no. 2, 148–158.

Monk, J.D. [11] Maximal free sequences in a Boolean algebra. Comment. Math.Univ. Carol. 52, 4 (2011), 593–611.

Monk, J.D. [12] Remarks on continuum cardinals on Boolean algebras. Math. Log.Quart. 58, no. 3 (2012), 159–167.

Monk, J.D.; Nyikos, P. [97] On cellularity in homomorphic images of Boolean alge-bras. Proceedings of the 12th Summer Conference on General Topology andits Applications (North Bay, ON, 1997). Topology Proc. 22 (1997), Summer,341–362.

Parovicenko, I.I. [67] The branching hypothesis and the correlation between localweight and power in topological spaces. Dokl. Akad. Nauk SSSR 174, 30–32(1967).

Peterson, D. [97] Cardinal functions on ultraproducts of Boolean algebras. J. Symb.Logic 62, no. 1 (1997), 43–59.

Peterson, D. [98] Reaping numbers and operations on Boolean algebras. Alg. Univ.39, no. 3-4 (1998), 103–119.

[email protected]

References 557

Rabus, M.; Shelah, S. [99] Topological density of ccc Boolean algebras – everycardinality occurs. Proc. Amer. Math. Soc. 127 (1999), 2573–2581. Publ. 631of Shelah.

Rosenstein, J. [82] Linear orderings. Acad. Press, xvi + 487 pp.

Roslanowski, A.; Shelah, S. [98] Cardinal invariants of ultraproducts of Booleanalgebras. Fundamenta Math. 155 (1998), 101–151. Publication 534 of Shelah.

Roslanowski, A.; Shelah, S. [00] More on cardinal invariants of Boolean algebras.Ann. Pure Appl. Logic 103 (2000), 1–37. Publication 599 of Shelah.

Roslanowski, A.; Shelah, S. [01] Historic forcing for Depth. Colloq. Math. 89(2001), 99–115. Publication 733 of Shelah.

Roslanowski, A.; Shelah, S. [01a] Forcing for hL and hd. Colloq. Math. 88 (2001),273–310. Publ. 651 of Shelah.

Rubin, M. [83] A Boolean algebra with few subalgebras, interval Boolean algebras,and retractiveness. Trans. Amer. Math. Soc. 278 (1983), 65–89.

Shapiro, L. [76a] The space of closed subsets of Dℵ2 is not a dyadic bicompact.(Russian) Dokl. Akad. Nauk SSSR 228, no. 6 (1976) English translation:Soviet Math. Dokl. 17, no. 3, 937–941.

Shapiro, L. [76b] On spaces of closed subsets of bicompacts. (Russian) Dokl. Akad.Nauk SSSR 231, no. 2 (1976); English translation: Soviet Math. Dokl. 17, no.6, 1567–1571.

Shelah, S. [80] Remarks on Boolean algebras. Alg. Univ. 11 (1980) 77–89. Publ.92.

Shelah, S. [88] Was Sierpinski right? I. Isr. J. Math. 62, 355–380. Publ. 276.

Shelah, S. [90] Products of regular cardinals and cardinal invariants of products ofBoolean algebras. Israel J. Math. 70, no. 2, 129–187. Publ. 345.

Shelah, S. [91] Strong negative partition relations below the continuum. Acta Math.Hung. 58, no. 1–2 (1991), 95–100. Publ. 327.

Shelah, S. [92] Factor = quotient, uncountable Boolean algebras, number of endo-morphism and width. Math. Japon. 37 (1992), 385–400. Publ. 397.

Shelah, S. [94] Cardinal arithmetic. Oxford Univ. Press, 481 pp.

Shelah, S. [94a] The number of independent elements in a product of intervalBoolean algebras. Math. Japon. 39 (1994), 1–5. Publ. 503.

Shelah, S. [96] On Monk’s questions. Fundamenta Math. 151 (1996), 1–19. Publ.479.

Shelah, S. [97] Colouring and non-productivity of ℵ2-cc. Ann. Pure Appl. Logic 84(1997), 153–174. Publ. 572.

Shelah, S. [97a] σ-entangled linear orders and narrowness of products of Booleanalgebras. Funda. Math. 153 (1997), 199–275. Publ. 462.

[email protected]

558 References

Shelah, S. [99] Special subsets of cf(μ)μ, Boolean algebras, and Maharam measurealgebras. Topol. Appl. 99 (1999), 135–235. Publ. 620.

Shelah, S. [01] Constructing Boolean algebras for cardinal invariants. Alg. Univ.45 (2001), 353–373. Publ. 641.

Shelah, S. [02] More constructions for Boolean algebras. Arch. Math. Logic 41(2002), no. 5, 401–441. Publication no. 652.

Shelah, S. [03] On ultraproducts of Boolean algebras and irr. Archive for Math.Logic 42 (2003), 569–581. Publication 703.

Shelah, S. [05] The depth of ultraproducts of Boolean algebras. Alg. Univ. 54, no.1 (2005), 91–96. Publication 853.

Shelah, S. [09] A comment on “p < t”. Canad. Math. Bull. 52 (2009), 303–314.Publication 885.

Shelah, S.; Spinas, O. [99] On tightness and depth in superatomic Boolean algebras.Proc. Amer. Math. Soc. 127 (1999), no. 12, 3475–3480. Publication 663 ofShelah.

Shelah, S.; Spinas, O. [00] On incomparability and related cardinal functions onultraproducts of Boolean algebras. Math. Japon. 52 (2000), 345–358. Publi-cation 677 of Shelah.

Sirota, S. [68] Spectral representation of spaces of closed subsets of bicompacta.(Russian) Dokl. Akad. Nauk SSSR 181, no. 5 (1968); English translation:Soviet Math. Dokl. 9, no. 4, 997–1000.

Solovay, R.; Tennenbaum, S. [71] Iterated Cohen extensions and Suslin’s problem.Ann. of Math. 94 (1971), 201–245.

Todorcevic, S. [81] Trees, subtrees and order types. Ann. Math. Logic 20 (1981),233–268.

Todorcevic, S. [85] Remarks on chain conditions in products. Compos. Math. 55,no. 3 (1985), 295–302.

Todorcevic, S. [86] Remarks on cellularity in products. Compos. Math. 57 (1986),357–372.

Todorcevic, S. [87] Partitioning pairs of countable ordinals. Acta math. 159, 261–294.

Todorcevic, S. [87a] On the cellularity of Boolean algebras. Handwritten notes.

Todorcevic, S. [89] Partition problems in topology. Contemp. Math. 84, Amer.Math. Soc. 1989, xii + 116 pp.

Todorcevic, S. [90] Free sequences. Topol. Appl. 35 (1990), 235–238.

Todorcevic, S. [93] Irredundant sets in Boolean algebras. Trans. Amer. Math. Soc.339, no. 1 (1993), 35–44.

[email protected]

References 559

de la Vega, R.; Kunen, K. [04] A compact homogeneous S-space. Topol. Appl. 136(2004), 123–127.

Weese, M. [80] A new product for Boolean algebras and a conjecture of Feiner.Wiss. Z. Humboldt-Univ. Berlin Math.-Natur. Reihe 29 (1980), 441–443.

Weiss, W. [84] Versions of Martin’s axiom. In Handbook of set-theoretic topology,North-Holland, 827–886.

Williams, N. [77] Combinatorial set theory. North-Holland, xi + 208 pp.

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Symbol Index

∏Bc A, 21∏Bi∈I Ai, 14

A, 52

A , 318

a, 121, 153

A∗n, 336A -dimA, 507

A-partition of a, 338

A ≤free B, 6

A ≤m B, 6, 54

A ≤mg B, 6, 59

A ≤π B, 6

A ≤proj B, 6

A ≤rc B, 6

A ≤reg B, 6

A ≤σ B, 6

A ≤s B, 6

A ≤u B, 6

alt, 311, 429

aspect, 121

Aut, 3

A � x, 52

b, 303

B[F ], 292

c, 2

c′, 84card, 2

Cardh+, 316

CardH−, 311, 443Cardh−, 316CardHs, 316

ccc, 83

cf, 311

cH+, 120

ch+, 121

cH−, 121ch−, 121cHr, 141

cHs, 127

cmm, 121

cS+, 121

cS−, 121cSr, 137

cSs, 136

d, 2

dCardS+, 316

dCardS−, 316dcS+, 121

dcS−, 121dd, 419

dDepthS+, 178

dDepthS−, 178, 289ddS+, 231

ddS−, 231Ded, 280

Depth, 2

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Basel 2

561

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562 Symbol Index

DepthH+, 175, 177

Depthh+, 178

DepthH−, 178Depthh−, 178DepthHr, 203

DepthHs, 199

DepthS+, 178

DepthS−, 178DepthSr, 199

DepthSs, 199

dH+, 231

dh+, 231

dH−, 231dh−, 231dhdS+, 454

dhLS+, 443

dHr, 232

dHs, 231

dIndS+ , 346

dIndS−, 347dIrrS+, 297

dj, 507

dj-dim, 508

dkS+, 7

dkS−, 7dLengthS+, 289

dLengthS−, 289dn, 227

dπS+, 247

dπS−, 247dπχS−, 382dS+, 231

dS−, 231dSr, 232

dSs, 232

dsS+, 413

dsS−, 413dtS−, 397Dup, 26, 36

End, 3

ends, 338

ends′, 338ess.supFi∈Iαi, 109

Exp , 28, 29, 397, 427

Exp ′, 31

Φ, 318

f, 398

f′, 419Finco, 13

{Finco(ω)}-dim, 508

finco-dim, 508

free-dim, 507

Freecal, 355

fsp, 398

h, 188

h, 13

h-cof, 3

h-cof2mm, 482

h-cof2spect, 482

h-cofmm, 482

h-cofspect, 482

hd, 3, 244

hdH+, 454

hdh+, 454

hdidmm, 454, 455

hdm, 460

hdmm, 454

hdS+, 454

hdS−, 454hdspect, 454

hL, 3

hLH+, 443

hLh+, 443

hLh−, 443hLHr, 446

hLHs, 446

hLm, 448

hLmm, 443

[email protected]

Symbol Index 563

hLS+, 443

hLS−, 443hLSr, 446

hwd, 264

i, 347

IAint, 74

IAat, 27

ic, 507

ic-dim, 508

icp, 507

Id, 3

Id′, 31IF , 292

Inc, 3

Incmm, 465

Incspec, 465

Inctreemm, 466

Inctreespect, 466

Ind, 2

IndH+, 346

Indh+, 346

IndH−, 346Indh−, 346IndHs, 355

Indmm, 347

Indn, 9

Indn, 372

IndS+, 346

IndS−, 346Indspect, 347

IndSs, 355

Init, 71

int-dim, 507

{Intalg(R)}-dim, 508

Irr, 2

Irrmm, 309

Irrmn, 310

IrrS+, 297

IrrS−, 297

isp, 347

k′, 7κ-cc, 83, 84

kH+, 7

kh+, 7

kH−, 7kh−, 7kHr, 9

kHs, 8

kmm, 7

kS+, 7

kS−, 7kspect, 7

kSr, 9

kSs, 8

kU , 318

len, 59

Length, 2

Length′, 285LengthH+, 284, 289

Lengthh+, 289

LengthH−, 289Lengthh−, 289LengthHr, 290

LengthHs, 289

Lengthmm, 290

LengthS+, 289

LengthS−, 289Lengthspect, 290

LengthSr, 290

LengthSs, 290

linf , 7

lsup, 7

m, 507

M c, 318

mg-dim, 508(mn

)�→(ji

)1,1k,q

, 250

Mo, 318

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564 Symbol Index

M ↓ p, 35M ↑ p, 35M r, 318

nw, 332

nwdc, 484

oi, 507

p, 188

π, 2

P(ω)/fin, 13, 121, 152, 178, 188, 215,220, 230, 242, 261, 347, 417, 429,467, 531, 532

p-alt, 311

π-basis, 383

Pend, 501

πH+, 246

πh+, 246

πH−, 246πh−, 246πHr, 276

P ×Q, 84

Pr1, 89

precal, 361

πS+, 247

πS−, 247pspect, 188

πSr, 276

πSs, 275

Psub, 499

pta, 506

ptree-dim, 508

πχ, 2

πχH+, 380

πχh+, 380

πχH−, 382πχh−, 382πχHr, 384

πχinf , 261, 382

πχS+, 382

πχS−, 382πχSr, 384

r, 247, 248

rel, 264

rm, 248

rmn, 249

s, 3

sa, 506

sa-dim, 507

sg-dim, 508

sH+, 413

sh+, 413

sH−, 413sh−, 413sm, 420

smm, 413

SmpAx , 52

spa, 507

spl, 188

sS+, 413

sS−, 413sspect, 413

Sub, 3

t, 3, 175

ta, 506

Tailalg, 69

tH+, 397

th+, 397

tH−, 397tHr, 400

tHs, 400

tla, 506

tm, 401

tmn, 401

tow, 178

towwspect, 179

towspect, 178

tree-dim, 507

[email protected]

Symbol Index 565

tS+, 397

tS−, 397tSr, 400

tSs, 400

u, 427

Ult, 3

ultrf , 318

utm, 401

utmn, 401

V , 28

w, 330

χ, 3, 427

x0, 9

x1, 9

Xfi, 318

χh+, 427

χH−, 427χh−, 427χHr, 432

χHs, 432

X id, 318

χinf , 427

χS+, 427

χS−, 427χSr, 432

χSs, 432

[email protected]

Subject Index

3-signature, 339

⊥-free, 81

A -dimension, 507

Alexsandroff duplicate, 26, 115, 175,230, 243, 283, 345, 378, 397, 413,421, 424, 426, 430, 442, 447, 453,511, 512, 531, 532, 536

Alexsandroff–Hausdorff theorem, 317

algebraic density, 2

altitude, 311

amalgamated free products, 106, 161,222, 239, 282

amalgamation, 168

Aronszajn subtree, 303

attained, 7, 158, 461

attainment, 83, 279, 335, 336, 375, 388,406, 422, 438, 449

automorphisms, 491

Balcar, Franek theorem, 316, 439, 519

Baumgartner-Komjath algebra, 512,531

branching point, 167

Brenner, Monk theorem, 515

caliber, 8, 136

cardinality, 311

cellularity, 2, 83

centered, 219

chain condition, 4

character, 3, 209, 422

closed domain, 318

co-complete, 66

co-interval algebra, 81

cofinality, 311

Cohen algebra, 81

compact Kunen line, 297

compatible, 84

configuration, 484

countable chain condition, 83

Dedekind cut, 285

dense, 2

density, 2

depth, 155, 392

derived functions, 7, 120, 175, 231, 284,297, 346, 380, 397, 413, 427, 443,454, 464

descendingly incomplete, 109

determined, 318

disjoint, 2

disjunctive, 35

disjunctively generated, 69

disjunctiveness, 507

double delta system lemma, 365

downward Lowenheim-Skolemtheorem, 102

Erdos–Rado theorem, 85, 106, 107, 112,152, 202, 204, 229, 283, 290, 317, 509

Erdos–Tarski theorem, 83, 142

, ,OI 10.1007/978-3- - -


014


Basel 2

567

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568 Subject Index

essential supremum, 109

exponential, 28, 38, 115, 175, 230, 243,283, 346, 379, 397, 413, 426, 442, 453

fan element, 154

finitely weakly dense, 260

free

caliber, 355

chain, 284

poset algebras, 82

products, 84, 158, 220, 238, 281,297, 343, 359, 376, 393, 410, 424,441, 452, 492

sequence, 175, 380, 388

Fusion lemma, 168

fusion sequence, 168

gap, 209

general h-cof sequence, 482

good, 468

good point, 474

h-cof sequence, 482

of the second kind, 482

Hajnal’s free set theorem, 409

hereditarily separable, 297

hereditary

cofinality, 3, 481

density, 3, 449

Lindelof degree, 3

homogeneous, 339

homogeneously weakly dense, 264

homomorphic k relation, 9

homomorphic spectrum, 8, 127

hwd-homogeneous, 271

hyper-rigid, 468

ideal independence, 120, 130

identity, 35

Inc, 462

incomparability, 3, 461

incomparable, 3

independence, 2, 335

independent, 2, 9

infinitely weakly dense, 261

initial chain algebra, 71

number, 507

for pseudo trees, 507

interval algebra, 36, 46, 63, 66, 74, 81,127, 130, 139, 150, 152, 178, 195,202, 204, 208, 209, 233, 235, 246,277, 432, 480, 483

irredundance, 2, 291

irredundant, 2

κ-centered, 219

κ-chain condition, 83

κ-dense, 468

κ, λ,P-caliber, 8

Kunen line, 430

left-separated, 244, 407

left-separated sequence, 449

length, 2, 279

limit-normal, 214

Lindelof BA, 81

Lindelof degree, 3

hereditary ∼, 3local π-base, 380

lower character, 209

m-dense, 248

m-free sequence, 401

m-ideal-independent, 420

m-independent, 372

m-left-separated, 460

m-right-separated sequence, 448

Martin’s axiom, 88, 144, 145, 149, 152,215, 235, 315, 447, 476, 532

maximal free sequence, 398

maximal right-separated sequences, 443

minimal extension, 54

minimal for (A,F ), 55

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Subject Index 569

minimality, 507

minimally generated, 59

(m,n)-dense, 249

m,n-free sequence, 401

mn-irredundant, 310

moderate, 81

moderate products, 13, 115, 174, 230,242, 283, 345, 378, 396, 412, 424,441, 453

n, 244

n-intersection property, 226

net weight, 332

network, 332, 411

nice antichain, 471

non-reflecting, 89

normal, 13

normal form theorem, 40, 73

nowhere dense for configurations, 484

nowhere relatively dense, 269

one-point gluing, 24, 115, 174, 230,242, 345, 378, 397, 413, 425, 442, 453

order-ideal number, 507

order-independence, 5, 392, 440

ordinary sup-function, 4, 107, 164, 282,345, 412

π-character, 2

π-weight, 2, 237

paracompact, 82

PCF theory, 102

perfect tree, 167

precaliber, 361

products, 11, 84, 156, 220, 238, 280,292, 335, 374, 387, 391, 409, 421,439, 451, 464, 491

projective, 81

projectively embedded, 6

pseudo

-altitude, 311

-tree, 39

-tree algebra number, 506

-tree algebras, 39

pure chain, 154

ramification set, 43

Ramsey ultrafilter, 170

rc-filtered, 81

realizes, 285

reaping number, 247, 248

refines, 115

reg, 414

regular, 109

closed, 318

filter, 318

ultrafilters, 164

relational, 102

relationships, 277, 297, 371, 382, 397,419, 430, 432, 446, 460, 483

representing chain, 59

retractive, 81

right-separated, 407, 437, 444

S-determined, 318

σ-embedded, 6

σ-filtered, 81

Sacks forcing, 168

Sacks reals, 167

saturation, 152

semigroup, 35

algebra, 36

number, 506

separate, 470

simple extension, 51

Skolem function, 102

special ♦-sequences, 455special classes, 150, 208, 233, 277, 384,

401, 448, 480

special subalgebras, 6, 74, 83, 121, 155,180, 189, 220, 237, 291, 335, 373,387, 405, 421, 439, 451, 464

splitting, 188

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570 Subject Index

spread, 3, 120, 405

strict club system, 89

strong p-set, 192

subalgebra k relation, 9

subalgebra spectrum, 8

subdirect product, 114, 174, 230, 242,283

sup-min function, 5, 240, 376, 424, 481

superatomic number, 507

support, 1

tail algebra, 69

number, 506

theorem

Alexsandroff–Hausdorff ∼, 317Balcar, Franek ∼, 316Brenner, Monk ∼, 515downward Lowenheim–Skolem ∼,

102

Erdos–Rado ∼, 85, 106, 107, 112,152, 202, 204, 229, 283, 290, 317,509

Erdos–Tarski ∼, 83, 142Hajnal’s free set ∼, 409

tightness, 3, 175

Todorcevic walks, 90

topological density, 2, 219

tower, 178

tree algebra, 36, 210

number, 506

type, 1

ultra-sup, 111, 297

function, 4, 282, 345, 410, 464

ultrafilters, 489

ultraproducts, 108, 164, 226, 239, 282,344, 376, 392, 410, 424, 440, 452,464, 481, 493

unions, 107, 151, 164, 222, 239, 282,345, 377, 395, 424

upper character, 209

weak

density, 247, 248

partition, 248

products, 1, 281, 296, 387, 440, 451

tower, 179

very ∼, 180weakly

dense, 248

dense in L, 280

homogeneous, 339

projective, 81

well met, 72

zero, 35

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Name Index

Abraham, U., 103, 551

Alexsandroff, P., 26, 115, 175, 230, 243,283, 317, 345, 378, 397, 413, 421,424, 426, 430, 442, 447, 453, 511,512, 531, 532, 536

Arhangelskiı, A., 1, 175, 430, 551

Balcar, B., 248, 252, 254, 260, 313, 316,439, 519, 551

Baumgartner, J., 152, 476, 483, 512,531, 536, 551

Baur, L., 71, 523, 524, 551

Bekkali, M., 480, 551

Bell, M., 297, 551

Blass, A., 69, 519, 551

Bonnet, R., 82, 468, 521, 523, 551

Bozeman, K., 248, 269, 384, 519, 552

Brenner, G., 69, 153, 403, 434, 515, 552

Brown, J., 153, 248, 265, 434, 552

Bruns, C., 81, 417, 552

Cech, E., 427

Chang, C.C., 109, 127, 552

Chestnut, R., 254, 552

Comfort, W.W., 1, 108, 114, 136, 165,223, 552

Corominas, E., 472

Cramer, T., 336, 552

Cummings, J., 504, 552

Day, G.W., 337, 552

Devlin, K., 106, 552

Donder, H.-D., 113, 166, 242, 282, 297,345, 376, 393, 410, 424, 440, 464,481, 552

van Douwen, E.K., 1, 245, 313, 315,552

Dow, A., 215, 248, 249, 385, 404, 519,522, 552

Engelking, R., 318, 329, 552

Erdos, P., 83, 85, 106, 107, 112, 142,152, 202, 204, 229, 283, 290, 317,406, 409, 509, 553

Fedorchuk, V., 115, 144, 145, 346, 455,460, 553

Fichtenholz, G, 204, 340

Fleissner, W., 86, 553

Foreman, M., 139, 147, 148, 553

Franek, F., 316, 439, 519

Gardner, R., 26, 553

Garti, S., 553

Ginsburg, J., 297, 551

Gratzer, G., 155, 553

Gurevich, Y., 13, 553

Hajnal, A., 85, 406, 409, 553

Hausdorff, F., 150, 191, 204, 209, 317,340, 511, 553

Heindorf, L., 6, 13, 35, 46, 71, 81, 297,302, 336, 489, 507, 551, 553

, ,OI 10.1007/978-3- - -


014


Basel 2

571

[email protected]

572 Name Index

Hodel, R., 1, 554

Hodges, W., 164

Jech, T., 169, 554

Juhasz, I., 1, 127, 144–147, 205, 207,233, 297, 317, 326, 330, 332, 334,406, 407, 420, 447, 460, 554

Just, W., 144, 146, 316, 554

Kantorovich, L., 204, 340

Keisler, H.J., 109, 127, 242, 552, 554

Komjath, P., 476, 483, 512, 531, 536,551

Koppelberg, S., 1, 6, 39, 54, 59, 69, 70,81, 82, 144–147, 149, 178, 205, 222,232, 239, 242, 276, 312, 315, 355,492, 497, 533, 554

Koszmider, P., 18, 141, 144–149, 316,334, 427, 460, 554, 555

Kunen, K., 86–88, 139, 147, 148, 167,215, 216, 235, 297, 299, 303, 313,417, 419, 429, 455, 511, 513, 531,535, 536, 554, 555, 559

Kuratowski, K., 317, 555

Kurepa, G., 233, 280, 555

Lakser, H., 155, 553

Laver, R., 88, 139, 143, 147–149, 167,170, 365, 440, 553, 555

Loats, J., 499

Lowenheim, L., 102

Mate, A., 85, 406, 409, 553

Magidor, M., 103, 113, 174, 283, 345,551, 555

Malyhin, V., 393, 555

Martınez, J.C., 390, 555

Martin, D.A., 88, 144, 145, 149, 152,215, 235, 315, 447, 476, 532, 559

McKenzie, R., 113, 121, 155, 158, 161,162, 167, 183, 189, 194, 281, 282,351, 400, 428, 527, 555

van Mill, J., 460, 532, 555

Milner, E., 462, 555

Mizokami, T., 555

Monk. J.D., passim

Negrepontis, S., 108, 114, 136, 165,223, 552

Nikiel, J., 46

Nyikos, P., 137, 143, 146–149, 556

Parovicenko, I.I., 422, 556

Peterson, D., 109, 111, 165, 240, 248,263, 268, 277, 282, 297, 345, 375,376, 385, 392, 410, 424, 440, 452,489, 556

Pfeffer, W., 26, 553

Pospısil, B., 427

Pouzet, M., 462, 555

Prikry, K., 109, 242, 554

Purisch, S., 46

Rabus, M., 233, 557

Rado, R., 85, 106, 107, 112, 152, 202,204, 229, 283, 290, 317, 406, 409,509, 553

Rosenstein, J., 519, 557

Roslanowski, A., 173, 199, 200, 226,292, 303, 345, 372, 393, 401, 410,420, 439, 441, 448, 450, 460, 481,521, 522, 557

Rubin, M., 81, 82, 297, 476, 484, 502,521, 531, 536, 538, 551, 557

Rudin, M.E., 297, 554

Selker, K., 125, 183, 191, 263, 345, 354,400, 428, 429

Shapiro, L., 6, 31, 81, 554, 557

Shapirovskiı, B., 175, 231, 246, 381

Shelah, S., passim

Simon, P., 248, 252, 254, 260, 313, 551

Sirota, S., 31, 558

Skolem, T., 102

[email protected]

Name Index 573

Solovay, R., 151, 558

Spinas, O., 215, 385, 404, 410, 424,441, 452, 464, 558

Steprans, J., 248, 249, 552

Szentmiklossy, Z., 326, 554

Tarski, A., 83, 102, 142

Taylor, A., 152, 551

Tennenbaum, S., 151, 558

Todorcevic, S., 90, 115, 127, 130, 139,

144, 147, 148, 297, 303, 393, 398,419, 420, 446, 461, 462, 551, 553, 558

de la Vega, R., 299, 559

Wagon, S., 152, 551

Watson, S., 248, 249, 552

Weese, M., 13, 559

Weiss, W., 460, 559

Wesep, R., 480

Williams, N., 559

[email protected]

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