saharon shelah- more constructions for boolean algebras

8/3/2019 Saharon Shelah- More Constructions for Boolean Algebras

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MORE CONSTRUCTIONSFOR BOOLEAN ALGEBRAS

SH652

Saharon Shelah

Institute of MathematicsThe Hebrew University

Jerusalem, Israel

Rutgers UniversityMathematics DepartmentNew Brunswick, NJ USA

Abstract. We construct Boolean Algebras with prescribed behaviour concerningdepth for the free product of two Boolean Algebras over a third, in ZFC using pcf;assuming squares we get results on ultraproducts. We also deal with the familyof cardinalities and topological density of homomorphic images of Boolean Algebras(you can translate it topology - on the cardinalities of closed subspaces); and lastly wedeal with inequalities between cardinal invariants, mainly d (B ) < |B | ind( B ) > Depth( B ) log( |B | ).

2000 Mathematics Subject Classication . 03E04.

Key words and phrases : set theory, Boolean Algebras, ZFC constructions, pcf application, forcing,depth of Boolean Algebra, cardinal invariants of Boolean Algebras.

This research was partially supported by the Israel Foundation for ScienceI would like to thank Alice Leonhardt for the beautiful typing.Latest Revision - 00/Nov/14

Typeset by AM S -TEX1


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2 SAHARON SHELAH

Annotated Content

1 The depth of free product may be bigger than the depths of those multiplied

[We prove, in ZFC, that for many Boolean Algebra B , for arbitrarily largecardinals we can construct Boolean Algebras B1 , B 2 extending B suchthat the depth of the free product of B1 , B 2 over B is strictly larger than> hence than the depths of B1 and of B2 which is ; using pcf. Thus, weanswer problem 10 of Monk [M2]. We give a condition ,, which impliesthat for some Boolean Algebra A = A of cardinality there are B1 =B 1,, , B 2 = B2 = B

2,, , satisfying Depth

+ (B t ) and Depth( B1A

B1) . We then start to investigate for a xed A, the existence of such B1 , B 2 ;gives sufficient conditions and necessary conditions, involving consistencyresults. Using a relative ,, we shed some light on problem 11 of [M2]dealing with sufficient and with necessary conditions on an innite A for:a class of for some extensions B1 , B 2 of B , Dp(B1

AB1) > Dp( B1) +

Dp( B2).]

2 On the family of homomorphic images of a Boolean Algebra

[We prove that e.g. if B is a Boolean Algebra of cardinality , and , are strong limit singular of the same conality, then B has a homomorphicimage of cardinality and with exactly ultralters.More generally if > cf() = cf( ) and B a Boolean Algebra of cardinality , then for some homomorphic image B of B we have |B | 2 or Depth( B ) log(|B |). For this we deal with a cardinal invariant si( B ).]

4 On omitting cardinals by compact spaces[We deal with the existence of Boolean Algebras B such that {|B | : B a homomorphic image of B } is restricted. We also deal with topologicalrelatives of the results of 2.]

5 Depth of ultraproducts of Boolean Algebras

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MORE CONSTRUCTIONS 3

[We show that if and = 0 then for some Boolean Algebras Bn (forn < ), we have Depth( Bn ) but for any uniform ultralter D on ,the ultraproduct

n

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4 SAHARON SHELAH

1 On the depth of free products

Monk [M2], Problem 10, 11 ask about the depth of BA

C (see there for the known

results, [M2]), we answer 10 and give some information on 11 here.We shall dene a spectrum SpDpFP (A) for a Boolean Algebra A (in 1.1) and phraseMonks question with it (1.2(2),(3)). We then phrase a combinatorial statement

,, and prove it gives examples of B, C extending A while BA

C has depth larger

than both (in 1.5), and note that it (provably in ZFC) holds for many cardinals(with = near a singular) (in 1.7). Later we note some variants of ,,and investigate when the construction in 1.5 works for a Boolean Algebra A, inparticular for many innite Boolean Algebra A, it holds for a class of s.

1.1 Denition. 1) For a Boolean Algebra A, we dene the spectrum of depth of free products over A,SpDpFP (A) as

(, ) : there are Boolean Algebras B, C extending A such that :

Depth( BA

C ) > Depth( B ) + Depth( A) .

We write for (+ , ).2) Similarly

SpDpFP + (A) = (, ) : there are Boolean Algebras B, C extending A such that

Depth + (BA

C ) > , Depth + (B ) + Depth + (A) .

We write for (, ).

1.2 Remark. 0) Recall that

Depth( B ) = {| X | : X B is well ordered by < B }

Depth + (B ) = {| X |+ : X B is well ordered by < B }

1) Note that


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6 SAHARON SHELAH

(d) = cf () = sup( Rang (c ))(e) in clause (c) we can demand i i( ) for any pregiven i( ) < .

Then for any [, ) we can nd Boolean Algebras A, B 1 , B 2 such that:

( ) |A| = , A depend just on and ( ) |B1 | = |B2 | = ( ) Depth + (B1

AB2) = + so A B1 , A B

() Depth + (B1) and Depth + (B2) ()+ moreover, Length + (B i ) for i = 1 , 2.

1.6 Remark. 1) Let ,, just means (a), (b), (c), i.e. omitting clause (d) butweakening (e) to (e) , which means i( ) Rang( c ). So by 1.9(3) below ,,

,, and there are obvious monotonicity properties.

2) Let (a) mean < . Note that if in the hypothesis of 1.7 below we omit < cf(), and in the conclusion replace (a) by (a) the proof still works.3) Let +,, means that in the denition of ,, in clause (c), when c { , k } j & c { k , } j we add c { , k } = c { k , }.In 1.7 below we can get those versions by working a little more.4) We could have omitted j in clause (c) of ,, , (replacing j by > i , so j = i + 1), but usually in our proofs we show that j can be arbitrarily larger thani, and in ( ii ) get i so this formulation makes it clearer.

We quote [Sh:g]

1.7 Observation. The demand on < is not hard, in fact assuming also < cf() we can nd c such that clauses (a), (b), (c), (d), (e) of ,, hold if thecardinals ,, satises at least one of the following statements:

( )1 for some (, ) we have 0 , = cf( ),

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( )3 = cf( ) = > > cf( ) = , i : i < is strictly increasing with limit, i < max pcf { j : j < i } < (or just < ) and

i

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8 SAHARON SHELAH

We can also demand

3 if v, m < n , then the truth value of , > m , , < m does

not depend on < +

4 if = m are from v and = m then < < + , < ,m .

For each < + let i = Max[ {c {, } : = w { : < n }\{ }} { i( )}],now clearly i < so as cf( + ) > and + (by clause (a) or more exactlyby our assumptions) without loss of generality i = i for < + . As , (i +1) : < n have only ( i +1 )n < possible values, without loss of generality < + , (i + 1) = , moreover < + , (i + 2) = because|{ (i + 2) : < }| .

Also without loss of generality

5 for every < + and nite u , for + ordinals < + we have

i u


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10 SAHARON SHELAH

Let I t be the ideal of A generated by {a ti : i < } for t = 1 , 2. Let I be the ideal of A generated by I 1 I 2 and I the ideal generated by {a ti : i < , t { 1, 2}} so (if = then I = I ) A/I is the trivial (= two elements) Boolean Algebra.

Clearly

a t (1)i (1) = at (2)i (2) (t(1) , i(1)) = ( t(2) , i(2))

t { 1, 2} & i < j < A |= a ti < a tj

[Why? Well, you may check it using ( ) of stage D below; for any t, i let h t,i :{a1j , a 2j : j < } {0, 1} be dened by h t,i (a sj ) is 1 if s = t & i j and h t,i (a sj ) iszero otherwise. Interpreting {0, 1} as the trivial Boolean Algebra, h t,i respect the

equations is ( )1 + ( )2 hence extend to a homomorphismh t,i from A into {0, 1}.Now assume t(1) , t (2) { 1, 2} and i(1) , i(2) < ; if t(1) = t(2) i(1) < i (2) then

h t (2) ,i (2) maps a t (2) ,i (2) to 1 and a t (1) ,i (1) to 0 hence A |= at (1)i (2) = a

t (2)i (2) . Also if

t { 1, 2} and i < j < then a ti a tj by ( )1 and h t,j (a tj ) = 1 , h t,j (a ti ) = 0 henceA |= a ti < a tj .]For t = 1 , 2 let B t be the extension of A by {x t : < } freely except that:

( )3 x t x t a tc {, } for < <

( )4 x t a3 ti = 0 for < , i < .

Clearly B t extends A.[Why? Well use ( ) of stage D, dene a function h from A { x t : < } to A byh(x) = x for x A and h(x t ) = 0. Clearly h respects the equations in the BooleanAlgebra A and the equations is ( )3 +( )4 hence the homomorphism h from B onto{0, 1} is well dened and it extends id A .]

Let B = B1A

B2 and let x = x1 x2 B . Let J 1 , J 2 , J be the ideals of B which

I 1 , I 2 , I generates (resp.), so J 1 J 2 J .Clearly |B | .

Stage B: Depth+

(B ) = +

.Clearly Depth + (B ) | B |+ + . Hence it suffices to prove that x : < isstrictly increasing. So let < . First we prove x x . Let D be an ultralteron B (so D t = D B t Ult( B t ) and D1 A = D2 A), and we shall provex D x D . This suffices for proving x x (as D was any ultralter).

Case 1: D A is disjoint to I .

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Now modulo J t in B t , x t x t (by ( )3 + a ti I ) hence modulo J , x t x thence mod J, x x hence x D x D.

Case 2: D A is not disjoint to I .So D I = hence for some t { 1, 2} and i < , a ti D, but x3 t is disjoint to

a ti by ( )4 , so x3 t / D hence x = x1 x2 / D so trivially x D x D.

We still have to prove x = x . Let D t be the ultralter on B t generated byx t ( ), x t ( < ) and the members of { x : x I } (check that it is okaytrivially). Now D 1 A = A\ I = D 2 A (check, trivial). So there is an ultralterD on B satisfying D B t = D t . So x t , x t D t for t = 1 , 2 so x , x D asrequired.

Stage C: Length + (B t ) .Assume not, so we can nd c : < a chain (so with no repetition). Let

c = (as , 1i , 1 , . . . , a

s ,k i ,k , x

t, 1 , . . . , x

t,n ) where i, 1 < . . . < i ,k < and , 1 0 , without loss of generality:

( )5 k = k , n = n , = , s, = s (and if |A| < cf() then i, = i).( )6 for some m k we have:

(i) 0 < m i, = i <

(ii ) (m , k ] i,

(iii ) for < < , for some v = v, (m , k ] we have

(a) v i, = i, and(b) (m , k ] & / v implies that i, = i,

and both are not in {i,k , i,k : k = }.

[why ( )6? we can assume clauses (i) + (ii) by the pigeon-hole principle.

For clause (iii) if is regular, by the -system lemma (so then v, = v)and if singular, apply it twice. More elaborately for singular as > , let =

< cf( )

, where 1 < 2 |A| < 1 < 2 < . For each < cf()

we can nd X [ , + ) of cardinality + such that [m , k ) i, =i, . Then apply the -system lemma on : i ( ) (see clause (e)). So c , c aredistinct members of a chain of B t .Now read Stage D below.

So let w =: { , , , : = 1 , . . . , n } []< 0 .As in Stage D, B t,w is the subalgebra of B t generated by A { x t : w} andclearly c , c B t,w , hence also in B t,w , c , c are distinct members of a chain.

By symmetry assume B t,w |= c < c , hence there is a homomorphism f fromB t,w to (the trivial Boolean algebra) {0, 1} such that f (c ) = 0 , f (c ) = 1 and byStage D we can extend f to a homomorphism from B t to {0, 1}; we denote it, too,by f . Let (f ) = Min { : if < then f (a t ) = 1 }.

As in the Boolean Algebra A Bt

, at

: < is increasing, clearly( )7 for < , f (a t ) = 1 (f )

( )8 for < , f (a3 t ) = 0.

[Why does ( )8 hold? Assume that not, so, by ( )3 , we have f (x t ) = 0 for < .By ( )6(iii )(b), if k, (m , k ]\ v then i, = i,k . Hence by the denition of Aand ( ) of Stage D there is an endomorphism g of A such that

g(a s ) = asi , if s = s and = i, for some (m , k ]\ v,

g(a s ) = asi , if s = s and = i, for some (m , k ]\ v,

g(a s ) = as otherwise .

So s { 1, 2} & < g(a s ) = a s and g permutes the {a s : < , s { 1, 2}}.By the form of ( )3 + ( )4 and the fact that c {, } < , there is a homomorphismg from B t into B t extending g and taking each x t to x t for < . Clearly theng (a s i , ) = a

s i , for all = 1 , . . . , k . Similarly, g

(a s i , ) = as i , for all = 1 , . . . , k .

Hence

f (g (c )) = (f (a s 1i , 1 ), . . . , f (as ki ,k ), 0, 0, . . . , 0) = f (c ) = 1 ,

and similarly f (g (c )) = 0, contradicting c < c . So ( )8 holds.]Now

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( )9 the function g : {a s : < , s { 1, 2}} {0, 1} induce a homomorphism gfrom A to {0, 1} where g is dened by (recall i, j were chosen together with, ):

(i) g(a t ) = f (a t ) for < j

(ii ) g(a t ) = 1 if j, <

(iii ) g(a3 t ) = f (a3 t ) = 0 for <

(iv) g(a s i , ) = f (as i , ), g(a

s i , ) = f (a

s i , ) for = 1 , . . . , k

(v) g(a s ) = f (a s ) if a s / { as,i , , a

s i , : = 1 , . . . , k } and [, )

[why? now g is well dened as, e.g. for contradiction concerns (iv), two

instances do not contradict by ( )6(iii ) and they do not contradict othersby ( )6(i) + ( ii ). By ( ) of Stage D below we should check the equationsappearing in the denition of A. For those in ( )1 , i.e. a s a s for < < ,if s = 3 t this is trivial by clause (iii), if s = t, > j , this is trivial byclause (ii) and if s = t, < j , then g(a s ) = f (a s ) f (a s ) = g(a s ).

As for the equations in ( )2 that is a1 a2 = 0 for , < they are preservedtrivially by ( )7 and clause (iii).]

Dene a function h from A { x t, , xt, : = 1 , . . . , n } to {0, 1} as follows: h A

is the homomorphism g to{

0, 1}

and dene h(x t,

) = f (x t,

) and h(x t,

) =f (x t, ). Now

( )10 h induces a homomorphism h from B t,w to {0, 1}[why? by clause (c)(i) of ,, , i.e. the choice of , the function h is welldened; we use ( ) of Stage D; now

(a) the equations in A are respected.By the choice of h A by g being a homomorphism from A to {0, 1}

(b) the equations x t , a3 ti = 0 for { , }, i < .

This is respected as h(a3 ti ) = g(a

3 ti ) = 0.

(c) for { , } the equation x t , xt ,m a

tc { , , ,m } for < m .

Let be such that {, } = {, } and remember that i c { , , ,m } =c { , , ,m } (by the choice of ,, i , j see ,, (c)( ii ) of 1.5) so h(a tc {, , ,m } ) =f (a tc { , , ,m } ) = f (a

tc { , , ,m } ) by the choice of h A and of g i.e.

( )9(i), hence

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h(x t , ) h(xt ,m ) = f (x

t , ) f (x

t ,m )

= f (x t , x t ,m ) f (a tc { , , ,m } )= h(a tc { , , ,m } )

as required.(d) If , m { 1, . . . , m } and = are from {, } and , < ,m the

equation x t , xt ,m a

tc { , , ,m } .

Now we have to look at clause (c) of ,, of 1.5, by it there are twopossibilities

possibility 1: c { , , ,m } j (but necessarily < ), theng(a tc { , , ,m } ) = 1 by clause (ii) of ( )9 so the equation isrespected.

possibility 2: c { , , ,m } = c { , , ,m } < j hence[ , < ,m , < ,m ](read (c) of ,, so both holds).

So h(x t, ) h(xt ,m ) = f (x

t, ) f (x

t ,m ) = f (x

t , x

t,m )

f (a tc {, , ,m } ) = f (atc {, , ,m } ) = g(a

tc { , , ,m } ) =

h(a tc { , , ,m } ).So we have proved ( )10 .]

Now we have two homomorphisms f, h from B t,w to {0, 1} and they satisfy:

( )11 (a) f (a s i , ) = h(as i , )

(b) f (a s i , ) = h(as i , )

(c) f (x t, ) = h(xt, )

(d) f (x t, ) = h(xt, )

[why? for (a) + (b) note that h A = g A and ( )9(iv) and for (c) + (d) just see the choice of h.]

So clearly f (c ) = h(c ), f (c ) = h(c ), but f (c ) = 0 < f (c ) = 1 so h(c ) = 0 , w, then the equation x t x t a tc {, }[why? the meaning of the demand is that we should check

h(x t ) h(x t ) h(a tc {, } )

that is

h(x t ) x t a tc {, }

that is w (x t a tc {, } ) x

t a tc {, }

for this it suffices to show

w B t,w |= x t x t a tc {, } atc {, }

but we know

<

&

w &

w B t,w |= xt x

t a

tc { , }

and apply this to ( , ) = ( , ); so B t,w |= x t x t a tc {, } . Now as ati : i <

is increasing and as by clause (b) of ,, we have

c {, } Max{c {, }, c {, }}

we get B t,w |= a tc {, } atc {, } a

tc {, } , so together we are done.] 1.5

1.8 Observation. 1) Assume

( ) = is weakly compact > .

If A is a Boolean Algebra, A B1 , A B2 , B = B1A

B1 , |A| and:

Depth + (B ) > or just Length + (B ) > , then 2

t =1Depth + (B t ) > .

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2) Similarly if ()2 and cf () > 2 .3) We can above replace = cf () > 2 by > .

Remark. Clearly part (2) generalizes part (1).

Proof. 1), 2) We can nd pairwise distinct c B (non-zero) for < such that < < c , c are comparable in B .

For each we can nd bt, B t ( < n , t = 1 , 2) such that c =n 1

=0

(b1, b2, ).

Without loss of generality b2, : < n are pairwise disjoint (in B2). Withoutloss of generality n = n( ).

For each < there is an ultralter D of B such that

c D , 2 2| A | , without loss of generality

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18 SAHARON SHELAH

constant so without loss of generality < D 0 A = D 0 . Let D be anultralter of A extending D 0 and let D be an ultralter of B extending D 0 D .Now continue as above. 1.8

Remark. If we deal only with 1.8(1), we could use the partition property twice:rst for the colouring {, } c {, } and second for the colouring , t {, } .

1.9 Claim. 1) In 1.5 we may replace in clauses (b), (c) of the assumption theusual (well) ordering of the ordinals < and of the ordinals < by linear orders< , < , but can retain the order on , this does not make a difference (call thisassumption ,, ), provided that we weaken clause ( ) of the conclusion by

( )

Length +

(B1 A B2) = +

2) In ,, of 1.5 if we add = sup Rang (c ), then we can omit clause (e) as it follows.3) If c , ,, satises (a),(b),(c) of ,, , then Rang (c ) has no last element.4) If ,, and 1 1 , then 1 , 1 , .

Proof. 1) Same proof as in 1.5.2) Add to w dummy members to increase i, possible as Rang( c ) is unbounded in

(and included in the next proof).3) Let = {c {, }+1 : , < }, and let = cf( ), now > 0 as Rang( c ) = ,if is a successor ordinal, say + 1 let 0 = 1 be such that c ( 0 , 1) = andlet , = for < 2, < ; applying clause (c) we get a contradiction. So isa limit ordinal, and let : < be increasing continuous with limit . Denec : []2 by c {, } = Min { < : c {, } < } so c {, } is always asuccessor ordinal.

Let us prove (c) + (e). So let ( ) < and let w = { , : < n }, , 0 < , 1 < . . . < ,n 1 < for < . Let = max {c { , , ,m } : < m < n } andas cf() > (so cf() > is an overkill) without loss of generality is constantso = max {( ) + 1 , + 1 : < } < , so as = sup(Rang c ), for some < < we have c { , } > , hence c { , } > . Without loss of generality for all s the truth value of , < , , > , , < , , > arethe same. Now apply the old clause (c) to w = w { , } and we can nd = and i, j as there. Now ,, i = Min { : j < }, j =: i + 1 are as required.4) Trivial. 1.9


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By Denition 1.10(6), Qs 2(,, ) mean that we are given a Boolean algebra A,of cardinality , an ideal I of A and a function f : []2 J (I ).

Let f : []2 I be such that < < f {, } f (, ). So f

is a function with domain [ ]2

and with range of cardinality | I | | A| .By the denition of []2,< 0 there is X [] and nite w such that

< X f (, ) w. Now let b = { a : a w}, as a nite union of membersof I clearly b is in I . Also for < from X, f (, ) is a member of w hence is b,but is also a member of f (, ) which is a subset of I upward closed, so b f (, ),is as required (we did not use f {, } is closed under intersection). 1.13

1.14 Remark. 1) So we have consistency results by [Sh 276], [Sh 546].2) In 1.15 below we can moreover get Qr 2(,, ).3) When does of 1.5 fail? See below (and 5.5).

1.15 Claim. 1) Assume = cf () < =

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22 SAHARON SHELAH

( Ai )[c { i , } < j i ]

( B i )[c { i , } j i ]

For i even we have no free choice.For i odd we ask

( )i is there < such that j and c {, } j } D + ?

If yes, choose such = i , so Ai 1 \ ( i + 1) =

j i , c { i , } < j } D +

and without loss of generality j i > i . Choose this set as Ai and B i as B i,j .If no, let for < , j ( ) < be a counterexample. So by the normality of the

lter, B =: { B i 1 : for every < we have c {, } j ( )} = mod D. SoB i 1 \ B D + but normality implies + -completeness hence for some j < wehave B = { B i 1 : / B and j ( ) = j } D + .

So B D + and for 1 < 2 in B we have c { 1 , 2} < j ( 1) = j ; butB D + | B | = (again normality).

This contradicts the choice of c , that is clause (c) of 1.5 so clearly ( )i holds forour odd i.So we succeed to choose Ai : i , B i : i , so we can nd A and B \ ( + 1) (as A , B D + ).So for each i <

( ) i i < < , c { i , } < j i , c { i , } j i so c { i , } < c { i , }.

But c { i , } Max{c { i , }, c {, }}. Hence c {, } c { i , } but the latteris j i and j i i, so c {, } i. As this holds for any i < we have gotten acontradiction. 1.15

1.16 Remark. 1) We can weaken the demand on the even player in the game(,D, ) for = to the demand

i

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( ) make A1 B1 =( ) if i = 4 j + 1 retain demand (c) in the proof but if i = 4 j + 3 uses a similar

demand interchanging Ai and B i .

2) Moreover, instead even has a winning strategy it is enough, that odd has nowinning strategy for winning at least one of two plays, played simultaneously.Now we can deal with other variants.

We may wonder what is required from A in Claim 1.5.

1.17 Claim. Assume ,, (see below) and A is a Boolean Algebra of cardinality < cf () (for simplicity) and

( ) there are ati A for i < , t { 1, 2} such that:

(i) i < j < A |= a ti a tj(ii ) a1i a2i = 0 for i <

(iii ) there are no disjoint a1 , a 2 B such that t { 1, 2} & i < a ti a t .

Then (, ) SpDpFP (A) where

,, is like ,, but we replace clause (c) by (c)

below where(c) like clause (c) but we are given also an unbounded Y and demand j Y .

Proof. Similar to the proof of 1.5.

We mark the changesStage A:

We let I t

be the ideal of A generated by {a ti

: i < } and let I t

= {a A: wehave a ti a = 0 A for every i < }. Clearly I 3 t I t . Let I be a maximal ideal of B which include I 1 I 2 (it exists by clause (iii) of the assumptions).

We let B t be the extension of A by {x t : < } freely except the equations:

( )3 x t x t a tc {, } for < <

( )4 x t a = 0 for a I t .


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we use ,, . Let < and f be a homomorphism from B t to {0, 1} such thatf (x ) = 0 , f (x ) = 1. Now

( )8 a I t f (a) = 0[why? if a I t is a counterexample then < w f (x t ) = f (x t

a ) = f (x ) 1 = 0 so clearly f (c ) = f (c ), contradiction]( )9 f (a ) = 1

[why? otherwise f (c ) = 0 = f (c ), contradiction]( )10 there is a homomorphism g : A {0, 1} such that

(i) g(a t ) = f (at ) for < j

(ii ) g(a t ) = 1 if j, < (iii ) g(a ) = 1

(iv) b I t g(b) = 0.[Why? If ( )( j < f (a t ) = 1), then g = f is O.K. If not, then

clause (i) means g(a t ) = 0 for < j ; hence for proving the existence of q itis enough to show

j < j a (a tj a tj ) / I t

this holds by the choice of ( ,, i , j ), i.e. j Y .]

We continue dening h as in the proof of ( )9 in the proof of 1.5. 1.17

1.18 Claim. In 1.7 we can also get ,, .

Proof. We repeat the proof but: proving clauses (c) + (e) we are also given anunbounded set Y { i < : i odd}, and choose j there. 1.18

1.19 Claim. Assume

(a) A is a Boolean Algebra and I 1 , I 2 are ideals of A and I 1 I 2 = {0}(b) for = 1 , 2 we have NQs 2(,,A,I ) and |A| < cf ()

(c) I 1 I 2 generates A (or less).Then there are Boolean Algebras B1 , B 2 extending A such that Depth + (B1

AB2) =

+ , Depth + (B1) , Depth + (B2) .

Remark. We can weaken clause (c). We may wonder on using more ideals.

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26 SAHARON SHELAH

Proof. Like the proofs of 1.5, 1.17 but: A is given let c : []2 J (I ) exempliesNQs 2(,,A,I ) and ( )3 , ( )4 are now

( )3 xt x

t a when a f t {, }, t { 1, 2}

( )4 x t a = 0 if a I t , < .

1.19

Note (not used here, just for background see more on clause (c) in 5)1.20 Claim. 1) Assume

(a) = cf () < , and (b) is strong limit singular, cf () = < (c) for any c : []2 satisfying < < c {, } Max {c {, }, c {, }},

there is X [] such that Rang (c [X ]2) is a bounded subset of .

then { < + : cf () = } I [].2) We can replace (b) by

(b) > > = cf () and ( < )[

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2 On the family of homomorphic images of a Boolean Algebra

Our best result is 2.6(2) but we rst deal with more specic cases. On backgroundsee [JuSh 612].2.1 Lemma. Assume < < and is a strong limit singular of conality and cf () = .

Then every Boolean Algebra of cardinality has a homomorphic image of car-dinality [, 2


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28 SAHARON SHELAH

, so |B | .So we are done. 2.2

2.3 Remark. If is regular, |B | , then B has a homomorphic image of cardinality[, 2


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Proof of 2.1. So by 2.2 without loss of generality = 2


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30 SAHARON SHELAH

|B | |{ a Y : a B }| i

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Q = {X :X is a subset of B of cardinality < with the FIP

and {D Ult( B ) : X D}| > }.

Now

(a) Q is not empty (as Q)(b) for X Q dene the two place relation E X on B : aE b iff the set {D

Ult( B ) : X D and a D b D} has cardinality .

Now E X is an equivalence relation; easily

|{D Ult( B ) : X D}| 2|B/E X |

+

hence necessarily |B/E X | . So by the Erdos Rado theorem, we have

(c) for any < there is a sequence a : < such that < X { a , a } Q moreover by the proof of Erdos-Rado theorem X { a : < } { a } Q.

We can let =n

n with n < n +1 . We then can easily get a tree a :


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32 SAHARON SHELAH

Algebra of cardinality .) The completion of B has as homomorphic imagesmany -complete Boolean Algebras of cardinality (though consistency not allby a result of Dow and Vermeer).

Proof. 1) Without loss of generality B is a Boolean Algebra of subsets of andlet =

i +

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[, 0.Let D be an ultralter on B c such that { x : < } D and for < letD be an ultralter on B c such that: D B c, = D B c, and a D .We dene a homomorphism f : B c P () by f (b) = { < : b D }.Clearly f is a homomorphism. Let B be the range of f . Now f (x ) = {}for < so B

contains all singletons from . Next suppose b B c

. Hence

for some cn B for n < we have b = sup {bn : n < }. Hence for some n < (for n < ) and (innite) Boolean term , b = (. . . , x n , . . . )n

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34 SAHARON SHELAH

3 If d(B ) is small, then depth or ind are not tiny

The following denition 3.1 inspires the main result of this section 3.3 though isnot used. The result says that e.g. d(B ) = < |B | & ind+ (B ) Depth( B ).

3.1 Denition. 1) We say a = a : < is semi-independent in B if: it isa sequence of distinct elements in a Boolean Algebra B and some ideal I on Bwitness the semi-independence of a in B , which means:

B,I, a for any < < and b a : 0)( )3 b / I, b a I b a I ( )4 b / I, b a I b a I .

2) si + (B ) = Min { : there is no a : < in B which is semi-independent }and we say a = a : < and I witness < si + (B ) if I witnesses the semi-independence of a in B . Let si (B ) = sup { : there is a semi-independent sequencea : < in B }.

3) si 1+ (B ) is dened similarly to si + (B ) except that we use a : < + 1 . Wesay I, a : witness < si 1+ (B ).

Remark. In fact as it is known that t+ (B ) ind + (B )+ Depth + (B ), below part (2)of 3.2 follows from part (3) of 3.2; proof included for familiarity with si .

3.2 Fact. 0) si (B ), si 1+ (B ) si + (B ) (si (B )) + .1) ind+ (B ) si + (B ).2) si + (B ) t+ (B ).3) si + (B ) ind+ (B ) + Depth + (B ).

Proof. 0) Read the denition.1) ind+ (B ) si + (B ) holds as independent implies semi-independent for the ideal{0B }.2) Let < si + (B ) and a : < , I witness it.

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Let D be an ultralter on B disjoint to I . Without loss of generality ( )(aD ) or ( )(a / D). As we can replace a : < by a : < ,without loss of generality

0, so wecan assume k < n . Clearly bk b : < k B holds and k < k +1 so inDenition 3.1(1) with ( bk , k , k+1 ) here standing for ( b,, ) there the demandshence the conclusions of ( )2 ( )4 holds. Now by trivial reasons we have bk+1 ( a k +1 ) = bk ( a k ) ( a k +1 ) = ( bk ( a k )) (bk ( a k +1 )); now by 3.1(1)( )2 this is > 0 and by the maximality of k, bk+1 I , so by 3.1(1) ( )1 we havebk+1 ( a k +1 ) = bk+1 ( a k +2 ) = . . . hence bk+1 ( a k ) = bk+2 = . . . = bn sobn = 0.3) Let a : < , I witness < si + (B ).

If a : < is independent we are done, so assume not, so let < beminimal such that a : is not independent modulo I ; so a : < isindependent modulo I and for some b a : 0, (so b / I by the assumption on ), we have b a I or b a I . Now by symmetry(as we can replace a : < by a : < ) without loss of generality the

former holds.Hence [ , ) b a I (by ( )3 of B,I, a from Denition 3.1(1)) hence < 1 < 2 < b a a 1 = b a a 2 (by ( )1 of B,I, a fromDenition 3.1(1) applied to b b ) hence ( ) < < b a a +1 =b a a +1 a = b a a +1 a a +1 b a a +1 .So ba a +1 : [ , ) is increasing. Now for from [ , ), b / I, b a I hence b (b a a +1 ) belongs to a j : j < + 2 B but does not belong toI hence by B,I, a ( )2 we have (b (b a a +1 )) a +2 a +3 > 0 henceb a a +1 = b a +2 a +3 hence b a2 b2 +1 : 2 [ , ) is strictlyincreasing in B , a required. 3.2

3.3 Claim. Assume B is an innite Boolean Algebra satisfying ind + (B ) and d(B ) = .

Remark. I think that:

( ) if in addition + ( + 1) 3 for every < , then Depth + (B ) > .

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36 SAHARON SHELAH

Proof. Let a : < + be a list of pairwise distinct elements of B . As d(B )without loss of generality B is a subalgebra of P (). Let B = { a : < } Band

E =: < + : B {a : < + } = {a : < }

clearly it is a club of + .For every S 0 =: { E : cf() } we let I =: {b B : a b B}, so

I is an ideal of the subalgebra B of B . Let S 0 , and J be an ideal on B andnow we try to choose by induction on i < , an ordinal ,J,i such that:

(a) ,J,j < ,J,i < for j < i(b) a ,J,j /J : j < i is independent in the Boolean Algebra B /J .

If we succeed, then a ,J,i : i < contradict the assumption ind+ (B ), so forsome i(, J ) < we have: ,J,i is dened iff i < i (, J ). This is true in particularfor J = I . So for some stationary S 1 S 0 and i( ) < and i : i < i ( ) , anincreasing sequence of ordinals < + , the set S 1 = { S 0 : i(, I ) = i( ) andi < i ( ) ,I ,i = i } is stationary.Let b : < ( ) list the non-zero Boolean combinations of {a i : i < i ( )}so ( ) < (as ind + (B ) 1) and for S 1 let Y ( ) be such that < ( ) & S 1 [b I Y ]. As B is a subalgebra of P () we canchoose a function H such that Dom( H ) = B \{ } , H (c) c and dene, for s { 0, 1}

the function H s

with domain {c B : c as

> 0B } by H s

(c) = H (c as

). Choosea function F such that Dom( F ) = I and c I F (c) = c a B . Againfor some x s : < ( ) and s = 0 , 1 and Y we have

S 2 = { S 1 :Y = Y and for ( )\ Y we haveH (b a

s

) = xs

for s = 0 , 1}

is a stationary subset of + .For each S 2 and t { 0, 1} and Y we try to choose by induction on

i < , B ,, t ,i and an ordinal ,, t ,i such that:

(a) ,, t ,j < ,, t ,i < for j < i(b) ,, t ,i > j for j < i ( )(c) a t ,, t ,i b I

(remember a t is a if t = 1 and is a if t = 0)

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(d) B ,, t ,i is the smallest subalgebra of B containing {b } { a j : j < i ( )}{a ,, t ,j : j < i }

(e) if c I B,, t ,i then a ,, t ,i c = F (c) = a c (in fact just c{b a t ,, t ,j : j < i } suffice)

(f ) if c B,, t ,i and s { 0, 1} then c a s = 0 H (c as

) as

,, t ,i (in fact just c { b a t ,, t ,j : j .

[Why? We just prove that a t b at

,, t ,i : i < is strictly increasing. Let

j < i < , so by clause (c) we know that b a t ,, t ,j I but b a t ,, t ,j B,, t ,iby clause ( d) hence a ,, t ,i b a

t

,, t ,j = F (b at

,, t ,j ) = a b at

,, t ,j byclause (e) and it follows that ( a ,, t ,i ) b a

t

,, t ,i = ( a ) b at

,, t ,j .

So as t { 0, 1} we always have a t ,, t ,i b at

,, t ,j = at

b at

,, t ,j .So

x a t ,, t ,i b at

,, t ,j x at

b at

,, t ,j .

So if x a t b then

x a t ,, t ,i at

,, t ,j x at

,, t ,j .

hence

x a t ,, t ,j x at

,, t ,i .

The above statement means

x a t b [x at

,, t ,j x at

,, t ,i ]

hence a t b at

,, t ,i : i < is -increasing. But b / I hence at

b / B , hencefor j < i, a t b = a

t

b at

,, t ,j (as by clause ( c) we know that b a t ,, t ,j I

so a t b at

,, t ,j B ).

So 0 < a t b at

b at

,, t ,j = ( b at

,, t ,j ) at

hence x =: H ((b at

,, t ,j ) a t ) a

t

is well dened and belongs to at

,, t ,i by clause ( f ) . So x belongs to

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38 SAHARON SHELAH

a t , b at

,, t ,j , at

,, t ,i so it exemplies at

b at

,, t ,j = at

b at

,, t ,i . Soa t b a

t

,, t ,i : i < is strictly increasing.]

We have proved ( ), so assume toward contradiction that for S 2 , ( )\ Y andt { 0, 1} the ordinal ,, t ,i is well dened iff i < j (,, t ) where j (,, t ) < . Soagain for some ordinals j (, t ) and , t ,i (for ( )\ Y, t { 0, 1} and i < j (, t ))and B, t ,i we have

S 3 = S 2 :for every ( )\ Y we have

j (,, t ) = j (, t ), [i < j (, t ) ,, t ,i = , t ,i ]

and [i j (,t) B ,,

t,i = B,

t,i ]

is stationary.So for some stationary S 4 S 3 we have: s { 0, 1} & 1 , 2 S 4 H s1

, t ,i

B, t ,i = H s

2, t ,i

B , t ,i and F 1 (y at

1 ) = F 2 (y at

1 ) for any y, t ,i

B , t ,i

belonging to I 2 (equivalently I 1 ) and I 1 , t ,i

B, t ,i = I 2 , t ,i

B , t ,i . Let 1 < 2

be in S 4 and we shall get a contradiction as follows.Recalling the choice of i(2 , I

2),

2,I

2,i : i , t ,i


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40 SAHARON SHELAH

4 On omitting cardinals by compact spaces

We continue Juhasz Shelah [JuSh 612]. We investigate what homomorphic imagessome Boolean Algebras may have, and (in 4.16) prove the topological analog of 2, showing the existence of some subspaces for Hausdorff spaces (not, necessarilycompact).Recall4.1 Denition. 1) U () = Min {|P | : P [] and ( X [] )( a P )

( |a X | = )}.2) Let a () = Min {|A | : A [] is -MAD} where A [] is called -ADif A = B A | A B | < and we have A is -MAD when in addition A ismaximal under those restrictions.

4.2 Remark. 1) In the case 2 , in which we are interested, U () = a () = |A |whenever A is -MAD for . (See more and connection of pcf theory [Sh 506],[Sh 589], but we do not use any non-trivial fact.)2) We could have used other variants like [, ] = Min {|P | : P [] , and everyA [] is the union of < members of P } or cov(, + , , ) = .3) Our main interest below is in regular .4) Recall that B |= b =

i U a i if b is the lub of {a i : i U } in B and then we

say i U

a i exists (in the sense of B ).

4.3 Denition. 1) For J 1 J 2 ideals of a Boolean Algebra B , we say J 2 is -fullover J 1 inside B , if ( ) or at least ( ) where

( ) is regular and for every X [J 1] there is b J 2 such that |{x X : B |=x b}| =

( ) not necessarily regular and for every x i J 1 for i < there is b J 2 suchthat |{i < : x i b}| = .

(When is regular, they are equivalent).2) The ideal J of B is -full if: J is -full over J inside B .3) J is -local inside B , if |{b B : B |= b x}| for x J .4) In part (1), J 2 is strongly -full over J 1 inside B if ( ) or just ( ) where

( ) is regular and for every X [J 1] for some Y [X ] , for every Z Y wehave

b Z

a exists in the sense of B and belongs to J 2

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( ) is not necessarily regular and for every x i J for i < , for some Y [] ,for every Z Y , the union

i Z

x i in the sense of B exists and belongs to

J 2 .(For regular ( ), ( ) are equivalent).4A) Similarly in part (2).

4.4 Fact: 1) If J 2 is -full over J 1 inside B , h is a homomorphism from B onto Band J = h(J ) for = 1 , 2, then J 2 is -full over J 1 inside B .2) If J 2 is -full over J 1 inside B and J 2 is -local inside B , then U (|J 1 |) + |J 2 |.3) If the ideal J of B is -full and -local inside B , then U ( |J |) + |J |.4) In parts 2) and 3), if = then we get U (|J 1 |) | J 2 | , U (|J |) | J | respectively.

Proof. 1) Trivial.2) Let J 1 = {a i : i < |J 1 |} , let P y = {X | J 1 | : |X | = and ( i X )(B |=a i y)} for each y J 2 , so |P y | . Lastly, let P = { P y : y J 2} so|P | | J 2 | sup

y J 2|P y | + |J 2 | . Easily P is as required in Denition 4.1.

3) Follows by part (2).4) Similar to the proof of part (2) only now P y = {{i < |J 1 | : B |= a i y}}.

4.4

4.5 Fact: 1) Assume

( ) < and , =: { : U () = } and = cf( ) +and for every we have cf() = , (clearly there is such cardinal: = + ).

Then there is a Boolean algebra B such that

(a) |B | = (b) B is atomic with exactly atoms, say [ ]< 0 B P ()

(c) B has a maximal ideal J which is 2 -local, moreover x J |{ y : B |=y x & y an atom (of B )}| ; in other words, J []

(d) J is -full (inside B ) for every (e) () if 2 then P () is isomorphic to some B {x : x a}

( ) If 2 > , B 0 P () has cardinality then in ( ) we canreplace P () by B 0 .

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42 SAHARON SHELAH

2) We can add in part (1)

(f ) if 2 , x J and y x, then y B(g) 2 & J is strongly -full inside

B(h) if 2 and we x B 0 P () which has cardinality then for somesubalgebra B 1 of P () extending B 0 we have() for every x J B satisfying |x| = we have B x = B P (x) is

isomorphic to B 1( ) x J y I (x y & |y| = )( ) if x, y B 1 and |x | = |y| then B 1 x = B 1 y.

Proof. 1), 2).Case A: 2 .

As we can nd pairwise distinct x i [] for i < , x 0 = {i : i < }. LetJ 0 be the ideal of P () generated by {x i : i < } {} : < , so J 0 []

and |J 0 | = as = |{x i :< }| | J 0 | i

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4.6 Claim. 1) If B , J (and ,,, ) are as in fact 4.5(1),(2), 2 and B isa homomorphic image of B and B 2 , then U ( B ) = B .2) Hence if , 2 < , ( < )( | | 2 & equality holds (in any case |Ult(B )| B ).

4.7 Remark. 1) Note that in 4.6(2), if = then the conclusion says nothing as[, ) = (this occurs e.g. if = 2 ).2) In second clause of 4.6(3), the assumption can be weaken to if X J, |X | = (2 )+ then for some y I we have = |{x X : x y}|, to get this weneed much less than U () = .

Proof. 1) 2), 3) Let h : B B be a homomorphism from B onto B , letJ = {h (x) : x J }. First assume 1 B J , this means that for some xJ, h (x) = 1 B , so B x =: B {y : B |= y x} has B as a homomorphicimage so B | P (x)| 2 , and the number of ultralters of B is 22

; also

if B 2 we get B = 2 hence U ( B ) = U (2 ) = 2 as (2 ) = 2 ;this nishes. So assume 1 B / J , hence J is a maximal ideal of B , also clearlyJ is -full inside B for every (by Fact 4.4(1)). As J is 2 -local (see Fact4.5(c)), clearly J is 2 -local, hence by 4.4(2), (letting =: |J |), we have U ( ) (2 ) + , but U ( ) 2 = (2 ) so U ( ) = . Also B = + = . So

we have gotten the conclusion of 4.6(1). Now 4.6(2) follows easily as U ( )

asP =: { : < } : is a -AD family of subsets of X =: { : > }now |X | = |. Hence it suffices to assume that A [> ]and X [> ] a A [|a X | = ], and show that |A | = . Let P = {{ : < } : }. Thus a A [|{ : a}| = ]. So

( ) =a A

{ : |{ < : a}| = }.

Now x a A and let T a = { : |{ < : a}| = }. For each T a let f () = { : a}. Clearly f is a one-to-one function. Hence|T a | 2 2 . Hence by ( ), | A | 2 . Since 2 < , it follows that|A | = .]

Lastly, for 4.6(3), if D is an ultralter of B then either D = B \ J or for somex J , x D but for each x J , B {y B : y x} has 2 members so

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the number of ultralters of B to which x belongs is 22

, that means

4.8 Fact: If the Boolean Algebra B has a -local maximal ideal, then|Ult( B )| 2 + B .Last point is the second sentence in part (3); that is we assume =: B > 2 ;we have to prove that B has 22

ultralters. Let x i J for i < be such

that i < j < h (x i ) = h (x j ) (possible as B = |J |). So by the -systemargument without loss of generality for some x , i < j < (2 )+ x i x j = x.Also i < j < x i = x j hence x i x : i < (2 )+ is a sequence of (2 )+pairwise disjoint non-zero (in the sense of B ) members of B . Now we know thatJ is -full (ideal of B , by 4.5(1), clause (d), J is -full for every ). Applythe denition of J is -full to the set x i x : i < so we can nd y Bsuch that w = {i < : x

i x y} has cardinality (remember by

assumption of the second inequality in 4.6(3)). Hence P ( { x i x : i w}) B .Now for every set u w, zu =

i u

(x i x) by clause (f) of 4.5(2) belongs to B and

[i u x i x zu ] and [i w\ u (x i x) zu = 0]. But h is a homomorphismso in B we have ( u w)( z B )([i u h (x i x) z] & [i w\ uh (x i x) z = 0]). Hence B has a homomorphic image isomorphic to P (), just let D i be an ultralter of B to which h(x i x) belongs and g : B P ()by g(a) = {i : a D i }. Hence B has 22

ultralters.

Together we nish. 4.6

By claims 4.5, 4.6 we really nish. Let me point some specic conclusions:conclusion 4.9 is the theorem of Juhasz Shelah [JuSh 612].

4.9 Conclusion. For every > 2 , there is a Boolean algebra B such that: B is atomic with atoms, B = and for every homomorphic image B of B of cardinality > 2 we have = and the number of ultralters of B is 22

+

in particular [, ) is impossible.

Proof. We apply fact 4.5 + claim 4.6 to = and our , so = { : }. Sofor we get B . Let B be a homomorphic image of B (equivalently, a quotient of B ) and = B > 2 . So U ( ) = , now if < let = Min { : > }, so = cf( ) and


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Proof. By 4.6 and 4.13.

Another example

4.15 Conclusion. If is strong limit singular, = cf( ) < cf(), + < 2 , (+ )

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By renaming without loss of generality Z i = [ i , i+1 ). Let Y = c{y : < }.It suffices to prove that |Y | = (i.e. clause ( ), noting that clause ( ) holdstrivially), for this it suffices to prove

( ) if x Y , then for some i < we have

x c({y : < i } { y j : j < }).

If x contradicts ( ), then for every i < there is i < such that x U i , U i ({y : < i } { y j : j < }) = .Now i < =

j

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W j,A = U Y : < j and [i [ j, ) A [ i , i + i )

y U ]

and [i [ j, )\ A [ i , i + i )

y / U ] .

1 W is a family of open subsets of Y .[Why? As each U W is open subset of X .]

2 W is a basis of Y .[Why? Let z Y , z U , U an open subset of X . As W = {U : < }is a basis of X , for some < we have z U U . As z Y ,for some i(0) < and i (0) and > i (1)and A = {i < : y i U }. Clearly satises the requirements in thedenition of W j,A so U Y W j,A , as W j,A W we are done.]

3 W has cardinality


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(remember w(X ) is the weight of the topological space X ).

4.18 Remark. Of course, we can use disjoint sums of Boolean Algebras to get moreexamples (and similarly for topological spaces) as

wcSp(i

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5 Depth of Ultraproducts of Boolean Algebras

5.1 Claim. Assume (i.e. there is C : < + limit such that C is a club of , C of order type < if cf() < and 1 acc(C 2 ) C 1 = C 2 1).Let = cf( ) < . Then there are Boolean Algebras B for < such that:

(a) Depth (B )

(b) for any uniform ultralter D on we have + Depth ( 1 , = 0the set S = { < + : cf() = 0 , and otp( C ) is divisible by 2} is stationary (if E is a club disjoint to S , choose acc(E ) of conality 1 and a club of C are as required). If = 1 , = 0 , use the construction in [Sh 351, 4].]So without loss of generality for every limit < + we have C S = (why?by deleting the rst + 1 elements from any C of greater order type) alsowithout loss of generality [ C \ acc(C ) non-limit]. Without loss of gen-erality S ( + 1) = and let = + .So

(a) < are regular(b) C : < limit is a square sequence

(c) S { < : cf() = } is stationary(d) S otp( C ) = so cf( ) = (e) C S =(f ) is divisible by 2 .

Done 7-8/97

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5.3 Fact: Assume ,,S, C : < limit satisfy above. Then there are setsA, (for < , < ) such that:

,,S

A , :

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Let A, = { } A , where is the -th member of acc( C ) (so necessarily acc( C ) and < C = C A ,0 ( )[(A , =A , )].

Check. 5.3

5.4 Remark. This may be relevant to a problem from [Sh 108], see 1.20.Continuation of the proof of 5.1. Let A = A, : < + , < be as in 5.3. Let< be the following two place relation on + : < A, . It is a partialorder (by clause (iii) of 5.3). Also

( )1 < /D : < + exemplies this by ( )1 above. Assume to-ward contradiction that < and Depth + (B ) > + so assume b = b : ( ) ( ) A (( ) ,) , and (( ), ) < ( ) (( ), ) A( ) , and let = max( {} { : < n ( ) }), so [(( ), ) ( ) A (( ) ,) , ( ) = A( ) , ]

and ( )

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Proof. Easy, e.g.

(a) let A, : < : < + be as in the proof of 5.1. Let for <

c {, } =: Min { : A, }.

Now to prove clause (c) of ,, from 1.5, use stationary and get possi-bility (i) there. 5.5

5.6 Remark. 1) If is (weakly) inaccessible > = cf( ), S { < : cf() = } isstationary non-reecting and we have a square as required in the beginning of theproof of 5.1 (e.g. if V = L, by Beller and Litman [BeLi80]) then for some A, the

statement,,SA is proved by 5.3. Then repeating the proof of 5.1 we get BooleanAlgebras B for < such that Depth + (B ) = , and for every uniform ultralter

D on , < Depth + ( are regular, S { < : cf() = } is stationary, clause (i) and (ii)

of 5.3 hold and(iii ) A, A, A,

(iv)+ S & < & < sup( A, ) < .

3) If = + , = cf( ) < and S = { < : cf() = } then for some A, ,,SA(as in [Sh 108], [Sh 88a]) but /D so this does not give an example about thedepth of the ultraproduct above the ultraproducts of the depths).

Clearly by 5.1:5.7 Conclusion: If e.g. V = L, > = cf( ) and = then for some sequenceB : < of Boolean Algebras for every uniform ultralter D on we have

Depth(


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REFERENCES.

[BeLi80] Aaron Beller and Ami Litman. A strengthening of Jensens squareprinciples. The Journal of Symbolic Logic , 45 :251264, 1980.

[JMMP] Thomas Jech, Menachem Magidor, William Mitchell, and Karel Prikry.On precipitous ideals. J. of Symb. Logic, 45 :18, 1980.

[Ju] Istvan Juhasz. Cardinal functions. II. In Handbook of set-theoretictopology , pages 63109. NorthHolland Publ. Co, 1984.

[Ju1] Istvan Juhasz. On the weight spectrum of a compact spaces. Israel Journal of Mathematics , 81 :369379, 1993.

[JuSh 612] Istv an Juh asz and Saharon Shelah. On the cardinality and weight spec-tra of compact spaces, II. Fundamenta Mathematicae , 155 :9194, 1998.math.LO/9703220.

[McMo82] Ralph McKenzie and J. Donald Monk. Chains in boolean algebras.Annals of Mathematical Logic , 22 :137175, 1982.

[M2] J. Donald Monk. Cardinal Invariants of Boolean Algebras , volume 142of Progress in Mathematics . Birkh auser Verlag, BaselBostonBerlin,1996.

[Sh:E12] Saharon Shelah. Analytical Guide and Corrections to [Sh:g].math.LO/9906022.

[Sh 108] Saharon Shelah. On successors of singular cardinals. In Logic Collo-quium 78 (Mons, 1978) , volume 97 of Stud. Logic Foundations Math ,pages 357380. North-Holland, Amsterdam-New York, 1979.

[Sh:b] Saharon Shelah. Proper forcing , volume 940 of Lecture Notes in Math-ematics . Springer-Verlag, Berlin-New York, xxix+496 pp, 1982.

[Sh 88a] Saharon Shelah. Appendix: on stationary sets (in Classication of nonelementary classes. II. Abstract elementary classes). In Classi- cation theory (Chicago, IL, 1985) , volume 1292 of Lecture Notes in Mathematics , pages 483495. Springer, Berlin, 1987. Proceedings of the

USAIsrael Conference on Classication Theory, Chicago, December1985; ed. Baldwin, J.T.

[Sh 276] Saharon Shelah. Was Sierpi nski right? I. Israel Journal of Mathematics ,62 :355380, 1988.

[Sh 351] Saharon Shelah. Reecting stationary sets and successors of singularcardinals. Archive for Mathematical Logic , 31 :2553, 1991.


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[Sh 420] Saharon Shelah. Advances in Cardinal Arithmetic. In Finite and In-nite Combinatorics in Sets and Logic , pages 355383. Kluwer AcademicPublishers, 1993. N.W. Sauer et al (eds.). 0708.1979.

[Sh:g] Saharon Shelah. Cardinal Arithmetic , volume 29 of Oxford LogicGuides . Oxford University Press, 1994.

[Sh 454a] Saharon Shelah. Cardinalities of topologies with small base. Annals of Pure and Applied Logic , 68 :95113, 1994. math.LO/9403219.

[Sh 506] Saharon Shelah. The pcf-theorem revisited. In The Mathematics of Paul Erd os, II , volume 14 of Algorithms and Combinatorics , pages 420459.Springer, 1997. Graham, Nesetril, eds.. math.LO/9502233.

[Sh:f] Saharon Shelah. Proper and improper forcing . Perspectives in Mathe-

matical Logic. Springer, 1998.[Sh 589] Saharon Shelah. Applications of PCF theory. Journal of Symbolic Logic ,

65 :16241674, 2000.

[Sh 460] Saharon Shelah. The Generalized Continuum Hypothesis revisited. Is-rael Journal of Mathematics , 116 :285321, 2000. math.LO/9809200.

[Sh 546] Saharon Shelah. Was Sierpi nski right? IV. Journal of Symbolic Logic ,65 :10311054, 2000. math.LO/9712282.

saharon shelah- more constructions for boolean algebras

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