maximal chains in interval algebras

Algebra Universalis, 27 (1990) 32-43 0002-5240/90/010032-!2501.5(] + 0.20/0 �9 1990 Birkfi~iuser Verlag, Basel

Maximal chains in interval algebras

SABINE KOPPELBERG

In Memory of Evelyn Nelson

A maximal chain in a Boolean algebra A is a subset of A which is maximal with respect to the property of being totally ordered under the Boolean (partial) order of A. And A is generated by some chain C ~_ A iff A is isomorphic to an interval algebra (cf. Section 1 for the relevant definitions). More precisely, if A is the interval algebra Intalg X of a linear order X, then A is generated by some maximal chain C ~_ A isomorphic (up to addition or deletion of endpoints) to X. Thus it seems to be a natural question which interval algebras have the property that all of their maximal chains are isomorphic. E.g.: are all maximal chains of the interval algebra of the reals isomorphic? This is easily answered by part of the following observation.

FACT 1. Let A be a Boolean algebra. Then each maximal chain of A contains 0A and la. A is atomless iff each maximal chain of A is densely ordered. A is complete (as a Boolean algebra) iff each maximal chain of A is complete (as a linear order).

Now the interval algebra of an infinite linear order is not even a-complete, so we find that the interval algebra of the reals has both complete and non-complete maximal chains.

Let us denote by [0, 1] the real unit interval, by [0, 1) the half-open interval [0, 1]\{1} etc., and by [0, 1]rat the rational unit interval [0, 1] fq Q. The interval algebra of [0, 1)rat is countable; hence by Fact 1 and Cantor's theorem on countable dense linear orders, all of its maximal chains are isomorphic to [0, 1]rat.

THEOREM 1. If X is a dense linear order and no interval of X is ~o!-saturated (in the sense of model theory), then Intalg X has a countable maximal chain.

In particular, Intalg X has a countable maximal chain if X is a complete linear order.

Presented by R. S. Pierce. Received July 6, 1987 and in final form January 27, t988.

32

Vol. 27, 1990 Maximal chains interval algebras 33

Only very few examples of Boolean algebras are known with the property that all of their maximal chains are isomorphic: if A is an infinite free Boolean algebra, then each maximal chain of A is isomorphic to [0, 1]rat- If A is the completion of an infinite free algebra or if A is complete, atomless and carries a strictly positive, finite, o-additive measure, then each maximal chain of A is isomorphic to [0, 1]; similarly if A is complete, atomless, satisfies the countable chain condition and no Souslin lines exist. Finally if A is atomless, x-saturated (in the sense of model theory) and has cardinality x, then each maximal chain of A is isomorphic to the unique dense x-saturated linear order (with end points) of cardinality x.

T H E O R E M 2. If X is a dense linear order and all maximal chains of Intalg X are isomorphic, then X is countable, hence isomorphic to [0, ])rat-

By Theorem 1, Intalg [0, 1) has a maximal chain isomorphic to the rational unit interval; can it have a maximal chain isomorphic to the irrationals, more precisely to the set I of irrational numbers between 0 and 1? No, since I is not isomorphic to a dense F~-subset of [0, 1]:

T H E O R E M 3. A linear order is isomorphic to a maximal chain of lntalg [0, 1) iff it is isomorphic to a subset F of [0, 1] satisfying

(i) 0 and 1 are in F (ii) F is dense in [0, 1]

(iii) F is an Fo-subset of [0, 1], i.e. a union of countably many compact subsets

of [0, 1].

After setting up the necessary machinery in Section 1, we prove Theorem 1 and 2 in Section 2. In Section 3 we show Theorem 3, the main result of the paper.

The paper is self-contained and no particular knowledge of Boolean algebras or interval algebras will be required. Cf. [Hall for general information on Boolean algebras and [Mo, Ta], [May, Pie] for basic facts on interval algebras. The most involved argument, i.e. the proof of Theorem 3 can be viewed as a study of topology and measure on the real line.

1. co-gaps and ~o-points in maximal chains

In this section, assume that

(X, <) is a dense linear order with a least element 0x and no greatest element.

34 SABINE KOPPELBERG ALGEBRA UNIV,

We recall the definition of the interval algebra of X: for x < y in X, put

[x, y ) = (z ~ X : x < - z < y } ,

a half-open interval. !ntalg X, the interval algebra of X, is the set of all finite

unions of half-open interval of X ; it is a Boolean algebra under set-theoretical

union, intersection, and complementat ion with respect to X. We sometimes write

-< for the Boolean partial order of Intalg X, i.e. a -< b means that a _~ b and a < b

means that a is a proper subset of b. Each element a of Intelg X has a unique

standard representation

a = I x , , y , ) U . o - U [x,, y,,) (1)

where Ox<--xl<yl<'''<X,, in X and either x n < y n in X or y , = { ~ } - i . e .

[Xn~, y,) = [x,, 2) = {z ~ X :xn <-z}.

D E F I N I T I O N . Let a ~ Intalg X have standard representat ion (1). Then Int a

is the finite set of intervals constituting a:

I n t a = {[xi, y i ) : 1 < - i < - n } ,

and le (a) = n = card (Int a) is the length of a. (In particular, le (a) = 0 iff a = Q.)

With these notions at our hands, we can prove the easy direction of Theorem

3. For M a subset of some linear order C or some Boolean algebra A, we denote by infCM, supCM, inf A M, sup A M the greatest lower bound resp. least upper

bound of M in C resp. A, if they happen to exist. Note that if C c A is a maximal

chain in a Boolean algebra A, then for M ~_ C, sup c M exists iff sup A M exists, and in this case, sup c M = sup a M. Similarly for inf c M and inf A M.

L E M M A 1. Assume X U {~} is complete, A = lntalg X, k ~ o), and M ~_ A is a chain in A such that le (m) <- k for all m e M. Then sup A M and inf A M exist and le (sup A M) <- k, le (inf A M) -< k.

Proof. Consider the equivalence relation - on U M c_ X satisfying, for x - y,

x - y iff [x, y ) ~_ m, for some m ~ M.

The equivalence classes are convex subsets of X, and there are at most k of them,

say Y 1 , . . . , Yr. Clearly

sup A M --- [inf x Y1, sup x Y1) U. �9 . U [inf x Yr, sup x Y~).

Similarly for inf A M.


C O R O L L A R Y . Assume X U {w} is complete and C ~ Intalg X is a maximal chain. C is densely ordered (by Fact 1); let D be its De&kind completion, equipped with the order topology, and assume C c_ D. Then C is an F~-subset o lD.

Proof. Letting Ck = {c e C: le (c) -< k}, we have C = (_,Jk~o Ck. By Lemma 1, each Ck is closed under sup and inf in D; so Ck is a compact subset of D.

Let us derive one direction of Theorem 3 from this corollary, assuming X = [0, 1), C is a maximal chain in Intalg X, and D its completion. By Fact 1 and the corollary, C is a dense Fo-subset of D containing 0D and 1D, so we are left

with showing that D ~ [0, 1]. Denote by Z(c) the Lebesgue measure of c e C; then Z : C---~ [0, 1] is an order-isomorphism from C to a subset of [0, 1]. Hence C has a countable dense subset and its completion is isomorphic to [0, 1].

DEFINITION. Let Y be a linear order. The pair (L, U) is a gap of Y if L is an initial segment of Y, U= Y \ L , both L and U are non-empty, L has no greatest element and U has no least element. (L, U) is an m-gap if L has a countable cofinal subset L0 and U has a countable coinitial subset Uo; we also call

(Lo, Uo) an m-gap. A point of Y is an (o-point it if is the limit of some strictly increasing or decreasing sequence of type ~o.

Thus X is m~-saturated iff it has no m-gaps, no m-points, and no countable subset of X is cofinal in X.

DEFINITION. Let a, b c Intalg X and a -< b. Define a map

conn~ : Int a ~ Int b

(the map connecting I n t a and I n t b ) by letting, for u e In ta , connob (u) the unique v E Int b including u, and let

d(a, b) = card (range conn~) ;

thus d(a, b) <- le (a) and d(a, b) <- le (b).

The subsequent lemma gives a technique for constructing ~o-gaps in maximal chains of Intalg X.

36 SABINE KOPPELBERG ALGEBRA UNIV.

L E M M A 2. Let M, N ~_ A = Intalg X be such that M U N is a chain in A and

m <- n holds for all m ~ M, n �9 N. I f there exists a ~ A such that m ~ a <- n for all

m �9 M, n ~ N, then

{d(m, n) :m �9 M, n ~ N}

is a bounded (=finite) subset o f to. The converse holds if X U {~}/s compiete.

Proof. Assume a separates M and N. Then for m e M and n �9 N, m -< a -< n

implies that connm, = conna, o conn,,a and thus d(m, n) = card (range connmn) -<

card (range conning) -< le (a), i.e. le (a) is an upper bound of {d(m, n) : m

M, n 6 N } .

Conversely assume X U {~} is complete and d(m, n) <- k for all m �9 M, n z N.

As in the proof of Lemma 1, define an equivalence relation on Y = U M by

x - y iff for all n e N, x and y are in the same interval of Int n.

The equivalence classes of - are convex in Y, and there are at most k of them.

For otherwise, pick k + 1 non-equivalent points y l < ' ' ' < y k + ~ in Y. Choose

m �9 M large enough such that y~ . . . . , Yk+l ~ m, say Yi �9 ui �9 Int m, and choose

n ~ N small enough such that y~ . . . . . Yk+~ are in pairwise distinct intervals

Vl . . . . . Vk+l of In tn . It follows that connmn(Ui)=vi and d ( m , n ) =

card (range connm,) -> k + 1, contradiction. Let Y ~ , . . . , Yr be the equivalence

classes of - ; then

a = [inf x Y1, sup x Y0 U. -- U [inf x Yr, sup x Yr)

separates M and N.

FACT 2. Assume p, q �9 A = Intalg X and p < q. Then there are Vk, Wk �9 A

for k �9 to such that

p < v o < v ~ < o . - < w l < w o < q , d(Vk, Wk)>--k.

Hence for each maximal chain C of A which happens to include Vo = {vk : k �9 m}

and W0 = {Wk:k �9 co}, (Vo, Wo) will define an to-gap in C.

Sketch o f proof. First choose Vo, Wo, then vl , Wl ~ etc. and make sure that

le (Vk) >-- k, le (Wk) >- k, and d(Vk, w~) >-- k.


F A C T 3. Assume no interval of X is co l - sa tu ra t ed - i.e. be tween any two

points of x, there is an m-gap or an w-point . Let p , q e A = Intalg X and p < q.

Let r e o). Then there are s~, tk ~ A such that

p < � 9 < s 1 < s o < t o < t 1 < � 9 < q ,

p = inf ASk, q = sup a tk, kern k~o)

d(so, to) >- r, d(Sk+l, Sk) ~ r, d(t~, tk+~) >- r.

In part icular for each maximal chain C of A which happens to contain all sk and

tk, p and q will be w-points of C.

S k e t c h o f p r o o f . For the sake of simplicity, let us consider the simple case that

r = 2, p = [a, b) and q = [a, c) where a < b < c in X, and that there are w-points

x, y in X such that b < x < y < c and sequences (x~), (Yk) such that

a < b < x o < x ~ < " " x < y o < y l < " . . < y < c ,

x = sup x Xk, Y = sup x Y~. k ~ o k ~ o

For k e w, put

= [a, xk) u Ix, u [y, c);

then p < to < tl < " �9 �9 < q = sup A tk and d( tk , tk+l) -- 2. Starting with the si tuation

p < to instead of p < q, the s~ can be cons t ruc ted similarly.

We conclude this section by showing that the addit ional assumpt ion on X in

Fact 3 and in T h e o r e m 1 (no interval of X is COl-saturated) cannot he dispensed

with:

P R O P O S I T I O N 1. I f X is ~Ol-saturated, then no m a x i m a l cha in o f X has an

w-po in t .

Proo f . This amoun t s to proving that for no decreasing sequence Co > cl > " �9 �9 in

A = Intalg X, inf a Cn exists; it is enough to show that each sequence ao > al > �9 �9 -

in A has a non-zero lower bound . Cons ider the t ree T = Un~o T~ with T, = Int an

as its n ' t h level and where the order ing of T is such that for u c T~+I = Int an+l,

the unique predecessor of u in Tn is conn . . . . . . . ( u ) - the interval of an including

u. Each level Tn of T is finite and non-empty ; by K6nig 's lemma, T has an infinite

38 SABINE KOPPELBERG A_L-2iEBR~ UNIV.

branch, i.e. a sequence ( u , , ) ~ such that u, ~ Int a , , u, 4=~ and u~ ~ u~ ~ u2 _~ -o - . By ~0a-saturatedness of X, there is a non-empty half-open interval a of X

included in each un; a is a non-zero lower bound of the an in A.

2. Proof of Theorem 1 and Theorem 2

Facts 2 and 3 from Section 1 are enough to prove Theorem i. For Theorem 2,

we shall have to establish another lemma.

D E F I N I T I O N . If Y is a linear order and p, q ~ Y, we say that p, q constitute

a jump in Y and write "p ~ q in Y" if p < q and no element of Y lies strictly

between p and q.

Proof of Theorem 1. Let X be densely ordered with no co~-saturated interval.

In A - - I n t a l g X , we construct by induction chains Cn for n s m such that

C o ~ _ C ~ ~ _ o . . .

Let Co = {0A, 1a} . After Cn has been constructed, let Cn+: arise from C~ by

inserting into each jump p ~ q of C~ elements Spqk, tpqk (k e co) such that

p = inf A Spqk, q = s u p A tpqk,

d(x ,y)>-n for x ~ y i n C n + l ;

this is possible by Fact 3. So C=[_)n~,oCn is a countable chain in A; we show that it is max ima l

Otherwise there exists a e A and L, U ~_ C such that L LJ U = C and I < a < u for all l e L, u E U. We prove by induction on n that C~ 71 L has a greatest e lement

In and Cn N U has a least e lement un. Clearly lo = 0a and u0 = l a ; assume p = In and q = u, exist. Then p ~ q in C,; since p = inf a Spq k and q = sup A tpqk, we have

x < a < y for some jump x ~ y in Cn+l, i.e. ln+~ and u~+~ exist and satisfy

ln+1<a<un+l, d(ln+!, un+l)~n. (2)

Now (2) holds for every n, contradicting L e m m a 2. Thus C is maximal.

P R O P O S I T I O N 2. Let X be a dense linear order. Then ~ntaigX has a maximal chain C such that between any two points of C, there is an m-gap of C.

VoI. 27, 1990 Maximal chains interval algebras 39

Theorem 2 follows readily from this proposition:

Proof of Theorem 2. Assume X is densely ordered and all maximal chains of A = Intalg X are isomorphic. In fact, they are isomorphic to X U {oz}, since ([0x, x ) : x e X O {m}} is a maximal chain. By Proposition 2, any two points of X are separated by an co-gap, hence no interval of X is m~-saturated. By Theorem

1, X is countable.

Proof of Proposition 2. We construct an increasing sequence (Co~)~<_o of chains in A = Intalg X. Here p will be an ordinal of the form p = 3. + 2 for some

limit ordinal 3., and C will be C o. Let Co = {OA, 1A}. If 3. is limit and C~ is defined for all o~ < 3., put Ca = [._Jo~<, CoL; the inductive construction will guarantee that Ca

is densely ordered. Suppose 3. is a limit ordinal and Ca is defined. Then Ca+l arises from Cz by

inserting into each gap (L, U) of Ca an element of A separating L and U, if such an element happens to exist. Note that Ca is dense in Ca+l and Ca+~ is again densely ordered, but possibly C~+a = Ca. Ca+2 arises from Ca+z by inserting, for each p e Ca+~, elements p+ and p - of A such that

p < p + < u for all u 6 C~+1 satisfying p < u

! < p - < p for all l e Ca+ 1 satisfying l < p

if such elements happen to exist. If Ca+2 = Ca+l (i.e. if for all p ~ Ca+i, neither p+ nor p - exist), put p = 3. + 2, finishing the construction. Otherwise note that

C~+2 contains jumps p - ~ p resp. p ~ p + . Finally suppose Co~ is defined, o~ < p, and o~ < m or o: = ,~ + n for some limit

ordinal 3. and n ~ m, n->-2. Our construction will guarantee that Ca contains jumps. Let then Co~+l arise from Ca by using Fact 2, i.e. by inserting an m-gap (in A) into each jump p ~ q of Ca.

Now for a < b in C = Cp, there exists an m-gap of C between a and b - this

follows by induction on o~ where oc is minimal such that a, b e Co, The inductive construction of the Ca shows that C is a maximal chain.

3. Proof of Theorem 3

One direction of Theorem 3 was proved after the corollary in Section 1. For the non-trivial one, assume F is an Fo-subset of the real unit interval [0, 1] such that 0 and 1 are in F and F is dense in [0, 1]. For the sake of clarity we shall formulate a number of claims whose proofs are given later.

40 SABINE KOPPELBERG ALGEBRA UNIV.

In the whole proof, fix a sequence of closed subsets F~ of [0, 1] such that

Given these Fn, we define for p e F

h(p) = the least n ~ co such that p ~ Fn

and for p -< q in F

6(p, q) = the least n E ~o such that p -< z <- q for some z e F,

(thus 6(p, q) <-- h(p) and 6(p, q) <- h(q)). The proof consists mainly in construct-

ing, for p e F, elements ap of A = Intalg [0, 1] satisfying

le (ap)(the length of ap, cf. Section 1) = h(p) (3)

~.(ae)(the Lebesgue measure of a e ) = p (4)

i fp --< q in F, then a e <-- aq in A and d(a e, aq)(Cf. Section 1) = 6(p, q). (5)

It then follows from (4) and (5) that the assignment p map is an order-

isomorphism from F onto the chain

C= {a e:p e e } .

Claim 1. (4) and (5) imply that C is a maximal chain. Let us say that (3)-(5) hold for a subset G of F if % has been constructed for

p �9 G and (3)-(5) holds for all p, q e G. We construct ap for p e Fn, by induction on n. since Fo = {0}, we put a0 = 0A (the empty subset of [0, 11); so (3)-(5) holds for Fo. Assume they hold for Fn. Fix sets X, Y c_ [0, 1] such that

X is a countable subset of Fn+I\F,, Y = (F,~+I\Fn)\X,

and for all y e Y:

y = s u p [~ {X C fn U X :x < y } = int 4~ {x e F, kJ X :y < x } .

g ~ _ 5 ~ - ' - , F= tO F., g = (o}, 1~5. n E o )


Claim 2. Suppose (3)- (5) hold for Fn U X ; for y c Y, define

ay = s u p z ( a x : x E F n L_J S ~ x < y } . (6)

Then (3)-(5) holds for Fn+l. So we are left with defining ap for p ~ X. Write

X = {xk :k < p}

where p - oJ and the xk are pairwise dinstinct. Suppose (3)-(5) holds for the

closed set

D = F,, U (Xo, �9 �9 �9 , x l , - 1 }

and consider

X "~'~X k

with the aim of defining ax.

The following general fact on Intalg [0, 1] is easily seen by drawing pictures.

F A C T 4. Let a, b c A = I n t a l g [ 0 , I] be such that a < b and conna,b (cf. Section 1) is o n e - o n e or onto. Let m c o) such that le (a), le (b) -< m; let y ~ [0, 1]

be such that i ( a ) < y < ,~(b). Then there exists c E A satisfying

a < c < b, le (c) = m, ,~(c) = y,

connac is o n e - o n e and conncb is onto.

Since the set D above is closed and x ~ D, we can define two element of D by

d * = m a x ( d 6 D : d < x } , d * * = m i n ( d 6 D : x < d } ;

note that Z(ad*) = d * < x < d * * = )~(ad**).

C l a i m 3. The assumptions of Fact 4 hold for a = ad*, b = ad*., m = n + 1 and y = x .

Thus let ax = c, where c c A is as guaranteed by Fact 4.

C l a i m 4. (3)-(5) holds for D t_J (x} = Fn U {Xo . . . . . xk).

42 SABINE KOPPELBERG ALGEBRA UNtV.

Claims 2 through 4 allow to construct ap for all p e Fn+~ such that (3)-(5) hold for Fn+l. This finishes the construction of ap for p e Fn+~ and thus of C.

Proof o f Claim 1. Assume for contradiction that there exists a ~ A and L, U c_ F such that L U U = F and a~ < a < au for all I e L, u E U. Consider x = ,t.(a);

then x ~ F since by (4), l < x < u holds for all I e L, u e U. Let n e ~o be arbitrary; we shall find I e L and u e U such that 6(l, u)>-n + 1; i.e. by (5) d(a), a,)>- n + 1, thus contradicting Lemma 2. Since x ~ F~, F~ is closed and F is dense in [0, 1], there are l, u e F such that the closed subinterval [l, u I of [0, !] is a neighbourhood of x and disjoint from F~. It follows that l e L, u e U and

6(1, u)-->n + 1.

Proof o f Claim 2. First note that ay as defined in (6) exists in A, by Lemma 1, and le (ay)_< n + 1. The definition shows, by (3)-(5) for Fn tAX, t h a t p < q in F,+t

implies ap < aq and that )~(ap) = p for all p e F,+~. To show le (ay)= n + 1, assume for contradiction that le (ay)---n. Pick an

increasing sequence (pk) and a decreasing sequence (q~) in F, U X converging to y. Then ap ~ ay <- aq for all k, hence

6(p~, qk) = d(ap~, aq~) <-le (ay) <-n;

thus there exists zk e Fn such that Pk <- Ze <-- qk" The sequence (Zx) converges to y and F, is closed which gives the contradiction y e F,.

Let us finally show that d(ay, ay,)= (~(y, y') holds for y < y ' in F,+I. We consider the case that both y and y ' are in Y, the remaining ones being similar but easier. Choose a decreasing sequence (Pk) in F, U X converging to y and an increasing sequence (qk) in F, U X converging to y ' such that P0 < q~. Since F~ is

closed and y ' ~ F,, we have that

6(qk, y ' ) = n + ! for almost every k; (7)

in particular, n + I = h(qk) = h (y ' ) and hence n + 1 = !e (aqk) = le (%~) for large k. Since a e, = sup z aq,, the map conn%,,y, must be onto and hence

connaq~ a~ is bijective for large k. (8)

Similar reasoning shows that

6(y, Pk) = n + 1 for large k,

conna,,% is bijective for large k.

Consider, for k large enough, the situation

(9)

Y <P~, < q~ < y ' .


Then 6(y, y ' )= 6(p~, qk) follows from (7) and (9) and d(ay, ay , )= d(ap~, aqk )

from (8) and (10). Now (5) holds for p~, q~ e X, proving d(ay, ay') = (~(y, y ' ) .

Proof of Claim 3. We have to show that conn~d.,~., is o n e - o n e or onto. Consider k =d(ad,, ad**); since (5) holds for D, we have k = 6(d*, d**). No element of F, lies strictly between d* and d**, so

6(d*, d**) = min (h(d*), h(d**)) = (by (3)) min (le (aa.), le (aa**)).

Thus either d(ae., ad**)= le (ado and conn,~.,ad., is o n e - o n e or d(ad., ad**)= le (aa**) and conn,,, a,.. is onto.

Proof of Claim 4. We have to show that (5) holds for arbitrary p, q e D U {x}. Without loss of generality consider p eD, q=x, so p<-d*<x. Then

h(x) = n + 1 and no element of Fn lies strictly between d* and x, so 6(d*, x) = h(d*) and 6(p, x)= 6(p, d*). By construction of ax, conn,~.,a, is o n e -o n e ; by con~p,~ x = conna~ o conn,~,a~., we have d(ap, ax) = d(ap, ad.). Thus

d(ap, ax) = d(ap, ado = 6(p, d*) = 6(p, x).

REFERENCES

[Hal] P .R . HALMOS, Lectures on Boolean algebras, van Nostrand, 1963. [May, Pie] R. D. MAYER and R. S. PIERCE, Boolean algebras with ordered bases, Pac. J. Math. 10

(1960), 925-942. [Mo, Ta I A. MOSTOWSKI and A. TARSKI, Boolesche Ringe mit geordneter Basis, Fund. Math. 32

(1939), 69-86.

2. Mathematisches Institut Freie Universitiit Berlin, West Germany

maximal chains in interval algebras

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