A bstract IS THE PRODUCT OF CCC SPACES A CCC SPACE ?

A ccc space is a topological space which satisfies the countable chain condition : Each family of (pairwise) disjoint nonempty open sets is countable . A separable space, for instance, is a ccc space . This is so because given a family .F of disjoint nonempty open sets in a separable space X, one can define an injection of F into a countable dense subset S of X by choosing one point of S in each member of .F . A ccc space, however, need not be separable . For example, if I is a set with cardinality greater than 1 f$ 1 (the cardinality of the set of real numbers) and for each i E I, Xi = {0,1} with the discrete topology, then the product space RiEIXi is ccc but not separable [10, p . 511 . A simple example of a topological space which is not ccc is any uncountable set with the discrete topology. In a 1947 paper [12], E. Marczewski gave a proof (which may also be found in [15]) of Pondiczery's theorem that the product of at most F8 1 separable spaces is separable ; and he raised the question as to whether the product of (just) two ccc spaces is a ccc space . It turns out that special axioms are needed to answer this question : it has a negative answer in the presence of the Continuum Hypothesis (CH) and an afirmative answer when Martin's Axiom and the negation of CH are assumed. The main goal of this paper is to give fairly self-contained proofs of these two assertions, which we do in Sections


. Introduction
A ccc space is a topological space which satisfies the countable chain condition : Each family of (pairwise) disjoint nonempty open sets is countable .A separable space, for instance, is a ccc space.This is so because given a family .F of disjoint nonempty open sets in a separable space X, one can define an injection of F into a countable dense subset S of X by choosing one point of S in each member of .F.A ccc space, however, need not be separable .For example, if I is a set with cardinality greater than 1 f$ 1 (the cardinality of the set of real numbers) and for each i E I, Xi = {0,1} with the discrete topology, then the product space RiEIXi is ccc but not separable [10, p. 511 .A simple example of a topological space which is not ccc is any uncountable set with the discrete topology.In a 1947 paper [12], E. Marczewski gave a proof (which may also be found in [15]) of Pondiczery's theorem that the product of at most F8 1 separable spaces is separable ; and he raised the question as to whether the product of (just) two ccc spaces is a ccc space.It turns out that special axioms are needed to answer this question : it has a negative answer in the presence of the Continuum Hypothesis (CH) and an afirmative answer when Martin's Axiom and the negation of CH are assumed.The main goal of this paper is to give fairly self-contained proofs of these two assertions, which we do in Sections 2 and 3. Related topics (Souslin's Hypothesis, Property (K)) are touched upon in Sections 4 and 5. None of the results and proofs are new .
First, some preliminaries .Let us recall that an ordinal number may be identified with the set of all smaller ordinal numbers ; thus for ordinals a and ,0, the statements a < /3, aEP, and a C P are all equivalent.The smallest infinite ordinal and the smallest uncountable ordinal will be denoted by w and wl respectively .We will need the fact that w and wl are also cardinal numbers.(A good reference for cardinal and ordinal numbers is [6] .)The cardinality of the set of all functions from w to 2 is 2°', and 2°' =l ld 1 .The Continuum Hypothesis (CH) is the statement wl = 2w.The negation of CH, denoted here by -CH, is therefore the statement wl < 2'.Godel and Cohen proved that CH is independent of the usual axioms of set theory, namely, the Zermelo-Fraenkel axioms together with the Axiom of Choice (usually denoted by ZFC) .
We review the definition of the Gleason space of a topological space X, for use several times in the sequel .An open subset U of X is called a regular open set if U is equal to the interior of its closure, Le., U = (U)°.The set R(X) of all regular open subsets of X forms a complete Boolean algebra with respect to the operations U n V = U n V U V V = (U U V)°, and U' -~I x (X\U)°.The Gleason space of X, denoted here by G(X), is the Stone space' of R(X) .The space G(X) is constructed as follows.A subset F of R(X) is -called a filter of R(X) if (i) 0 ~F, (ii) U n Ve.F for all U, V e.F, and (iii) Ve .F if U C V and Ue,F .An ultrafilier of 1Z(X) is a filter which is not properly contained in any filter of 7Z(X ) .Then G(X) is the set of all ultrafilters of R(X ) .For each open set U in X, let U* = {.FeG(X) : (U)°6F} .
With the topology determined by the base {U* : U is open in X}, the Gleason space G(X) is compact, Hausdorff and extremally disconnected; and if X is a ccc space, then so is G(X).
The cardinality of a set X will be denoted by 1 X ~. .We will need the following delta system lemma several times .It is so called because the family 13 is a delta system .
Lemma.Let G be an uncountable family of finite sets .Then there is an uncountable subfamily B of 9 and a fixed set R such that A n B = R whenever A and B are distinet members of B .
Proof.. [8, p . 225] .Since G is uncountable and the sets are finite, there must be uncountably many members of 9 with the same number of elements; therefore we may assume that for some n, 1 X 1= n for,all X in g.We proceed by induction on n.For n = 1, we may take R = 0. Assume that the lemma holds .for n = k and let G be an uncountable family of sets each of which has k + 1 elements .
If there is some point a which belongs to each set in an uncountable subfamily C of 9, then the induction hypothesis may be applied to the family {X\{a} XeC } to yield an uncountable subfamily 13 of C and a finite set R such that (X\{a}) n (Y\{a}) = R for any distinct X and Y in 13.Then X fl Y = R U {a} for any distinct X and Y in B.
Otherwise, each element a belongs to only countably many members of ~, and we construct a disjoint subfamily 13 = {X,, : a < wl } of 9 by transfinite induction on a, as follows .Assume that we have constructed X,, for all a < ,0.Then each element of the countable set U,,,<#X,,, belongs to at most countably many members of 9, so there is some X in 9 which is disjoint from U,,<flXa .Let Xp = X and the proof is complete.

. Martin's Axiom and products of ccc spaces
Martin's Axiom [13] states that no compact Hausdorff ccc space is the union of fewer than 2' nowhere dense sets .Observe that Martin's Axiom (henceforth denoted by MA) is implied by the Continuum Hypothesis because under CH, MA is an immediate consequence of the Baire category theorem.However, MA :P* CH because, as is shown in [19], there is a model of ZFC in which MA holds and wl < 2 " .
In this section we prove that (MA+-CH) implies that every product of ccc spaces is a ccc space, and we do it with the help of an interesting theorem about products of ccc spaces (Theorem 2 .2) whose proof requires no special axioms.
Lemma.[17, p. 16] .(MA+~CH) implies that if X is a compact Hausdorff ccc space, then any uncountable family g = {U,, : a < wl } of nonempty opera sets has a cardinality wl subfamily with nonempty intersection .
Proof.: Consider the set S of all families of disjoint nonempty open subsets of X with the property that if FES, then each member of F meets only countable many members of G.If S is empty, we are through .So suppóse that S is not empty.Then by Zorn's lemma, S has a maximal member, say F. Since X is a ccc space, F is countable .Hence there is a member V of G which does not meet any member of F.Then, because F is maximal, any open subset of V must intersect wl members of G.For each /i < wl , define Hp = V\(U .>aU.) .
Then Hp is nowhere dense in the compact Hausdorff ccc space V. (V is ccc because V is.) Hence (MA+ ~CH) implies that V :~Up<,,,1 Hp .Thus there is v in V such that v 1 U# <,1 Hp, and therefore v belongs to wl Ua 's .
N .M .Roy Theorem 2.1 .(MA -i--CH) implies that if X and Y are ccc spaces, then X x Y is a ccc space .
Proof-Suppose that {Ua : a < wl} is an uncountable family of nonempty open subsets of X x Y.We will show that there axe two members of this family which intersect.By shrinking if neccessary, we may assume that each U« is basic.Then there are open sets V« C X asid W« C Y such that Ua = Va x Wa .The Gleason spaces, G(X) and G(Y), of X and Y are compact, Hausdorf and ccc.Hence by the lemma above, there is an uncountable subset D of w1 such that n.EDV .,*qÉ 0 and naeD W« :~0-(Recall from Section 1 that U* is the set of all ultra,filters which contain (U) °.) Let /P, yED with P qÉ y.Then The ccc has the interesting property that if the product of any two ccc spaces is a ecc space, then the product of any number of ccc spaces is a ccc space .To see this, assume that ccc is preserved by products of two spaces .Then by induction, it is preserved by products with a finite numbers of factors .Then by the following theorem (which seems to have originated in [14]), it is preserved by arbitrary products.Theorem 2 .2 .Suppose that {Xi : ieI} is a family of topological spaces such that IIi, jXi is a ccc space for every ftinite J C I. Then IIi,IXi is a ccc space.
Proof.[10, p. 51] .Let X = IIj,IXi and suppose that there exists an uncountable family {U,, : a < w1 } of disjoint nonempty open subsets of X .As before, we may assume that each Ua is basic .Then by definition of the product topology, for each a < w1 there is a finite subset Fa of I such that Ua = n {P¡-1 (Vi) : iEFa }, where Pi is the projection of X onto Xi, and Vi is open in Xi .By the delta system lemma of Section 1, there is an uncountable subset D of w1 and a set R , such that Fa n Fp = R whenever a, f6D and Fa 7É Fp.Note that R cannot be empty because Fa n Fp = 0 implies Ua n Up = 0.For each a in D, let P(Ua ) = IIiERPi(Ua) .It is not hard to verify that {P(Ua ) : acD} is an uncountable family of disjoint nonempty open subsets of IIi,RXi, which is a contradiction .
Since a finite product of separable spaces is separable, the following corollary is an immediate consequence of Theorem 2.2 .Other consequences of (MA + -CH), not all topological, may be found in [4] and [18] .
In this section we describe the example due to R. Laver and F .Galvin [5], which assumes CH, of ccc spaces Xo arid Xl such that Xo x Xl is not a ccc space .This will lead (via Gleason spaces) to a compact Hausdorff extremally disconnected ccc space X whose square is not a ccc space.The material in this section is adapted from [2] and [5] .
The following lemma can be proved by induction, A proof of a somewhat more general result may be found in [2, p. 190] .Lemma 3.1 .Let A be a set and for each n < w, let {Fi, n : i6In } be a family of disjoint finite subsets of A with 1 In ~= w .Then there are two subsets Ao and Al of A such that I {2eln : Fi n C At} I= w for all n < w and t = 0,1 .
3. The Laver-Galvin example Lemma 3 .2 .Assume that wl = 2°' .Then there are two families {KO (a) a < wl }, { Ki (a) : a < wl } of subsets of wl such that (i) Ko(a) U K, (a) = a for all a < wl , and (ii) If .F = {Fi : i < w} is a countably infinite family of disjoint finite subsets of wl , then there is an ordinal A (depending on .F) with A < wl , such that if A < a < wl , U.F C a, X is a finite subset of a, te{0,1} and then Hence I {i < w : Fi C a\Kt(l) dP¿X} I= w, Proof. .We begin by computing, under CH, the cardinality of the set S of all countably infinite families of disjoint finite subsets of wl .Clearly ~S ~> wl since for each ordinal a such that w < a < wl , the family of a singletons is countably infinite and disjoint .To see that ~S ~< w1 , first note that the set of all finite subsets of wl has cardinality wl + (wl )2 + (wl )3 .{-. . .= wl + wi + wi + . . .= wl w = wl .Thus 1 S 1= wl .Let {.Fa : .\< w,} be a well-ordering of S in such a way that for each A < w, the members of Fa áre subsets of w.
In what follows, Fi,a denotes the ith member of the family .Fa .We now define the sets Ko(a),Kj(a) by recursion .We set Kt(a) = .afor t = 0,1 and a <_ w.Let w < a < wl and suppose that Kt(ce') has been defined for t = 0,1 and a' < a .Consider the set of all triples (t, A, X) such that te{0,1}, A < a, U,-"',X C a, X is a finite subset of a, and This set of triples is countably infinite because X can be the empty set, and if A < w then certainly UF>, C a. Let {(tn,A .,X.) : n < w} be an enumeration of this set, and for each n < w let Then 1 In 1= w, hence we may apply Lemma 3.1 with A and Fi, n replaced by a and Fj,an respectively ; then there are disjoint subsets Lo(a) and Ll (a) of a such that for n < w and t = 0,1 .We set Let us verify that properties (i) and (ii) are satisfied .
(i) Clearly Ko (a)UKj (a) = a because Lo(a) and Lj (a) are disjoint subsets of a. (ii) Let .'F= {F; : i < w} be a countably infinite family of disjoint finite subsets of wl .Then .F = .Fa for some A < wl .Suppose that A < a < wl, U.Fa C a, X is a finite subset of a, te{0,1}, and I {i < w : Fi ,>, C ,P\Kt(a) dpcX} I= w .
Then (t, A, X) = (tn, An, Xn) for some n < w.Hence if we let Our goal is to show that 1 H 1= w, and we will do this by proving that J C H . Let ieJ .Then Fi,a C Lt (a), hence Fi,a C a\Kt(ca).Because ¡EIn we have Proof. .For t6{0, l}, let {Kt (a) : a < wl} be the family of Lemma 3.2 and let Xt be the set of all functions on wl to {0,1} .Set Vt (a) = {xcX, : x 1 K,(a) -0, x(a) = 1} for a < wl, and give Xt the topology determined by the subbase {Vj(a) : a < wl }.The space X o x XI is not ccc because the uncountable family Then since {Vo(a) x Vi (c,) : a < wl} consists of disjoint nonempty open sets.To see that they are disjoint, suppose that there are a, a' with a < a' < wl and a6Kt (a') for sume t in {0,1} ; for this t, we have xt(a) = 0 and xt(a) = 1, a contradiction .
We now verify that X o and Xl are ccc spaces .Let t6{0, 1} and let {U¡ : i < w,} be an uncountable family of nonempty basic open subsets of Xt.For each i < w there is a finite subset Gi of wl such that Then This is so because if we choose x in U¡, then xEVt (a) for all a in Gi, hence x 1 Gi -1 and x 1 Ua,G ;Kt(a) -0.Thus the intersection is empty .Applying now the delta system lemma of Section 1 to the family {Gi}, we may assume without loss of generality that there is a set R such that Gi n Gi, = R whenever i < i' < wl .Let F = G¡\R for each i < wl .Then if i < i' < wl and y 1 Kt(a) for all y6F¡ and Q6F, , , it follows that Ui n U;, ~.We outline the verification .First show that Gi n (Ua,G;, K,(a)) _ and it then follows that Ui = n«EG ;Vt(a) .
(Gi U Gi,) n (Ua,G,UG ;, Kt(a)) = 0 .Now choose x in Xt such that x 1 Gi U Gi, = 1 and Then x6U¡ n U¡, .Thus to prove that {Ui : i < wl } is not a disjoint family, it is enough to show that there are i, i' such that i < i' < wl and "y 0 Kt(a) for all yeFi and fEFi, .Set .F = {Fi : i < w} and let A < wl be an ordinal satisfying condition (ii) of Lemma 3 .2for F. We may assume that UF C A. Because A + 1 is countable and {Fi : w < i < wl } is uncountable, there is an ordinal i' such that w < i' < wl and .1 < a for all a in F¡, .Then UF C a for all a in F¡ , .
If Fi , = ~, then for any i < i' we have that y 1 Kt(fl) for all yeFi and PEFi, .
Therefore we assume that Fi , :~0 and we let {al, a2, . ..,a,,} be the elements of Fi, in the order inherited from wl .We note that and correspondingly.Hence In particular, there is i < w such that y 1 Kt(P) for all yeFi and PEFi, .Thus U i n U;, :~0 and so Xt is a ccc space .
Corollary.Assuming CH, Mere is a compact Hausdorf extremally disconnected ccc space X such that X x X is not a ccc space.
Proof.Let Xo and Xl be the ccc spaces of the above theorem and let X be the disjoint union of the Gleason spaces G(Xo) and G(X1 ).Then X is compact, Hausdorff, extremally disconnected and ccc Because both G(Xo) and G(X1) are .To prove that X x X is not a ccc space, it is enough to show that G(Xo) x G(XI) is not a ccc space Because G(Xo) x G(XI) is homeomorphic to an open (and closed) subset of X x X.Let {Ui x Vi : i < wl } be an uncountable, family of disjoint nonempty basic open subsets of X o x XI .Then, as is not too difficult to verify, {U; x Vi* : i < wl } is an uncountable family of disjoint nonempty open subsets of G(Xo) x G(X 1 ).Thus G(Xo) x G(X1) is not a ccc space .
Before leaving this section we remark that S .Argyros, S. Mercourakis and S. Negrepontis have proved the existence, under CH, of a Corson-compact ccc space whose square is not ccc [1] .

. The Souslin problem
Suppose that X is a totally ordered set satisfying (a) X has no first or last element, (b) X is connected in the order topology, and (c) X is separable in the order topology.Then X is linearly isomorphic to R .(A proof is in [3, p.8].)In 1920, the Russian mathematician M .Souslin [20] asked whether (c) can be replaced by (c') X is ccc in the order topology.A positive answer to Souslin's question has come to be known as Souslin's Hye ,othesis (SH).The axiom SH is undecidable in ZFC and is implied by (MA+-CH) ( [7], [21], [19]).Indeed, to quote D .Fremlin [4, p. 184], Souslin's problem was the original stimulus for the invention of Martin's axiom.
A Souslin line is a totally ordered set satisfying properties (a), (b) and (c'), but not (c) .Thus the existente of a Souslin line is equivalent to the negation of SH .G .Kurepa proved [11] that if X is a Souslin line, then X x X is not a ccc space.(A proof may be found in [10, p . 66].)There is also the notion of a Souslin tree; it is shown in [16] that the existente of a Souslin tree is equivalent to the existente of a Souslin line .

. Property (K)
A topological space is said to have property (K) (after Knaster [9]) if each uncountable family of open sets contains an uncountable subfamily in which every two sets have nonempty intersection .Clearly every space with property (K) is a ccc space.The converse is true under (MA + ~CH) [4, Theorem 41 A] and is false under CH.To verify the latter, we make use of Marczewski's theorem [12] that a product of spaces has property (K) if (and only if) each space has property (K) .Recall from Section 3 that there exists, under CH, a ccc space X such that X x X is not a ccc space .Then X does not have property (K) because otherwise X x X would have property (K) and thus be a ccc space.
The reader has probably noticed that another approach (in Section 2) to proving that (MA + -CH) implies that a product of ccc spaces is a ccc space would be to combine Fremlin's theorem that under (MA + -CH) every ccc space has property (K), with Marczewski's theorem quoted above .
For open problems and additional referentes involving ccc spaces, the reader may consult [2, Chapter 7] and [4, f 44].

Corollary 2 . 3 .
If {Xi : ieI} is a family of separable spaces, then IIi,IXi is a ccc space.

Corollary 2 .
4.  (MA + -CH) implies that if {Xi : ieI} is a family of ccc spaces, then IIi,,Xi is a ccc space .