]]> 2001 American Mathematical Society Transactions of the American Mathematical Society Trans. Amer. Math. Soc. tran 354 05 20011121 November 27, 2000 July 30, 2001 November 21, 2001 1783 1783-1791 9 S0002-9947-01-02936-1 10.1090/S0002-9947-01-02936-1 0002-9947 1088-6850 English 54E35 03E35 03E65 54E45 2000 http://www.ams.org/jourcgi/jour-getitem?pii=S0002-9947-01-02936-1 Introduction In his 1948 paper , Miroslav Katetov proved that if the cube of a compact space is T 5 (also called completely normal or hereditarily normal) Recall that the separation axiom T 5 says that every two separated sets A and B (i.e., B A ) can be separated by disjoint open sets. then the space must be metrizable. He asked if the cube can be replaced by the square in this statement. It has been known for some time that MA 1 and CH each give counterexamples to the implication, and more recently, a considerable weakening of CH was proved to be sufficient for a counterexample (see Pr2,GN,Moo ). In this note we show the consistency of Katetov's metrization statement: Every compact space with T 5 square is metrizable. Our research in this area centers around a combinatorial analysis of forcing axioms compatible with the existence of a particular type of Souslin tree (related work appears in Fa,Mi,Mi2,Lar,LarT,ShZ,W ). It turns out that for solving Katetov's problem it suffices to analyze the maximal amount of MA 1 compatible with the existence of this type of tree. Since MA 1 is by far the most familiar forcing axiom (see Fr ), in order to make this paper accessible to a wider audience we have decided to extract only this part of our analysis from the more general picture that will be dealt with in subsequent papers. It has been known for some time that the metrizability of a compactum X whose square is T 5 is closely tied with two seemingly contradictory statements, the assumption that there are more sets of countable ordinals than reals (which has been used quite successfully in analyzing the class of separable first countable normal spaces ever since Burton Jones's work Jo in 1937), and the assumption that a sufficient fragment of MA 1 holds, in particular the fragment that would ensure the separability of perfectly normal compacta (see KuTa,GN,T00 ). Finding the right form of Jones's hypothesis and the right form of MA 1 , consistent with each other yet strong enough to give enough information about T 5 spaces to solve Katetov's problem, is the difficulty which has kept the problem open for so long. It is reasonable to expect that a continuation of this work will eventually result in a much better understanding of the whole class of compact T 5 spaces. For example, the knowledge already accumulated suggests that separable --- or even c.c.c. --- spaces from this class must be closely tied to the unit interval, almost as closely tied, for instance, as the split interval 0,1 0,1 . We also hope that this work will motivate further study of fragments of forcing axioms surviving mild forcing extensions. In our case, the extension is by a Souslin tree. Equally attractive are forcing extensions by measure algebras, as these are also quite relevant to the structure of compact T 5 spaces, as shown for example in T00 , 4. We refer the reader to the works R,Lav,T96,Mooth,Hir for more on the preservation of forcing axioms under measure algebra extensions. Notation We consider 1 -trees in this paper as subsets of 1 , ordered by extension. If S is an 1 -tree and s,t S , then s t denotes that s is an initial segment of t , which corresponds to s S t in the corresponding forcing order. The length of s S , equivalently its level in S , is denoted by l(s) . If a is a finite subset of S , then inf (a) is the longest common initial segment of the members of a , and if the members of a are all pairwise compatible, then max (a) is the node given by a . The set of unordered n -tuples (finite sets) from a set X is denoted X n ( X ). For a function f and a subset a of the domain of f , f a is the function with domain a which agrees with f on all points in a . If A is a subset of a topological space X , is the closure of A in X . We use the usual forcing terminology (see, e.g., Ku ). Every object in an extension by a forcing is represented by a -name in the ground model. The class of -names is defined recursively, each name being a set of pairs (p, ) where p is a condition in and is a -name which has already been defined. The pair (p, ) being a member of a name means that the condition p forces that the realization of will be a member of the realization of . For each set x in the ground model, is the canonical name for x , i.e., the set of pairs (1 , ) for a x , where 1 is the empty condition in . The names that we refer to in this paper are all names for subsets of sets from the ground model, and as such are composed of pairs of the form (p, ) . All topological spaces are assumed to be T 3 . Souslin's axiom A coherent tree is a downward closed subtree S of 1 with the property that equation s t dom (s) dom (t) : s() t() equation is finite for all s, t S . In this paper we work with coherent Souslin trees, which are Souslin trees given by a coherent family of functions in 1 closed under finite modifications. For S a coherent Souslin tree and s,t on the same ( th) level of S , there is a canonical isomorphism S st between the cones above s and t , defined by letting S st (v)() be t() if and v() otherwise, for each v s . We let SA 1 (where SA stands for Souslin's Axiom") denote the statement that there is a coherent Souslin tree S such that for all posets with S c.c.c., and any collection D ( 1) of dense open subsets of , there is a filter G such that G D for all 1 . In our discussion of SA 1 we let S refer to such a coherent tree. To obtain a model of SA 1 , start with a coherent Souslin tree S and a cardinal such that 1 , and build a finite support iteration of length of c.c.c. posets of size 1 . This is worked out in Fa,Lar . Stronger versions of the axiom appear in Mi,Mi2,Lar,LarT,ShZ . In the following section we shall prove that under SA 1 a considerable amount of MA 1 is true after forcing with S . It turns out that along with a certain failure of MA 1 which necessarily holds after forcing with a Souslin tree, this amount is sufficient to give the metrizability of any compact space whose square is T 5 . Rectangularly refinable partitions Recall the notion of a c.c.c. partition 1 2 K 0 K 1 from T89 : the partition is said to be c.c.c. if for all sequences a 1 ( 1) either some a 2 K 0 or there is a pair 0 1 such that a 0 a 1 2 K 0 . Let 2 be the statement that every c.c.c. partition of the pairs from 1 has an uncountable homogeneous subset. The statement 2 is clearly a consequence of MA 1 , and it is well known that 2 implies many of the consequences of MA 1 . In fact, it is still unknown whether 2 implies MA 1 , and it is this question which has motivated us to investigate this subject (see T89 , 7; LarT ). In this section we show that forcing with the coherent Souslin tree S over a model of SA 1 gives 2 for a class of c.c.c. partitions on 1 2 which is especially relevant to Katetov's problem. definition A partition 1 2 K 0 K 1 satisfies the rectangle refining property if for all uncountable A, B 1 there are uncountable A' A , B' B such that , : A', B', K 0 . Equivalently, for all uncountable families , 1 of pairwise disjoint subsets of 1 there are uncountable ' , ' such that , : a, b K 0 for all a ' , b ' . definition Partitions satisfying the rectangle refining property are easily seen to be c.c.c. We let 2(rec) denote the statement that every partition of 1 2 satisfying the rectangle refining property has an uncountable homogeneous set. theorem SA 1 tworec The Souslin tree S forces 2(rec) . theorem Given an 1 -tree S and a set C 1 , let S C denote the set of s S such that l(s) C , with the inherited ordering. Note that an uncountable C 1 and a subset K of the set of pairs from S C induce an S -name K for a subset of 1 2 , letting t , K if and only if s, r K , where s and r are the predecessors of t on the th and th levels of S C respectively. Conversely, if S is a Souslin tree and is an S -name for a set of pairs from 1 , we can define an uncountable C 1 and a subset K of the pairs from S C such that K gives rise to the same set as . To define K , first note that we can find a strictly increasing function f 1 1 such that for all 1 , every node on the f() th level of S decides 1 2 . Let C f 1 , and define K S C 2 by letting s, t K if and only if s and t are incompatible or t s and s , , where l(s) f() and l(t) f() . Then K is as desired. So now let C 1 be uncountable, and let K be a subset of S C 2 such that the partition associated to K is forced to have the rectangle refining property. Let K be the set of finite K -homogeneous subsets of S C , ordered by inclusion. We would like to see that S K is c.c.c., since by SA 1 this would give an uncountable homogeneous set for K , giving in turn an S -name for an uncountable homogeneous set for K . Now if S is a coherent Souslin tree, then the induced ordering on S C also gives a coherent Souslin tree, and so to establish Theorem tworec it suffices to prove the following lemma. lemma Say that S is a coherent Souslin tree and K S 2 is such that s, t K for all incomparable s,t , and such that the partition associated to K is forced to have the rectangle refining property. Then if K is the set of finite K -homogeneous subsets of S , ordered by inclusion, then S K is c.c.c. lemma proof Let (t , a ) : 1 be an uncountable subset of S K . We wish to find a pair such that t and t are compatible in S and a a K . We may fix an integer n and assume that each a has size n . Since moving any given t further up in S makes our job harder, we may assume that for all 1 , l(t ) l(s) for all s a . Since S is coherent, the sets t s , s a , are finite, so by n applications of the -system lemma we may fix an 1 and an increasing function f : 1 1 such that, for all 1 , equation l(t ) l(s) : s a (f(), f( 1)), equation equation t s : s a (f(), f( 1). equation Note that this reduction involves throwing out those s a with l(s) . We can fix distinct nodes v 0,...,v m-1 , for some integer m n , on the th level of S such that t : 1 is a dense subset of the cone above above v 0 , and such that every member of every a is above some v i . We may also assume that the relationships among the nodes inf (b) b t a are the same for all 1 . For each 1 , let t i S v 0, v i (t ) , and let a i s a v i s . Our application of the -system lemma gives us the following density property: for any uncountable set A 1 and any s S , if t : A is dense above s , then so is equation inf ( i m v iv 0 S a i ) : A . equation For each 1 , let c i,j i m, j k i enumerate the maximal pairwise compatible subsets of a , with each c i,j a i and each c i,0 s a s t i . Note that if i m c i,0 is empty for any (all) 1 , then we are done. Let A,B 1 . We will now refine the families equation (t , a ) : A , (t , a ) : B equation successively in i m k i many steps, where in each step we replace A and B with uncountable subsets A' and B' , and we replace each remaining t ( A' ) with a t' t . Each refinement corresponds to a pair (i, j) , i m , j k i , after which for all A' , B' , if t' t' and equation S v 0v i (t' ) c i,0 c i,j equation is an S -chain, then c i,0 c i,j K . Each refinement also maintains the fact that the families equation t : A , t : B equation are both dense above v 0 . Now to do one step, fix A,B,i,j . By the density condition above, the sets c i,0 ( A ) and c i,j ( B ) define S -names , for uncountable sets of finite pairwise disjoint subsets of 1 , where each t i forces l(s) : s c i,0 into and each max (c i,j ) forces l(s) : s c i,j into . By the rectangle refining property for K , there exist names 0 , 0 for uncountable subsets of , such that a b 2 K for all a 0 , b 0 . Now let equation A' A t t i s.t. t l(s) : s i,0 0 equation and equation B' B max (c i,j ) l(s) : s i,j 0 . equation For each A' , replace t with t' v iv 0 S(t ) , where t is any extention of t i forcing l(s) : s c i,0 into 0 . By the choice of 0 and 0 , we see that A',B' and the t' 's are as desired. Having completed our refinements, we pick any A . By the density property we maintained, we may pick a B such that equation t inf ( i m v iv 0 S a i ). equation Then by the property of our refinements, t t and a a K , so and are as desired. proof The motivating question for our research in this area remains the following. question SA 1 Does S force 2 question Subspaces of first countable compact spaces The purpose of this section is to prove a result which shows that 2(rec) imposes considerable structure on compact spaces with T 5 squares. It is this result that will move the difficulty of Katetov's problem to some questions about the real line which we consider in the next section. theorem 2(rec) teefae The following are equivalent for every space X which has a first countable compactification X . enumerate (1) For every family of open subsets of X there is a countable subfamily 0 such that 0 . (2) For every subset Y of X there is a countable Y 0 Y such that 0 . enumerate theorem proof To prove the implication from (1) to (2), let Y be a given uncountable subset of X . If there is no countable Y 0 Y such that 0 , then by shrinking Y we may assume that Y comes with a wellordering w of ordertype 1 such that each yY is not in x Y x w y . For each y Y pick an open set U y such that y U y and y x Y x w y . Consider the following partition Y 2 K 0 K 1 : equation x,y K 0 iff y U x x U y. equation Since we have 2(rec) at our disposal, the proof of the implication (1) (2) is finished once we show that this partition satisfies the rectangle refining property. So let A and B be uncountable subsets of Y . By (1), there is a complete accumulation point z of B such that z B . Since z has a countable neighborhood base and since z x for all x Y with x w z , there is a neighborhood V of z such that equation A 0 x A : V x equation is uncountable. Let B 0 B V . By the assumption that z is a complete accumulation point of B , B 0 is uncountable. Then x,y K 0 for all x A 0 and y B 0 with x w y . To prove the implication from (2) to (1), let be a given collection of open subsets of X such that 0 for all countable 0 . Then we can fix a set Y X , a wellordering w of Y of ordertype 1 , and for each x Y a member U x such that x U x but y U x for all y w x . Let X be a first countable compact space containing X . We can replace U x with an open subset of X whose trace on Y is the same as the trace of our original U x , thus assuring that U x is open in X . For each x Y choose a set V x open in X such that x V x x U x , and consider Y 2 K 0 K 1 as follows: equation x,y K 0 iff x V yy V x. equation Again it remains only to check that this partition satisfies the rectangle refining property, so let A and B be uncountable subsets of Y . Let z X be a condensation point of A . Note that z y for any y Y , so, since by our assumption z has a countable neighborhood base in X , we can find an open neighborhood V of z such that equation B 0 y B V y equation is uncountable. Let A 0 V A . By the assumption that z is a complete accumulation point of A , the set A 0 is uncountable, and equation x,y K 0 for all x A 0, y B 0, x w y. equation This completes the proof. proof remark Note that in the above proof all we need to assume is that the ambient space X is first countable and Lindelof. remark Subspaces of the real line Katetov showed that if the square of a compact space X is T 5 then X is perfect , i.e., closed subsets of X are G . Note that every such X is first countable and satisfies hypothesis (1) of Theorem teefae . So under 2(rec) every compactum X whose square is T 5 satisfies (2). In particular, X is separable. We shall need the following classical facts. lemma Sn , see also Ga A compact space is metrizable if and only its diagonal is a G subset of its square. lemma proof To prove the nontrivial direction, suppose we are given a compact space X 2 whose diagonal can be written as the intersection n 0 G n of open subsets of X 2 . Since X 2 is normal, we may assume that in fact n 0 n . For each n choose f n X 2 0,1 such that f n 0 and f n X 2 G n 1 . Finally, define d X 2 0,1 by equation d(x,y) n 0 f n(x,z) - f n(y,z) : z X 2 n . equation It is easily checked that d is a metric on X which generates the original topology on X . proof lemma Jo,Hea johea If a separable, first countable normal space has an uncountable closed discrete subspace, then there is an uncountable set A of reals such that every subset of A is relatively G . lemma proof Let Z be a normal first countable space that can be written as C D , where C is a countable dense subset of Z , D is a closed discrete subset of Z and C D . For each x D , choose a subset a x of C which converges to x . It suffces to show that the almost disjoint family A a x : x D has the property that for every B A there is a b C such that for every x D , x B if and only if a x b is infinite (see Remark okrem ). Given such a B A , since Z is a normal space there is a continuous f Z 0,1 such that f(x) 0 for all xD with a x B , and f(x) 1 for all x D with a x B . Let b x C f(x) 1 2 . proof Let X be a given compact first countable space and assume that 2(rec) holds. Note that if the diagonal of X is not G , then X 2 fails to satisfy property (1) of Theorem teefae . To see this, consider the family of all sets of the form X 2 F , where F is a closed G superset of the diagonal. Applying 2(rec) to X 2 , we conclude that there is a Y X 2 such that 0 for all countable Y 0 Y . Shrinking Y , we may assume that it comes with a wellordering w of ordertype 1 such that y x Y x w y for all y Y . Applying 2(rec) to Y , via Theorem teefae we conclude that Y fails to satisfy (1). Let be a family of open subsets of Y such that 0 for all countable 0 . Then we can find uncountable Z Y such that for each x Z there is an element U x such that y U x whenever x w y . It follows that Z is a discrete subspace of X 2 . So if X 2 were T 5 the space X would be perfect by Katetov's result , and so it would satisfy the hypothesis (1), and therefore the hypothesis (2), of Theorem teefae , as 2(rec) holds. In particular, X , and therefore X 2 , would be separable. Choosing a countable dense set C in X 2 disjoint from and applying Lemma johea to the subspace C Z , we obtain an uncountable set of reals A satisfying the conclusion of the lemma. As customary, let 1 A B A G -sets G (B G A) . denote the statement that such a set of reals cannot be found, i.e., the statement that for every uncountable A there is a B A such that B G A for any G -set G . So we have established the following fact. The referee points out that the argument for the proof of Theorem sixthree is essentially the argument presented in Theorem 6.4 of GN . theorem 2(rec) 1 sixthree A compact space is metrizable if and only if its square is 5 . theorem remark okrem The proof of Lemma johea shows that if there is a separable, first countable normal space with an uncountable closed discrete subspace, then there is a family y ( 1) of pairwise almost disjoint subset of such that for every set A 1 there is a z such that equation A 1 z y is infinite . equation The existence of such a family is an equivalent form of 1 . This is the form used most often in set theory JenSolo , and it is commonly used in topology as well (see, e.g., Ta ). For example, it is essentially shown in GN that 1 implies that there is an uncountable E 0,1 such that the square of S(E) is T 5 . Here S(E) ( 0,1 0 ) (E 1 ) is considered as an ordered compactum with the lexicographical ordering. Thus 1 implies the negative answer to Katetov's question. remark Now the consistency of the positive answer to Katetov's question follows from the following fact. lemma LarT 1 holds after forcing with a Souslin tree. lemma proof Let S be a Souslin tree, and let ( 1) be S -names for an 1 -sequence of pairwise almost disjoint subsets of . Since S is c.c.c. and has height 1 , and since forcing with S doesn't add reals, we may assume that there is a club set C of levels of S such that for C , the value of is decided before the th level of S , where is the least member of C above ; in other words, every member of the th level decides the value of . There is an S -name for a subset of 1 such that for all C , " is never decided until level 1 . For instance, we can let be the set of pairs (t, ) T C such that dom (t) and t( ) 0 . Then is forced to be a subset of C , and t forces if and only if is in the domain of t and t( ) 0 . Then no real x from the ground model can code the realization of , since for each C , the statement is infinite" is decided before the statement ." Since forcing with S adds no reals, we are done. proof corollary SA 1 The Souslin tree S forces that every compact space whose square is 5 is metrizable. corollary