The independence of katetovs problem

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(1)UNIVERSIDAD DE LOS ANDES. The independence of Katětov’s Problem. by Daniel Rodriguez. A work submitted as a requirement for the degree of Mathematician in the Mathematics Department. Advisor: Ramiro Hernando de la Vega Sinisterra January 19, 2010.

(2) UNIVERSIDAD DE LOS ANDES. Abstract Mathematics Department Student of Mathematics by Daniel Rodrı́guez. During my early years as a mathematician candidate I got amazed by various things I heard about, but, particularly, the most amazing thing I ever heard was the existence of independent statements. Even though I was incapable of understanding how it was possible, my interest for a broader knowledge on that matter grew, in order to understand how those things happen. Few years later I took the basic course in Set Theory. It was there where I had my first approach to independence proofs indeed, and where I found how amazing Set Theory was, making this part of Mathematics of great interest for me. Therefore, in this work I present an approach to one independent statement that talks only about topology. The statement was first proposed by Miroslav Katětov in a classical paper [1] where he asked whether the statement was true or not. When I first saw it many questions popped out, and it became of particular interest to me. In his classical paper (1948) Katětov showed that a compact space whose cube is completely normal is metrizable [1]. Then he asked the natural question arising from this theorem: whether there could appear the square of the space instead of its cube in the hypothesis of the theorem. Before reading Katětov’s proof, the first question that I had was: the metric being a function from the space squared to the reals (satisfying some conditions), why is it important the cube and not the square for a metrization result? We expect to answer this and other questions in a certain way. Katětov’s question was open for many years, then Nyikos announced in 1977 that under Martin’s Axiom and the negation of the Continuum Hypothesis (MA+¬CH) there was a counterexample, and so the negative answer to the question resulted to be consistent with ZFC. More than a decade later Gruenhage and Nyikos published a paper [2] in which they showed the details of the first counterexample announced by Nyikos. In addition, they constructed another counterexample under the Continuum Hypothesis (CH). It follows then that there are counterexamples in the two most familiar extensions of ZFC. Thinking about this, it seemed that there must be a counterexample in ZFC or that a consistency result must be very tricky. Finally, in 2002 Larson and Todorčevic̀.

(3) ii [3] proved the consistency of a positive answer to Katětov’s question. It results then that this question is independent of ZFC, and so the question that was open for a little more than 50 years was answered. In this work, I intend to show these results in a very detailed way, so that every student having taken basic courses in Topology and Set Theory should have no problem following it. A minimal notion of forcing is assumed from the reader. In the first chapter we present the preliminary results necessary for the main part of the work. The second chapter presents Katětov’s Theorem. The counterexample under MA+¬CH is treated in the third chapter. Since it is very technical we omit the presentation of the counterexample under CH. Additionally, the consistency result is presented in the fourth chapter..

(4) Acknowledgements Foremost, I would like to thank my advisor, Professor Ramiro de la Vega, who guided me during this process, and that brought me back to the path when I was lost. Thanks to Moncheese who helped me, not only with some technical aspects, but also in the emotional ones. Camacho thanks for being a brother to me. Julia and Lombardo thanks for being always there for me and for making me the human being that I am. Marı́a more than a sibling, you are my companion, my friend, thanks for all. Betty, Ángulo, Edgar and Nicolás thanks for all those great moments that we spent together after class. I would also like to thank everyone else who made this stage in my life a very enjoyable one.. iii.

(5) Contents Abstract. i. Acknowledgements. iii. 1 Preliminaries 1.1 Notation and Conventions . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.2 Separation Axioms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.3 Martin’s Axiom and Q-sets . . . . . . . . . . . . . . . . . . . . . . . . . . 2 Katětov’s Metrization Theorem. 1 1 1 7 11. 3 Katětov’s Question Counterexample under 3.1 Alexandroff’s Double Arrow . . . . . . . . . 3.2 Modifying Alexandroff’s Double Arrow . . . 3.3 Construction of the Counterexample . . . .. MA+¬CH 14 . . . . . . . . . . . . . . . . . 14 . . . . . . . . . . . . . . . . . 17 . . . . . . . . . . . . . . . . . 18. 4 Katětov’s Question Consistency 4.1 Preliminary Definitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.2 Sufficient conditions for Katětov’s Question to have an affirmative answer 4.3 Construction of the model where Katětov’s Question has a positive answer. 30 30 33 38. Bibliography. 49. iv.

(6) Chapter 1. Preliminaries 1.1. Notation and Conventions. Throughout this paper we use standard notation, so (X, τ ) will denote a topological space and τ will denote the topology of X, if there is no ambiguity we will abuse the notation by referring to “the topological space X”, assuming that it is implicitly clear what is the topology of X. If A is a subset of a topological space its clousure will be denoted by A and its interior by int (A). If it is the case that A ⊆ Y and A ⊆ X we will specify where the interiors and clousures are taken by putting as superindex or subindex Y. of the operation the space where the operation is taken; for example A is the clousure of A in Y and intX (A) is the interior of A in X, if Y ⊆ X then the operation whitout index indicates the operation with respect to X. Along the paper it is assumed that all the spaces are regular and Hausdorff (i.e. T3 ).. 1.2. Separation Axioms. For the study of Katětov’s theorem it is essential to understand some separation axioms, the concepts of a completely normal and a perfectly normal space are the most important ones for this work. In this chapter there is a presentation of these concepts and some preliminary results. It is important to recall that if X is a topological space and A ⊆ X then U is called a neighborhood of A if it is an open subset of X and A ⊆ U . Recall that a local base for a point p is a family B of neighborhoods of p such that for every open V containing p there is an element U of B between p and V (p ∈ U ⊆ V ). We extend this definition to arbitrary subsets of X, so if A ⊂ X, a family B of neighborhoods of A is called a local base for A if for every open set U containing A there is an element V of B between A and the open set. A property is said to be hereditary for a topological 1.

(7) Chapter 1. Preliminaries. 2. space X if every subspace Y of X has the property. In addition given a point p of X a local pseudobase for p is a family of neighborhoods of p whose intersection is {p}. This definition is extended for arbitrary subsets of X in the same way as we did for T local bases (i.e. for A ⊆ X, B ⊆ τ is called a local pseudobase if B = A). It is of. interest to know how ‘big” a subset of the topology must be in order to be a local base or a local pseudobase for a given set. For that purpose, the concepts of character and pseudocharacter (as well as other cardinal functions) have been studied for many years. Definition 1.2.1. Let X be a topological space and A ⊆ X a closed subset of X. The character of A in X, denoted χX (A) and the pseudocharacter of A in X, denoted ψX (A) are defined as follows: χX (A) = min{|B| : B is local base for A}. ψX (A) = min{|B| : B is local pseudobase for A} Observe that the character is defined for every closed set in every topological space. In opposition the pseudocharacter could not exist. Although, if the space we are dealing with is T3 then the pseudocharacter is well defined for every closed set. Since every space is assumed here to be regular and Hausdorff we will be able to use these notions in order to deduce some results (for further documentation about cardinal functions see [4]). Now we define the concept of a completely normal (or T5 ) space. It is known that T is the initial of the Trennungsaxioms that in German means axioms of separation. It is expected for T5 to be stronger than the usual normality T4 , we will show this fact. Definition 1.2.2. (X, τ ) is completely normal or T5 if every two separated subsets of X can be separated by open sets , i.e. ∀A, B ⊆ X(A ∩ B = A ∩ B = ∅ → ∃U, V ∈ τ ((U ⊇ A) ∧ (V ⊇ B) ∧ (V ∩ U = ∅))). A characterization of completely normal spaces is given in the next lemma. To compute and to use the complete normality this charaterization is very useful. The following Lemmas and Theorems are to be found in a basic Topology book such as [5] or [6]. Most of them are excercises that we prove in the following lines. Lemma 1.2.3. X is hereditarily normal iff it is T5 . Proof. (⇒): Let X be hereditary normal, assume that A and B are separated, so by definition A∩B = A∩B = ∅. Consider Y = X\A ∩ B as a subspace of X. By assumption Y is normal. Since A ∩ B = ∅, then A ⊆ Y , in the same manner B ⊆ Y . In addition we Y. Y. know that in Y the clousures of A and B are disjoint (as A = A ∩ Y and B = B ∩ Y , Y. aditionally by definition of Y it follows that A ∩ B. Y. = ∅). Thus by normality of Y.

(8) Chapter 1. Preliminaries there are U and V open subsets of Y such that A. 3 Y. = A ∩ Y ⊆ U, B. Y. = B∩Y ⊂ V. and U ∩ V = ∅, since Y was considered as a subspace of X, then there are open subsets Ũ and Ṽ of X such that U = Ũ ∩ Y and V = Ṽ ∩ Y . Clearly Ũ ∩ Ṽ ∩ Y = ∅, thus Ũ ∩ Ṽ ⊆ A ∩ B and it follows that Ũ ∩ Ṽ ⊆ A ∩ B. We also know that A, B⊆ Y ; in consequence A ∩ Ũ ∩ Ṽ = ∅, and in the same way B ∩ Ũ ∩ Ṽ = ∅. Notice that A ⊆ Ũ and B ⊆ Ṽ , thus A ⊆ Ũ \ Ũ ∩ Ṽ and B ⊆ Ṽ \ Ũ ∩ Ṽ . Now, if we let U1 = Ũ \ Ũ ∩ Ṽ and V1 = Ṽ \ Ũ ∩ Ṽ , evidently U1 and V1 are disjoint and open, and they separate A and B, then X is T5 .. (⇐): Let X be T5 and consider a subset Y of X as a subspace. Let A, B ⊆ Y be two disjoint closed subsets in Y , it follows that A ∩ Y = A and B ∩ Y = B, thus A ∩ B = A ∩ B ∩ Y = (A ∩ Y ) ∩ B = A ∩ B = ∅. In the same way A ∩ B = ∅. Since X is T5 there are U and V , open in X, that separate A and B. Let U1 = U ∩ Y , V1 = V ∩ Y , then they are open in Y and clearly separate A and B, so Y is normal and consequently X is hereditarily normal. Corollary 1.2.4. If X is T5 then it is normal. Now we introduce a stronger version of normality called perfect normality. Perfectly normal spaces will result to be completely normal. Thus the property of being perfectly normal is designated T6 . Definition 1.2.5. X is said to be perfectly normal or T6 if every two disjoint closed subsets of X can be precisely separated by a function, that is, given A and B two closed disjoint subsets of X there is a continuous function f : X → R such that f −1 ({0}) = A and f −1 ({1}) = B. We give a characterization of perfectly normal spaces. Intuitively this is easier to verify than the original definition of perfect normality. Lemma 1.2.6. X is T6 iff it is normal and every closed subset of X is a Gδ . Proof. (⇒): Let A ⊆ X be closed (if A = X then clearly A is Gδ ), then assume that there is a b ∈ X \ A. Since we assume X to be T2 then {b} is closed, it is also disjoint from A, thus by assumption there is an f : X → R continuous s.t. f −1 ({0}) = A and f −1 ({1}) = {b}. So {0} is clearly Gδ in R, then A is also a Gδ . It is clear that perfectly normal implies normal. (⇐): We first prove the following preliminary result: If X is normal, A is Gδ and closed iff there exists f : X → [0, 1] continuous such.

(9) Chapter 1. Preliminaries. 4. that f −1 ({0}) = A, we give a short proof for that: T If A is a closed Gδ , then A = n∈ω An for some An , open subsets of X, thus X \ A = S S n∈ω Fn with Fn closed, additionally n∈ω X \ An . Now, let X \ An = Fn then X \ A = ∀n ∈ ω, Fn ∩ A = ∅, so, by normality of X and Urysohn’s Lemma, for each n ∈ ω. there is a continuous fn : X → [0, 1] such that fn (Fn ) = {1} and fn (A) = {0}. Let P 1 f = n∈ω n+1 fn , since f is the limit of a uniformly convergent sequence of continuous 2 functions, f is contiuous. Furthermore if x ∈ A, then clearly f (x) = 0 and if x ∈ / A then x ∈ Fn for some n ∈ ω, hence f (x) 6= 0. Now if such f exists then A is the preimage of 0, since f is continuous and {0} is Gδ then A is also a Gδ .. Now Let A and B be two closed disjoint sets. By our preliminary result there are continuous functions f and g from X into [0, 1] such that A = f −1 ({0}) and B = f (x) g −1 ({0}). Let h(x) = , thus h is a function over X into [0, 1], it is well f (x) + g(x) defined because A and B are disjoint, so f and g are never both 0 at the same point. It is easy to see that h precisely separates A and B. Note that if X is perfectly normal then every open set is Fσ . As we said before T6 spaces are also T5 . We show this in the following lemma. Lemma 1.2.7. If X is T6 then it is T5 . Proof. Let X be T6 and let A and B be two separated subsets of X, since X is T6 , then there are continuous f, g : X → [0, 1] such that f −1 ({0}) = A and g −1 ({0}) = B Now define h : X → [−1, 1] by h(x) = f (x) − g(x), then h is continuous and if x ∈ A then x∈ / B so h(x) < 0, then A ⊆ h−1 ([−1, 0)) and similary B ⊆ h−1 ((0, 1]). Clearly this is a separation of A and B, and we can conclude that X is T5 . Metric spaces satisfy various separation axioms; for example every metric space is normal (the strongest axiom of separation that is studied in a basic course of topology). And since a metric gives us more flexibility when separating sets with neighborhoods, it is expected that metric spaces satisfy stronger versions of normality. Actually they do. We prove this result in our next lemma. Lemma 1.2.8. If X is metrizable then it is T6 ..

(10) Chapter 1. Preliminaries. 5. Proof. Let X be metrizable and let d be a metric for X. Let A be a closed set in X then A = {x ∈ X : d(A, x) = 0} 1 . Consequently A = f −1 (0) where f : X → R is defined by f (x) = d(A, x). Evidently f is continuous and {0} is Gδ , therefore A is also a Gδ . Additionally it is well known that every metric space is normal. Therefore we can conclude that X is T6 .. Theorem 1.2.9. Let A and B be subsets of a space X, and assume that there are countable open covers U and V of A and B respectively, such that for every U ∈ U, U does not meet B and viceversa (for every V ∈ V, V does not meet A). Then A and B can be separated by open sets. Proof. Let A and B as above. Since U and V are countable we can assume that U = {Un |n ∈ ω}, V = {Vn |n ∈ ω} such that Un ⊆ Um and Vn ⊆ Vm whenever n ≤ m. Let S S S S Ũn = Un \ i≤n Vi and Ṽn = Vn \ i≤n Ui . Then WA = n∈ω Ũn and WB = n∈ω Ṽn are open sets that separate A and B:. For every n ∈ ω, Ũn is clearly open (since Un is open and. S. i≤n Vi. is closed). The same. argument shows that Ṽn is open also. Thus WA and WB are open. If a ∈ A then if follows that there is n ∈ ω such that a ∈ Un (since U is an open cover of A). Besides, by hypothesis we have that for every i ∈ ω, a ∈ / Vi , thus a ∈ Ũn . It follows then that A ⊆ WA , similarly B ⊆ WB . We will finish our proof when we show that WA ∩ WB = ∅. To reach a contradiction, suppose that WA ∩ WB 6= ∅; therefore there are n and m in ω such that Ũn ∩ Ṽm 6= ∅. Without loss of generality suppose n ≤ m, then Ṽm ∩ U n = ∅ and additionally U n ⊇ Un ⊇ Ũn , it follows that Ṽm ∩ Ũn = ∅, a contradiction. When proving that two sets can be separated by open sets, this last theorem is very useful, since it shows that it is enough to find a countable open cover for each set such that the clousures of the cover’s elements do not meet the other set. Now we present a theorem very useful to prove that a given space is perfectly normal. It somehow links enumerability axioms with separation ones. Recall that a space is said to be Lindelöf if every open cover of the space has a countable subcover. Theorem 1.2.10. Let X be a space that is T3 and hereditarily Lindelöf, then it is T6 . Proof. Let X be T3 and hereditarily Lindelöf we will use Lemma 1.2.6 in order to prove that X is T6 . Consequently, we will first show that X is in fact T4 and then that every closed subset is a Gδ . Let A and B be two closed disjoint sets in X. We will use the last Theorem (1.2.9) in 1. Recall that if a B ⊆ X then d(B, x) = inf{d(b, x) : b ∈ B}, it is easy to see that if A ⊂ X is closed then A = {x ∈ X : d(A, x) = 0}..

(11) Chapter 1. Preliminaries. 6. order to prove that A and B can be separated by open sets. Thus it is enough to find open covers U and V of A and B respectively such that they are countable and for every U ∈ U its clousure misses B and viceversa. For every a ∈ A there is an open neighborhood Ua of a that does not intersect B (since a∈ / B), additionally X is T3 , thus we can choose Ua such that U a ∩ B = ∅, it follows that Ũ = {Ua |a ∈ A} is an open cover of A, since X is Lindelöf and A is closed, there is U countable subcover of Ũ , by construction every neighborhood in U has a clousure that misses B. Similarly, for B there is an open countable cover V, such that every element in V has a clousure that misses A, so by Theorem 1.2.9, A and B can be separated by open sets, and thus X is normal (note that to prove that X is normal we only use that it is T3 and Lindelöf). Let A be closed in X, then for every b ∈ X \ A there is an open set Ub such that b ∈ Ub and U b misses A (due to the fact that X is T3 ). Consider now U = {Ub |b ∈ X \ A}, as it is an open cover of X \ A and X is hereditarily Lindelöf it follows that there is A ⊆ U that is countable and still covers X \ A. Remember that for every U ∈ A we T have U ∩ A = ∅, so X \ U ⊇ A, put VU = X \ U . Now we verify that A = U ∈A VU. and this will bring our proof to an end. Since for all U ∈ A we have VU ⊇ A, then T A ⊆ U ∈A VU . If b ∈ X \ A then there is a U ∈ A such that b ∈ U , so b ∈ U . It follows T T then that b ∈ / VU and thus b ∈ / U ∈A VU , thus A = U ∈A VU . The following theorem links also separation axioms with enumerability ones. We use now separation properties to get enumerability ones. Theorem 1.2.11. If X is a Lindelöf perfectly normal space then it is hereditarily Lindelöf. Proof. It is enough to show that for every family of open sets U there is a countableU0 ⊆ S S S S S U such that U = U0 . Since X is T6 then U is a Fσ so U = n∈ω Cn where every Cn is closed (in X), and thus Lindelöf. Additionally U is an open cover for every Cn ,. so for each n ∈ ω there is Un ⊆ U at most countable that stills covers Cn . Therefore, S S S U0 = n∈ω Un it is at most countable, and obviously U = U0 . It follows then that X is hereditarily Lindelöf.. Theorem 1.2.12. If X is a second countable space and Y a hereditarily Lindelöf one then X × Y is hereditarily Lindelöf. Proof. Let X and Y as above. Let B = {Bn |n ∈ ω} be a countable base for X. Consider A ⊆ X × Y and let U = {Ba × Uα |α ∈ κ} a family of basic open sets that covers A. Let S Un = {B × U ∈ U |B = Bn }. Thus U = n∈ω Un . But since Y is hereditarily Lindelöf if. Vn = {U |Bn × U ∈ Un } then it has a countable subcover, say Ṽ , so {Bn × U |U ∈ Ṽ } is a countable subcover of Un , what finish the proof..

(12) Chapter 1. Preliminaries. 7. Further documentation about separation and enumerability axioms can be found in [5] and [6]. These are the most important results about separation axioms. We will use them in order to proof Katětov’s metrization theorem and in the construction of the counterexample under MA+¬CH. In order to construct this counterexample we will need Martin’s Axiom, so we now continue with a discussion of this Axiom.. 1.3. Martin’s Axiom and Q-sets. When working in consistency proofs it is common to use additional axioms to the standard ZFC axioms to derive results. A very powerful axiom is Martin’s Axiom, MA, it is often used to force the cardinals strictly between ω and 2ω to have similar properties as ω. MA is related to partial orders (recall that a partial order is a pair (P, ≤) where P is a set and ≤ a binary relation on P that is transitive and reflexive). We will often abuse notation by referring to the “partial order P” instead of referring to the “partial order (P, ≤)” . If p and q are in P and p ≤ q it is said that p extends q. If P is a partial order, then p and q elements of P are said to be compatible if there exists r ∈ P such that r extends both p and q, and they are incompatible (p ⊥ q) just if they are not compatible. An antichain in P is a set A ⊆ P such that ∀p, q ∈ A(p 6= q → p ⊥ q). P is said to have the countable chain condition (or is c.c.c.) if every antichain in P is at most countable. A subset D of P is dense in P if ∀p ∈ P(∃q ∈ D(q ≤ p)). G ⊆ P is a filter in P if the next two conditions are satisfied: 1. ∀p, q ∈ G∃r ∈ G(r ≤ p, r ≤ q), i.e. the elements in G are pairwise compatible, and G has a witness of that. 2. ∀p ∈ G∀q ∈ P(p ≤ q → q ∈ G). Now, we are ready to say what MA is. Definition 1.3.1. MA(κ) is the assertion: If P is a non-empty c.c.c. partial order, and D is a family of at most κ dense sets in P, then there exists a filter in P that intersects every element of D. MA is the statment: ∀κ < 2ω (M A(κ)). The construction of the counterexample to Katětov’s question under MA+¬CH lies essentially in the existence of a subset of reals having a very particular property, such.

(13) Chapter 1. Preliminaries. 8. sets are called Q-sets. We now introduce the notion of Q-set. In the following lines we use partial functions, therefore it is important to recall that F n(A, B) is the set of all finite partial functions from A to B. Definition 1.3.2. A Hausdorff space X is called a Q-set space if every Y ⊂ X is a Gδ in X. We use MA to construct Q-sets that later will be very important in order to construct a counterexample to Katětov’s question.The next theorem states that under MA every Hausdorff space of size less than the continuum that has a countable base is a Q-set space. M.E. Rudin gave a proof for a special case of this Theorem in [7]. This fact is well known, though I could not found a paper containing a proof of it. Rudin’s proof uses some technical consequences of Martin’s Axiom. In opposition to Rudin’s proof, the following proof uses just the definition of MA. It is a result of an idea that Prof. De la Vega gave to me. Theorem 1.3.3. Let ω ≤ κ. MA(κ) implies that every second-countable Hausdorff space (X, τ ) of cardinality at most κ is a Q-set. Proof. : Let B be a countable base for X and let Y ⊂ X. We will show that Y is a Gδ in X. For this, let P = {hp, Ai : p ∈ F n(ω × ω, B) and A ∈ [X \ Y ]<ω } ordered by hp, Ai ≤ hq, Bi iff: 1. p ⊇ q and A ⊇ B. 2. ∀(i, j) ∈ dom(p) \ dom(q)(p(i, j) ∩ B = ∅). First observe that P is clearly reflexive. If hp, Ai ≤ hq, Bi ≤ hr, Ci then obviously A ⊇ C and p ⊇ r, and if (i, n) ∈ dom(p) \ dom(r), we have two cases: either (i, n) ∈ dom(q) or (i, n) ∈ / dom(q), in the last case since hp, Ai ≤ hq, Bi, then we have p(i, n) ∩ B = ∅, therefore p(i, n) ∩ C = ∅; in the first case since p ⊇ q we have that p(i, n) = q(i, n), therefore since hq, Bi ≤ hr, Ci, then p(i, n) ∩ C = q(i, n) ∩ C = ∅. In both cases we see that the second condition of the relation is also satisfied, then hp, Ai ≤ hr, Ci and we can conclude that P is a partial order. In order to use Martin’s Axiom we need also to see that P is c.c.c.. Note that ∀p ∈ F n(ω × ω, B) and ∀A, C ⊂ X \ Y , we have that hp, Ai and hp, Ci are compatible: it is obvious that A ⊂ A ∪ C, p ⊆ p, since dom(p) \ dom(p) = ∅, then we have that hp, A ∪ Ci ≤ hp, Ai, similarly hp, A ∪ Ci ≤ hp, Ci..

(14) Chapter 1. Preliminaries. 9. Consider A = {hpα , Aα i : α ∈ ω1 }, a subset of P, since |F n(ω × ω, B)| = ω there are α and β in ω1 such that α 6= β, pα = pβ , therefore hpα , Aα i and hpβ , Aβ i are compatible, in consequence A cannot be an antichain, then every antichain in P is at most countable and so P is c.c.c.. The idea is to construct a total function g from ω × ω to B, that gives us the (countable) family of open sets, whose intersection is precisely Y . To do this we force g to satisfy T S for every n ∈ ω, that Un = i∈ω g(n, i) ⊇ Y and that n∈ω Un = Y . So, consider the following collections of sets: • For y ∈ Y and i ∈ ω, we let Dyi = {hp, Ai : (∃n ∈ ω)((i, n) ∈ dom(p)∧p(i, n) 3 y)}. • For a ∈ X \ Y , we put Da = {hp, Ai : a ∈ A}. • For (i, j) ∈ ω × ω, we let Cij = {hp, Ai : (i, j) ∈ dom(p)}. We prove that every set of every collection described above is dense:. • First we prove that every set on the first collection is dense. Let y ∈ Y , i ∈ ω be arbitrary and let hp, Ai ∈ P. Suppose that hp, Ai ∈ / Dyi , since y ∈ / A and A is finite, there is a U basic neighborhood of y that does not intersect A (we can found such U because X is Hausdorff). Since p is finite there is an n ∈ ω such that (i, n) ∈ / dom(p), therefore hp ∪ {((i, n), U )}, Ai is clearly an element of Dyi and it extends hp, Ai, for this reason Dyi is dense. • Now we prove the density for the sets of the second collection. Let a ∈ X \ Y and hp, Ai ∈ P then clearly hp, A ∪ {a}i ≤ hp, Ai, thus Da is dense. • Let us prove the density of the sets in the last collection. Let (i, n) ∈ ω × ω and hp, Ai ∈ P. Suppose that hp, Ai ∈ / Cij , recall that A is finite, then there exists U ∈ B that does not intersect A (just let x ∈ / A, and as in the first case, there exists a neighborhood U ∈ B of x that does not intersect A), therefore hp ∪ {((i, j), U )}, Ai is in Cij and extends hp, Ai, and so Cij is also dense for every (i, j) ∈ ω × ω. Now consider D = {Dyi : y ∈ Y ∧ i ∈ ω} ∪ {Da : a ∈ X \ Y } ∪ {Cij : (i, j) ∈ ω × ω}. D is a collection of dense sets and since |X| ≤ κ, D has size at most κ, then by M A(κ) there is G a filter in P that intersects every element of D. Notice that since every two elements in G are S compatible, then g = {p : ∃A such that hp, Ai ∈ G} is a function. Since G intersects.

(15) Chapter 1. Preliminaries. 10. every Cij , g is a total function from ω × ω to B. Put Ui =. S. n∈ω. g(i, n). Our proof will. be complete when we show that these open sets are the witnesses for Y to be Gδ (i.e. T i∈ω Ui = Y ). So let us prove that the Ui satisfy the following properties: 1. ∀i ∈ ω we have Ui ⊇ Y : Fix i ∈ ω and let y ∈ Y then since G intersects Dyi then there is hp, Ai ∈ G such that there is n ∈ ω with p(i, n) 3 y. Then g(i, n) 3 y and so y ∈ Ui , thus Ui ⊇ Y . 2. ∀a ∈ X \ Y there is a j ∈ ω such that a ∈ / Uj : Fix a ∈ X \ Y then G intersects Da and so there is hp, Ai ∈ G with a ∈ A. Now there is j ∈ ω such that ∀n ∈ ω(j, n) ∈ / dom(p) (if such j did not exist then p would not be finite). Fix such j and let n be in ω, then there is hq, Bi ∈ G such that (j, n) ∈ dom(q), since G is a filter then there is hr, Ci ∈ G that extends both hp, Ai and hq, Bi. Since (j, n) ∈ dom(r) and (j, n) ∈ / dom(p), A ∩ r(j, n) = ∅, therefore a∈ / r(j, n), and in consequence a ∈ / g(j, n), since n was arbitrary, it follows that S a∈ / n∈ω g(j, n) = Uj . It follows that Y =. T. i∈ω. Ui . Note that every Ui is open because it is a union of open. sets, for this reason Y is a Gδ . Since Y was arbitrary, then we conclude that X is a Q-set..

(16) Chapter 2. Katětov’s Metrization Theorem We begin this chapter with some preliminary results that are used in the main metrization theorem. We will give a little modified version of Katětov’s original proof [1].. Our first result uses the notion of pseudocharacter that we defined in Chapter 1. The theorem is due to Katětov [1] and is very general. Therefore we will use only a particular case of it in order to prove the main metrization theorem. We follow a part of Katětov’s original proof, but simplify it using some theorems proved in the previous chapter. Theorem 2.0.1. Let κ be an infinite cardinal, X and Y topological spaces s.t. X × Y is T5 , then either every A ⊆ Y of cardinality at most κ is closed, or every closed set in X has pseudocharacter at most κ. Proof. Let X, Y and κ be as above. Assume that there is a non-closed A ⊆ Y s.t. |A| ≤ κ, and let C be a closed subset of X. We must see that the pseudocharacter of C is at most κ. Define D = C×A, fix a ∈ A\A and put E = X\C×{a}, then D = C×A = C×A ⊆ C×Y and E = X \ C × {a} ⊆ X × {a}. Therefore E ∩ D = E ∩ D = ∅ and since X × Y is completely normal we can separate E and D by open sets V and U , so that D ⊆ U and U ∩ E = ∅. For every y ∈ A let Uy = {x ∈ X : hx, yi ∈ U }. Since f : X × {y} → X given by hx, yi 7→ x is a homeomorphism, for every y ∈ A, Uy is an open subset of X. Observe also that for every y ∈ A we have Uy ⊇ C. T Now, consider B = {Uy : y ∈ A}, we show that B = C: T Suppose that there is z ∈ B \ C, observe that hz, ai ∈ U (to see this let H × W be. a basic neighborhood of hz, ai, by assumption a ∈ A and W is a neighborhood of a so there is y ∈ A ∩ W and since z ∈ Uy then we have hz, yi ∈ U and hz, yi ∈ H × W , so. H × W ∩ U 6= ∅, as wanted). Now, hz, ai ∈ X \ C × {a} = E so hz, ai ∈ E ∩ U and this 11.

(17) Chapter 2. Katětov’s Metrization Theorem is a contradiction. Thus,. T. 12. B = C.. Now |B| ≤ κ because |A| ≤ κ, for this reason B is a local pseudobase for C of size at most κ. We conclude that ψX (C) ≤ κ as we wanted. The particular case of the theorem above we will use in the proof of Katětov’s metrization theorem is the following one. It is is also due to Katětov and is proved in [1]. Corollary 2.0.2. If X and Y are such that X × Y is T5 , then every A ⊆ Y of size at most ω is closed, or X is T6 . Proof. X is normal because it is a subspace of X × Y and the last one is hereditarily normal. Now, by Theorem 2.0.1 letting κ = ω, either every A ⊆ Y of size at most ω is closed or every closed subset of X is Gδ and since X is normal this is equivalent to saying that either every A ⊆ Y of size at most ω is closed or X is T6 . The next lemma is the last result we need in order to prove Katětov’s metrization theorem. It is a classical well-known metrization theorem which states that for compact spaces it is sufficent to look at the diagonal as a subspace of X 2 in order to prove metrizability. Recall that the diagonal ∆ of a space X is ∆ = {hx, xi : x ∈ X}. Lemma 2.0.3. If X is a compact space whose diagonal ∆ is a Gδ then X is metrizable. Proof. Let ∆ =. T. n∈ω. An where each An is open in X 2 . We will use Urysohn’s metriza-. tion theorem in order to conclude the proof. Therefore we will show that X has a countable base. Let n be fixed, since An is open and ∆ ⊆ An , then for every x ∈ X we have that there is a basic neighborhood Ux × Vx of hx, xi that is contained in An . Let Hx = Ux ∩ Vx , obviously (Hx )2 ⊆ An and by regularity of X we can find Wx neighborhood of x such that Wx ⊆ Hx , therefore (Wx )2 ⊆ An . Now, consider Bn = {(Wx )2 : x ∈ X}, then Bn is an open cover of ∆ and by compactness of ∆ (it is compact since it is closed S and X 2 is compact) Bn has a finite subcover, say B̃n . Let D = n∈ω B̃n . Observe. that D consists of open squares B 2 , note also that for every x ∈ X we have that T {B|B 2 ∈ D and x ∈ B} = {x} (suppose that there is a y ∈ X such that y 6= x and for every B 2 ∈ D, such that x ∈ B, y is also in B. Let n ∈ ω, then there is a B 2 ∈ B̃n. such that x ∈ B, in consequence y ∈ B. Thus hx, yi ∈ B 2 , therefore hx, yi ∈ An . T Since n was arbitrary hx, yi ∈ n∈ω An and this is a contradiction). Additionally D is countable, since it is the countable union of finite sets. Put S = {B|B 2 ∈ D}, thus S. is countable. We claim that E = {U : U is a finite intersection of elements of S} is a base for X, note that E is also countable (because |E| ≤ | [S]<ω | = |S|). Let x ∈ X and.

(18) Chapter 2. Katětov’s Metrization Theorem. 13. let U be an open set such that x ∈ U , then X \ U is closed and hence compact. Now, for every y ∈ X \ U there is a Vy ∈ S such that x ∈ Vy and y ∈ / Vy , then y ∈ X \ Vy and {X \ Vy : y ∈ X \ U } is an open cover of X \ U . It follows that this set has a finite subcover, X \ Vy1 , . . . , X \ Vyn , then X \ Vy1 ∩, . . . , ∩Vyn ⊇ X \ U , consequently Tn Tn i=1 Vyi ⊆ U and then V = i=1 Vyi ⊆ U . Additionally, since for every y we had that x ∈ Vy , then x ∈ V ⊆ U and, by definition of E, we have that V ∈ E, so E is a base for. X. Thus X is second-countable and, since X is assumed to be regular, by Urysohn’s Theorem X is metrizable. Actually the last lemma is an equivalence, since every metrizable space is T6 and so the diagonal is a Gδ . We are now able to prove Katětov’s metrization theorem.. Katětov’s Metrization Theorem If X is compact then X 3 is T5 iff X is metrizable. Proof. If X is metrizable then X 3 is also metrizable, then it is completely normal (since every metrizable space is T6 , as it is shown in Lemma 1.2.8). So let us consider the other direction. Suppose that X is compact and X 3 is T5 , if X is finite, it is discrete, hence metrizable. So let us consider the case that X is infinite. Since X × (X × X) is T5 , corollary 2.0.2 tells us that either every subset of X of size at most ω is closed or X × X is T6 , but the first condition is imposible (since X is infinite then it has to have a subset A of size ω and by compactness of X then A will have a limit point a. Consider B = A \ {a}, clearly |B| = ω but B is not closed), therefore X 2 has to be perfectly normal. Now, X is T2 (since X 3 is Hausdorff), so the diagonal ∆ is closed in X 2 and since X 2 is T6 then ∆ is a Gδ , and by the previous lemma X is metrizable. Observe that the only consequence we use of the complete normality of X 3 is the property that this plus compactness implies that X 2 is perfectly normal. Now, a counterexample to Katětov’s question has to be a compact non-metrizable X such that X 2 is T5 , consequently X has to be at least T6 . Such kind of topological space is not easy to imagine. One example of a T6 non-metrizable space is Alexandorff’s double arrow D (see [5]). Nevertheless D2 is not T5 , so it cannot be a counterexample to Katětov’s question. We will later present a discussion in depth of this idea..

(19) Chapter 3. Katětov’s Question Counterexample under MA+¬CH In the present chapter we confine our attention to the counterexample to Katětov’s question that Nyikos constructed under MA+¬CH. It is not a surprise that this counterexample is alike to Alexandroff’s Double Arrow (the only well known example of a T6 non-metrizable compact space). Firstly, we look at this space in order to get more familiarized with the construction.. 3.1. Alexandroff ’s Double Arrow. Alexandroff’s double arrow D is also called the “split interval” because it consists of just two disjoint copies of the interval. One copy is declared to be the + copy and the other one the - copy, since D = [0, 1]+ ∪ [0, 1]− we denote the elements of D by x+ or x− (with x ∈ [0, 1]) being clear which copy of the interval it comes from. If we do not want to specify the copy where the element comes from we will write x∗ . Additionally there is the natural projection π : D → [0, 1] given by π(x∗ ) = x for all x ∈ [0, 1]. The order in D is the following: if a = x∗ and b = y ∗ then a < b if and only if x < y or, x = y and a = x− , b = x+ . It is easily seen that this is a total order. We consider D with the order topology, and show that it is compact and T6 .(see Figure below, 3.1).. 14.

(20) Chapter 3. Katětov’s Question Counterexample under MA+¬CH. 15. Figure 3.1: Alexandroff’s Double Arrow. Theorem 3.1.1. D is compact. Proof. It is well known that if an ordered space has the supremum property1 then it has to be compact with the order topology. Consequently it will be enough just to show that D has the supremum property: Let A ⊆ D and let c = sup π (A) (this supremum is taken in the usual order of the interval). If c ∈ π(A), then let d = max π −1 ({c}) ∩ A clearly this is going to be the supremum of A. If c ∈ / π(A) then c− is the supremum of A. In order to prove that D is perfectly normal we will first show that Rl the Sorgenfrey line is hereditarily Lindelöf. Additionally, D is the union of two copies of the interval seen as a subspace of the Sorgenfray line. Remember that in the first chapter we proved that a regular and hereditarily Lindelöf space is T6 . Since D is an ordered space, then it is regular, and since the union of two hereditarily Lindelöf spaces is hereditarily Lindelöf, it then follows that D is perfectly normal. We present these ideas in the following results. Theorem 3.1.2. [0, 1]+ and [0, 1]− are homeomorphic to the interval seen as a subspace of the Sorgenfrey line. Proof. Observe that (x− , y ∗ ) = [x+ , y ∗ ) (see Figure 3.1). Note also that [x+ , y ∗ ) ∩ [0, 1]+ = [x+ , y + ) ∩ [0, 1]+ . It follows that π [0, 1]+ is a homeomorphism. Similarly we have that (x∗ , y + ) = (x∗ , y − ], thus (x∗ , y + ) ∩ [0, 1]− = (x− , y − ] ∩ [0, 1]− . Hence the assingment x− → 1−x makes [0, 1]− homeomorphic to [0, 1]l , the Sorgenfrey interval. Now we prove that the Sorgenfrey line Rl is hereditarily Lindelöf. Theorem 3.1.3. Rl is hereditarily Lindelöf. 1. Here the supremum property is taken as the property that every subset has a least upper bound..

(21) Chapter 3. Katětov’s Question Counterexample under MA+¬CH. 16. Proof. Let X ⊆ Rl and let U = {[aα , bα )|α ∈ κ} be a family of basic open sets covering X. Consider U1 = {(aα , bα )|α ∈ κ}. Since R is hereditarily Lindelöf with the usual S S topology there is U2 ⊆ U1 at most countable, such that U1 = U2 . Now, we just have S to cover B = X \ U1 with countable elements of U , so we will finish our proof when we prove that B is countable. To see this let x ∈ B, so [x, ax ) is in U for some ax > x,. then for x ∈ B we can choose qx ∈ [x, ax ) ∩ Q. Now, consider x → qx , this assignment is injective. To see this, suppose that there are x, y ∈ B such that qx = qy = q. Without loss of generality suppose that x < y, we also know that q > y so y ∈ (x, q) ⊆ (x, ax ) and then y cannot be in B and this is a contradiction. Thus Rl is hereditarily Lindelöf. Based on the last two results we can conclude that D is hereditarily Lindelöf. And as we mentioned D is also T3 , and so it follows that D is perfectly normal (see Theorem 1.2.10). Recall that D is also compact, thus it is a candidate to be a counterexample to Katětov’s Question. We show that this is not the case. It is clear that D is not metrizable (since [0, 1] with the Sorgenfrey topology is a subspace). We will show that D2 is not T5 . Observe that (0, 1)+ × (0, 1)+ is homeomorphic to (0, 1)2 with the Sorgenfrey topology, and so (0, 1)+ × (0, 1)+ is homeomorphic to R2l . Hence, to prove that D2 is not T5 it is just enough to show that R2l is not normal. We show this fact in the following lines. Theorem 3.1.4. The Sorgenfrey plane, R2l , is not normal Proof. To get a contadiction suppose that R2l is normal. Since ∆o = {hx, −xi|x ∈ R} is closed in the usual topology of R2 and the Sorgenfrey plane has a finer topology, it is also closed in R2l . Note that ∆o is also discrete ([x, x + 1) × [−x, −x + 1) ∩ ∆o = {hx, −xi}) Therefore, every subset of ∆o is closed in R2l , hence if R2l is normal then for every two disjoint subsets of ∆o there are open subsets of R2l that separate them. And so for every A ⊆ ∆o we can choose UA and VA open (in R2l ) and disjoint, such that A ⊆ UA and ∆o \ A ⊆ VA , for A = ∆o choose UA = R2 and VA = ∅ (and for A = ∅ the other ω. way around). Define f : P(∆o ) → P(Q2 ) by f (A) = UA ∩ Q2 . Since |P(∆o )| = 22. and |P(Q2 )| = 2ω we will get our contradiction when we show that f is injective. Let A 6= ∅ ⊂ ∆o , therefore f (A) is not empty (due to the fact Q2 is dense in R2l ) and is not Q2 (since VA ∩ Q2 is not empty). It remains only to prove that if ∅ 6= B ⊂ ∆o , B 6= A then f (A) 6= f (B). In this case there is one point that is in one set but not in the other, without loss of generality suppose that x ∈ A \ B, then x ∈ ∆o \ B, hence x ∈ VB ∩ UA , it follows that VB ∩ UA 6= ∅, so it contains at least one point of Q2 . This point is in UA but is not in UB , and so f is injective, giving us the desired contradiction. Thus R2l cannot be normal..

(22) Chapter 3. Katětov’s Question Counterexample under MA+¬CH. 17. Even though D is not a counterexample to Katětov’s Question it is very close. D satisfies two properties that a counterexample to Katětov’s Question has to fulfill (to be compact and T6 ). Our goal is to modify in certain degree Alexandroff’s double arrow, D, to find a counterexample.. 3.2. Modifying Alexandroff ’s Double Arrow. The main idea to modify D is the following one, introduced by Nyikos and Gruenhage in [2]. For A ⊆ [0, 1] call D(A) the interval where we only split the elements of A (just as with the Alexandroff’s Double Arrow). So one copy (A+ ) is declared to be the + copy and the other copy (A− ) is declared to be the - copy. In other words, D(A) = ([0, 1] \ A)(∪ A− ∪ A+ . Let πA be the natural projection πA : D(A) → [0, 1] a, if x = a− o x = a+ given by: πA (x) = x, if x ∈ / A− and x ∈ / A+ For ease of use we adopt the following notation. Given q ∈ D(A), the image of q will be denoted by q ∗ , the inverse image of q (where q ∈ [0, 1]) will be denoted by q̂, so q̂ is a two-element set iff q ∈ A. The order in D(A) is given by: x < y if and only if πA (x) < πA (y) or x = a− and y = a+ . As we did with D we consider the order topology for D(A). It is straightforward to prove that D(A) is compact (the same argument we used when proving that D is compact works, see Theorem 3.1.1). That D(A) is perfectly normal is also straightforward (as it is the union of two subspaces of Rl and one subspace of [0, 1] with the usual topology). Note that there are three types of neighborhoods in D(A). Each a+ has a local base in −1 D(A) consisting of the open intervals In (a+ ) = πA [a, a + 1/n) \ {a− }. In the same way −1 for a− a local base is the family of open intervals In (a− ) = πA [a − 1/n, a) \ {a+ }, while −1 for p ∈ (D(A) \ Â) the intervals In (p) = πA (p − 1/n, p + 1/n) form a local base for p.. Observe that if κ = |A| and 2κ > 2ω , then D(A)2 is not T5 , this is a consecuence of a small adaptation of Theorem 3.1.4 of our own authory that we present in the following lemma. Lemma 3.2.1. If κ = |A| and 2κ > 2ω , then D(A)2 is not T5 . −1 Proof. First of all observe that D(A) is separable. To do this consider E = πA (Q ∩. [0, 1]).Observe that the projection over [0, 1] of every open set in D(A) has a non-empty interior (in the usual topology), thus E is dense in D(A). Consider B = {ha− , a+ i|a ∈ A} ⊂ D(A)2 . If ha− , a+ i ∈ B then I1 (a− ) × I1 (a+ ) ∩ B = {ha− , a+ i}, so B is discrete; it follows then that all H, K ⊆ B disjoint are actually separated. Then if D(A)2 is T5 , for every H ⊆ B we can separate H and B \ H with.

(23) Chapter 3. Katětov’s Question Counterexample under MA+¬CH. 18. open sets in D(A). Following the same idea of Theorem 3.1.4 (E plays the role of Q) we get that 2κ ≤ 2ω . Now we have seen that for D(A) to be a counterexample A has to verify the inequality 2|A| ≤ 2ω . Besides, Przymusinski [8] showed in 1973 that if Y ⊂ R2 is a Q-set then it is normal as a subspace of the Sorgenfrey Plane. Since A+ × A+ as subspace of D(A) is homeomorphic to A2 seen as a subspace of the Sorgenfrey Plane (look at Theorem 3.1.2), Przymusinski’s result suggests that we are in the correct way to find out our counterexample if we choose A such that A2 is a Q-set of cardinaltity κ such that κ < 2ω . Using Martin’s Axiom, 2κ = 2ω . Note also that if A is countable then D(A) has a countable base and then it is metrizable. Thus the counterexample has to be D(A) (A a Q-set) with ω < |A| < 2ω . We give a look to these ideas in the following section.. 3.3. Construction of the Counterexample. As we said in the previous section, the counterexample to Katětov’s Question in MA+¬CH is D(A) with A2 a Q-set and ω < |A| < 2ω . We will follow Gruenhage and Nyikos’ construction in [2], so to make the proof easier we will assume that A is dense in [0, 1], this assumption causes no problem since MA implies that every subset of R2 of cardinality strictly smaller than 2ω is a Q-set (see Theorem 1.3.2). Thus if we also assume ¬CH, there is an Ã ⊂ R of cardinality strictly between ω and 2ω . Then A = Ã ∪ (Q ∩ [0, 1]) satisfies the desired properties. Fix A with these properties from here on. Since D(A) is compact and not metrizable (|A| > ω and so {ha+ , a− i|a ∈ A} is an uncountable discrete subspace of D(A)2 , but D(A) is separable, thus D(A)2 is separable and thus it cannot be metrizable, thus D(A) is not metrizable), the only thing we have to prove in order to show that D(A) is a Katětov’s counterexample is that D(A)2 is completely normal. Before we enter in the main proof, we will prove some preliminary results. For simplicity we denote D(A) by P . We begin with some Lemmas stated and proved by Nyikos and Gruenhage in [2]. Lemma 3.3.1. P 2 \ Â2 is hereditarily Lindelöf. Proof. Clearly, P 2 \ Â2 = (P \ Â)2 ∪ [(P \ Â) × Â] ∪ [Â × (P \ Â)]. Since (P \ Â) is a subspace of [0, 1] it is second countable, thus (P \ Â)2 is hereditarily Lindelöf. Observe that (P \ Â) × Â is the disjoint union of (P \ Â) × A+ and (P \ Â) × A− , also note that since A+ and A− are homeomorphic to subspaces of Rl , then they are hereditarily Lindelöf. Theorem 1.2.12 implies now that (P \ Â)×A+ and (P \ Â)×A− are hereditarily Lindelöf, and consequently (P \ Â) × Â is hereditarily Lindelöf. Similary Â × (P \ Â) is.

(24) Chapter 3. Katětov’s Question Counterexample under MA+¬CH. 19. hereditarily Lindelöf. Then P 2 \ Â2 is the union of three hereditarily Lindelöf spaces, so it is also hereditarily Lindelöf. Remember that we proved in the first chapter (Theorem 1.2.10) that every hereditarily Lindelöf regular space is T6 , indeed. Since P is an ordered space it is T3 , thus P 2 is T3 (due to the fact that being regular is a productive property), it follows that P 2 \ Â2 is regular (remember that being T3 is a hereditary property), therefore the previous lemma implies that P 2 \ Â2 is, in fact, T6 . Corollary 3.3.2. P 2 \ Â2 is T6 . We will also use the following result, which was stated in [2] and we prove here. Lemma 3.3.3. Let A be dense in [0,1]. then (A+ )2 is dense in P 2 . Proof. Note that it is enough to prove that A+ is dense in P . To prove this let x be in P and let In (x) be a basic neighborhood. Observe that πA [In (x)] has a non empty interior. Thus there is an a in A that is also in the interior of (πA [In (x)]). It follows that a+ ∈ In (x), and so A+ is dense in P . Another lemma that we will use is the following. It is a general fact that we prove here. Lemma 3.3.4. If Y is a dense subset of a space X and U is open in Y then U ⊆ int(U ). Additionally int(U ) = U (where the interiors and closures are taken in X). Proof. Let X, Y and U be as above. Then U = Y ∩ V where V is open in X. Thus U = V (since Y = X). Then if x ∈ U we have that x ∈ V ⊆ U and so x ∈ int(U ). It follows that U ⊆ int(U ). Now, U ⊆ int(U ) implies that U ⊆ int(U ). And since U ⊇ int(U ), taking clousures we see that U ⊇ int(U ). In order to prove that P is completely normal we will use Theorem 1.2.9. Thus we will consider two separated sets H and K and find a countable cover for each one such that the closures of its elements do not intersect the other set. Now define the basic sets for hx, yi element of P 2 , let Bn (hx, yi) be In (x) × In (y).. For ease of use we will introduce the following notation. Let B be a subset of a space X, we will say that a family of open sets U is B-good if no element of U has a clousure that intersects B. In an analogous way, we say that an open set U is B-good if U ∩ B = ∅. Since we will use Przymusinski’s Result, we state it here..

(25) Chapter 3. Katětov’s Question Counterexample under MA+¬CH. 20. Theorem 3.3.5. (Przymusinski [8]) if Y ⊂ R2 is a Q-set then it is normal as a subspace of the Sorgenfrey Plane.. Now, We give a detailed proof of Nyikos and Gruenhage’s Theorem in [2], filling some gaps of that proof. Main Theorem 3.3.6. If A ⊂ [0, 1], ω < |A| < 2ω , A is dense in [0, 1], and A2 is a Q-set, then P 2 is completely normal. Proof. Let H and K be two separated sets in P 2 . We will construct a countable K-good cover of H. The construction of a H-good cover for K will be analogous. By Theorem 1.2.9 this will prove that P 2 is T5 . Let us call H o = H ∩ (P 2 \ Â2 ). Since P 2 is regular, for every h ∈ H o there is an open neighborhood Uh , such that Uh ∩ K = ∅, so Ã = {Uh |h ∈ H o } is an open cover for H o . But H o is Lindelöf (since P 2 \ Â2 is hereditarily Lindelöf, by Lemma 3.3.1) so there is a countable subcover A of Ã. Thus it is enough to construct a countable K-good cover of H \ H o . Let us put H1 = H ∩ (A+ )2 , H2 = H ∩ (A− × A+ ), H3 = H ∩ (A− )2 and H4 = H ∩ (A+ × A− ). Note that elements of Hi have the ith quadrantric “orientation”. For example if p = ha+ , b+ i is an element of H1 then Bn (p) looks like a “first quadrantic” neighborhood (see Figure 3.2) . Additionally, note that H \ H o = H1 ∪ H2 ∪ H3 ∪ H4 . Let us define in the same way K1 , K2 , K3 and K4 . Thus we need to show that for every Hi there is a countable K-good cover. We will show how to construct one for H1 , the construction for the other Hi ’s is analogous.. Since the construction of such cover is rather long, it will be convenient to divide it into several steps.. Step I. There is a K1 -good neighborhood WK1 of H1 : Theorem 3.1.2 implies that (A+ )2 is homeomorphic to A2 seen as a subspace of R2l , also we now that A2 is a Q-set, so by Przymusinski’s result (Theorem 3.3.5) (A+ )2 is normal. In addition every subset of A2 is a Gδ in the usual topology of R. Since A2 seen as subspace of the Sorgenfrey plane has a finer topology, it follows that every subset of (A+ )2 is a Gδ , so by Lemma 1.2.6 (A+ )2 is perfectly normal and particularly it is T5 . Observe that H1 and K1 are separated as subsets of (A+ )2 (because H and K are separated in P 2 ). Thus there is a V open subset of (A+ )2 containing H1 such that V ∩ K1 = ∅. We now call W = int V . By Lemmas 3.3.3 and 1.2.8 we have that V ⊆ W , so H1 ⊆ W ..

(26) Chapter 3. Katětov’s Question Counterexample under MA+¬CH. A+. 21. A+. Bn ( a- , b+ ). Bn ( a+ , b+ ). a - , b+. +. b. a+ , b +. +. b -. a-. A. a+. A-. A+. A-. Bn ( a- , b- ). Bn ( a+ , b- ) a- , b -. b-. a+ , b b-. a-. A-. a+. A+. Figure 3.2: Orientation of the neighborhoods. Additionally Lemma 1.2.8 says that W = V and so W es a K1 -good neighborhood of H1 .. Now we show that there is also a K2 ∪ K3 ∪ K4 -good countable cover for H1 . We first write H1 as a countable union of closed sets. And then we find for each of these closed sets the desired cover. Step II. Write H1 as a countable union of closed sets as follows: For each p ∈ H1 let n(p) be min{n|Bn (p) ⊆ W and Bn (p)∩K = ∅}. Let H1m = {p|n(p) = m}. Since every subset of (A+ )2 is a relative Gδ it is also true that every subset of (A+ )2 S n such that every F n is a closed set in (A+ )2 . is an Fσ in (A+ )2 . So let H1m = n∈ω Fm m S S i . Observe that each F is also closed in (A+ )2 (since it is F Now let Fn = k≤n n i≤n k.

(27) Chapter 3. Katětov’s Question Counterexample under MA+¬CH a union of finitely many closed sets). Additionally H1 = Fki ,. S. n Fn .. 22. Fix n and let x ∈ Fn ,. then there are k and i both at most n such that x ∈ thus x ∈ H1k and so n(x) ≤ n. S So we can write H1 = n Fn such that if x ∈ Fn then n(x) ≤ n and Fn is closed in (A+ )2 . We will finish our proof when we find a K-good countable cover for each Fn . We now write each Fn as a finite union of sets, so that we reduce our problem to finding countable K-good covers for smaller sets.. Step III. Reduce the problem by writting each Fn as a finite union of sets as follows: Fix Fn and let S be a finite cover of I 2 = [0, 1]2 , such that each S ∈ S is an open square (in R2 ) of side strictly less than 1/n (this can be done, because I 2 is compact). For each S S ∈ S define U (S) = {Bn (p)|p ∈ Ŝ∩Fn }∩ Ŝ, note that U (S) is open since Ŝ is open and S S every Bn (p) is also open. Since Fn ∩ Ŝ ⊆ U (S), then Fn = S∈S Fn ∩ Ŝ ⊆ S∈S U (S).. Thus we just need to show that for every S there is a countable K-good cover for Fn ∩ Ŝ.. Additionally, we construct the desired cover for Fn ∩ Ŝ with open subsets of U (S). Fix an S ∈ S from here on.. Let ∂ = F r(U (S)∗ ) ∩ S, and define two functions f and g in S as follows: f (x) = inf{y|hx, yi ∈ U (S)∗ } and g(y) = inf{x|hx, yi ∈ U (S)∗ } (define them just where the infimum can be taken). We will show that ∂ = graph(f ) ∪ graph(g), where the last clousures (and boundary) are taken in S.. Step IV. If hx, yi ∈ S then hx, yi ∈ U (S)∗ iff there is a hx̃, ỹi ∈ Fn∗ ∩ S such that x̃ ≤ x and ỹ ≤ y: We know that hx, yi ∈ U (S)∗ iff there are a and b such that ha, bi∗ = hx, yi and ha, bi ∈ U (S) iff there is hc, di ∈ Ŝ ∩ Fn and ha, bi ∈ Bn (hc, di) iff c∗ ≤ a∗ < (c∗ + 1/n) and d∗ ≤ b∗ < (d∗ + 1/n) iff c∗ ≤ a∗ , d∗ ≤ b∗ and ha∗ , b∗ i ∈ S (because S has side less than 1/n) and this is iff c∗ ≤ x, d∗ ≤ y and hx, yi ∈ S, put x̃ = c∗ and ỹ = d∗ .. Step V. f and g are decreasing: Let x1 < x2 then hx1 , yi ∈ U (S)∗ iff there is hx̃, ỹi ∈ Fn∗ ∩ S such that x̃ ≤ x1 and ỹ ≤ y (by the previous step). It follows that x̃ ≤ x2 and ỹ ≤ y, so hx2 , yi ∈ U (S)∗ , thus {y|hx1 , yi ∈ U (S)∗ } ⊆ {y|hx2 , yi ∈ U (S)∗ }, and hence f (x2 ) ≤ f (x1 ). A similar argument works for g..

(28) Chapter 3. Katětov’s Question Counterexample under MA+¬CH. 23. Step VI. ∂ = graph(f ) ∪ graph(g): (⊆): Let p ∈ ∂ and p ∈ / graph(f ), then there is U open subset of S such that p ∈ U and U ∩graph(f ) = ∅. We want to show that p ∈ graph(g), so let V be an open set containing p. Then V ∩ U is an open neighborhood of p, thus we can choose a basic open neighborhood Ṽ = (a, b) × (c, d) of p contained in V ∩ U . Since p ∈ ∂ then Ṽ intersects U (S)∗ and S \ U (S)∗ . We choose hx1 , y1 i ∈ Ṽ ∩ U (S)∗ and hx2 , y2 i ∈ Ṽ ∩ (S \ U (S)∗ ). Now, since hx1 , y1 i ∈ U (S)∗ then f (x1 ) ≤ y1 and since Ṽ ∩graph(f ) is empty, then hx1 , f (x1 )i is not in Ṽ , so f (x1 ) ≤ c. Additionally hx2 , y2 i ∈ / U (S)∗ implies that y2 ≤ f (x2 ) 2 , additionally in this case we have that hx2 , f (x2 )i ∈ / Ṽ , thus d ≤ f (x2 ). Thus f (x2 ) > f (x1 ) so Step V implies that x1 > x2 . Now, since hx1 , y1 i ∈ U (S)∗ then g(y1 ) ≤ x1 , additionally we have that hx2 , y1 i ∈ / U (S)∗3 . It follows that a < g(y1 ) ≤ x1 (if g(y1 ) ≤ a, then g(y1 ) < x2 , so there is ã < x2 such that hã, y1 i ∈ U (S)∗ and so Step IV says that hx2 , y1 i ∈ U (S)∗ and this is a contradiction), so hg(y1 ), y1 i ∈ Ṽ , thus V ∩ graph(g) is not empty. Since V was arbitrary then p ∈ graph(g). (⊇): Let p ∈ graph(f ), then for all U open set containing p we have that U intersects graph(f ). Fix U open set containing p, we want to show that U intersects U (S)∗ and S \ U (S)∗ . Since U meets graph(f ) we can choose hx, f (x)i in U , without loss of generality we can assume that U = (a, b) × (c, d), thus f (x) < d, so there is y such that f (x) ≤ y < d and hx, yi ∈ U (S)∗ so U ∩U (S)∗ is not empty. On the other hand c < f (x), so there is ỹ such that c < ỹ < f (x) and hx, ỹi ∈ / U (S)∗ (in fact every z between c and f (x) satisfies this property) so U meets S \ U (S)∗ , thus p ∈ ∂. For p ∈ graph(g) the proof is essentially the same, so we omit it.. Observe that the steps IV, V and VI imply that U (S)∗ is the part of S that is above and to the right of ∂ (observe this in Figure 3.3).. ˆ Step VII. Contruction of a countable K-good cover for Fn \ ∂: Let E = int(U (S)∗ ), so, if p ∈ E there is an open neighborhood Up of p such that Up ⊆ E ⊆ S (because I 2 is regular). Take B a countable subcover of {Up |p ∈ E} (we can do this because R2 is heredetitarily Lindelöf). Define A = {Û |U ∈ B}, then A is countable. We want to show that if V ∈ A, then V ∩ K = ∅, this will be done if we prove that V ⊆ U (S). So let V ∈ A, thus V = Û for some U ∈ B, it follows that U ⊆ S, by definition if x ∈ Û then for every W open set containing x, we have that W ∩ Û 6= ∅. 2 If f (x2 ) < y2 then there is z < y2 such that hx2 , zi ∈ U (S)∗ by Step I. we have that there is ha, bi ∈ Fn∗ ∩ S such that a ≤ x2 and a ≤ z < y2 so one more time by Step IV hx2 , y2 i ∈ U (S)∗ , but this is a contradiction, thus y2 ≤ f (x2 ). 3 If hx2 , y1 i ∈ U (S)∗ then f (x2 ) ≤ y1 < d giving us a contradiction..

(29) Chapter 3. Katětov’s Question Counterexample under MA+¬CH. 24. Figure 3.3: U (S)∗. We show that x∗ ∈ U . Let M open set containing x∗ then M̂ intersects Û so there is a ∈ Û ∩ M̂ , thus a∗ ∈ U ∩ M , and x∗ ∈ U . Thus x∗ ∈ S. It follows that x ∈ Ŝ, in addition x∗ ∈ E so it is above and to the right of ∂, thus x ∈ U (S). In consequence ˆ If x ∈ U (S) \ ∂ˆ Û ⊆ U (S) as we wanted. We now want to show that A covers U (S) \ ∂. then x∗ ∈ U (S)∗ \ ∂ thus x∗ ∈ E and so there is U ∈ B such that U 3 x∗ , it follows that ˆ x ∈ Û ∈ A. This proves that A is a countable K-good cover for Fn \ ∂.. ˆ Besides, in Step I we Observe that we have found a countable K-good cover for Fn \ ∂. found a neighborhood that separates H1 from K1 . Thus in order to complete the proof ˆ we need to find a countable K-good cover for Fn ∩ ∂. ˆ Step VIII. Covering Fn ∩ ∂: Note that not all points of ∂ˆ have to be in U (S). For example if a segment of a vertical line x = a that lies in ∂, its preimage in P 2 consists of two vertical lines, namely x = a+ and x = a− . Since every point of the line x = a− will have neighbordhoods facing to the left of this line (like the point p3 in Figure 3.4), these neighborhoods are never going to meet U (S). Thus only the line x = a+ has points in ∂ˆ ∩ U (S). Similary if y = a has a segment in ∂ only the “upward facing” part of its preimage, y = a+ , will be in U (S) (as an example, the point p2 in the Figure 3.4 lies in y = a− and thus is not in U (S)). Additionally, since the points of (A− )2 have neighborhoods facing down and to the left,.

(30) Chapter 3. Katětov’s Question Counterexample under MA+¬CH. 25. Figure 3.4: U (S). the points that are in (A− )2 ∩ ∂ˆ will never meet U (S). Hence U (S) does not meet K3 . In addition we know that as U (S) ⊂ WK1 (constructed in the first step), it misses K1 .. Observe that points of U (S) ∩ K can only be in horizontal and vertical segments contained in ∂: These points can only be in K2 , K4 or in K ∩ P 2 \ (Â)2 . Then, if there is a point x ∈ U (S) ∩ K that is not in a vertical or a horizontal line, it is a decreasing point of f and g (like points p1 and p4 in figure 3.4), so by the way its neighborhoods face it would be in Fn ⊆ H, and then H ∩ K would not be empty, a contradiction.. It is also easy to see that vertical and horizontal segments contained in ∂ correspond to discontinuites of f and g respectively. It is a fact that these segments are countable4 .. Let Q be the set of the endpoints of the segments of ∂. Define D = Q ∪ (Fn∗ ∩ ∂) and S ∗ is closed in the relative usual topology of D (this can let Fn ∩ ∂ˆ = m∈ω Cm , where Cm be done since Fn∗ ∩ ∂ is a subset of D, D ⊆ A2 and A2 is a Q-set so Fn∗ ∩ ∂ is an Fσ. 4 Every segment l of ∂ contains a segment of the form (a, b) × {c} or of the form {a} × (b, c). Consider just the segments of the first type, so for every l segment of the first type l = (a, b) × {c}, choose ql ∈ (a, b), now consider the assignment l → ql , this is one to one since f is decreasing so the segments of the first type are countable; a similar argument works for the segments of the second type..

(31) Chapter 3. Katětov’s Question Counterexample under MA+¬CH relative to A2 and so to D) . For each m, let U (S)m = clousure of any U (S)m does not meet K:. S. 26. {Bn (q)|q ∈ Cm } ∩ Ŝ. Then the. Note that this last fact will bring to an end our whole proof (since U (S)m is open then ˆ Let U = U (S)m for some m, then {U (S)m |m ∈ ω} is a countable open cover of Fn ∩ ∂). U ⊆ U (S). Thus to show that U ∩ K = ∅ it is enough to show that if p ∈ U (S) ∩ K then there is a neighborhood of p that misses U ; such p verifies that p∗ can be only in vertical and horizontal segments of ∂.. Case 1. p∗ is in a horizontal segment, y = a, but it is not the right endpoint. Let q ∗ be the left end point of this segment. Now, we prove that there cannot be a point of Fn∗ between q ∗ and p∗ (except possibly p∗ ). If there is r∗ ∈ Fn∗ between q ∗ and p∗ , observe that p = hx, a+ i (because p ∈ U (S)) so if r = hz, a+ i is in the segment and z < x and r ∈ Fn then p ∈ Bn (r) and so p ∈ U (S) (a contradiction). In particular q ∗ cannot be ∗ . Thus there is V open square containing q ∗ that does not meet in Fn∗ , thus q ∗ ∈ / Cm ∗ and does not contain p∗ (remember that C ∗ is closed in D). Since q ∗ = hq ∗ , q ∗ i is Cm m x y. the left endpoint then for hx0 , y0 i ∈ V ∩ ∂ and x0 strictly smaller than qx∗ , y0 > qy∗ ; take such hx0 , y0 i and consider R0 = {hx, yi|y < y0 } ∩ V . Additionally R0 does not intersect U ∗ (remember that the neighborhoods of the elements of Cm have a first quadrantic ∗ such that B (r)∗ ∩ R 6= ∅, then since r ∗ ∈ orientation, so if there is a point r∗ of Cm /V n 0. it has to be below or to the left of R0 in the first case we conclude that p∗ ∈ int(U (S)∗ ) in the last that hx0 , y0 i is in int(U (S)∗ ), and both are contradictions). Now look at R̂0 , ∗ between q ∗ and p∗ our goal is to extend R̂ to the right since there are no points of Cm 0. without picking points of Cm and reaching p, to do it so we present the two cases below (recall that if p = hx, a+ i is in K ∩ U (S) then there are two possibilities, one that x = b− for some b ∈ A, and the other one that x ∈ P \ Â). Subcase 1. p = hx, a+ i and x = b− for some b ∈ A. Then since there are no elements of Cm between the right side of R̂0 and p then extending R̂0 to the right up to x ≤ b− (i.e. x < b+ ) we pick p and no elements of Cm , call this extension R, thus R is open does not meet U and contains p (see Figure 3.5)..

(32) Chapter 3. Katětov’s Question Counterexample under MA+¬CH. 27. Figure 3.5: Case1/Subcase1. Figure 3.6: Case1/Subcase2. Subcase 2. p = hx, a+ i and x ∈ P \ Â since hx, a+ i ∈ / H there is a k such that Bk (hx, a+ i) misses H, in particular, it misses Cm . Remember that Bk (hx, a+ i) = Ik (x) × Ik (a+ ), so Ik (x) × {a+ } does not meet Cm . Thus we can extend R̂0 up to the right edge of Ik (x) × a+ and call these new rectangle R; since Ik (x) is open R results also open and then R contains p but no point of Cm as desired5 (see Figure 3.6) . 5. Note that we did not consider the case where p∗ lies in the segment y = t∗ with t∗ ∈ / A. This case is ∗ easily solved as follows: since t∗ ∈ / A then if q ∗ is the left endpoint then q ∗ ∈ / Cm thus consider an open ∗ take hx0 , y0 i to the left of q ∗ in V ∩ ∂ consider R0 = {hx, yi|y < square V containing q ∗ that misses Cm ∗ / A we can extend R0 up to the right endpoint (without including it, so that the y0 } ∩ V , since t ∈ ∗ remaning rectangle remains open) of the segment containing p∗ without catching any element of Cm , ˆ call the last open set R̃, then R̃ does not meet U and contains p..

(33) Chapter 3. Katětov’s Question Counterexample under MA+¬CH. 28. Figure 3.7: Case 3. Case 2. p∗ is in a vertical segment but is not the upper endpoint. Let q ∗ be the lower endpoint of this segment. In this case we can argue in the same manner as we did in the previous case, having on mind that here the horizontal direction behaves in the same manner as the vertical direction does in Case1 and viceversa. Case 3. p∗ is a right endpoint but it is not a vertical endpoint. Since p ∈ P × A+ and p∈ / K1 , then there are two subcases, p ∈ K2 or p ∈ (P \ Â) × A+ . If p ∈ K2 we can argue exactly as in case 1. If p ∈ (P \ Â) × A+ , then p = hx, a+ i with x ∈ P \ Â. Since ∗ then there is a rectangle (c, d) × (e, f ) containing p∗ that does not intersect p∗ ∈ / Cm ∗ , in adition x ∈ (c, d). Thus if we construct the R (like in case 1 ) and, let e be an Cm 0 0. element of ê, consider then just the part above y = e0 of R̂0 (i.e. R̂0 ∩ {hx, yi|y > e0 }) and then extend this resulting rectangle up to the right of (c,ˆd) then it follows that this last (open) rectangle contains p and misses U .6 (see Figure 3.7). Case 4. p∗ is an upper endpoint but it is not a right endpoint. It can be argued essentially as we did in the previous case. Case 5. p∗ is an upper endpoint as well as a right endpoint. If p is in K2 treat it like Case 1. If p ∈ K4 treat it like the case 2. Clearly p cannot be in K1 or in K3 . Thus the remaning cases treat them as follows: Let q1∗ be the left endpoint and q2∗ the lower ∗ then there are V and V that contain q ∗ and q ∗ endpoint since these are not in Cm 1 2 1 2. respectively. Take hx1 , y1 i in V1 ∩ ∂ and on the left of q1∗ and hx2 , y2 i in V2 ∩ ∂ and below 6. One more time if y = t∗ with t∗ ∈ / A argue exactly as the last subcase of this case, we omit the details..

(34) Chapter 3. Katětov’s Question Counterexample under MA+¬CH. 29. Figure 3.8: Case 4. q2∗ . Alike, define R01 = {hx, yi|y < y1 } ∩ V1 and R02 {hx, yi|x < x2 } ∩ V2 , like in the past cases they do not intersect U ∗ thus we can extend R̂01 to the right up to the right side of R̂02 without intersecting U this last extension contains p and does not intersect U . (see Figure 3.8).. With this result we see that the negative answer to Katětov’s Question is consistent with ZFC, the usual axioms of the set theory. Observe also that the only consequence we use of Martin’s Axiom is the fact there is an uncountable subset of the real line that is a Q-set. In the following chapter we will see that the positive answer to this question is also consistent with ZFC..

(35) Chapter 4. Katětov’s Question Consistency In the present chapter we will show which conditions a model has to satisfy in order to give Katětov’s Question a positive answer. Additionally following Todorčevic̀ and Larson’s construction [3], we show how to construct one.. 4.1. Preliminary Definitions. In order to understand the properties a model has to satisfy to give a positive answer to Katětov’s Question we need to introduce some notions about partitions. Definition 4.1.1. If [A]2 = K0 ∪ K1 is a partition of [A]2 then it is said to be c.c.c. for K0 (or K0 -c.c.c.) if and only if for all ω1 -sequence haδ iδ∈ω1 of distinct elements of [A]<ω , there is δ such that [aδ ]2 * K0 or there are δ1 and δ2 different such that [aδ1 ∪aδ2 ]2 ⊆ K0 . Observe that if we call P = {a ∈ [A]<ω |[a]2 ⊆ K0 } then hP, ⊇i is c.c.c. if and only if the partition induced by K0 is K0 -c.c.c. in the sense of our last definition. We continue our discussion with a property that is true under Martin’s Axiom and ¬ CH. Definition 4.1.2. K2 is the assertion: Every partition of [ω1 ]2 that is c.c.c. has an uncountable homogeneous subset (i.e. there is an uncountable A ⊂ ω1 such that [A]2 ⊂ K0 or [A]2 ⊂ K1 ). As we said, K2 is true under Martin’s Axiom assuming ω1 < 2ω , see Theorem 5.23 of [9] . But, under MA Katětov’s Question has a negative answer so we want to have a property that does not need the full power of MA to be true. It is still an open question 30.