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DISJOINT UNIONS OF TOPOLOGICAL SPACES AND CHOICE

Paul Howard, Kyriakos Keremedis, Herman Rubin, and Jean E. Rubin

October 20, 1997

Abstract. We find properties of topological spaces which are not shared by disjoint unions in the absence of some form of the of Choice.

Introduction and Terminology This is a continuation of the study of the roll the plays in general . See also [vd], [gt], [wgt], and [hkrr]. Our primary concern will be the use of the axiom of choice in proving properties of disjoint unions of topological spaces (See Definition 1, part 11.) For example, in theory with choice the disjoint of metrizable topological spaces is a metrizable topological space. The usual proof of this fact begins with the choice of metrics for the component spaces. We will show that the use of some form of choice cannot be avoided in this proof and in fact without choice the of metrizable spaces may not even be metacompact. In section 1 we show that many assertions about disjoint unions of topological spaces are equivalent to the axiom of multiple choice. Models of and corresponding independence results are described in section 2. In section 3, we study the roll the Axiom of Choice plays in the properties of disjoint unions of collectionwise Hausdorff and collectionwise normal spaces. We begin with the definitions of the symbols and terms we will be using. Definition 1. 1. A family K of of a topological space (X, T)isl.f.(locally finite) iff each point of X has a neighborhood meeting a finite number of elements of K. 2. X is paracompact iff X is T2 and every open cover U of X has a l.f.o.r. (locally finite open refinement) V.Thatis,V is a locally finite open cover of X and every member of V is included in a member of U. 3. A family K of subsets of X is p.f. (point finite) iff each of X belongs to only finitely many members of K. 4. X is metacompact iff each open cover U of X has an o.p.f.r. (open point finite refinement). 5. An open cover U = {Ui : i ∈ k} of X is shrinkable iffthereexistsanopencover V = {Vi : i ∈ k} of non-empty sets such that V i ⊆ Ui for all i ∈ k. V is also called a shrinking of U. 6. X is a PFCS space iff every p.f. open cover of X is shrinkable.

1991 Mathematics Subject Classification. 03E25, 04A25, 54D10, 54D15.

Typeset by AMS-TEX 1 2 HOWARD, KEREMEDIS, RUBIN, AND RUBIN

7. D ⊆ X is discrete if every x ∈ X has a neighborhood U such that |U ∩ D| =1. AsetC ⊆P(X)isdiscrete if ∀x ∈ X, there is a neighborhood U of x such that U ∩ A =6 ∅ for at most one element A ∈ C. (Thus, D ⊆ X is discrete iff {{d} : d ∈ D} is discrete.) 8. X is cwH (collectionwise Hausdorff)iffifX is T2 and for every closed and discrete D of X there exists a family U = {Ud : d ∈ D} of disjoint open sets such that d ∈ Ud for all d ∈ D. 9. X is cwH(B)(collectionwise Hausdorff with respect to the base B)incaseU ⊆ B, where U is defined as in 8. 10. X is cwN (collectionwise normal)ifX is T2 and for every discrete C of pairwise disjoint closed subsets of X, there exists a family U = {UA : A ∈ C} of disjoint open sets such that A ⊆ UA for all A ∈ C.(cwN(B) is defined similarly to cwH(B)in9.) { ∈ } 11. The disjoint union of theS pairwise disjoint family (Xi,Ti):i k of topological spaces is the space X = {Xi : i ∈ k} such that O is open in X iff O ∩ Xi is open in Xi for all i ∈ k. It is clear from Definition 1 that

(1) T2 +compact→ paracompact → metacompact.

None of the implications in (1) are reversible. It is provable in ZF0,(ZFmi- nus Foundation) that paracompact spaces are normal. (For example, the proof of Theorem 20.10 p. 147 given in [w] goes through in ZF0 with some minor changes.) However, this conclusion does not hold for metacompact spaces. Dieudonn´e’s Plank (Example 89, in [ss] p. 108) is an example of a metacompact, non-. Any infinite set X endowed with the discrete topology is an example of a non- compact, paracompact space. Below we give our notation for the principles we use and their precise statements: Definition 2.

1. AC: The Axiom of Choice. For every family A = {Ai : i ∈ k} of non-empty pairwise there exists a set C which consists of one and only one element from each element of A. 2. MC: The Multiple Choice Axiom: For every family A = {Ai : i ∈ k} of non- empty pairwise disjoint sets there exists a family F = {Fi : i ∈ k} of finite non-empty sets such that for every i ∈ k, Fi ⊆ Ai. 3. CMC: The Countable Multiple Choice Axiom. MC restricted to a countable . 4. ω-MC: For every family A = {Ai : i ∈ k} of non-empty pairwise disjoint sets there exists a family F = {Fi : i ∈ k} of countable non-empty sets such that for every i ∈ k, Fi ⊆ Ai. 5. CMCω: (Form 350 in [rh]) CMC restricted to countable sets. 6. MP: Every is paracompact. 7. MM: Every metric space is metacompact. 8. DUM: The disjoint union of metrizable spaces is metrizable. 9. DUP: The disjoint union of paracompact spaces is paracompact. 10. DUMET: The disjoint union of metacompact spaces is metacompact. 11. DUPFCS: The disjoint union of PFCS spaces is PFCS. DISJOINT UNIONS AND CHOICE 3

12. DUPN: The disjoint union of paracompact spaces is normal. 13. DUMN: The disjoint union of metrizable spaces is normal. 14. vDCP(ω): (van Douwen’s Choice Principle.) If A = {Ai : i ∈ ω} is a family of non-empty disjoint sets and f a function such that for each i ∈ ω,f(i)isan ordering of Ai of type Z (= the integers), then A has a choice function. 15. DUN: The disjoint union of normal spaces is normal. 16. DUcwH: The disjoint union of cwH spaces is cwH. 17. DUcwN: The disjoint union of cwN spaces is cwN. 18. DUcwNN: The disjoint union of cwN spaces is normal. All the proofs below are in ZF0 and wlog abbreviates “without loss of generality”.

1. Disjoint Unions and Multiple Choice

In this section we show that many “disjoint union” theorems are equivalent to MC. As a consequence of the following well known lemma, these theorems are equivalent to AC in ZF. Lemma 1. [j]. In ZF, AC is equivalent to MC. It is also known, [j], that in ZF0, MC does not imply AC. Theorem 1. Each of the following are equivalent to MC : (i) DUP. (ii) DUMET. (iii) DUPFCS. (iv) If X is the disjoint union of a family {Xi : i ∈ k} of compact T2 spaces, then every open cover U of X has an open refinement V such that for every i ∈ k only a finite number of elements of V meet Xi non-trivially. (v) Every open covering of a topological space X which is the disjoint union of spaces Xi, i ∈ k, each of which is the one point compactification of a discrete space, can be expressed as a well ordered union of sets Uα ⊆P(X) where each Uα is locally finite (point finite). (vi) Every open covering of a topological space can be expressed as the well ordered union of locally finite (point finite) sets. If we restrict (i) through (v) to countable families of topological spaces, then the resulting statements are equivalent to CMC.

Proof. MC→(i). Fix a family X = {Xi : i ∈ k} of pairwise disjoint paracompact spaces and let X be their disjoint union. Fix U an open cover of X. Wlog we assume that each member of U meets just one member of X .PutUi = {u ∈U: u∩Xi =6 ∅} and Ai = {V : V is a l.f.o.r. of Ui}.AsXi is paracompact, Ai =6 ∅.UseMCto obtain a family F = {Fi : i ∈ k} of finite sets satisfying Fi ⊆ Ai, i ∈ k. Then, V = ∪{∪Fi : i ∈ k} is a l.f.o.r. of U. (i)→MC. Fix A = {Ai : i ∈ k} a family of pairwise disjoint non-empty sets. Wlog { ∈ } we assume thatS each Ai is infinite. Let yi : i k be distinct sets so that for each ∈ ∈ ∗ i k, yi / A. Put the discrete topology on Ai and let Ai denote the one point compactification of Ai by adjoining the point yi to Ai. Let X be the disjoint union { ∗ ∈ } ∗ ∗ of the family Ai : i k .AsAi is paracompact, (Ai is a compact T2 space) our hypothesis implies that X is of the same kind. Let V be a l.f.o.r of the open cover U { ∗ 6 ∗ ∈ } V = U : U is a neighborhood of yi in Ai , U = Ai for some i k . Since is a l.f. 4 HOWARD, KEREMEDIS, RUBIN, AND RUBIN S open cover, it follows that Fi = {Ai\v : v ∈V∧yi ∈ v} is finite and non-empty. Thus, F = {Fi : i ∈ k} satisfies MC for A. MC↔(ii). This can be proved exactly as in MC↔(i). MC→(iii) and MC→(iv) can be proved as in MC→(i). (iii)→MC. Fix A = {Ai : i ∈ k} a family of infinite pairwise disjoint sets and let ∗ → ∗ Ai and X be as in the proof of (i) MC. Clearly, Ai is a PFCS space. Thus, by (iii), X is also a PFCS space. Let V be a shrinking of the p.f. open cover U { ∗ ∈ } ∈V ⊂ = u : u = Ai or u = Ai ,i k of X. Clearly Fi = v, v and v Ai is a finite non-empty subset of Ai. Hence, F = {Fi : i ∈ k} satisfies MC for A. (iv)→MC. Let A and X be as in (iii) →MC. Let V be a refinement of the open U { ∗\{ } ∈ ∈ } cover = Ai a : a Ai,i k of X which is guaranteed by (iv). Clearly, ∪{ ∗\ ∈V ∈ } { Fi = Ai v : v ,yi v is a non-empty finite set included in Ai and F = Fi : i ∈ k} satisfies MC for A. MC→(vi). Since MC is equivalent to the statement “Every set is the union of a well ordered family of finite sets” ([le]), it is clear that MC implies (vi). It is also clear that (vi) implies (v). (v)→MC. Let A = {Ai : i ∈ k} a family of disjoint non-empty infinite sets. Let X be as in (i)→MC and let G be the set of proper neighborhoods of yi,forsomei ∈ k. ∈ ∈ ( ∗ (If g G, then there is an i k such that gSis a neighborhood of yi and g Ai .) G { ∈ } is an open covering of X,soby(v),G = Uα : α γ ,whereeachSUα is locally (point) finite. For i ∈ k,letαi be the smallest α such that yi ∈ Uα. (Notice that yi ∈ g,forg ∈ G iff every neighborhood of yi has a non-empty intersection with g, so it does not matter whether the Uα’s are locally finite or point finite.) It ··· follows that there are only a finite numberT of sets, g1,g2, ,gn,inUαi such that ∈ ··· n yi gj,forj =1, 2 ,n. Let hi = j=1 gj,thenhi is a neighborhood of yi and ( ∗ \ hi Ai .Consequently,Ai hi is a finite non-empty subset of Ai so that it follows that {Ai \ hi : i ∈ k} is a multiple choice set for A.  Remark 1. We remark here that the space X in (i)→MC is normal (also completely normal (every subspace of it is normal)), but if MC is false it is neither paracompact ∗ → nor metacompact. Also, since Ai in the proof of (i) MC is a compact T2 space, it follows that MC is equivalent to the statement: (1) The disjoint union of compact T2 spaces is paracompact. If in (1) we replace disjoint union with product and paracompact with compact,then the resulting statement, Tychonoff’s Compactness Theorem for compact Hausdorff spaces, (2) The product of compact T2 spaces is compact. is equivalent to the Boolean Prime Ideal Theorem, ([ln]) and (2) clearly implies: (3) The product of compact T2 spaces is paracompact. We do not know if (3) implies (2). Similarly, the statements: (4) The disjoint union of compact T2 spaces is metacompact. and (5) The disjoint union of compact T2 spaces is a PFCS space. are equivalent to MC and (3) implies (6) The product of compact T2 spaces is metacompact. We do not know the relationship between (3), (6) and (7) The product of compact T2 spaces is PFCS. DISJOINT UNIONS AND CHOICE 5

(However, the product of paracompact spaces may not be normal, and therefore, may not be paracompact. See [so].) Remark 2. Working as in Theorem 2.2 from [vd] one can easily establish that DUPN implies vDCP(ω). Hence DUPN cannot be proved in ZF without AC. Since paracompact spaces are normal (without appealing to AC) it follows that DUP implies DUPN and consequently, MC implies DUPN. In the next theorem using ideas from [hkrr] we show that the converse is also true. In fact we could deduce it as corollary to Theorem 1 of [hkrr], but we won’t do it. We prefer to give a new proof and exploit some new ideas.

Theorem 2. DUPN can be added to the list of Theorem 1. Proof. In view of Remark 2, it suffices to show that DUPN implies MC. Let A = {Ai : i ∈ k} be a family of infinite pairwise disjoint sets. Let A = {a : i ∈ k} and B = {b : i ∈ k} i i S be any two disjoint sets, each of which is disjoint from A. Let Ti be the topology <ω <ω on Xi = {ai,bi}∪[Ai] ,(where[Ai] is the set of all finite subsets of Ai) which <ω <ω is generated by Bi = {Cy(ai):y ∈ [Ai] }∪{Dw(bi):w ∈ [Ai] }∪{{x} : x ∈ <ω <ω [Ai] },where,Cy(ai)={ai}∪{x ∈ [Ai] : x ⊇ y} and Dw(bi)={bi}∪{x ∈ <ω [Ai] : x ∩ w = ∅}.

Claim 1. Xi is paracompact.

ProofofClaim1. First we show that Xi is a T2 space. Fix x, y ∈ Xi. We consider the following cases. <ω (i) x, y ∈{ai,bi}. Assume that x = ai and y = bi. Then for any w ∈ [Ai] , Cw(ai)andDw(bi) are disjoint neighborhoods of x and y respectively. <ω <ω (ii) Assume x = ai and y ∈ [Ai] . Let w ∈ [Ai] be disjoint from y.Then{y} and Cw(ai) are the required disjoint neighborhoods. If x = bi then {y} and Dy(bi) are the required disjoint neighborhoods. <ω (iii) x, y ∈ [Ai] then {x}, {y} are disjoint neighborhoods of x and y respectively. Thus, Xi is a T2 space as required. To see that (Xi,Ti) is paracompact, we fix U an open cover of Xi.IfE(ai),O(bi) ∈ U are neighborhoods of ai and bi respectively, then it follows easily that <ω V = {E(ai),O(bi)}∪({{x} : x ∈ [Ai] } is a locally finite open refinement of U.

Claim 2. If Cy(ai) and Dw(bi) are disjoint neighborhoods of ai and bi,thenw∩y =6 ∅

ProofofClaim2. If y ∩w = ∅,theny is in Cy(ai)∩Dw(bi) which is a contradiction.

Let X be the disjoint union of the family {Xi : i ∈ k}.ThenX,inviewof the hypothesis, is a normal space and A and B are closed and disjoint sets in X. Hence, there exists disjoint open sets OA ⊇ A and QB ⊇ B. For every i ∈ k put ∩ ∩ EAi (ai)=OA Xi,OBi (bi)=QB Xi, Z { ∈ <ω ⊆ } i = y [Ai] : Cy(ai) EAi (ai) , W { ∈ <ω ⊆ } i = w [Ai] : Dw(bi) OBi (bi) , {| | ∈Z} Z { ∈Z | | } ni = min y : y i , ini = y i : y = ni . 6 HOWARD, KEREMEDIS, RUBIN, AND RUBIN

Z Claim 3. Either ini is finiteS or, ∗ ( ) there exists a finite Q ⊆ Wi satisfying ∀w ∈Wi,Q∩ w =6 ∅. Z ∗ Proof of Claim 3. Assume, to the contrary, that ini is infinite and ( ) fails. We shall construct inductivelyS a pairwise disjoint set of finite non-empty sets { } ⊆ W { ∈Z ⊆ w0,w1, ..., wni+1 ,wj i, such that the set P = y ini : wj y for ≤ } ∈ | |≥ allS j ni +1 is infinite. This contradiction (any y P will satisfy ni = y | {wv : v ≤ ni +1}| >ni) will establish the claim. 0 ∈W 0 ∈Z Fix w i. Since Dw (bi) is disjoint from every Cy(ai), y ini , itS follows ∈Z 0 Z {{ ∈ from Claim 2, that each y ini meets non-trivially w . Hence, ini = y Z 0 ∩ } ⊆ 0 ∧ 6 ∅} ini : w y = w : w w w = and, consequently, there exists a non-empty ⊆ 0 { ∈Z ⊆ } w w such that H1 = y ini : w y is infinite. Put w0 = w. Assume that pairwise disjoint and non-empty sets w0,w1, ..., wn have been chosen so that the { ∈Z ⊆ } ∗ set y ini : w0,w1, ..., wn y is infinite. By the negation of ( ), there exists n n w ∈Wi such that (w0 ∪ w1 ∪ ... ∪ wn) ∩ w = ∅. By Claim 2 again, there exists ⊆ n { ∈Z ⊆ } anon-emptyw w such that Hn = y ini : w0,w1, ..., wn,w y is infinite. Put wn+1 = w. This terminates the induction and the proof of the claim. Z Claim 4.S Assume ini is infinite and let l0 be the least integer for which there is ∗ ∈ W l0 |Q | Q { ∈ ∪W l0 a Q [ i] satisfying ( ). Then il0 <ω,where il0 = Q [ i] : Q satisfies (∗)}. Q Proof of Claim 4. Assume, to the contrary, that il0 is infinite. We construct inductively a finite sequence of finite,S pairwise disjoint sets w0,w1,... ,wr,forsome ∈ r r ω, with the property that W = i=0 wi has l0 and infinitely many Q elements of il0 include W . This contradiction will establish the claim. 0 ∈W 0 Q 0 Fix w i.Asw and each member of il0 meets w non-trivially, it follows ⊆ 0 { ∈Q ⊆ } that there exists a non-empty w0 w such that K0 = Q il0 : w0 Q is infinite. If |w0| = l0,thenr = 0 and the induction terminates. Assume that pairwise disjoint finite non-empty sets w0,w1, ..., wn have been cho- sen so that |w0 ∪ w1 ∪ ... ∪ wn|

Kn+1 = {Q ∈ Kn : w0,w1, ..., wn,wn+1 ⊆ Q} is infinite, terminating the induction and the proof of the claim.

By Claims 3 and 4, it follows that F = {Fi : i ∈ k} where,  S Zin if Zin is finite, F = S i i i Q Z il0 if ini is infinite. satisfies MC for A, finishing the proof of the theorem. 

Since the space Xi of Theorem 2 is also normal, cwH, and cwN, we get as a corollary to Theorem 2. DISJOINT UNIONS AND CHOICE 7

Corollary. [hkrr] The following can be added to the list of Theorem 1: (i) DUN. (ii) DUcwH. (iii) DUcwN. (iv) DUcwNN.

It is easy to see that closed subspaces of paracompact (metacompact) spaces are also paracompact (metacompact) without appealing to AC. The next result shows that CMC is needed for Fσ sets (countable unions of closed subsets).

Theorem 3. The following are equivalent: (i) CMC. (ii) Fσ subsets of paracompact spaces are paracompact. (iii) Fσ subsets of metacompact spaces are metacompact. Proof. (i)→(ii). Follow the proof of Theorem 20.12 ([w], p. 148). ∗ (ii)→(i). Fix A = {Ai : i ∈ ω} a family of disjoint non-empty sets. Note that if X is the one point compactification of X as given in (i)→MC of Theorem 1 then X ∗ is Fσ in X and thus, paracompact. Continue theS proof as in Theorem 1. (i)→(iii). Let (X, T) be metacompact, let G = {Gn : n ∈ ω} be an Fσ set in X, and let U be an open cover of G. For every u ∈Upick vu ∈ T such that u = vu ∩G. (Note that we do not need AC in order to pick the element vu. We can pick all such elements and then take their union. Clearly the union is such an element.) For every n ∈ ω put Un = {vu : u ∈U∧vu ∩ Gn =6 ∅} ∪ {X\Gn}. Clearly Un is an open cover of X.AsX is metacompact An = {W : W is an o.p.f.r. of Un}6= ∅. { ⊆ ∈ } By CMC, there is a family F S= Fn An : nS ω of finite non-empty sets. For ∈ { ∩ \ ∈ ∧ ∩ 6 ∅} each n ω,letBn = G s m

→ ∗ Theorem 1, (i) MC. Since each Ai is metrizable, it follows by DUM that X is also metrizable. Let d be a metric on X producing its topology. For every i ∈ k let ni = min{n : Ai =6 D(yi, 1/n) ∩ Ai},whereD(yi, 1/n) is the open ball of radius 1/n centered at yi. Clearly, Fi = Ai\D(yi, 1/ni) is a finite non-empty subset of Ai and F = {Fi : i ∈ k} satisfies MC for A. 

Corollary. DUM implies CMCω. Proof. The proof is similar to the second part of the proof of Theorem 4. Theorem 5. DUMN + ω-MC iff MC. Proof. (←) It is shown in Theorem 4 that MC implies DUM. The result follows because DUM implies DUMN and MC implies ω-MC. (→) Let A = {Ai : i ∈ k} be a family of pairwise disjoint non-empty sets. By ω-MC we may assume that the members of A are countably infinite. If Ai is countable so <ω is [Ai] . Let (Xi,Ti) be the corresponding topological space as defined in Theorem 2. The topology, Ti as defined in Theorem 2, clearly has a countable base. It is shown in Theorem 2 that Xi is paracompact, and, therefore, regular. Thus, it follows from [gt] Corollary 4.8, that Xi is metrizable. Using DUMN it follows that the disjoint union of the Xi’s is normal. Using the same ideas as in Theorem 2, we can construct a multiple choice function for A. 

Remark 3. In the Cohen-Pincus model M1(hω1i), see [rh] and [pi], ω-MC holds. Since MC is false in this model, it follows from Theorem 4 that DUM is also false. Thus, ω-MC does not imply DUM.

2. Examples and Models

Example 1. (A non-metrizable, non-paracompact countable disjoint union of paracompact spaces). The space L of Theorem 2.2 from [vd] is clearly a non-normal (hence, non-paracompact) countable disjoint union of paracompact spaces.

Example 2. (A metrizable non-paracompact countable disjoint union ofS para- compact spaces). Let N be the permutation model with set of atoms: A = {Qn : n ∈ ω},whereQn = {an,q : q ∈ Q}. Suppose < is the lexicographic ordering on A, i.e.

an,q

The group of permutations G, is the group of all permutations on A which are ∈ | a rational translation on Qn, i.e. if φ G,thenφ Qn(an,q)=an,q+rn for some rn ∈ Q, and supports are finite. Let d be the metric on A given by:

d(an,q,am,p)=1ifn =6 m and d(an,q,am,p)=|q − p|/(1 + |q − p|)ifn = m. Good/Tree/Watson ([wgt]) prove that (A, d) is a linearly ordered, zero dimensional, metric space, but is not paracompact (MP is false). Each Qn, being a second countable metric space, is paracompact, and the disjoint union topology on A DISJOINT UNIONS AND CHOICE 9

coincides with the metric topology which is normal. Hence, A can be considered as a normal non-paracompact disjoint union of paracompact spaces. The disjoint union of the Qn’s is also not metacompact because V = {(a, b): (∃n ∈ ω)(a, b ∈ Qn ∧ a

Thus, we have shown that N|= “Every infinite set has a countably infinite subset.” + ¬ MM + ¬ DUMN + ¬C(∞, 2).

Example 3. In the Mostowski linearly ordered model, the disjoint union of metrizable spaces is metrizable. Proof. Let (A, <) be the set of atoms in the the Mostowski linearly ordered model M.(A is countable and < is a dense linear ordering on A without first or last elements.) For notational convenience we add first and last elements −∞ and ∞ to the ordering (A, <). Let G denote the group of order automorphisms of (A, <)and for each E ⊆ A let GE = {φ ∈ G : φ fixes E pointwise}. Supports in this model are finite subsets of A.Foreachx ∈Mwe denote the minimal support of x by sup(x). The crucial lemma is Lemma 2. If (X, T) is a metrizable topological space in M, then there is a metric m on X such that (a) m is in M, (b) T is the topology induced by m,and (c) sup(m) = sup(X, T). We shall show first that it follows from Lemma 2 that DUM is true in M. Assume that {(Xi,Ti):i ∈ K} is a collection of disjoint metrizable topological spaces in M with support D.Foreachi ∈ K choose a metric mi on Xi so that mi is in M, mi induces the topology Ti and sup(mi) = sup(Xi,Ti). (We may assume wlog that each mi is bounded by one.) It follows from the last of these properties that for all φ ∈ GD, φ(mi)=mi if and only if φ(Xi,Ti)=(Xi,Ti). (This would not be true, for example, in the basic Fraenkel model where “φ fixes sup(x) pointwise” 7→ implies “φ fixes x”, but not conversely.) Therefore, the function (Xi,Ti) Smi has M support D so it is in . We can use this function to define a metric m on i∈K Xi which induces the disjoint union topology. (m(x, y)=mi(x, y)ifx, y ∈ Xi for some i ∈ K;andm(x, y)=1otherwise.) To prove Lemma 2, let (X, T) be a metrizable topological space in M and let d be a metric on X which is in M and which induces the topology T .Ifφ ∈ G fixes d then φ fixes X and T and therefore sup(X, T) ⊆ sup(d). Assuming that sup(X, T) =6 sup(d), it suffices to construct a metric m satisfying (a) and (b) from the statement of Lemma 2 and (c0) sup(m) ( sup(d). By our assumption that sup(X, T) =6 sup(d) there is an element t ∈ sup(d) \ sup(X, T). Let e0 and e1 be the unique elements of sup(d) ∪{−∞, ∞} such that e0

Lemma 3. If (C, <1) and (D, <2) are countable dense linear orders without first or last elements and C0 and D0 are finite subsets of C and D respectively with |C0| = |D0| then there is an order isomorphism f from C onto D such that f(C0)= D0. 0 Using Lemma 3, let ∆ be an order isomorphism from (e0,e1)onto(e0,t)and DISJOINT UNIONS AND CHOICE 11

define ∆ : A → A \ [t, e1)by

 0 ∆ (a)ifa ∈ (e0,e1) ∆(a)= . a otherwise

We first note that for any x ∈M, there is a ψ ∈ GE such that ψ agrees with ∆ on sup(x). (This is a consequence of Lemma 3.) Further, if ψ and φ are in GE and ψ and φ agree with ∆ on sup(x)thenψ(x)=φ(x). This allows us to define a ∗ ∗ function ∆ from the topological space X to itself by ∆ (x)=ψ(x)whereψ ∈ GE and ψ agrees with ∆ on sup(x). We leave to the reader the proof that ∆∗ is one to ∗ one and onto the set X = {x ∈ X : sup(x) ∩ [t, e1)=∅}. There are two natural ways of putting a topology on the set X∗:

∗ ∗ ∗ T1 = {U : U ∈ T } where U = {∆ (x):x ∈ U} and ∗ T2 = {U ∩ X : U ∈ T }. ∗ ∗ The topology T1 is the topology under which the function ∆ : X → X is a homeomorphism (between the spaces (X, T)and(X, T1)). The topology T2 is the relative topology on X∗ inherited from (X, T) and is therefore the topology induced by the metric d restricted to X∗. We will call this restricted metric d∗. We are now able to outline our plan for finding a metric m for the topology T with support contained in E: We will show that T1 = T2. It follows that if we “pull the metric d∗ back to X using (∆∗)−1” we get a metric which induces the topology T on X. (This “pull back” of d∗ is the metric m defined below.) The metric d∗ ∗ has support E ∪{t}, however in pulling d back to m, t is identified with e1 and therefore m will have support E. The most difficult part of the argument is the proof that T1 = T2. We postpone this part and begin with the definition of m.

m(x, y)=d(∆∗(x), ∆∗(y)) The fact that m is a metric follows from the fact that d is a metric. To prove the triangle inequality, for example, choose x, y and z ∈ X and a ψ ∈ GE such that ψ agrees with ∆ on sup(x) ∪ sup(y) ∪ sup(z). Then m(x, y)+m(y,z)= d(ψ(x),ψ(y)) + d(ψ(y),ψ(z)) ≥ d(ψ(x),ψ(z)) = m(x, z). We now argue that ∀η ∈ GE, η fixes m. This can be accomplished by showing (∀x, y ∈ X)(m(η(x),η(y)) = m(x, y)). Assume η ∈ GE and x, y ∈ X.Choose φ ∈ GE such that φ agrees with ∆ on sup(x)∪sup(y)∪sup(η(x))∪sup(η(y)). By the definition of m, m(x, y)=d(φ(x),φ(y)) and m(η(x),η(y)) = d(φ(η(x)),φ(η(y))). The permutation φηφ−1 takes φ(x)toφ(η(x)) and φ(y)toφ(η(y)). Further, since the range of ∆ is A \ [t, e1), the four sets sup(φ(x)) = φ(sup(x)), sup(φ(y)) = φ(sup(y)), sup(φ(η(x))) = φ(sup(η(x))), and sup(φ(η(y))) = φ(sup(η(y))) are all subsets of A\[t, e1). Therefore, we can find a permutation σ ∈ GE such that −1 σ agrees with φηφ on sup(φ(x))∪sup(φ(y)) and such that σ fixes [t, e1) pointwise. It follows that σ(φ(x)) = φ(η(x)) and σ(φ(y)) = φ(η(y)). Also, since σ fixes E∪{t}, 12 HOWARD, KEREMEDIS, RUBIN, AND RUBIN

σ(d)=d. Therefore d(φ(x),φ(y)) = d(σ(φ(x)),σ(φ(y))) = d(φ(η(x)),φ(η(y))). Hence m(x, y)=m(η(x),η(y)). We conclude that (a) and (c0) are true of m. We shall complete the proof of Lemma 2 and also show that T1 = T2 by proving part (b) of Lemma 2, T is the topology induced by m.Wefirstshow (i) Every U ∈ T is in the topology induced by m. Assume U ∈ T and x ∈ U.Chooseφ ∈ GE so that φ agrees with ∆ on sup(x) ∪ sup(U). Since φ fixes T , φ(x) ∈ φ(U) ∈ T . Since the metric d induces T , there is an >0 such that the open ball B1 = {y ∈ X : d(φ(x),y) <}⊆φ(U). We claim that B2 = {y ∈ X : m(x, y) <}⊆U. For suppose y ∈ B2, then (by definition of m) d(φ0(x),φ0(y)) <,whereφ0 agrees with ∆ on sup(x) ∪ sup(y) ∪ sup(U). Since 0 0 0 0 φ (x)=φ(x), φ (y) ∈ B1. Therefore, φ (y) ∈ φ(U)=φ (U) and hence y ∈ U. Nowweshow (ii) If U is open in the topology induced by m then U ∈ T . It suffices to show that for every x ∈ X and every >0,

(∗) ∃δ>0 such that {z ∈ X : d(x, z) <δ}⊆{z ∈ X : m(x, z) <}.

Assume that there is an x ∈ X and an >0 for which (∗) is false. Fix an s in the interval (t, e1) so that sup(x) ∩ [s, e1)=∅. The failure of (∗) gives, for each positive ∈  n ω,anelementzn of X such that d(x, zn) < n and m(x, zn) >. (The function n 7→ zn may not be in M.)

Claim. We may assume that sup(zn) ∩ [s, e1)=∅. If this is not the case, choose an s0 in the interval (t, s) which also satisfies 0 0 0 0 sup(x)∩[s ,e1)=∅. Let γ be an order isomorphism of (s ,e1) for which γ (sup(zn)∩ 0 0 0 0 (s ,e1)) ⊆ (s ,s). The permutation γ which agrees with γ on (s ,e1) and is 0 the identity outside of (s ,e1) then has the property that sup(γ(zn)) ∩ [s, e1)= γ(sup(zn)) ∩[s, e1)=∅. In , γ fixes E ∪{t}∪sup(x) pointwise. This means  that γ fixes d, m,andx. Hence d(x, γ(zn)) < n and m(x, γ(zn)) >.Wemay therefore replace zn by γ(zn). This proves the claim.

Choose φ ∈ GE which agrees with ∆ on (e0,s). (This can be accomplished by using Lemma 2 to obtain an order isomorphism β from [s, e1)onto[∆(s),e1). Then define φ(a)=β(a)fora ∈ [s, e1)andφ(a)=∆(a) otherwise.) For n ∈ ω \{0}, define Zn = {ψ(zn):ψ fixes sup(d) ∪ sup(x) ∪{s} pointwise} Then (A) Zn has support sup(d) ∪ sup(x) ∪{s}. ∀ ∈  (B) ( w Zn)(d(x, w) < n ).  (Since w = ψ(zn)whereψ fixes x and d and d(x, zn) < n .) Similarly, (C) (∀w ∈ Zn)(m(x, w) >)and (D) (∀w ∈ Zn)(sup(w) ∩ [s, e1)=∅).

By (D), for all w ∈ Zn, φ (which is in GE ) agrees with ∆ on sup(w) ∪ sup(x). Hence, using (C) and the definition of m we conclude that (E) d(φ(x),φ(w)) >. S∞ S∞ By (A), Zn ∈M.By(B),x is in the T of Zn since d n=1 S∞ n=1 induces T . Similarly, by (E), φ(x)isnotintheT closure of φ( n=1 Zn). This is a DISJOINT UNIONS AND CHOICE 13

contradiction since φ ∈ GE and therefore fixes T . This completes the proof of (b) and the theorem.  3. Disjoint Unions of Collectionwise Hausdorff, Collectionwise Normal Spaces and the Axiom of Choice

Theorem 6. The following are equivalent: (i) AC. (ii) If a topological space (X, T) is cwH, then it is cwH(B) for every basis B.. (iii) The disjoint union of cwH spaces is cwH(B) for any basis B. (iv) If a topological space (X, T) is cwN, then it is cwH(B) for any basis B.

Proof. (i)→(ii). Let (X, T) be cwH, G = {gi : i ∈ k} a closed discrete subset of X and B abasisforX.AsX is cwH, there exists a disjoint family {Oi : i ∈ k}⊆T such that gi ∈ Oi for all i ∈ k. Since B is a basis it follows that Qi = {b ∈ B : gi ∈ b ⊆ Oi}6= ∅.PutQ = {Qi : i ∈ k} and let v be a choice function on Q. Clearly V = {v(i):i ∈ k}⊆B is the required family of open sets. The proofs (ii)→(iii) and (iii)→(iv) are straight forward. It therefore suffices to prove (iv)→(i). A { ∈ } ∈A Fix = Ai : i ω a familyS of infinite pairwise disjoint sets. For each A , let xA be a set not in A∪ A and chosen so that the function A 7→ xA is one-to- one. (To be specific, we could take xA to be the (A, A).) For each A ∈A, we define a topological space (YA, TA) as follows. YA = {xA}∪(ω × A)and TA is the topology with basis sets

BA = {{(n, a)} :(n, a) ∈ ω × A}∪

{{xA}∪Z :(ω \ m) × A ⊆ Z ⊆ ω × A for some m ∈ ω}. S T Let Y = A∈A YA and let be the disjoint union topology on Y . We claim that Y is cwN. Indeed, let Z be a discrete collection of closed sets. It can be readily verified that, ∗ ∗ (1) For every A ∈Aeither xA ∈/ z for all z ∈ Z or if xA ∈ z for some z ∈ Z, ∗ then there exists nz∗ ∈ ω such that for every z ∈ Z, z =6 z , z ⊆ nz∗ × A. For every z ∈ Z,let S S S (2) Uz = {(ω ×A)∩z : A ∈A,xA ∈/ z}∪( {z ∪(YA \ Z):A ∈A,xA ∈ z}).

In view of (1) and (2), it follows that U = {Uz : z ∈ Z} is a disjoint family of open sets separating the members of Z. It is easy to see that each YA is T2,sothat Y is also T2.Thus,(Y,T ) is cwN as claimed. Now we construct another basis B for theS topology T . The basis B consists of the sets {(n, a)} such that n ∈ ω and a ∈ A and the sets BA,a,n = {xA}∪ {(n, a)}∪{(m, b):m>n∧ b ∈ A} for each triple (A, a, n) such that A ∈A, a ∈ A, and n ∈ ω. Applying cwNH(B) to this basis and the discrete set {xA : A ∈A}gives for each A ∈Aa unique set BA,a,n. The function f defined by f(A)=a is a choice function for A.  Theorem 7. The following are equivalent (i) MC. (ii) Every space (X, T) which is cwH, is also cwH(B) for any basis B closed under 14 HOWARD, KEREMEDIS, RUBIN, AND RUBIN

finite intersections. (iii) The disjoint union of cwH spaces is cwH(B) for any basis B closed under finite intersections. (iv) Every space (X, T) which is cwN, is also cwH(B) for any basis B closed under finite intersections.

Proof. The proof of (i)→(ii) is similar to the proofT of (i)→(ii) in Theorem 6. (Here v would be a multiple choice function, and V = { v(i):i ∈ k}⊆B is the required family of open sets.) The proofs that (ii)→(iii) and (iii)→(iv) are again straight forward.

(iv)→(i). Let A = {Ai : i ∈ k} be a family of infinite pairwise disjoint sets and let X be as in Theorem 1 (i)→ MC. We claim that X is cwN. Indeed, let D = {dj : j ∈ J} be a discrete collection of closed sets. Clearly, for every i ∈ k either yi ∈/ dj for all ∗ ∗ j ∈ J or if yi ∈ dj∗ for some j ∈ J,(j is unique). Then, |Ai ∩ dj| <ωfor all j ∈ J, j =6 j∗ and

|{j ∈ J : Ai ∩ dj =6 ∅}| <ω.

For every j ∈ J, put

∪{ ∩ ∈ ∈ }∪ ∪{ ∗\ ∈ 6 } Oj = Ai dj : i k, yi / dj ( Ai dv : yi dj ,v = j ).

Clearly O = {Oj : j ∈ J} is a disjoint family separating the members of D.Itis easy to see that D = {{yi} : i ∈ k} is a discrete family of closed subsets of X. If B is defined as follows:

{ ∗\ ∈ <ω\{∅} ∈ }∪{{ } ∈∪A} B = Ai a : a [Ai] ,i k x : x ,

then B is a basis for X closed under finite intersections and D = {yi : i ∈ k} is a closed discrete subspace of X. It follows from cwH(B) that there is a pairwise disjoint set V = {Vi : i ∈ k}⊆B such that for each i ∈ k, yi ∈ Vi. Putting F = {Fi = Ai\Vi : i ∈ k}, we see that each Fi, i ∈ k, is a finite non-empty subset of Ai, which proves MC. 

Summary

We summarize our results in the following diagram. (The forms cwH(B), DUcwH(B), and cwNH(B) in the diagram are abbreviations for (ii), (iii), and (iv), respectively, in Theorem 6.) DISJOINT UNIONS AND CHOICE 15

AC l cwH(B) ↔ cwNH(B) ↔ DUcwH(B) ↓↓ ↓ DUP ↔ MC ↔ DUcwNN ll l DUPFCS DUPN DUMET l↓ l DUcwH DUM DUcwH l↓ l DUcwN DUMN DUcwNN

We have shown that DUM does not imply ω-MC or MC in ZF0. (See Example 3 above. DUM is true in M, but ω-MC and MC are false.) However, we still have the following questions.

Questions: (i) Does DUM imply ω-MCorMCinZF? (ii) Does DUMN imply DUM?

BIBLIOGRAPHY [vd] E. K. van Douwen, Horrors of topology without AC: a non normal orderable space, Proc. Amer. Math. Soc., 95 (1985), 101-105. [gt]C.GoodandI.J.Tree,Continuing horrors of topology without choice,Topology and its Applications, 63 (1995), 79-90. [wgt] C. Good, I. J. Tree and S. Watson, On Stone’s theorem and the axiom of choice, preprint 1996. [hkrr] P. Howard, K. Keremedis, H. Rubin and J. E. Rubin, Versions of normality and some weak forms of the axiom of choice. preprint (1996) [rh] P. Howard and J. E.Rubin, Weak forms of the Axiom of Choice, in preparation. [j] T. J. Jech, The Axiom of Choice, North Holland, Amsterdam, 1973. [le] A. Levy, of multiple choice, Fund. Math. 50 (1962) 475-483. [ln] J.Lo´ s and C. Ryll-Nardzewski, Effectiveness of the representation theory for Boolean algebras, Fund. Math. 41 (1954), 49-56. [pi] D. Pincus, Adding dependent choice, Ann. Math. Logic 11 (1977), 105-145. [so] R. H. Sorgenfrey, On the topological product of paracompact spaces, Bull. AMS 53 (1947), 631-632. [ss] L. A. Steen and J. A. Seebach, Jr, Counterexamples in Topology, Springer- Verlang, 1986. [w] S. Willard, , Addison-Wesley Publ. co., 1968.

Paul Howard Department of Mathematics, Eastern Michigan University, Ypsilanti, MI 48197 USA 16 HOWARD, KEREMEDIS, RUBIN, AND RUBIN email: [email protected]

Kyriakos Keremedis Department of Mathematics, University of the Aegean, Karlovasi, Samos 83200 GREECE email: [email protected].

Herman Rubin Department of Statistics, Purdue University, West Lafayette, IN 47907 USA email: [email protected].

Jean E. Rubin Department of Mathematics, Purdue University, West Lafayette, IN 47907 USA email: [email protected].