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On Stone's Theorem and the Axiom of Choice PROCEEDINGS OF THE AMERICAN MATHEMATICAL SOCIETY Volume 126, Number 4, April 1998, Pages 1211{1218 S 0002-9939(98)04163-X ON STONE'S THEOREM AND THE AXIOM OF CHOICE C. GOOD, I. J. TREE, AND W. S. WATSON (Communicated by Andreas R. Blass) Abstract. It is a well established fact that in Zermelo-Fraenkel set theory, Tychonoff's Theorem, the statement that the product of compact topological spaces is compact, is equivalent to the Axiom of Choice. On the other hand, Urysohn's Metrization Theorem, that every regular second countable space is metrizable, is provable from just the ZF axioms alone. A. H. Stone's The- orem, that every metric space is paracompact, is considered here from this perspective. Stone's Theorem is shown not to be a theorem in ZF by a forc- ing argument. The construction also shows that Stone's Theorem cannot be proved by additionally assuming the Principle of Dependent Choice. Introduction Given an infinite set X, is it possible to define a Hausdorff topology on X such that X has at least two non-isolated points? In ZFC, the answer is easily shown to be yes. However, models of ZF exist that contain infinite sets that cannot be expressed as the union of two disjoint infinite sets (amorphous sets—see [6]). For any model of ZF containing an amorphous set, the answer is no. So, as we see, even the most innocent of topological questions may be undecidable from the Zermelo- Fraenkel axioms alone. Further examples of the counter-intuitive behaviour of ‘choiceless topology’ can be found in [3] and [4]. The concept of paracompactness in a topological space was first defined by Dieudonn´e in [2], in which he proved that every metrizable space that is second countable or locally compact is paracompact. The importance of paracompactness in general topology was raised when A. H. Stone proved the theorem of the title, namely that every metric space is paracompact [11]. Mary Ellen Rudin later im- proved the proof in [10]. One notable point about Rudin’s proof is the very first line: ‘Let = Uα be an open cover indexed by ordinals’, an immediate use of the AxiomU of Choice.{ } Stone’s original proof also uses Choice, but in a less obvious manner. We are therefore prompted to ask whether this is an essential part of Stone’s Theorem: just how heavily does Stone’s Theorem depend on the Axiom of Choice? We address this question here. The fact that every separable metrizable space is paracompact can be proved from ZF [4]. That every second countable metric space is paracompact, Dieudonn´e’s original result, can also be so proved (but recall from [4] that there are models of ZF containing second countable metric spaces that are not separable). However, we show that this is not true of the general theorem: we construct symmetric models of ZF in which there are metric spaces that are not paracompact. Received by the editors March 27, 1996 and, in revised form, September 17, 1996. 1991 Mathematics Subject Classification. Primary 54D20; Secondary 03E25. 1211 c 1998 American Mathematical Society License or copyright restrictions may apply to redistribution; see https://www.ams.org/journal-terms-of-use 1212 C. GOOD, I. J. TREE, AND W. S. WATSON Symmetric models of ZF We review some facts about symmetric models, referring the reader to [6] for further details. Our notation follows that of [6] and [7]. Let be a transitive model of ZFC and B a complete Boolean algebra in . For anM automorphism π of B,weextendπto B by induction on the rankM of x B: M ∈M(1) π(0) = 0; (2) dom(πx)= πy : y dom(x) and (πx)(πy)=π(x(y)). It follows that π{is a one-to-one∈ } function from B onto itself, and πxˇ =ˇxfor every x .Let be a group of automorphismsM of B. A non-empty set of subgroups∈M of is calledG a normal filter on if and only if for all subgroups H;F K of , G G G(i) if K and K H then H , (ii) if H∈F and K⊆ then H ∈FK , ∈F ∈F ∩ 1 ∈F (iii) if π and H then πHπ− . Let be∈G a fixed normal∈F filter. For each∈Fx B, define sym(x)= π : πx = xF.Wesaythatx B is symmetric if ∈Msym(x) .TheclassHS{ ∈FB of all hereditarily} symmetric∈M names is defined by recursion:∈F ⊆M (a) 0 HS; (b) if ∈dom(x) HS and x is symmetric, then x HS. Now let G be an⊆ -generic ultrafilter on B and i∈ be the interpretation of B M G M by G. Define = iG(x):x HS .Then is a symmetric extension of and N[G] .{ More importantly,∈ } is aN model of ZF. M M⊆N ⊆M N The partial order We use one partial order as the basis for our symmetric model constructions, namely P = Fn(! R !1 !1;2;!1), the set of partial functions p with dom(p) < ! , dom(p) ! ×R ×! ×! and ran(p)= 0;1 . We define the ordering| on| P 1 ⊆ × × 1 × 1 { } by p 6 q if and only if q p.LetB=RO(P)in , the complete Boolean algebra of regular open sets of P⊆in . M Define the following elementsM of [G] together with their canonical names: M x = δ ! : p Gp(n; r; α, δ)=1 ; nrα { ∈ 1 ∃ ∈ } dom(x )= δˇ:δ x ; nrα { ∈ nrα} x (δˇ)=sup p P : p(n; r; α, δ)=1 =u ; nrα { ∈ } nrαδ X = x : α ! ; nr { nrα ∈ 1} dom(X )= x : α ! ; nr { nrα ∈ 1} Xnr(xnrα)=1; R = X : r R ; n { nr ∈ } dom(R )= X : r R ; n { nr ∈ } Rn(Xnr)=1; M= R :n ! ; { n ∈ } dom(M)= R :n ! ; { n ∈ } M(Rn)=1: License or copyright restrictions may apply to redistribution; see https://www.ams.org/journal-terms-of-use ON STONE'S THEOREM AND THE AXIOM OF CHOICE 1213 Let d(x; y)= x y be the usual distance function on R that generates the Eu- | − | clidean topology (in ). Recall that if G is P-generic over then (R)M = [G] M M (R)N =(R)M ,where [G]—see [7, 7.6.14]. M⊆N ⊆M Theorem 1. Let be a transitive model of ZFC. There is a symmetric extension of which containsM a collection of sets R for which there is no set S N M n ∈N with ? = S Rn ( Rn for all n and where Rn can be topologized as a metric 6 ∩ space with each Rn a connected subspace. S Proof. We define a group and a filter such that the names x ;X ;R and G F nrα nr n M are all symmetric. Observe that every permutation π on ! R !1 induces an order-preserving 1-1 mapping on P,by(πp)(π(n; r; α),δ)=p×(n; r;× α, δ), and an automorphism of B,byπu = sup πp : p u . One can check that { 6 } ( ) π(u )=u and π(x )=x :::: † nrαδ π(n;r;α)δ nrα π(n;r;α) Let be the group of all automorphisms of B induced by permutations of ! R ! G × × 1 satisfying π(n; r; α)=(n; ρ(r),α0), where π(n; r; :) is a permutation on !1 for fixed π n; r and ρ = ρ : R R is either a reflection about some point xπ R,orthe identity map. → ∈ By ( ), dom(πX )= x : α ! = dom(X ). It follows that, as † nt { π(n;t;α) ∈ 1} nρ(t) π is an automorphism of B, π(R )=R and π(M)=M. ∈G n n For each finite subset e ! R !1,letfix(e)= π : s eπ(s)=s . Let be the filter on generated⊆ × by× fix(e):e [! {R∈G! ]<!∀ ∈. } F G { ∈ × × 1 } Claim 1.1. is a normal filter on . F G Proof. Omitted. Let HS be the set of all hereditarily symmetric names in B.Let be the symmetric extension of given by the interpretation of HSMby G. N M Claim 1.2. For all n; r and α,thesetsx ;X ;R ;M are in the model . nrα nr n N ˇ Proof. By ( ), sym(xnrα)=fix((n; r; α)) . Inductively, δ HS for all δ !1, so dom(x †) HS. Hence x HS.Fix∈Fα !.Then ∈ ∈ nrα ⊆ nrα ∈ 0∈ 1 sym(X )= π :πX = X nr { ∈G nr nr} = π : dom(πX )=dom(X ) { nr nr and πX (πx )=π(X (x )) = π(1) = 1 nr nrα nr nrα } fix(n; r; α ): ⊇ 0 So sym(X ) and dom(X ) HS.ThuseachX is in HS. It follows nr ∈F nr ⊆ nr that dom(Rn) HS and sym(Rn)= ,sodom(M) HS and sym(M)= . ⊆ G ⊆ GB Therefore each Rn and M is in HS. By the definition of the interpretation of by G,wehavex ;X ;R ;M . M nrα nr n ∈N Claim 1.3. There is no function f such that dom(f)=Mand f(R )isa ∈N n proper non-empty subset of Rn for each n. Proof. Assume there is such an f .Letfbe a symmetric name for f and let p G be such that ∈N 0 ∈ p0 (f is a function) and ( nˇ ? = f(Rn) ( Rn): ∀ 6 License or copyright restrictions may apply to redistribution; see https://www.ams.org/journal-terms-of-use 1214 C. GOOD, I. J. TREE, AND W. S. WATSON Let e be a finite subset of ! R ! such that fix(e) sym(f). Pick n ! such × × 1 ⊆ ∈ that e ( n R !1)=?.
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