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On Locally Compact Hausdorff Spaces with Finite Metrizability Number

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On Locally Compact Hausdorff Spaces with Finite Metrizability Number CORE Metadata, citation and similar papers at core.ac.uk Provided by Elsevier - Publisher Connector Topology and its Applications 114 (2001) 285–293 On locally compact Hausdorff spaces with finite metrizability number M. Ismail ∗, A. Szymanski Department of Mathematics, Slippery Rock University, Slippery Rock, PA 16057, USA Received 9 September 1999; received in revised form 9 February 2000 Abstract The metrizability number m(X) of a space X is the smallest cardinal number κ such that X can be represented as a union of κ many metrizable subspaces. In this paper, we study compact Hausdorff spaces with finite metrizability number. Our main result is the following representation theorem: If X is a locally compact Hausdorff space with m(X) = n<ω, then for each k,1 k<n, X can be represented as X = G ∪ F ,whereG is an open dense subspace, F = X \ G, m(G) = k,and m(F ) = n − k. 2001 Elsevier Science B.V. All rights reserved. AMS classification: 54A25; 54D30 Keywords: Compact space; Metrizability number The metrizability number m(X) of a space X is the smallest cardinal number κ such that X can be represented as a union of κ many metrizable subspaces. In [5–7], we studied the metrizability number and related cardinal invariants for the class of compact Hausdorff spaces. The aim of this paper is to study compact (in general, locally compact) Hausdorff spaces with finite metrizability number. Typical examples of nonmetrizable compact Hausdorff spaces with finite metrizability number are the one-point compactification of an uncountable discrete space, the Alexandroff duplicate of the unit segment, and the one- point compactification of the space Ψ (cf. [3, 5I]). Our main goal is to prove Theorem 6, below, and some of its consequences. We first prove the following lemmas. Lemma 1. A locally compact Hausdorff space which is a union of countably many dense (or open) metrizable subspaces is metrizable. * Corresponding author. E-mail addresses: [email protected] (M. Ismail), [email protected] (A. Szymanski). 0166-8641/01/$ – see front matter 2001 Elsevier Science B.V. All rights reserved. PII:S0166-8641(00)00043-2 286 M. Ismail, A. Szymanski / Topology and its Applications 114 (2001) 285–293 Proof. Any such space has a σ -disjoint (hence, point-countable) base. Furthermore, any locally compact Hausdorff space with a point-countable base is metrizable (see [4, Section 7]). ✷ Lemma 2. A locally compact Hausdorff space with finite metrizability number contains an open dense metrizable subspace. Proof. Let X be a locally compact Hausdorff space such that m(X) < ω. We first prove, by induction on m(X),thatX contains a non-empty open metrizable subspace. Clearly,thisistrueifm(X) = 1. Let m(X) = n,where1<n<ω, and suppose that all locally compact Hausdorff spaces of metrizability number less than n contain non-empty open metrizable subspaces. Let X = M1 ∪ M2 ∪···∪Mn, where each Mi is metrizable. Since X is not metrizable, by Lemma 1, for some j, Mj is not dense in X.LetU be a non-empty open subset of X such that U ∩ Mj =∅.Since m(U) n−1andU is a locally compact Hausdorff subspace, by the inductive hypothesis, U contains a non-empty open metrizable subspace. D Let be a maximal disjoint family of non-empty open metrizable subspaces of X.Then D is the required dense open metrizable subspace of X. ✷ = { } Lemma 3. Let X be a locally compact Hausdorff space, and let X Mi: i<ω ,where each Mi is metrizable. Then {cl Mi: i<ω} is metrizable. Proof. Let F = {cl Mi: i<ω}. For each i,letBi be a σ -discrete base of Mi .For each V ∈ Bi, fix an open subset U(V) of clMi such that V = U(V) ∩ Mi .IfBi = { ∈ B } B B U(V): V i ,then i isabasein cl Mi at each point of Mi .Also, i is σ -disjoint. Let B = {Bi: i<ω}.SinceF ⊆ {Mi : i<ω}, the family {U ∩ F : U ∈ B} is a σ - disjoint (hence point-countable) base of F .SinceF is locally compact, F is metrizable (see [4, Section 7]). ✷ = LetX be a locally compact Hausdorff space, and let m(X) n,where2 n<ω.Let X = {Mi : i = 1,...,n}, where each Mi is metrizable. We may assume, without loss of generality, that Mi ∩ Mj =∅whenever i = j. For each non empty subset A of the set {1,...,n},let Y(A)= {cl Mi: i ∈ A}. For each k ∈{1,...,n},let Zk = Y(A): |A|=n − k + 1 . = { = } = Clearly, each Zk is a closed subspace of X.Also,Z1 clMi : i 1,...,n , Zn {clMi: i = 1,...,n}=X,andZ1 ⊆ Z2 ⊆···⊆Zn. Lemma 4. For each k = 1, 2,...,n− 1, Zk+1 \ Zk is metrizable. M. Ismail, A. Szymanski / Topology and its Applications 114 (2001) 285–293 287 Proof. Clearly, Zk+1 \ Zk = {Y(A)\ Zk: |A|=n − k}. First, let us show that for each ⊆{ } | |= − \ A 1,...,n , with A n k, Y(A) Zk is metrizable. Towards this end, let us notice \ = \ { ∈ } = \ { ∈ } that Y(A) Zk Y(A) cl Mi: i/ A .IfW X clMi : i/A,thenW is a locally = { ∩ ∈ } { ∩ ∈ compact Hausdorff space and W W Mi : i A . By Lemma 3, clW (W Mi ): i A} is metrizable. Since Y(A)\ Zk ⊆ {clW (W ∩ Mi ): i ∈ A},Y(A)\ Zk is metrizable. Now, let A and B be distinct subsets of {1,...,n} such that |A|=|B|=n − k.Then A \ B =∅and B \ A =∅.Leti ∈ A \ B.Thencl(Y (A) \ Zk) ⊆ clMi and (Y (B) \ Zk) ∩ =∅ \ ∩ \ =∅ \ ∩ \ = clMi . Thus cl(Y (A) Zk) (Y (B) Zk ) . Similarly, cl(Y (B) Zk) (Y (A) Zk) ∅. This shows that Zk+1 \Zk = {Y(A)\Zk: |A|=n−k}. As a disjoint sum of metrizable spaces, Zk+1 \ Zk is metrizable. ✷ Lemma 5. For each k = 1,...,n, m(Zk) = k and m(X \ Zk) = n − k. Proof. (i) First, we show by induction on k that m(Zk) k. Since Z1 = {cl Mi: i = 1,...,n}, by Lemma 3, Z1 is metrizable, i.e., m(Z1) = 1. Suppose that k<nand that m(Zk) k. By Lemma 4, Zk+1 \ Zk is metrizable and since Zk+1 = Zk ∪ (Zk+1 \ Zk), m(Zk+1) k + 1. (ii) Next, we show by induction on k that m(Zk) k. Since Zn = X, m(Zn) = n. Suppose that k<n− 1andthatm(Zn−k) n − k.By Lemma 4, Zn−k \ Zn−(k+1) is metrizable and since Zn−k = Zn−(k+1) ∪ (Zn−k \ Zn−(k+1)), m(Zn−(k+1)) n − (k + 1). It follows from (i) and (ii), above, that m(Zk) = k for each k = 1,...,n. (iii) For each k = 1,...,n, X \ Zk = (Zn \ Zn−1) ∪ (Zn−1 \ Zn−2) ∪···∪(Zk+1 \ Zk). Hence, by Lemma 4, m(X \ Zk) n − k.Sincem(X) = n and m(Zk) = k, m(X \ Zk) n − k. Thus m(X \ Zk) = n − k. ✷ Theorem 6. If X is a locally compact Hausdorff space with m(X) = n, 2 n<ω,then for each k, 1 k<n, X can be represented as X = G ∪ F ,whereG is an open dense subspace of X, F ∩ G =∅, m(G) = k, and m(F ) = n − k. Proof. Let W be a dense open metrizable subspace of X (such a subspace exists because of Lemma 2). We set G = W ∪ (X \ Zn−k). Since (X \ Zn−k) ⊆ G and, by Lemma 5, m(X \ Zn−k) = k, m(G) k. Since G = W ∪ (X \ Zn−1) ∪ (Zn−1 \ Zn−2) ∪···∪(Zn−k+1 \ Zn−k) = (W ∪ (X \ Zn−1)) ∪ (Zn−1 \ Zn−2) ∪···∪(Zn−k+1 \ Zn−k) and, by Lemma 1, W ∪ (X \ Zn−1) is metrizable, m(G) k. Thus m(G) = k. Let F = X \ G.SinceF ⊆ Zn−k, m(F ) n − k.Sincem(X) = n and m(G) = k, m(F ) = n − k. ✷ Corollary 7. If X is a locally compact Hausdorff space with m(X) = n, 2 n<ω,then X can be represented as X = G ∪ F ,whereG is an open dense metrizable subspace of X, F ∩ G =∅, and m(F ) = n − 1. 288 M. Ismail, A. Szymanski / Topology and its Applications 114 (2001) 285–293 Corollary 8. If X is a locally compact Hausdorff space with m(X) = n, 2 n<ω,then X can be represented as X = G ∪ F ,whereG is an open dense subspace of X with m(G) = n − 1, F ∩ G =∅, and F is metrizable. Definition 9 (cf. [8]). Given a cardinal function ϕ and a space X,theϕ-spectrum of X, denoted by Sp(ϕ, X),isdefinedasSp(ϕ, X) ={ϕ(F): F = clF ⊆ X and |F | ω}. Corollary 10. If X is a locally compact Hausdorff space with m(X) = n, n<ω,then Sp(m, X) ={1,...,n}. Proof. This is an immediate consequence of Theorem 6. ✷ Let us observe that if m(X) > ω, then the above result may not hold even for compact 2ω Hausdorff spaces. For example, Sp(m, ω1 + 1) ={1,ω1}, Sp(m, βN) ={2 },andifX is the top and the bottom of the lexicographic square, then Sp(m, X) ={1, 2ω} (see [6, Examples 4, 5, 8]). Theorem 11. If X is a locally compact Hausdorff space with m(X) = n, n<ω, and Y is a perfect image of X,thenm(Y ) m(X). Proof. The proof is by induction on n. If n = 1(i.e.,ifX is metrizable), then any perfect image of X is metrizable (cf. [2, Theorem 4.4.15]).
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