Lobachevskii Journal of Mathematics http://ljm.ksu.ru ISSN 1818-9962 Vol. 21, 2006, 57–63

Ying Ge
ON CLOSED INVERSE IMAGES OF BASE-PARACOMPACT SPACES
(submitted by M. A. Malakhaltsev)

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2000 Mathematical Subject Classiﬁcation. 54C10, 54D20, 54D70, 54E35.

Key words and phrases. Base-paracompactness, base-paracompact mapping, closed Lindelöf mapping, weight, regular cardinality .

This project was supported by NSFC(No.10571151).

Abstract. In this paper, we prove that every base-paracompact mapping $f : X \to Y$ inversely preserves base-paracompactness if $w (X) \geq w (Y)$ , where $w (X)$ and $w (Y)$ denote the weight of $X$ and the weight of $Y$ respectively. As an application of this result, we prove that every closed $L i n d e l \ddot{o} f$ mapping $f : X \to Y$ inversely preserves base-paracompactness if $X$ is a regular space and $w (X)$ is a regular cardinality, where “ $X$ is a regular space” cannot be relaxed to “ $X$ is a Hausdorff space”, which give some answers for a question on inverse images of base-paracompact spaces posed by L.Wu.

1. Introduction

In his paper [7], J.E.Porter introduced base-paracompactness, and obtained some analogous results of base-paracompactness to paracompactness. In particular, he proved that perfect mappings inversely preserve base-paracompactness [7, Theorem 3.6]. It is known that closed $L i n d e l \ddot{o} f$ mappings with regular domain inversely preserve paracompactness [2, Theorem 7.1], which is obtained by a “nice characterization” of paracompactness: a regular space is paracompact if and only if its every open cover has a $σ$ -locally ﬁnite open reﬁnement [6]. Naturally, it is interesting to investigate closed $L i n d e l \ddot{o} f$ inverse images of base-paracompact spaces. Contrary to what one might hope or expect, we do not know whether the analogous “nice characterization” of base-paracompactness is true. Thus our investigation from paracompactness to base-paracompactness case is not straightforward. Based on the discussion mentioned above, L.Wu [9] raised the following question.

Question 1.1. Is base-paracompactness inversely preserved under closed $L i n d e l \ddot{o} f$ mappings?

In this paper, we investigate Question 1.1 for domains are Hausdorff spaces and regular spaces, respectively. We denote Axiom of Choice and Generalized Continuum Hypothesis by AC and GCH respectively. The weight $w (X)$ of a space $X$ is the minimal cardinality of bases for $X$ . Let $κ$ be a cardinality. We denote the coﬁnality of $κ$ by $c f (κ)$ . A cardinality $κ$ is called regular if $c f (κ) = κ$ . We prove that every base-paracompact mapping $f : X \to Y$ inversely preserve base-paracompactness if $w (X) \geq w (Y)$ , which improves [7, Theorem 3.6]. As an application of this result, we prove that (AC+GCH) every closed $L i n d e l \ddot{o} f$ mapping $f : X \to Y$ inversely preserves base-paracompactness if $X$ is a regular space and $w (X)$ is a regular cardinality, where “ $X$ is a regular space” can not be relaxed to “ $X$ is a Hausdorff space”. By these results, we give some answers for Question 1.1.

Throughout this paper, all spaces are assumed to be Hausdorff and all mappings are continuous and onto. $N$ denotes the set of all natural numbers, $ω_{0}$ denotes the ﬁrst inﬁnite cardinality. The cardinality of a set $A$ is denoted $∣ A ∣$ . Without loss of generality, in this paper we can assume that $w (X) \geq ω_{0}$ . If $f : X \to Y$ is a mapping, $U$ and $V$ are families of subsets of $X$ and $Y$ respectively, then $f (U) = {f (U) : U \in U}$ and $f^{- 1} (V) = {f^{- 1} (V) : V \in V}$ . For a set $A$ , families $U$ and $V$ of sets, $⋃ U$ = $⋃ {U : U \in U}$ , $U \land A = {U ⋂ A : U \in U}$ , and $U \land V = {U ⋂ V : U \in U a n d V \in V}$ . We say that $V$ is a partial reﬁnement of $U$ , if for every $V \in V$ there exists $U \in U$ such that $V \subset U$ ; moreover, we say that $V$ is a reﬁnement of $U$ , if in addition $⋃ V = ⋃ U$ is also satisﬁed. One may refer to [2], [3] and [5] for undeﬁned notations and terminology.

Deﬁnition 1.2. A space $X$ is called base-paracompact [7] if there exists a base $ℬ$ with $∣ ℬ ∣ = w (X)$ such that every open cover of $X$ has a locally ﬁnite reﬁnement $ℬ^{'} \subset ℬ$ .

Remark 1.3. Base-paracompact $\Rightarrow$ paracompact $\Rightarrow$ normal.

Let $A$ be a subset of a space $X$ and let $U$ be a family of subsets of $X$ . We call that $U$ is locally ﬁnite at $A$ in $X$ [8], if for every $x \in A$ there exists an open (in $X$ ) neighborhood of $x$ that intersects at most ﬁnite members of $U$ .

Deﬁnition 1.4. A mapping $f : X \to Y$ is called base-paracompact, if there exists a base $ℬ$ for $X$ with $∣ ℬ ∣ = w (X)$ such that for every $y \in Y$ and every family $U$ of open subsets of $X$ which covers $f^{- 1} (y)$ , there exist an open neighborhood $O_{y}$ of $y$ and a partial reﬁnement $ℬ_{y}$ of $U$ , where $ℬ_{y} \subset ℬ$ , such that $f^{- 1} (O_{y}) \subset ⋃ ℬ_{y}$ and $ℬ_{y}$ is locally ﬁnite at $f^{- 1} (O_{y})$ in $X$ .

Recall a subset $F$ of a space $X$ is called a $L i n d e l \ddot{o} f$ subset if every open cover of $F$ has a countable subcover.

Deﬁnition 1.5. A closed mapping $f : X \to Y$ is called perfect (closed $L i n d e l \ddot{o} f$ ), if $f^{- 1} (y)$ is a compact subset ( $L i n d e l \ddot{o} f$ subset) of $X$ for every $y \in Y$ .

2. Inverse Images of Base-paracompact Spaces for Hausdorff Domains

Throughout section, all domains need not to be regular.

Theorem 2.1. Let $f : X \to Y$ be a base-paracompact mapping and $w (X) \geq w (Y)$ . If $Y$ is base-paracompact, then $X$ is base-paracompact.

Proof. Let $ℬ_{Y}$ be a base for $Y$ which witnesses base-paracompactness for $Y$ and let $ℬ_{X}$ be a base for $X$ with $∣ ℬ_{X} ∣ = w (X)$ which witnesses base-paracompactness for $f$ . Put $ℬ = ℬ_{X} \land f^{- 1} (ℬ_{Y})$ . Since $w (X) \geq w (Y)$ , $∣ ℬ ∣ = ∣ ℬ_{X} ∣ = w (X)$ and $ℬ$ is a base for $X$ . We prove that $ℬ$ witnesses base-paracompactness for $X$ as follows.

Let $U$ be an open cover of $X$ . Since $ℬ_{X}$ witnesses base-paracompactness for $f$ , for every $y \in Y$ there exist an open neighborhood $O_{y}$ of $y$ and a partial reﬁnement $ℬ_{y}$ of $U$ , where $ℬ_{y} \subset ℬ_{X}$ , such that $f^{- 1} (O_{y}) \subset ⋃ ℬ_{y}$ and $ℬ_{y}$ is locally ﬁnite at $f^{- 1} (O_{y})$ in $X$ . Notice that $Y$ is regular from Remark 1.3. There exists an open neighborhood $G_{y}$ of $y$ such that ${\bar{G}}_{y} \subset O_{y}$ , thus $f^{- 1} (y) \subset f^{- 1} (G_{y}) \subset f^{- 1} ({\bar{G}}_{y}) \subset f^{- 1} (O_{y}) \subset ⋃ ℬ_{y}$ . Put $V = {G_{y} : y \in Y}$ . By base-paracompactness for $Y$ , $V$ has a locally ﬁnite reﬁnement $ℬ_{Y}^{'} \subset ℬ_{Y}$ . We write $ℬ_{Y}^{'} = {B_{α}^{'} : α \in Γ}$ . For every $α \in Γ$ , pick $y_{α} \in Y$ such that $B_{α}^{'} \subset G_{y_{α}}$ . Put $ℬ^{'} = ⋃ {ℬ_{y_{α}} \land f^{- 1} (B_{α}^{'}) : α \in Γ}$ . Then $ℬ^{'}$ is a reﬁnement of $U$ and $ℬ^{'} \subset ℬ$ . To complete the proof, it suffices to show that $ℬ^{'}$ is locally ﬁnite.

Let $x \in X$ . Notice that $f^{- 1} (ℬ_{Y}^{'}) = {f^{- 1} (B_{α}^{'}) : α \in Γ}$ is locally ﬁnite in $X$ . There exist an open neighborhood $U$ of $x$ and a ﬁnite subset $Γ_{x}$ of $Γ$ such that $U ⋂ f^{- 1} (B_{α}^{'}) = \emptyset$ for every $α \in Γ - Γ_{x}$ .

Let $α \in Γ_{x}$ . If $x \notin f^{- 1} (\bar{B_{α}^{'}})$ , then there exists an open neighborhood $V_{α}$ of $x$ such that $V_{α} ⋂ f^{- 1} (\bar{B_{α}^{'}}) = \emptyset$ . Thus $V_{α} ⋂ f^{- 1} (B_{α}^{'}) = \emptyset$ . If $x \in f^{- 1} (\bar{B_{α}^{'}})$ , then $x \in f^{- 1} ({\bar{G}}_{y_{α}}) \subset f^{- 1} (O_{y_{α}})$ . Since $ℬ_{y_{α}}$ is locally ﬁnite at $f^{- 1} (O_{y_{α}})$ in $X$ , there exists an open neighborhood $V_{α}$ of $x$ that intersects at most ﬁnite members of $ℬ_{y_{α}}$ . Thus we can obtain $V_{α}$ for every $α \in Γ_{x}$ as above.

Put $W = U ⋂ (⋂ {V_{α} : α \in Γ_{x}})$ . Then $W$ is an open neighborhood of $x$ . It is not difficult to check that $W$ intersects at most ﬁnite members of $ℬ^{'}$ . □

Remark 2.2. We do not know if the condition “ $w (X) \geq w (Y)$ ” in Theorem 2.1 can be omitted.

Lemma 2.3. [3]. A mapping $f : X \to Y$ is closed if and only if for every $y \in Y$ and every open subset $U$ in $X$ which contains $f^{- 1} (y)$ , there exists an open neighborhood $V$ of $y$ such that $f^{- 1} (V) \subset U$ .

Remark 2.4. Base-paracompact mappings are closed mappings from Lemma 2.3.

Proposition 2.5. Let $f : X \to Y$ be a perfect mapping. Then $f$ is base-paracompact.

Proof. Let $ℬ$ be a base for $X$ with $∣ ℬ ∣ = w (X)$ . Let $y \in Y$ and let $U$ be a family of open subsets of $X$ which covers $f^{- 1} (y)$ . For every $x \in f^{- 1} (y)$ , there exists $B_{x} \in ℬ$ such that $x \in B_{x} \subset U$ for some $U \in U$ . Since $f^{- 1} (y)$ is compact, the family ${B_{x} : x \in f^{- 1} (y)}$ has a ﬁnite subfamily $ℬ_{y} \subset ℬ$ such that $f^{- 1} (y) \subset ⋃ ℬ_{y}$ . By Lemma 2.3, there exists an open neighborhood $O_{y}$ of $y$ such that $f^{- 1} (O_{y}) \subset ⋃ ℬ_{y}$ . Notice that $ℬ_{y}$ is ﬁnite, so $ℬ_{y}$ is locally ﬁnite at $f^{- 1} (O_{y})$ in $X$ . Thus $f$ is base-paracompact. □

Corollary 2.6. [7]. Let $f : X \to Y$ be a perfect mapping. If $Y$ is base-paracompact, then $X$ is base-paracompact.

Proof. It is straight from Proposition 2.5 , [3, Theorem 3.7.19] and Theorem 2.1. □

Now we use [4, The Counterexample] to answer Question 1.1 negatively.

Example 2.7. Closed $L i n d e l \ddot{o} f$ inverse images of Base-paracompact spaces need not to be base-paracompact.

Proof. Let $X$ , $Q$ and $I$ be the set of all real numbers, the set of all rational numbers and the set of all irrational numbers respectively. Deﬁne a base $ℬ$ of $X$ as follows.

$ℬ = {{x} : x \in I} ⋃ {G (x, n) : x \in Q, n \in N}$ , here $G (x, n) = {y \in I : - 1 ∕ n < y - x < 1 ∕ n} ⋃ {x}$ .

That is, $X$ is a Bennett and Lutzer’s space [1]. Deﬁne an equivalence relation $R$ on $X$ as follows: $x R y$ if and only if either $x, y \in Q$ or $x = y$ . Put $Y$ is the quotient space $X ∕ R$ and put $f : X \to Y$ is a natural mapping.

Fact 1. $f$ is a closed $L i n d e l \ddot{o} f$ mapping.

Fact 2. $X$ is Hausdorff, but $X$ is not regular.

Fact 3. $X$ is not paracompact, and so $X$ is not base-paracompact.

Fact 4. $Y$ is normal.

Fact 5. $Y$ is base-paracompact.

We only need to prove Fact 5, other facts hold from [4, The Counterexample].

Let $ℬ$ be a base for $Y$ with $∣ ℬ ∣ = w (Y)$ . Pick $x_{0} \in Q$ and put $y_{0} = f (x_{0})$ . Note that ${y}$ is open in $Y$ for every $y \in Y - {y_{0}}$ . So ${y} \in ℬ$ for every $y \in Y - {y_{0}}$ . Let $U$ be any open cover of $Y$ . There exists $B_{0} \in ℬ$ such that $y_{0} \in B_{0} \subset U$ for some $U \in U$ . Put $ℬ^{'} = {B_{0}} ⋃ {{y} : y \in Y - B_{0}}$ . Then $ℬ^{'} \subset ℬ$ . It is clear that elements of $ℬ^{'}$ are mutually disjoint. So $ℬ^{'}$ is a locally ﬁnite reﬁnement of $U$ . Consequently, $Y$ is base-paracompact. □

3. Inverse Images of Base-paracompact Spaces for Regular Domains

Throughout this section, all domains assume to be regular.

Proposition 3.1. If $f : X \to Y$ is a closed $L i n d e l \ddot{o} f$ mapping, then $f$ is base-paracompact.

Proof. Let $ℬ$ be a base for $X$ with $∣ ℬ ∣ = w (X)$ . We can assume that $ℬ$ is closed under ﬁnite unions, ﬁnite intersections and complements of closures from [7, Theorem 3.4]. Let $y \in Y$ and let $U$ be a family of open subsets of $X$ , which covers $f^{- 1} (y)$ . For every $x \in f^{- 1} (y)$ , there exist $B_{x}^{'}, B_{x}^{''} \in ℬ$ such that $x \in B_{x}^{'} \subset \bar{B_{x}^{'}} \subset B_{x}^{''} \subset U$ for some $U \in U$ . Since $f^{- 1} (y)$ is $L i n d e l \ddot{o} f$ , the family ${B_{x}^{'} : x \in f^{- 1} (y)}$ has a countable subfamily ${B_{x_{n}}^{'} : n \in N}$ covering $f^{- 1} (y)$ . By Lemma 2.3, there exists an open neighborhood $O_{y}$ of $y$ such that $f^{- 1} (O_{y}) \subset ⋃ {B_{x_{n}}^{'} : n \in N}$ .

Put $B_{1} = B_{x_{1}}^{''}$ and $B_{n} = B_{x_{n}}^{''} - ⋃ {\bar{B_{x_{i}}^{'}} : i < n}$ for every $n \geq 2$ . Put $ℬ_{y} = {B_{n} : n \in N}$ .

Claim 1. $ℬ_{y} \subset ℬ$ : It follows from that $ℬ$ is closed under ﬁnite unions, ﬁnite intersections and complements of closures.

Claim 2. $ℬ_{y}$ is a partial reﬁnement of $U$ : It is clear.

Claim 3. $f^{- 1} (O_{y}) \subset ⋃ ℬ_{y}$ : Let $x \in f^{- 1} (O_{y})$ . Put $n_{x} = m i n {i \in N : x \in \bar{B_{x_{i}}^{'}}}$ . Then $x \in B_{n_{x}}$ .

Claim 4. $ℬ_{y}$ is locally ﬁnite at $f^{- 1} (O_{y})$ in $X$ : Let $x \in f^{- 1} (O_{y})$ . Then there exists $i \in N$ such that $x \in B_{x_{i}}^{'}$ , thus $B_{x_{i}}^{'}$ is an open neighborhood of $x$ which misses $B_{n}$ for every $n > i$ .

This proves that $f$ is base-paracompact. □

Corollary 3.2. Let $f : X \to Y$ be a closed $L i n d e l \ddot{o} f$ mapping and $w (X) \geq w (Y)$ . If $Y$ is base-paracompact, then $X$ is base-paracompact.

Remark 3.3. Let $X$ and $Y$ be spaces stated in Example 2.7. It is clear that $w (X) = w (Y) = c$ , where $c$ is the cardinal number of the continuum. So Proposition 3.1 and Corollary 3.2 do not hold for Hausdorff domains by Theorem 2.1 and Example 2.7.

Lemma 3.4. [5]. (AC+GCH) Let $κ$ be a cardinality. If $c f (κ) > ω_{0}$ , then $κ^{ω_{0}} = κ$ .

Lemma 3.5. (AC+GCH) Let $f : X \to Y$ be a closed $L i n d e l \ddot{o} f$ mapping. If $c f (w (X)) > ω_{0}$ , then $w (X) \geq w (Y)$ .

Proof. Let $ℬ$ be a base for $X$ such that $∣ ℬ ∣ = w (X)$ and let $A = {⋃ ℬ^{'} : ℬ^{'} \subset ℬ a n d ∣ ℬ^{'} ∣ \leq ω_{0}}$ . Since $c f (w (X)) > ω_{0}$ , $∣ A ∣ = w (X)$ by Lemma 3.4. Put $C = {Y - f (X - A) : A \in A}$ . Then $∣ C ∣ = ∣ A ∣ = w (X)$ . It suffices to show that $C$ is a base for $Y$ . It follows from the deﬁnition that every member of $C$ is open. Let $y \in Y$ and $W$ be a neighborhood (in $Y$ ) of $y$ . Then $f^{- 1} (y) \subset f^{- 1} (W)$ and $f^{- 1} (y)$ is a $L i n d e l \ddot{o} f$ subset of $X$ , thus there exists a $A \in A$ such that $f^{- 1} (y) \subset A \subset f^{- 1} (W)$ . It is not difficulty to prove that $y \in Y - f (X - A) \in C$ and $Y - f (X - A) \subset W$ . This proves that $C$ is a base for $Y$ . □

Theorem 3.6. (AC+GCH) Let $f : X \to Y$ be a closed $L i n d e l \ddot{o} f$ mapping and $w (X)$ be a regular cardinality. If $Y$ is base-paracompact, then $X$ is base-paracompact.

Proof. If $w (X) = ω_{0}$ , then $X$ is metrizable. So $X$ is base-paracompact from [7, Theorem 3.3].

If $w (X) > ω_{0}$ , then $c f (w (X)) = w (X) > ω_{0}$ because $w (X)$ is a regular cardinality. Thus $w (X) \geq w (Y)$ from Lemma 3.5. So $X$ is base-paracompact from Corollary 3.2. □

Remark 3.7. Theorem 3.6 does not hold for Hausdorff domains by Example 2.7. But we do not know if the condition “ $w (X)$ be a regular cardinality” in Theorem 3.6 can be omitted.

The author would like to thank the referee for his/her valuable amendments and suggestions very much.

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Department of Mathematics, Suzhou University, Suzhou 215006, P.R.China

E-mail address: geying@pub.sz.jsinfo.net

Received March 31, 2006; Revised version April 24, 2006