The Finite Intersection Property in Compact Spaces and Countably Compact Spaces

Posted on November 30, 2009 by Dan Ma

This is a discussion of a basic result about compact spaces, namely the characterization of compactness in terms of the finite intersection property. All spaces are at least Hausdorff. A space $X$ is compact if and only if every collection of close subsets of $X$ satisfying the finite intersection property has non-empty intersection. We also present a version of this theorem for countably compactness. We also show that in countably compact space, any locally finite collection of non-empty subsets is finite. As a corollary, in the class of paracompact spaces, countably compactness is equivalent to compactness.

A collection $\mathcal{C}$ of subsets of $X$ has the finite interection property if for every finite $F \subset \mathcal{C}$ , $\bigcap F \neq \phi$ .

Theorem 1. A space $X$ is compact if and only if every collection of closed subsets of $X$ satisfying the finite intersection property has non-empty intersection.

Proof. $\Rightarrow$ Let $X$ be compact. Let $\mathcal{C}$ be a collection of closed subsets of $X$ . We show that if $\mathcal{C}$ has the finite intersection property, then it has non-empty intersection. Suppose that $\bigcap \mathcal{C}=\phi$ . Then $\mathcal{U}=\lbrace{X-C:C \in \mathcal{C}}\rbrace$ is an open cover of $X$ . By the compactness of $X$ , $\mathcal{U}$ has a finite subcover $\lbrace{X-C_0,X-C_1,...,X-C_n}\rbrace$ . it follows that $\bigcap_{i \leq n} C_i = \phi$ .

$\Leftarrow$ Let $\mathcal{U}$ be an open cover of $X$ such that it has no finite subcover. Note that $\mathcal{C}=\lbrace{X-U:U \in \mathcal{U}}\rbrace$ has the finite intersection property and $\bigcap \mathcal{C}=\phi$ .

Theorem 2. Let $X$ be a space. The following conditions are equivalent.

$X$ is countably compact.
Every countable collection of closed subsets of $X$ satisfying the finite intersection property has non-empty intersection.
For every sequence $\lbrace{C_0,C_1,C_2,...}\rbrace$ of non-empty closed subsets of $X$ where $C_0 \supset C_1 \supset C_2 \supset ...$ , we have $\bigcap_{n<\omega} C_n \neq \phi$ .

Proof. The direction of $1 \rightarrow 2$ is analogous to the direction of $\Rightarrow$ in Theorem 1. The direction of $2 \rightarrow 3$ is obvious.

$3 \rightarrow 1$ . Let $\mathcal{U}=\lbrace{U_0,U_1,U_2,...}\rbrace$ be an open cover of $X$ . For each $n$ , define $V_n=\bigcup_{i \leq n} U_i$ and $C_n=X-V_n$ . Note that $C_0 \supset C_1 \supset C_2 \supset ...$ . If $X=V_n$ for some $n$ , then we are done. If not, each $C_{n}$ is non-empty and we have $\bigcap_{n<\omega} C_n \neq \phi$ . Then we have a point that does not belong to all $V_n$ , which is a contradiction. Thus we must have $X=V_n$ for some $n$ .

Theorem 3. Let $X$ be a space. Then the following conditions are equivalent.

$X$ is countably compact.
Every locally finite collection of subsets of $X$ is finite.
Every infinite subset of $X$ has an accumulation point.

Proof. $1 \rightarrow 2$ . Suppose we have an infinite locally finite collection $\lbrace{A_n \subset X:n \in \omega}\rbrace$ . For each $n$ , let $C_n=\bigcup_{i \ge n} \overline{A_i}$ , which is a closed set. Note that $C_0 \supset C_1 \supset ...$ and $\bigcap_{n<\omega} C_n = \phi$ . By theorem 2, $X$ is not countably compact.

$2 \rightarrow 3$ . Suppose we have $A \subset X$ such that $A$ has no accumulation point. Then for each $x \in X$ , there is an open set $O_x$ such that $x \in O_x$ and $O_x$ contains no point of $A$ other than $x$ . This means that the singletons $\lbrace{x}\rbrace$ , where $x \in A$ , form a discrete collection of subsets of $X$ (thus a locally finite collection). By (2), $A$ must be finite.

$3 \rightarrow 1$ . Suppose we have a sequence $\lbrace{C_0,C_1,C_2,...}\rbrace$ of closed subsets of $X$ where $C_0 \supset C_1 \supset C_2 \supset ...$ . We want to show that $\bigcap_{n<\omega} C_n \neq \phi$ . Then by Theorem 2, $X$ is countably compact. If there is some $n$ such that $C_n=C_m$ for all $m \ge n$ , then we are done. So assume that the sequence of $C_n$ is strictly decreasing. So there is an increasing sequence of integers $m(0)<m(1)<m(2)<...$ such that we can choose $x_i \in C_{m(i)}-C_{m(i)+1}$ . Then $A=\lbrace{x_0,x_1,x_2,...}\rbrace$ is an infinite set. Let $x$ be an accumulation point of $A$ . It follows that $x \in \bigcap_{n<\omega} C_n$ .

Corollary. If $X$ is countably compact and paracompact, then $X$ is compact.

Countably Compact Spaces with G-delta Diagonals

Posted on November 29, 2009 by Dan Ma

It is a classic result in general topology that any compact space with a $G_\delta-$ diagonal is metrizable ([3]). This theorem also holds for countably compact spaces (due to Chaber in [2]). The goal of this post is to present a proof of this theorem. We prove that if $X$ is countably compact and has a $G_\delta-$ diagonal, then $X$ is compact and thus metrizable. All spaces are at least Hausdorff. This post has a discussion on the theorem on compact spaces with $G_\delta-$ diagonal. This post has a discussion on some metrizaton theorems for compact spaces.

If $\mathcal{T}$ is a collection of subsets of a space $X$ , then for each $x \in X$ , define $st(x,\mathcal{T})=\bigcup\lbrace{T \in \mathcal{T}:x \in T}\rbrace$ . A sequence of open covers $\lbrace{\mathcal{T}_n:n \in \omega}\rbrace$ of the space $X$ is a $G_\delta-$ diagonal sequence for $X$ if for each $x \in X$ , we have $\lbrace{x}\rbrace=\bigcap_{n<\omega} st(x,\mathcal{T}_n)$ . We use the following lemma (due to Ceder, [1]). This lemma was proved in this previous post.

Lemma. The space $X$ has a $G_\delta-$ diagonal if and only if it has a $G_\delta-$ diagonal sequence.

Theorem. Let $X$ be a countably compact space that has a $G_\delta-$ diagonal. Then $X$ is compact.

Proof. Let $X$ be a countably compact space. Let $\lbrace{\mathcal{T}_n:n \in \omega}\rbrace$ be a $G_\delta-$ diagonal sequence for $X$ . If $X$ is Lindelof, then we are done. Suppose we have an open cover $\mathcal{V}$ of $X$ that has no countable subcover. From this open cover $\mathcal{V}$ , we derive a contradiction.

We inductively, for each $\alpha < \omega_1$ , choose a point $x_\alpha \in X$ and an integer $m(\alpha) \in \omega$ with the following properties:

For each $\alpha < \omega_1$ ,

$x_\alpha \in X-\bigcup\lbrace{st(x_\beta,\mathcal{T}_{m(\beta)}): \beta < \alpha}\rbrace$ , and
the open cover $\mathcal{V}$ does not have a countable subcollection that covers $X-\bigcup_{\beta \leq \alpha} st(x_\beta,\mathcal{T}_{m(\beta)})$ .

To start off, choose $x_0 \in X$ . There is an integer $m(0) \in \omega$ such that no countable subcollection of $\mathcal{V}$ covers $X-st(x_0,\mathcal{T}_{m(0)})$ . Suppose this integer $m(0)$ does not exist. Then for each $n \in \omega$ , we have a countable $\mathcal{V}_n \subset \mathcal{V}$ such that $\mathcal{V}_n$ covers $X-st(x_0,\mathcal{T}_n)$ . Then $\bigcup_{n<\omega} \mathcal{V}_n$ would be a countable subcollection of $\mathcal{V}$ that covers $X-\lbrace{x_0}\rbrace$ . This would mean that $\mathcal{V}$ has a countable subcover of $X$ .

Suppose that $\lbrace{x_\beta:\beta<\alpha}\rbrace$ and $\lbrace{m(\beta):\beta<\alpha}\rbrace$ have been chosen such that conditions (1) and (2) are satisfied for each $\beta<\alpha$ . We have the following claim. Proving this claim allows us to choose $x_\alpha$ and $m(\alpha)$ .

Claim. No countable subcollection of $\mathcal{V}$ covers $X-\bigcup_{\beta<\alpha} st(x_\beta,\mathcal{T}_{m(\beta)})$ .

Suppose we do have a countable $\mathcal{W} \subset \mathcal{V}$ such that $\mathcal{W}$ covers $X-\bigcup_{\beta<\alpha} st(x_\beta,\mathcal{T}_{m(\beta)})$ . Then $\mathcal{S}=\lbrace{st(x_\beta,\mathcal{T}_m(\beta)):\beta < \alpha}\rbrace \cup \mathcal{W}$ is a countable open cover of $X$ and thus has a finite subcover $\mathcal{F}$ . Let $\delta$ be the largest ordinal $<\alpha$ such that $st(x_\delta,\mathcal{T}_m(\delta))$ is in this finite subcover $\mathcal{F}$ . Then $\mathcal{W}$ is a counntable subcollection of $\mathcal{V}$ that covers $X-\bigcup_{\beta \leq \delta} st(x_\beta,\mathcal{T}_{m(\beta)})$ . This violates condition (2) above for the ordinal $\delta$ . This proves the claim.

Now, pick $x_\alpha \in X-\bigcup\lbrace{st(x_\beta,\mathcal{T}_{m(\beta)}): \beta < \alpha}\rbrace$ . There must be some integer $m(\alpha) \in \omega$ such that conditon (2) above is satisfied for $\alpha$ . If not, for each $n \in \omega$ , there is some countable $\mathcal{V}_n \subset \mathcal{V}$ such that $\mathcal{V}_n$ covers $X-\bigcup_{\beta \leq \alpha} st(x_\beta,\mathcal{T}_n)$ . Then $\bigcup_{n<\omega} \mathcal{V}_n$ would be a countable subcollection of $\mathcal{V}$ that covers $X-\biggl(\bigcup_{\beta<\alpha} st(x_\beta,\mathcal{T}_{m(\beta)}) \biggr) \bigcup \lbrace{x_\alpha}\rbrace$ . This would mean that $\mathcal{V}$ has a countable subcover of $X-\bigcup_{\beta<\alpha} st(x_\beta,\mathcal{T}_{m(\beta)})$ . This violates the above claim. Now the induction process is completed.

To conclude the proof of the theorem, note that there is some $n \in \omega$ and there is some uncountable $D \subset \omega_1$ such that for each $\alpha \in D$ , $n=m(\alpha)$ . Then $Y=\lbrace{x_\alpha:\alpha \in D}\rbrace$ is an uncountable closed and discrete set in $X$ . Note that each open set in $\mathcal{T}_n$ contains at most one point of $Y$ . Thus $X$ must be Lindelof. With $X$ being countably compact, $X$ is compact.

Reference

Ceder, J. G. Some generalizations of metric spaces, Pacific J. Math., 11 (1961), 105-125.
Chaber, Conditions which imply compactness in countably compact spaces, Bull. Acad. Pol. Sci. Ser. Math., 24 (1976), 993-998.
Sneider, V., Continuous images of Souslin and Borel sets: metrization theorems, Dokl. Acad. Nauk USSR, 50 (1945), 77-79.

Ψ-Spaces – spaces from almost disjoint families

Posted on November 27, 2009 by Dan Ma

As the title suggests the $\Psi-$ spaces are defined using almost disjoint families, in our case, of subsets of $\omega$ . This is a classic example of a pseudocompact space that is not countably compact. This example is due to Mrowka ([3]) and Isbell (credited in [2]), It is sometimes called the Mrowka space in the literature. This is another example that is a useful counterexample and set-theoretic construction. This being an introduction, I prove that the $\Psi-$ space, when it is defined using a maximal almost disjoint family of subsets of $\omega$ , is pseudocompact and not countably compact. On the other hand, I show that if a normal space is pseudocompact space, then it is countably compact. All spaces in this note is at least Hausdorff.

A space $X$ is countably compact if every countable open cover of $X$ has a finite subcover. According to Theorem 3.10.3 in [1], a space $X$ is countably compact if and only if every infinite subset of $X$ has an accumulation point. A space $X$ is pseudocompact if every real-valued continuous function defined on $X$ is bounded. It is clear that if $X$ is a countably compact space, then it is pseudocompact. We show that the converse does not hold by using the example of $\Psi-$ space. We also show that the converse does hold for normal spaces.

Let $\mathcal{A}$ be a family of infinite subsets of $\omega$ . The family $\mathcal{A}$ is said to be an almost disjoint family if for each two distinct $A,B \in \mathcal{A}$ , $A \cap B$ is finite. The almost disjoint family $\mathcal{A}$ is said to be a maximal almost disjoint family if $B$ is an infinite subset of $\omega$ such that $B \notin \mathcal{A}$ , then $B \cap A$ is infinite for some $A \in \mathcal{A}$ .

There is an almost disjoint family $\mathcal{A}$ of subsets of $\omega$ such that $\lvert \mathcal{A} \lvert=\text{continuum}$ . To see this, identify $\omega$ (the set of all natural numbers) with $\mathbb{Q}=\lbrace{r_0,r_1,r_2,...}\rbrace$ (the set of all rational numbers). For each real number $x$ , choose a subsequence of $\mathbb{Q}$ consisting of distinct elements that converges to $x$ . Then the family of all such sequences of rational numbers would be an almost disjoint family. By a Zorn’s Lemma argument, this almost disjoint family is contained within a maximal almost disjoint family. Thus we also have a maximal almost disjoint family of cardinality continuum. On the other hand, there is no countably infinite maximal almost disjoint family of subsets of $\omega$ . See comment below Theorem 2.

Let $\mathcal{A}$ be an infinite almost disjoint family of subsets of $\omega$ . Let’s define the space $\Psi(\mathcal{A})$ . The underlying set is $\Psi(\mathcal{A})=\mathcal{A} \cup \omega$ . Points in $\omega$ are isolated. For $A \in \mathcal{A}$ , a basic open set of of the form $\lbrace{A}\rbrace \cup (A-F)$ where $F \subset \omega$ is finite. It is straightforward to verify that $\Psi(\mathcal{A})$ is Hausdorff, first countable and locally compact. It has a countable dense set of isolated points. Note that $\mathcal{A}$ is an infinite discrete and closed set in the space $\Psi(\mathcal{A})$ . Thus $\Psi(\mathcal{A})$ is not countably compact. We have the following theorems.

Theorem 1. Let $\mathcal{A}$ be an infinite maximal almost disjoint family of subsets of $\omega$ . Then $\Psi(\mathcal{A})$ is pseudocompact.

Proof. Suppose we have a continuous $f:\Psi(\mathcal{A}) \rightarrow \mathbb{R}$ that is unbounded. We can find an infinite $B \subset \omega$ such that $f$ is unbounded on $B$ . If $B \in \mathcal{A}$ , then we have a contradiction since $\lbrace{f(n):n \in B}\rbrace$ is a sequence that does not converge to $f(B)$ . So we have $B \notin \mathcal{A}$ . By the maximality of $\mathcal{A}$ , $C=B \cap A$ is infinite for some $A \in \mathcal{A}$ . Then $\lbrace{f(n):n \in C}\rbrace$ is a sequence that does not converge to $f(A)$ , another contradiction. So $\Psi(\mathcal{A})$ is pseudocompact.

Theorem 2. For a normal space $X$ , if $X$ is pseudocompact, then $X$ is countably compact.

Proof. Suppose $X$ is not countably compact. Then we have an infinite closed and discrete set $A=\lbrace{a_0,a_1,a_2,...}\rbrace$ in $X$ . Define $f:A \rightarrow \mathbb{R}$ by $f(a_n)=n$ . According to the Tietze-Urysohn Theorem, in a normal space, any continuous function defined on a closed subset of $X$ can be extended to a continuous function defined on all of $X$ . Then $f:A \rightarrow \mathbb{R}$ can be extended to a continuous $f^*:X \rightarrow \mathbb{R}$ , making $X$ not pseudocompact.

Comment
If there is a countably infinite maximal almost disjoint family $\mathcal{B}$ of subsets of $\omega$ , then $\Psi(\mathcal{B})$ is a countable first countable space and is thus has a countable base (hence is normal). By Theorem 1, it is pseudocompact. By Theorem 2, it is countably compact. Yet $\mathcal{B}$ is an infinite closed and discrete subset of $\Psi(\mathcal{B})$ , contradicting that it is countably compact. Thus there is no countably infinite maximal almost disjoint family $\mathcal{B}$ of subsets of $\omega$ . In fact, we have the following corollary.

Corollary. If $\mathcal{A}$ is an infinite maximal almost disjoint family of subsets of $\omega$ , then $\Psi(\mathcal{A})$ cannot be normal.

Reference

Engelking, R., General Topology, Revised and Completed Edition, 1988, Heldermann Verlag Berlin.
Gillman, L., and Jerison, M., Rings of Continuous Functions, 1960, Van Nostrand, Princeton, NJ.
Mrowka, S., On completely regular spaces, Fund. Math., 41, (1954) 105-106.

Perfect Image of Separable Metric Spaces

Posted on November 23, 2009 by Dan Ma

This article was written on November 23, 2009 and is replaced with a better article with the same contents (see here).

$\text{ }$

Dan Ma Bow-Tie space
Daniel Ma Bow-Tie space

Dan Ma perfect map
Daniel Ma perfect map

Dan Ma separable metric space
Daniel Ma separable metric space

Dan Ma topology
Daniel Ma topology

$\copyright$ 2009-2023 – Dan Ma

Revised: March 28, 2023

Metrization Theorems for Compact Spaces

Posted on November 22, 2009 by Dan Ma

In this blog I have already presented two metrization theorems for compact spaces: (1) any compact space with a countable network is metrizable (see the post), (2) any compact space with a $G_\delta-$ diagonal is metrizable (see the post). I now present another classic theorem: any countably compact space with a point-countable base is metrizable. This theorem is a classic result of Miscenko ([1]). All spaces are at least Hausdorff and regular. We have the following three metrization theorems for compact spaces. In subsequent posts, I will discuss generalizations of these theorems and discuss related concepts.

Thoerem 1. Any compact space with a countable network is metrizable.
The proof is in this post.

Thoerem 2. Any compact space with a $G_\delta-\text{diagonal}$ is metrizable.
The proof is in this post.

Thoerem 3. Any countably compact space with a point-countable base is metrizable.

A base $\mathcal{B}$ for a space $X$ is a point-countabe base if every point in $X$ belongs to at most countably elements of $\mathcal{B}$ .

Proof of Theorem 3. Let $\mathcal{B}$ be a point-countable base for the countably compact space $X$ . We show that $X$ is separable. Once we have a countable dense subset, the base $\mathcal{B}$ has to be a countable base. So we inductively define a sequence of countable sets $\lbrace{D_0,D_1,...}\rbrace$ such that $D=\bigcup_{n<\omega}D_n$ is dense in $X$ .

Let $D_0=\lbrace{x_0}\rbrace$ be a one-point set to start with. For $n>0$ , let $E_n=\bigcup_{i<n}D_n$ . Let $\mathcal{B}_n=\lbrace{B \in \mathcal{B}:B \cap E_n \neq \phi}\rbrace$ . For each finite $T \subset \mathcal{B}_n$ such that $X - \bigcap T \neq \phi$ , choose a point $x(T) \in X - \bigcup T$ . Let $D_n$ be the union of $E_n$ and the set of all points $x(T)$ . Let $D=\bigcup_{n<\omega}D_n$ .

We claim that $\overline{D}=X$ . Suppose we have $x \in X-\overline{D}$ . Let $\mathcal{A}=\lbrace{B \in \mathcal{B}:B \cap D \neq \phi \phantom{X} \text{and} \thinspace x \notin B}\rbrace$ . We know that $\mathcal{A}$ is countable since every element of $\mathcal{A}$ contains points of the countable set $D$ . We also know that $\mathcal{A}$ is an open cover of $\overline{D}$ . By the countably compactness of $\overline{D}$ , we can find a finite $T \subset \mathcal{A}$ such that $\overline{D} \subset \bigcup T$ . The finite set $T$ must have appeared during the induction process of selecting points for $D_n$ for some $n$ (i.e. $T \subset \mathcal{B}_n$ ). So a point $x(T)$ has been chosen such that $x(T) \notin \bigcup T$ (thus we have $x(T) \in D_n \subset \overline{D}$ ). On the other hand, since $\overline{D} \subset \bigcup T$ , we observe that $x(T) \notin \overline{D}$ , producing a contradiction. Thus the countable set $D$ is dense in $X$ , making the point-countable base $\mathcal{B}$ a countable base.

Reference

Miscenko, A., Spaces with a point-countable base, Dokl. Acad. Nauk SSSR, 144 (1962), 985-988. (English translation: Soviet Math. Dokl. 3 (1962), 1199-1202).

Compact Spaces With G-delta Diagonals

Posted on November 18, 2009 by Dan Ma

In a previous post, I showed that any compact space with a countable network is metrizable. Another classic metrization theorem for compact spaces is that any compact space with a $G_\delta-$ diagonal is metrizable ([6]). The theroem I try to prove is: for a compact space $X$ , $X^2$ is perfectly normal if and only if $X$ has a $G_\delta-$ diagonal if and only if $X$ is metrizable. My proof is based on the notion of $G^*_\delta-$ diagonal. Every compact space with a $G_\delta-$ diagonal has a $G^*_\delta-$ diagonal, which allows us to define a countable base. The theorem discussed here had been generalized (see the comment at the end of this post). All spaces are at least Hausdorff.

Let $X$ be a space. The set $\Delta=\lbrace{(x,x):x \in X}\rbrace$ is called the diagonal of the space $X$ . The space $X$ has a $G_\delta-$ diagonal if $\Delta$ is a $G_\delta-$ set in $X^2$ .

Let $\mathcal{G}$ be a collection of subsets of $X$ and let $x \in X$ . Define $st(x,\mathcal{G})=\bigcup \lbrace{G \in \mathcal{G}:x \in G}\rbrace$ . A sequence $\lbrace{\mathcal{G}_n}\rbrace_{n<\omega}$ of open covers of $X$ is called a $G_\delta-$ diagonal sequence of $X$ if for each $x \in X$ , $\lbrace{x}\rbrace=\bigcap_{n<\omega} st(x,\mathcal{G}_n)$ . Lemma 1 shows that a space has a $G_\delta-$ diagonal if and only if it has a $G_\delta-$ diagonal sequence. This lemma is due to Ceder ([2]).

Another notion we need is that of the $G^*_\delta-$ diagonal. The space $X$ has a $G^*_\delta-$ diagonal if there is a $G_\delta-$ diagonal sequence $\lbrace{\mathcal{G}_n}\rbrace_{n<\omega}$ such that for each $x \in X$ , $\lbrace{x}\rbrace=\bigcap_{n<\omega} \overline{st(x,\mathcal{G}_n)}$ . Such a $G_\delta-$ diagonal sequence is called a $G^*_\delta-$ diagonal sequence. The notion of $G^*_\delta-$ diagonal is due to R. E. Hodel ([4]). Lemma 2 below shows that any compact space with a $G_\delta-$ diagonal has a $G^*_\delta-$ diagonal.

Lemma 1. The space $X$ has a $G_\delta-$ diagonal if and only if it has a $G_\delta-$ diagonal sequence.

Proof. $\Rightarrow$ Suppose that $\Delta=\bigcap_{n<\omega}U_n$ where each $U_n$ is open in $X^2$ . Let $\tau$ denote the topology on $X$ . For each $n$ , let $\mathcal{G}_n=\lbrace{V \in \tau:V \times V \subset U_n}\rbrace$ . We claim that for each $x \in X$ , $\lbrace{x}\rbrace=\bigcap_{n<\omega} st(x,\mathcal{G}_n)$ . Obviously $\lbrace{x}\rbrace \subset \bigcap_{n<\omega} st(x,\mathcal{G}_n)$ . Let $y \in \bigcap_{n<\omega} st(x,\mathcal{G}_n)$ . For each $n$ , $y \in V_n$ where $V_n \in \mathcal{G}_n$ and $x \in V_n$ . Thus $(x,y) \in V_n \times V_n \subset U_n$ for each $n$ . This implies $(x,y) \in \Delta$ and $x=y$ . Thus $\lbrace{x}\rbrace=\bigcap_{n<\omega} st(x,\mathcal{G}_n)$ . We have established that $\{ \mathcal{G}_n \}$ is a $G_\delta$ -diagonal sequence of $X$ .

$\Leftarrow$ Suppose $\lbrace{\mathcal{G}_n}\rbrace$ is a $G_\delta-$ diagonal sequence of $X$ . For each $n$ , let $U_n=\bigcup \lbrace{V \times V:V \in \mathcal{G}_n}\rbrace$ . Since $\bigcap_n st(x,\mathcal{G}_n)=\{ x \}$ , $\Delta \subset \bigcap_{n<\omega} U_n$ . To show the set inclusion for the other direction, let $(x,y) \in \bigcap_{n<\omega} U_n$ . For each $n$ , $(x,y) \in V_n \times V_n$ for some $V_n \in \mathcal{G}_n$ . This implies that $y \in st(x,\mathcal{G}_n)$ for each $n$ . It follows that $y=x$ . Thus $\Delta = \bigcap_{n<\omega} U_n$ .

Lemma 2. If $X$ is compact and has a $G_\delta-$ diagonal, then $X$ has a $G^*_\delta-$ diagonal. Furthermore, each open cover in the $G_\delta-$ diagonal sequence is finite.

Proof. Let $\lbrace{\mathcal{G}_n}\rbrace_{n<\omega}$ be the $G_\delta-$ diagonal sequence obtained in Lemma 1. We inductively define $\lbrace{\mathcal{H}_n}\rbrace_{n<\omega}$ , another $G_\delta-$ diagonal sequence.

Using the compactness of $X$ , obtain a finite subcollection $\mathcal{H}_0$ of $\mathcal{G}_0$ such that $\mathcal{H}_0$ is a cover of $X$ . Here’s how I obtain $\mathcal{H}_1$ . For each $x \in X$ , choose an open set $G_x \in \mathcal{G}_1$ and an open set $H_x \in \mathcal{H}_0$ such that $x \in G_x$ and $x \in H_x$ . Choose open set $V_x$ such that $x \in V_x$ and $\overline{V_x} \subset G_x \cap H_x$ . Let $\mathcal{H}_1$ be a finite subcollection of $\lbrace{V_x:x \in X}\rbrace$ such that $\mathcal{H}_1$ is a cover of $X$ . Continue the inductive process and we produce a sequence of open covers $\lbrace{\mathcal{H}_n}\rbrace_{n<\omega}$ satisfying the following two claims.

Claim 1
For each $x \in X$ , $\lbrace{x}\rbrace=\bigcap_{n<\omega} st(x,\mathcal{H}_n)$ .

Because each open cover $\mathcal{H}_n$ is chosen to be a subcover or a refinement of the open cover $\mathcal{G}_n$ , we have $st(x,\mathcal{H}_n) \subset st(x,\mathcal{G}_n)$ . Since $\lbrace{x}\rbrace=\bigcap_{n<\omega} st(x,\mathcal{G}_n)$ (from the definition of $G_\delta-$ diagonal sequence), we have $\lbrace{x}\rbrace=\bigcap_{n<\omega} st(x,\mathcal{H}_n)$ .

Claim 2
For each $x \in X$ , $\lbrace{x}\rbrace=\bigcap_{n<\omega} \overline{st(x,\mathcal{H}_n)}$ .

We only need to show $\bigcap_{n<\omega} \overline{st(x,\mathcal{H}_n)} \subset \lbrace{x}\rbrace$ . Let $y \in \overline{st(x,\mathcal{H}_n)}$ for each $n$ . Because $\mathcal{H}_n$ is finite, $y \in \overline{V}$ for some $V \in \mathcal{H}_n$ with $x \in V$ . Each such $\overline{V} \subset U$ for some $U \in \mathcal{H}_{n-1}$ . Thus $y \in \overline{st(x,\mathcal{H}_n)}$ implies $y \in st(x,\mathcal{H}_{n-1})$ . By Claim 1, $y=x$ .

We have shown that $X$ has a $G^*_\delta-$ diagonal by producing a $G^*_\delta-$ diagonal sequence $\lbrace{\mathcal{H}_n}\rbrace_{n<\omega}$ .

Theorem. If $X$ is compact and has a $G_\delta-$ diagonal, then $X$ is metrizable.

Proof. Let $\lbrace{\mathcal{H}_n}\rbrace_{n<\omega}$ be the $G^*_\delta-$ diagonal sequence obtained in Lemma 2. Furthermore, each $\mathcal{H}_n$ is a finite open cover. Let $\mathcal{H}=\bigcup_{n<\omega} \mathcal{H}_n$ . The collection $\mathcal{H}$ satisfies the properties stated in the following two claims.

Claim 3
For each $x,y \in X$ with $x \neq y$ , there is a $U \in \mathcal{H}$ such that $x \in U$ and $y \notin \overline{U}$ .

Since $\lbrace{x}\rbrace=\bigcap_{n<\omega} \overline{st(x,\mathcal{H}_n)}$ , $y \notin \overline{st(x,\mathcal{H}_n)}$ for some $n$ . Then there is some $U \in \mathcal{H}_n$ such that $x \in U$ and $y \notin \overline{U}$ . In fact, for any $U \in \mathcal{H}_n$ with $x \in U$ , it must be the case that $y \notin \overline{U}$ .

Claim 4
Let $\mathcal{B}=\lbrace{X-\overline{\bigcup F}: F \subset \mathcal{H} \phantom{x} and \phantom{x} \vert F \lvert < \omega}\rbrace$ . Then $\mathcal{B}$ is a countable base for $X$ .

To see Claim 4, note that $\mathcal{B}$ is a cover of $X$ and is closed under finite intersections. This makes $\mathcal{B}$ is base for a topology. To show that this base generate the same topology on $X$ , let $y \in U \subset X$ where $U$ is open. Then $X-U$ is compact. For each $x \in X-U$ , let $V_x \in \mathcal{H}$ such that $x \in V_x$ and $y \notin \overline{V_x}$ . We can choose $F=\lbrace{V_{x(0)},...,V_{x(n)}}\rbrace$ such that $F$ is a cover of $X-U$ . Then $y \in X-\overline{\bigcup F} \subset U$ .

With Claims 3 and 4, the theorem is established.

Corollary. Let $X$ be a compact space. The following conditions are equivalent.

$X^2$ is perfectly normal.
$X$ has a $G_\delta-$ diagonal.
$X$ has a countable base.

Proof. $1 \rightarrow 2$ and $3 \rightarrow 1$ are obvious. $2 \rightarrow 3$ follows from the theorem.

Examples. Based on the corollary, any non-metrizable compact Hausdorff space does not have a $G_\delta-$ diagonal. One handy example is the uncountable product of the unit interval $I^{\omega_1}$ where $I=[0,1]$ . Both $I \times I$ with the lexicographic order and the double arrow space are compact and non-metrizable (thus have no $G_\delta-$ diagonal). I discussed these two spaces in a previous post.

Comment. The notion of $G_\delta-$ diagonal plays an important role in metrization theorems. The theorem for compact space with $G_\delta-$ diagonal had long been generalized. For example, in [3] Chaber had shown that any countably compact space with a $G_\delta-$ diagonal is metrizable. In [1] and [5], it was shown that any paracompact space with a $G_\delta-$ diagonal is submetrizable. The theorem proved in this post would simply be a corollary of this result. In upcoming posts, I plan to discuss some of these theorems as well as explore the connection of submetrizability and various $G_\delta-$ diagonal properties.

Reference

Borges, C. R. On stratifiable spaces, Pacific J. Math., 17 (1966), 1-16.
Ceder, J. G. Some generalizations of metric spaces, Pacific J. Math., 11 (1961), 105-125.
Chaber, Conditions which imply compactness in countably compact spaces, Bull. Acad. Pol. Sci. Ser. Math., 24 (1976), 993-998.
Hodel, R., E., Moore spaces and $w \Delta-$ spaces, Pacific J. Math., 38, (1971), 641-652.
Okuyama, A., On metrizability of M-spaces, Proc. Japan. Acad., 40, 176-179.
Sneider, V., Continuous images of Souslin and Borel sets: metrization theorems, Dokl. Acad. Nauk USSR, 50 (1945), 77-79.

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Network Weight of Topological Spaces – II

Posted on November 17, 2009 by Dan Ma

This is a continuation of the discussion on network. In the previous post, I showed that the network weight (the minimum cardinality of a network) coincides with the weight for both metrizable spaces and locally compact spaces. In another post, I showed that this is true for compact spaces. I now show that this is also true for the class of Moore spaces. First, some definitions. A sequence $\lbrace{\mathcal{D}_n}\rbrace_{n<\omega}$ of open covers of a space $X$ is a development for $X$ if for each $x \in X$ and each open set $U \subset X$ with $x \in U$ , there is some $n$ such that any open set in $\mathcal{D}_n$ containing the point $x$ is contained in $U$ . A developable space is one that has a development. A Moore space is a regular developable space.

For a collection of $\mathcal{G}$ of subsets of a space $X$ and for $x \in X$ , define $st(x,\mathcal{G})=\bigcup\lbrace{U \in \mathcal{G}:x \in U}\rbrace$ . An equivalent way of defining a development: A sequence $\lbrace{\mathcal{D}_n}\rbrace_{n<\omega}$ of open covers of a space $X$ is a development for $X$ if for each $x \in X$ , $\lbrace{st(x,\mathcal{G}_n):n \in \omega}\rbrace$ is a local base at $x$ . For a basic introduction to Moore space and the Moore space conjecture, there are numerous places to look in the literature ([1] being one of them).

Theorem. If $X$ is a Moore space, then $nw(X)=w(X)$ .

Proof. Since $nw(X) \leq w(X)$ always holds, we only need to show $w(X) \leq nw(X)$ . To this end, we exhibit a base $\mathcal{B}$ with $\vert \mathcal{B} \lvert \leq nw(X)$ . Let $\lbrace{\mathcal{D}_n}\rbrace_{n<\omega}$ be a development for $X$ . Let $\mathcal{N}$ be a network with cardinality $nw(X)$ .

For each $N \in \mathcal{N}$ , choose open set $O(n,N) \in \mathcal{D}_n$ such that $N \subset O(n,N)$ . Let $\mathcal{B}_n=\lbrace{O(n,N):N \in \mathcal{N}}\rbrace$ and $\mathcal{B}=\bigcup_{n<\omega}\mathcal{B}_n$ . Note that $\lvert \mathcal{B} \lvert \leq nw(X)$ . Because $\mathcal{N}$ is a network, each $\mathcal{B}_n$ is a cover of $X$ . To see this, let $x \in X$ . Choose some $V \in \mathcal{D}_n$ such that $x \in V$ . There is some $N \in \mathcal{N}$ such that $x \in N \subset V$ . Then $x \in O(n,N)$ . For each $n$ , $\mathcal{B}_n \subset \mathcal{D}_n$ . The sequence $\lbrace{\mathcal{B}_n}\rbrace$ works like a development. We have just shown that $\mathcal{B}$ is a base for $X$ .

Corollary. The example of Butterfly space is not a Moore space.

The example of the Butterfly (or Bow-tie) space is defined in this previous post. This space has a countable network and the weight of this space is continuum. Thus this space cannot be a Moore space.

Reference
[1] Steen, L. A. & Seebach, J. A. [1995] Counterexamples in Topology, Dover Books.

Network Weight of Topological Spaces – I

Posted on November 16, 2009 by Dan Ma

In the previous post, I discussed the notion of network of a topological space. It was noted that for any space $X$ , the network weight (the least cardinality of a network for $X$ ) is always $\leq$ the weight (the least cardinality of a base for $X$ ). When $X$ is compact, the network weight and weight would coincide. Are there other classes of spaces for which network weight = weight? I would like to discuss two other classes of spaces where network weight and weight coincide, namely metrizable spaces and locally compact spaces. The following two theorems are proved. For a basic discussion on network, see the previous post.

Theorem 1. If $X$ is metrizable, then $nw(X)=w(X)$ .

Proof. For the case of $w(X)=\omega$ , we have $nw(X)=\omega$ . Now consider the case that $w(X)$ is an uncountable cardinal. Based on the Bing-Nagata-Smirnov metrization theorem, any metrizable space has a $\sigma-$ discrete base. Let $\mathcal{B}=\bigcup_{n<\omega} \mathcal{B}_n$ be a $\sigma-$ discrete base for the metrizable space $X$ . Now let $\mathcal{K}=\lvert \mathcal{B} \lvert$ . Because each $\mathcal{B}_n$ is a discrete collection of open sets, any network woulld have cardinality at least as big as $\lvert \mathcal{B}_n \lvert$ for each $n$ . If $\mathcal{K}=\lvert \mathcal{B}_n \lvert$ for some $n$ , then $\mathcal{K} \leq nw(X)$ . If $\mathcal{K}$ is the least upper bound of $\lvert \mathcal{B}_n \lvert$ , then $\mathcal{K} \leq nw(X)$ . Both cases imply $w(X) \leq nw(X)$ . Since $nw(X) \leq w(X)$ always hold, we have $w(X)=nw(X)$ .

Theorem 2. If $X$ is a locally compact space, then $nw(X)=w(X)$ .

Proof. Let $X$ be locally compact where $nw(X)=\mathcal{K}$ . The idea is that we can obtain a base for $X$ of cardinality $\leq \mathcal{K}$ (i.e. $w(X) \leq nw(X)$ ). Let $\mathcal{N}$ be a network whose cardinality is $\mathcal{K}$ . Here’s a sketch of the proof. Each point in $X$ has an open neighborhood whose closure is compact. For the compact closure of such open neighborhood, the weight would coincide with the network weight. Thus we can find a base of size $\leq \mathcal{K}$ within such open neighborhood. Because $\lvert \mathcal{N} \lvert=\mathcal{K}$ , we only need to consider $\mathcal{K}$ many such open neighborhoods with compact closure. Thus we can obtain a base for $X$ of cardinality $\leq \mathcal{K}$ . To make this sketch more precise, consider the following three claims.

Claim 1
The collection of all $N \in \mathcal{N}$ , where $\overline{N}$ is compact, is a cover of the space $X$ .

Claim 2
For every compact set $A \subset X$ , there is an open set $U$ such that $A \subset U$ and $\overline{U}$ is compact.

Claim 3
If $U \subset X$ is open with $\overline{U}$ compact, then we can obtain a base $\mathcal{B}_U$ for the open subspace $U$ with $\vert \mathcal{B}_U \vert \leq \mathcal{K}$ .

For each $N \in \mathcal{N}$ in Claim 1, we can select an open $U$ (as in Claim 2)such that $\overline{N} \subset U$ and $\overline{U}$ is compact. Let $\mathcal{B}$ be the union of all the $\mathcal{B}_U$ in Claim 3 over all such $U$ . There are $\leq \mathcal{K}$ many $N$ in Claim 1. Thus $\vert \mathcal{B} \lvert \leq \mathcal{K}$ . Note that $\mathcal{B}$ would form a base for the whole space $X$ .

Both Claim 1 and Claim 2 are direct consequence of locally compactness. To see Claim 3, let $U$ be open such that $\overline{U}$ is compact. We have $nw(\overline{U}) \leq nw(X)=\mathcal{K}$ (the network weight of a subspace cannot exceed the original network weight). By the result in the previous post, we have $w(\overline{U})=nw(\overline{U})$ . We now have $w(U) \leq w(\overline{U})$ (the weight of a subspace cannot exceed the weight of the space containing it). So the weight of any open subspace with compact closure cannot exceed $\mathcal{K}$ .

Corollary. For both metrizable spaces and locally compact spaces $X$ , $w(X) \leq \lvert X \lvert$ .

Spaces With Countable Network

Posted on November 14, 2009 by Dan Ma

The concept of network is a useful tool in working with generalized metric spaces. A network is like a base for a topology, but the members of a network do not have to be open. After a brief discussion on network, the focus here is on the spaces with networks that are countably infinite in size. The following facts are presented:

Any space with a countable network is separable and Lindelof.
The property of having a countable network is hereditary. Thus any space with a countable network is hereditarily separable and hereditarily Lindelof.
The property of having a countable network is preserved by taking countable product.
The Sorgenfrey Line is an example of a hereditarily separable and hereditarily Lindelof space that has no countable network.
For any compact space $X$ , $nw(X)=w(X)$ . In particular, any compact space with a countable network is metrizable.
As a corollary to 5, $w(X) \leq \vert X \vert$ for any compact $X$ .
A space $X$ has a countable network if and only if it is the continuous impage of a separable metric space (hence such a space is sometimes called cosmic).
Any continuous image of a cosmic space is cosmic.
Any continuous image of a compact metric space is a compact metric space.
As a corollary to 2, any space with countable network is perfectly normal.
An example is given to show that the continuous image of a separable metric space needs not be metric (i.e. an example of a cosmic space that is not metrizable).

All spaces in this discussion are at least $T_3$ (Hausdorff and regular). Let $X$ be a space. A collection $\mathcal{N}$ of subsets of $X$ is said to be a network for $X$ if for each $x \in X$ and for each open $U \subset X$ with $x \in U$ , then we have $x \in N \subset U$ for some $N \in \mathcal{N}$ . The network weight of a space $X$ , denoted by $nw(X)$ , is defined as the minimum cardinality of all the possible $\vert \mathcal{N} \vert$ where $\mathcal{N}$ is a network for $X$ . The weight of a space $X$ , denoted by $w(X)$ , is defined as the minimum cardinality of all possible $\vert \mathcal{B} \vert$ where $\mathcal{B}$ is a base for $X$ . Obviously any base is also a network. Thus $nw(X) \leq w(X)$ . For any compact space $X$ , $nw(X)=w(X)$ . On the other hand, the set of singleton sets is a network. Thus $nw(X) \leq \vert X \vert$ .

Our discussion is based on an important observation. Let $\mathcal{T}$ be the topology for the space $X$ . Let $\mathcal{K}=nw(X)$ . We can find a base $\mathcal{B}_0$ that generates a weaker (coarser) topology such that $\lvert \mathcal{B}_0 \lvert=\mathcal{K}$ . We can also find a base $\mathcal{B}_1$ that generates a finer topology such that $\lvert \mathcal{B}_1 \lvert=\mathcal{K}$ . These are restated as lemmas.

Lemma 1. We can define base $\mathcal{B}_0$ that generates a weaker (coarser) topology $\mathcal{S}_0$ on $X$ such that $\lvert \mathcal{B}_0 \lvert=\mathcal{K}$ . Thus $w(X,\mathcal{S}_0) \leq nw(X)$ .

Proof. Let $\mathcal{N}$ be a network for $(X,\mathcal{T})$ such that $\vert \mathcal{N} \vert=nw(X,\mathcal{T})$ . Consider all pairs $N_0,N_1 \in \mathcal{N}$ such that there exist disjoint $O_0,O_1 \in \mathcal{T}$ with $N_0 \subset O_0$ and $N_1 \subset O_1$ . Such pairs exist because we are working in a Hausdorff space. Let $\mathcal{B}_0$ be the collection of all such open sets $O_0,O_1$ and their finite interections. This is a base for a topology and let $\mathcal{S}_0$ be the topology generated by $\mathcal{B}_0$ . Clearly, $\mathcal{S}_0 \subset \mathcal{T}$ and this is a Hausdorff topology. Note that $w(X,\mathcal{S}_0) \leq \vert \mathcal{B}_0 \vert =\vert \mathcal{N} \vert$ .

Lemma 2. We can define base $\mathcal{B}_1$ that generates a finer topology $\mathcal{S}_1$ on $X$ such that $\lvert \mathcal{B}_1 \lvert=\mathcal{K}$ . Thus $w(X,\mathcal{S}_1) \leq nw(X)$ .

Proof. As before, let $\mathcal{N}$ be a network for $(X,\mathcal{T})$ such that $\vert \mathcal{N} \vert=nw(X,\mathcal{T})$ . Since we are working in a regular space, we can assume that the sets in $\mathcal{N}$ are closed. If not, take closures of the elements of $\mathcal{N}$ and we still have a network. Consider $\mathcal{B}_1$ to be the set of all finite intersections of elements in $\mathcal{N}$ . This is a base for a topology on $X$ . Let $\mathcal{S}_1$ be the topology generated by this base. Clearly, $\mathcal{T} \subset \mathcal{S}_1$ . It is also clear that $w(X,\mathcal{S}_1) \leq nw(X)$ . The only thing left to show is that the finer topology is regular. Note that the network $\mathcal{N}$ consists of closed sets in the topology $\mathcal{T}$ . Thus the sets in the base $\mathcal{B}_1$ also consists of closed sets with respect to $\mathcal{T}$ and the sets in $\mathcal{B}_1$ are thus closed in the finer topology. Since $\mathcal{B}_1$ is a base consisting of cloased and open sets, the topology $\mathcal{S}_1$ regular.

Discussion of 1, 2, and 3
Points 1, 2 and 3 are basic facts about countable network and they are easily verified based on definitions. They are called out for the sake of having a record.

Discussion of 4
The Sorgenfrey Line does not have a countable network for the same reason that the Sorgenfrey Plane is not Lindelof. If the Sorgenfrey Line has a countable netowrk, then the Sorgenfrey plane would have a countable network and hence Lindelof.

Discussion of 5
In general, $nw(X) \leq w(X)$ . In a compact Hausdorff space, any weaker Hausdorff topology must conincide with the original topology. So the weaker topology produced in Lemma 1 must coincide with the original topology. In the countable case, any compact space with a countable network has a weaker topology with a countable base. This weaker topology must coincide with the original topology.

Discussion of 6
Note that $nw(X) \leq \lvert X \lvert$ always holds. For compact spaces, we have $w(X)=nw(X) \leq \lvert X \lvert$ .

Discussion of 7
Let $X$ be a space with a countable network. By Lemma 2, $X$ has a finer topology that has a countable base. Let $Y$ denote $X$ with this finer second countable topology. Then the identity map from $Y$ onto $X$ is continuous.

For the other direction, let $f:Y \rightarrow X$ be a continuous function mapping a separable metric space $Y$ onto $X$ . Let $\mathcal{B}$ be a countable base for $Y$ . Then $\lbrace{f(B):B \in \mathcal{B}}\rbrace$ is a network for $X$ .

Discussion of 8
This is easily verified. Let $X$ is the continuous image of a cosmic space $Y$ . Then $Y$ is the continuous image of some separable metric space $Z$ . It follows that $X$ is the continuous image of $Z$ .

Discussion of 9
Let $X$ be compact metrizable and let $Y$ be a continuous image of $X$ . Then $Y$ is compact. By point 7, $Y$ has a countable network. By point 5, $Y$ is metrizable.

Discussion of 10
A space is perfectly normal if it is normal and that every closed subset is a $G_\delta-$ set. Let $X$ be a space with a countable network. The normality of $X$ comes from the fact that it is regular and Lindelof. Note that $X$ is also hereditarily Lindelof. In a hereditarily Lindelof and regular space, every open subspace is an $F_\sigma-$ set (thus every closed set is a $G_\delta-$ set.

Discussion of 11 (Example of cosmic but not separable metrizable space)
This is the “Butterfly” space or “Bow-tie” space due to L. F. McAuley. I found this example in [Michael]. Let $Y=T \cup S$ where
$T=\lbrace{(x,y) \in \mathbb{R}^2:y>0}\rbrace$ and
$S=\lbrace{(x,y) \in \mathbb{R}^2:y=0}\rbrace$ .

Points in $T$ have the usual plane open neighborhoods. A basic open set at $p \in S$ is of the form $B_c(p)$ where $B_c(p)$ consists of $p$ and all points $q \in Y$ having distance $<c$ from $p$ and lying underneath either one of the two straight lines in $Y$ which emanate from $p$ and have slopes $+c$ and $-c$ , respectively.

It is clear that $Y$ is a Hausdorff and regular space. The relative “Bow-tie” topologies on $T$ and $S$ coincide with the usual topology on $T$ and $S$ , respectively. Thus the union of the usual countable bases on $T$ and $S$ would be a countable network for $Y$ . On the other hand, $Y$ is separable but cannot have a countable base (hence not metrizable).

Reference
[Michael]
Michael, E., $\aleph_0-$ spaces, J. Math. Mech. 15, 983-1002.

Perfectly Normal Spaces Can Never Be Dowker Spaces

Posted on November 12, 2009 by Dan Ma

The Dowker’s theorem states that for a normal space $X$ , $X \times [0,1]$ is normal if and only if $X$ is countably paracompact. Since this theorem was published, any normal space that is not countably paracompact became known as Dowker space. There are classes of spaces that can never be Dowker spaces (e.g. metrizable spaces, paracompact spaces, compact spaces and Lindelof spaces). In [Katetov], it was shown that there are no perfectly normal Dowker spaces. My blog has a proof of the Dowker’s theorem (see the proof here). For more background on Dowker’s spaces, see the survey article [Rudin]. Dowker’s theorem was published in [Dowker].

Theorem. If $X$ is perfectly normal, then $X$ is countably paracompact.

To prove this theorem, we use the following characterization of countably paracompactness (you can find a proof here).

Lemma. Let $X$ be a normal space. Then $X$ is countably paracompact if and only if for each sequence $\lbrace{A_n:n \in \omega}\rbrace$ of closed subsets of $X$ such that $A_0 \supset A_1 \supset ...$ and $\bigcap_n A_n=\phi$ , there exist open sets $B_n \supset A_n$ such that $\bigcap_n B_n=\phi$ .

Proof of Theorem. Suppose $X$ is perfectly normal. Let $A_0 \supset A_1 \supset ...$ be a sequence of closed sets such that $\bigcap_n A_n=\phi$ . For each $n$ , let $A_n=\bigcap_{i<\omega} U_{n,i}$ where each $U_{n,i}$ is open in $X$ . For each $n$ , define $B_n=\bigcap_{i, j \leq n}U_{i,j}$ . Clearly, $B_n \supset A_n$ . It is easy to see that $\bigcap_n B_n=\phi$ . Note that all the open sets $U_{n,j}$ are used in defining the sequence $B_0,B_1,B_2,\cdots$ . Thus $\bigcap_n B_n \neq \phi$ would imply $\bigcap_n A_n \neq \phi$ .

Comment. As a consequence of this theorem and the Dowker’s theorem, if $X$ is perfectly normal, then $X \times Y$ is normal for any compact metric space $Y$ .

Reference
[Dowker]
Dowker, C. H. [1951], On Countably Paracompact Spaces, Canad. J. Math. 3, 219-224.

[Katetov]
Katetov, M., On real-valued functions in topological spaces, Fund. Math. 38 (1951), pp. 85-91.

[Rudin]
Rudin, M. E., [1984], Dowker Spaces, Handbook of Set-Theoretic Topology (K. Kunen and J. E. Vaughan, eds), Elsevier Science Publishers B. V., Amsterdam, 761-780.

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$\copyright \ 2009-2016 \text{ by Dan Ma}$
Revised November 23, 2016

Dan Ma's Topology Blog

Expository Articles In General Topology

Monthly Archives: November 2009

The Finite Intersection Property in Compact Spaces and Countably Compact Spaces

Countably Compact Spaces with G-delta Diagonals

Ψ-Spaces – spaces from almost disjoint families

Perfect Image of Separable Metric Spaces

Metrization Theorems for Compact Spaces

Compact Spaces With G-delta Diagonals

Network Weight of Topological Spaces – II

Network Weight of Topological Spaces – I

Spaces With Countable Network

Perfectly Normal Spaces Can Never Be Dowker Spaces