# The Finite Intersection Property in Compact Spaces and Countably Compact Spaces

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.

1. $X$ is countably compact.
2. Every countable collection of closed subsets of $X$ satisfying the finite intersection property has non-empty intersection.
3. 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.

1. $X$ is countably compact.
2. Every locally finite collection of subsets of $X$ is finite.
3. 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) 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

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 $\alpha < \omega_1$,

1. $x_\alpha \in X-\bigcup\lbrace{st(x_\beta,\mathcal{T}_{m(\beta)}): \beta < \alpha}\rbrace$, and
2. 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

1. Ceder, J. G. Some generalizations of metric spaces, Pacific J. Math., 11 (1961), 105-125.
2. Chaber, Conditions which imply compactness in countably compact spaces, Bull. Acad. Pol. Sci. Ser. Math., 24 (1976), 993-998.
3. 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

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

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

# Perfect Image of Separable Metric Spaces

In a previous post on countable network, it was shown that having a countable network is equivalent to being the continuous image of a separable metric space. Since there is an example of a non-metrizable space with countable netowrk, the continuous image of a separable metric space needs not be a separable metric space. However, the perfect image of a separable metrizable space is separable metrizable. First some definitions. A continuous mapping $f:X \rightarrow Y$ is a closed mapping if $f(H)$ is closed in $Y$ for any closed set $H \subset X$. A continuous surjection $f:X \rightarrow Y$ is a perfect mapping if $f$ is closed and $f^{-1}(y)$ is compact for each $y \in Y$.

Let $f:X \rightarrow Y$ be a perfect mapping where $X$ has a countable base $\mathcal{B}$. Assume $\mathcal{B}$ is closed under finite unions. Because $f$ is a closed mapping, $f(X-B)$ is closed and $f(B)$ is open in $Y$ for each $B \in \mathcal{B}$. We show that $\mathcal{B}_f=\lbrace{f(B):B \in \mathcal{B}}\rbrace$ is a base for $Y$. Let $y \in Y$ and $U \subset Y$ be open with $y \in U$. For each $x \in f^{-1}(y)$, choose $B_x \in \mathcal{B}$ such that $f(B_x) \subset U$. Since $f^{-1}(y)$ is compact, we can choose $B_{x(0)},...,B_{x(n)}$ that cover $f^{-1}(y)$. Let $B=B_{x(0)} \cup ... \cup B_{x(n)}$, which is in $\mathcal{B}$. We have $y \in f(B) \subset U$. Thus the topology on $Y$ can be generated by $\mathcal{B}_f$.

Update (11/24/2009):
The proof in the above paragraph is faulty. Thanks to Dave Milovich for pointing this out. Here’s the corrected proof.

Let me first prove a lemma.

Lemma. Let $f: X \rightarrow Y$ be a closed mapping and let $V \subset X$ be open. Then $f_*(V)=\lbrace{y \in Y:f^{-1}(y) \subset V}\rbrace$ is open in $Y$. Furthermore, $f_*(V) \subset f(V)$.

Proof of Lemma. Since $f$ is a closed mapping, $f(X-V)$ is closed. We claim that $f(X-V)=Y-f_*(V)$. It is clear that $f(X-V) \subset Y-f_*(V)$. To show that $Y-f_*(V) \subset f(X-V)$, let $z \in Y-f_*(V)$. Then $f^{-1}(z)$ cannot be a subset of $V$. Choose $x \in f^{-1}(z)-V$. Then we have $z=f(x) \in f(X-V)$. Thus $f(X-V)=Y-f_*(V)$ and $f_*(V)$ is open. It is straitforward to verify that $f_*(V) \subset f(V)$.

Now I prove that the perfect image of a separable metric space is a separable metric space. Let $f:X \rightarrow Y$ be a perfect mapping where $X$ has a countable base $\mathcal{B}$. Assume $\mathcal{B}$ is closed under finite unions. We show that $\mathcal{B}_f=\lbrace{f_*(B):B \in \mathcal{B}}\rbrace$ is a base for $Y$.

Let $y \in Y$ and $U \subset Y$ be open with $y \in U$. For each $x \in f^{-1}(y)$, choose $B_x \in \mathcal{B}$ such that $x \in B_x$ and $f(B_x) \subset U$. Since $f^{-1}(y)$ is compact, we can choose $B_{x(0)},...,B_{x(n)}$ that cover $f^{-1}(y)$. Let $B=B_{x(0)} \cup ... \cup B_{x(n)}$, which is in $\mathcal{B}$. Since $f^{-1}(y) \subset B$, we have $y \in f_*(B)$. We also have $f_*(B) \subset f(B) \subset U$. Thus the topology on $Y$ can be generated by the countable base $\mathcal{B}_f$.

# Metrization Theorems for Compact Spaces

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. 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

1. 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

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.

1. $X^2$ is perfectly normal.
2. $X$ has a $G_\delta-$diagonal.
3. $X$ 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

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

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$\copyright$ 2009 – Dan Ma, Revised February 1, 2018

# Network Weight of Topological Spaces – II

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.