When is a Pseudocompact Space Metrizable?

Compactness, countably compactness and pseudocompactness are three successively weaker properties. It follows easily from definitions that

$(A) \ \ \ \ \ \text{compact} \Rightarrow \text{countably compact} \Rightarrow \text{pseudocompact}$

None of these arrows can be reversed. It is well known that either compactness or countably complactness plus having a $G_\delta$-diagonal implies metrizability. We have:

$(B) \ \ \ \ \ \text{compact} + \text{having a } G_\delta \text{-diagonal} \Rightarrow \text{metrizable}$

$(C) \ \ \ \ \ \text{countably compact} + \text{having a } G_\delta \text{-diagonal} \Rightarrow \text{metrizable}$

A question can be asked whether these results can be extended to pseudocompactness.

Question $(D) \ \ \ \ \ \text{pseudocompact compact} + \text{having a } G_\delta \text{-diagonal} \Rightarrow \text{metrizable?}$

The answer to this question is no. The space defined using a maximal almost disjoint family of subsets of $\omega$ is an example of a non-metrizable pseudocompact space with a $G_\delta$-diagonal (discussed in this post). In this post we show that if we strengthen “having a $G_\delta$-diagonal” to being submetrizable, we have a theorem. Specifically, we show:

$(E) \ \ \ \ \ \text{pseudocompact} + \text{submetrizable} \Rightarrow \text{metrizable}$

For the result of $(B)$, see this post. For the result of $(C)$, see this post. In this post, we discuss the basic properties of pseudocompactness that build up to the result of $(E)$. All spaces considered here are at least Tychonoff (i.e. completely regular). For any basic notions not defined here, see [1] or [2].

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Pseudocompact Spaces

A space $X$ is said to be pseudocompact if every real-valued continuous function defined on $X$ is a bounded function. Any real-valued continuous function defined on a compact space must be bounded (and is thus pseudocomppact). If there were an unbounded real-valued continuous function defined on a space $X$, then $X$ would have a countably infinite discrete set (thus not countably compact). Thus countably compact implies pseudocompact, as indicated by $(A)$.

A space $X$ is submetrizable if there is a coarser (i.e. weaker) topology that is a metrizable topology. Specifically the topological space $(X,\tau)$ is submetrizable if there is another topology $\tau^*$ that can be defined on $X$ such that $\tau^* \subset \tau$ and $(X,\tau^*)$ is metrizable. The Sorgenfrey line is non-metrizable and yet the Sorgenfrey topology has a weaker topology that is metrizable, namely the Euclidean topology of the real line.

The following two theorems characterizes pseudocompact spaces in terms of locally finite open family of open sets (Theorem 1) and the finite intersection property (Theorem 2). Both theorems are found in Engelking (Theorem 3.10.22 and Theorem 3.10.23 in page 207 of [1]). Theorem 3 states that in a pseudocompact space, closed domains are pseudocompact (the definition of closed domain is stated before the theorem). Theorem 4 is the main theorem (result $E$ stated above).

Theorem 1
Let $X$ be a space. The following conditions are equivalent:

1. The space $X$ is pseudocompact.
2. If $\mathcal{V}$ is a locally finite family of non-empty open subsets of $X$, then $\mathcal{V}$ is finite.
3. If $\mathcal{V}$ is a locally finite open cover of $X$, then $\mathcal{V}$ is finite.
4. If $\mathcal{V}$ is a locally finite open cover of $X$, then $\mathcal{V}$ has a finite subcover.

Proof
$1 \Rightarrow 2$
Suppose that condition $2$ does not hold. Then there is an infinite locally finite family of non-empty open sets $\mathcal{V}$ such that $\mathcal{V}=\left\{V_1,V_2,V_3,\cdots \right\}$. We wish to define an unbounded continuous function using $\mathcal{V}$.

This is where we need to invoke the assumption of complete regularity. For each $n$ choose a point $x_n \in V_n$. Then for each $n$, there is a continuous function $f_n:X \rightarrow [0,n]$ such that $f_n(x_n)=n$ and $f_n(X-V_n) \subset \left\{ 0 \right\}$. Define $f:X \rightarrow [0,\infty)$ by $f(x)=f_1(x)+f_2(x)+f_3(x)+\cdots$.

Because $\mathcal{V}$ is locally finite, the function $f$ is essentially pointwise the sum of finitely many $f_n$. In other words, for each $x \in X$, for some positive integer $N$, $f_j(x)=0$ for all $j \ge N$. Thus the function $f$ is well defined and is continuous at each $x \in X$. Note that for each $x_n$, $f(x_n) \ge n$, showing that it is unbounded.

The directions $2 \Rightarrow 3$ and $3 \Rightarrow 4$ are clear.

$4 \Rightarrow 1$
Let $g:X \rightarrow \mathbb{R}$ be a continuous function. We want to show that $g$ is a bounded function. Consider the open family $\mathcal{O}=\left\{\cdots,O_{-3},O_{-2},O_{-1},O_0,O_1,O_2,O_3,\cdots \right\}$ where each $O_n=g^{-1}((n,n+2))$. Note that $\mathcal{O}$ is a locally finite family in $X$ since its members $O_n=g^{-1}((n,n+2))$ are inverse images of members of a locally finite family in the range space $\mathbb{R}$. By condition $4$, $\mathcal{O}$ has a finite subcover, leading to the conclusion that $g$ is a bounded function. $\blacksquare$

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

1. The space $X$ is pseudocompact.
2. If $\mathcal{O}=\left\{O_1,O_2,O_3,\cdots \right\}$ is a family of non-empty open subsets of $X$ such that $O_n \supset O_{n+1}$ for each $n$, then $\bigcap \limits_{n=1}^\infty \overline{O_n} \ne \varnothing$.
3. If $\mathcal{V}=\left\{V_1,V_2,V_3,\cdots \right\}$ is a family of non-empty open subsets of $X$ such that $\mathcal{V}$ has the finite intersection property, then $\bigcap \limits_{n=1}^\infty \overline{V_n} \ne \varnothing$.

Proof
$1 \Rightarrow 2$
Suppose that $X$ is pseudocompact. Suppose $\mathcal{O}=\left\{O_1,O_2,O_3,\cdots \right\}$ satisfies the hypothesis of condition $2$. If there is some positive integer $m$ such that $O_n=O_m$ for all $n \ge m$, then we are done. So assume that $O_n$ are distinct for infinitely many $n$. According to condition $2$ in Theorem 1, $\mathcal{O}$ must not be a locally finite family. Then there exists a point $x \in X$ such that every open set containing $x$ must meet infinitely many $O_n$. This implies that $x \in \overline{O_n}$ for infinitely many $n$. Thus $x \in \bigcap \limits_{n=1}^\infty \overline{O_n}$.

$2 \Rightarrow 3$
Suppose $\mathcal{V}=\left\{V_1,V_2,V_3,\cdots \right\}$ is a family of non-empty open sets with the finite intersection property as in the hypothesis of $3$. Then let $O_1=V_1$, $O_2=V_1 \cap V_2$, $O_3=V_1 \cap V_2 \cap V_3$, and so on. By condition $2$, we have $\bigcap \limits_{n=1}^\infty \overline{O_n} \ne \varnothing$, which implies $\bigcap \limits_{n=1}^\infty \overline{V_n} \ne \varnothing$.

$3 \Rightarrow 1$
Let $g:X \rightarrow \mathbb{R}$ be a continuous function such that $g$ is unbounded. For each positive integer $n$, let $V_n=\left\{x \in X: \lvert g(x) \lvert > n \right\}$. Clearly the open sets $V_n$ have the finite intersection property. Because $g$ is unbounded, it follows that $\bigcap \limits_{n=1}^\infty \overline{V_n} = \varnothing$. $\blacksquare$

Let $X$ be a space. Let $A \subset X$. The interior of $A$, denoted by $\text{int}(A)$, is the set of all points $x \in X$ such that there exists an open set $O$ with $x \in O \subset A$. Points of $\text{int}(A)$ are called the interior points of $A$. A subset $C \subset X$ is said to be a closed domain if $C=\overline{\text{int}(C)}$. It is clear that $C$ is a closed domain if and only if $C$ is the closure of an open set.

Theorem 3
The property of being a pseudocompact space is hereditary with respect to subsets that are closed domains.

Proof
Let $X$ be a pseudocompact space. We show that $\overline{U}$ is pseudocompact for any nonempty open set $U \subset X$. Let $Y=\overline{U}$ where $U$ is a non-empty open subset of $X$. Let $S_1 \supset S_2 \supset S_3 \supset \cdots$ be a decreasing sequence of open subsets of $Y$. Note that each $S_i$ contains points of the open set $U$. Let $O_i=S_i \cap U$ for each $i$. Note that the open sets $O_i$ form a decreasing sequence of open sets in the pseudocompact space $X$. By Theorem 2, we have $\bigcap \limits_{n=1}^\infty \overline{O_n} \ne \varnothing$ (closure here is with respect to $X$). Note that points in $\bigcap \limits_{n=1}^\infty \overline{O_n}$ are also points in $\bigcap \limits_{n=1}^\infty \overline{S_n}$ (closure with respect to $Y$). By Theorem 2, $Y=\overline{U}$ is pseudocompact. $\blacksquare$

Theorem 4 (Statement $E$ above)
Let $X$ be a pseudocompact submetrizable space. Then $X$ is metrizable.

Proof
Let $(X,\tau)$ be a pseudocompact submetrizable space. Then there exists topology $\tau^*$ on $X$ such $(X,\tau^*)$ is metrizable and $\tau^* \subset \tau$. We show that $\tau \subset \tau^*$, leading to the conclusion that $(X,\tau)$ is also metrizable. If $A \subset X$, we denote the closure of $A$ in $(X,\tau)$ by $cl_{\tau}(A)$ and the closure of $A$ in $(X,\tau^*)$ by $cl_{\tau^*}(A)$.

To show that $\tau \subset \tau^*$, we show any closed set with respect to the topology $\tau$ is also a closed set with respect to the topology $\tau^*$. Let $C$ be a closed set in $(X,\tau)$. Consider the family $\mathcal{W}=\left\{cl_{\tau}(U): U \in \tau \text{ and } C \subset U \right\}$. We make the following claims.

Claim 1. $C=\bigcap \left\{W: W \in \mathcal{W} \right\}$.

Claim 2. Each $W \in \mathcal{W}$ is pseudocompact in $(X,\tau)$.

Claim 3. Each $W \in \mathcal{W}$ is pseudocompact in $(X,\tau^*)$.

Claim 4. Each $W \in \mathcal{W}$ is compact in $(X,\tau^*)$.

We now discuss each of these four claims. For Claim 1, it is clear that $C \subset \bigcap \left\{W: W \in \mathcal{W} \right\}$. The reverse set inclusion follows from the fact that $X$ is a regular space. Claim 2 follows from Theorem 3. Note that sets in $\mathcal{W}$ are closed domains in the pseudocompact space $(X,\tau)$.

If sets in $\mathcal{W}$ are pseudocompact in the larger topology $\tau$, they would be pseudocompact in the weaker topology $\tau^*$ too. Thus Claim 3 is established. In a metrizable space, compactness and weaker notions such as countably compactness and pseudocompactness coincide. Because they are pseudocompact subsets, sets in $\mathcal{W}$ are compact in the metrizable space $(X,\tau^*)$. Thus Claim 4 is established.

It follows that $C$ is closed in $(X,\tau^*)$ since it is the intersection of compact sets in $(X,\tau^*)$. Thus $(X,\tau)$ is identical to $(X,\tau^*)$, implying that $(X,\tau)$ is metrizable. $\blacksquare$

Reference

1. Engelking, R., General Topology, Revised and Completed edition, 1989, Heldermann Verlag, Berlin.
2. Willard, S., General Topology, 1970, Addison-Wesley Publishing Company.