# An exercise gleaned from the proof of a theorem on pseudocompact space

Filling in the gap is something that is done often when following a proof in a research paper or other published work. In fact this is necessary since it is not feasible for authors to prove or justify every statement or assertion in a proof (or define every term). The gap could be a basic result or could be an older result from another source. If the gap is a basic result or a basic fact that is considered folklore, it may be OK to put it on hold in the interest of pursuing the main point. Then come back later to fill the gap. In any case, filling in gaps is a great learning opportunity. In this post, we focus on one such example of filling in the gap. The example is from the book called Topological Function Spaces by A. V. Arkhangelskii [1].

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Pseudocompactness

The exercise we wish to highlight deals with continuous one-to-one functions defined on pseudocompact spaces. We first give a brief backgrounder on pseudocompact spaces with links to earlier posts.

All spaces considered are Hausdorff spaces. A space $X$ is a pseudocompact space if every continuous real-valued function defined on $X$ is bounded, i.e., if $f:X \rightarrow \mathbb{R}$ is a continuous function, then $f(X)$ is a bounded set in the real line. Compact spaces are pseudocompact. In fact, it is clear from definitions that

$\text{compact} \Longrightarrow \text{countably compact} \Longrightarrow \text{pseudocompact}$

None of the implications can be reversed. An example of a pseudocompact space that is not countably compact is the space $\Psi(\mathcal{A})$ where $\mathcal{A}$ is a maximal almost disjoint family of subsets of $\omega$ (see here for the details). Some basic results on pseudocompactness focus on the conditions to add in order to turn a pseudocompact space into countably compact or even compact. For example, for normal spaces, pseudocompact implies countably compact. This tells us that when looking for pseudocompact space that is not countably compact, do not look among normal spaces. Another interesting result is that pseudocompact + metacompact implies compact. Likewise, when looking for pseudocompact space that is not compact, look among non-metacompact spaces. On the other hand, this previous post discusses when a pseudocompact space is metrizable. Another two previous posts also discuss pseudocompactness (see here and here).

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The exercise

Consider Theorem II.6.2 part (c) in pp. 76-77 in [1]. We do not state the theorem because it is not the focus here. Instead, we focus on an assertion in the proof of Theorem II.6.2.

The exercise that we wish to highlight is stated in Theorem 2 below. Theorem 1 is a standard result about continuous one-to-one functions defined on compact spaces and is stated here to contrast with Theorem 2.

Theorem 1
Let $Y$ be a compact space. Let $g: Y \rightarrow Z$ be a one-to-one continuous function from $Y$ onto a space $Z$. Then $g$ is a homeomorphism.

Theorem 2
Let $Y$ be a pseudocompact space. Let $g: Y \rightarrow Z$ be a one-to-one continuous function from $Y$ onto $Z$ where $Z$ is a separable and metrizable space. Then $g$ is a homeomorphism.

Theorem 1 says that any continuous one-to-one map from a compact space onto another compact space is a homeomorphism. To show a given map between two compact spaces is a homeomorphism, we only need to show that it is continuous in one direction. Theorem 2, the statement used in the proof of Theorem II.6.2 in [1], says that the standard result for compact spaces can be generalized to pseudocompactness if the range space is nice.

The proof of Theorem II.6.2 part (c) in [1] quoted [2] as a source for the assertion in our Theorem 2. Here, we leave both Theorem 1 and Theorem 2 as exercise. One way to prove Theorem 2 is to show that whenever there exists a map $g$ as described in Theorem 2, the domain $Y$ must be compact. Then Theorem 1 will finish the job.

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Reference

1. Arkhangelskii A. V., Topological Function Spaces, Mathematics and Its Applications Series, Kluwer Academic Publishers, Dordrecht, 1992.
2. Arkhangelskii A. V., Ponomarev V. I., Fundamental of general topology: problems and exercises, Reidel, 1984. (Translated from the Russian).

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$\copyright \ 2015 \text{ by Dan Ma}$

# C*-Embedding Property and Stone-Cech Compactification

This is a continuation of an introduction of Stone-Cech compactification started in two previous posts (first post: A Beginning Look at Stone-Cech Compactification; second post: Two Characterizations of Stone-Cech Compactification). In this post, we present another characterization of the Stone-Cech compactification, that is, for any completely regular space $X$, $X$ is $C^*$-embedded in its Stone-Cech compactification $\beta X$ and that any compactification of $X$ in which $X$ is $C^*$-embedded must be $\beta X$. In other words, this property of $C^*$-embedding is unique to Stone-Cech compactification. We prove the following two theorems (U3 has two versions).

The links for other posts on Stone-Cech compactification can be found toward the end of this post.

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Definition. Let $Y$ be a space. Let $A \subset Y$. The subspace $A$ is $C^*$-embedded in $Y$ if every bounded continuous function $f:A \rightarrow \mathbb{R}$ is extendable to a continuous $\hat{f}:Y \rightarrow \mathbb{R}$.

Theorem C3
Let $X$ be a completely regular space. The space $X$ is $C^*$-embedded in its Stone-Cech compactification $\beta X$.

$\text{ }$

Theorem U3.1
Let $X$ be a completely regular space. Let $I=[0,1]$. Let $\alpha X$ be a compactification of $X$ such that each continuous $f:X \rightarrow I$ can be extended to a continuous $\hat{f}:\alpha X \rightarrow I$. Then $\alpha X$ must be $\beta X$.

$\text{ }$

Theorem U3.2
If $\alpha X$ is any compactification of $X$ that satisfies the property in Theorem C3 (i.e., $X$ is $C^*$-embedded in $\alpha X$), then $\alpha X$ must be $\beta X$.
$\text{ }$

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Other Characterizations

Two other characterizations of $\beta X$ are proved in the previous post (Two Characterizations of Stone-Cech Compactification).

Theorem C1
Let $X$ be a completely regular space. Let $f:X \rightarrow Y$ be a continuous function from $X$ into a compact Hausdorff space $Y$. Then there is a continuous $F: \beta X \rightarrow Y$ such that $F \circ \beta=f$.

$\text{ }$

Theorem U1
If $K$ is any compactification of $X$ that satisfies condition in Theorem C1, then $K$ must be equivalent to $\beta X$.
$\text{ }$

Theorem C2
Let $X$ be a completely regular space. Among all compactifications of the space $X$, the Stone-Cech compactification $\beta X$ of the space $X$ is the largest compactification.

$\text{ }$

Theorem U2
The property in Theorem C2 is unique to $\beta X$. That is, if $\alpha X$ is a compactification of $X$, then $\alpha X$ must be equivalent to $\beta X$.

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Remark

The C theorems and the U theorems are a great tool to determine whether a given compactification is $\beta X$. Whenever a compactification $\alpha X$ of a space $X$ satisfies the property belonging to a C theorem, based on the corresponding U theorem, we know that this compactification $\alpha X$ must be $\beta X$. For example, any compactification $\alpha X$ that satisfies the function extension property in Theorem C1 must be $\beta X$. Th $C^*$-embedding property in Theorem C3 and Theorem U3 (both versions) is also a function extension property much like that in Theorems C1 and U1, but is easier to use. The reason being that we only need to extend a smaller class of continuous functions (i.e., to check whether functions from $X$ into $I=[0,1]$ can be extended), rather than checking all continuous functions from $X$ to arbitrary compact spaces. As the following example below about $\beta \omega_1$ illustrates that the $C^*$-embedding in Theorem C3 and U3.1 can be used to describe $\beta X$ explicitly.

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Proving Theorem U3.1 and Theorem U3.2

Let $Y$ be a space. Let $A$ be a subspace of $X$. Recall that $A$ is $C^*$-embedded in $Y$ if every bounded continuous function $f:A \rightarrow \mathbb{R}$ can be extended to a continuous $\hat{f}:Y \rightarrow \mathbb{R}$.

Any bounded continuous function $f: X \rightarrow \mathbb{R}$ can be regarded as $f: X \rightarrow I_f$ where $I_f$ is some closed and bounded interval. The $C^*$-embedding property in Theorem C3 is a function extension property like the one in Theorem C1, except that it deals with function from $X$ into a specific type of compact spaces $Y$, namely the closed and bounded intervals in $\mathbb{R}$. Theorem C3 is a corollary of Theorem C1 (see below). So we only need to prove Theorem U3.1 and Theorem U3.2. Theorem U3.2 is a corollary of Theorem U3.1.

Proof of Theorem U3.1
By Theorem C2, we have $\alpha X \le \beta X$. So we only need to show $\beta X \le \alpha X$. To this end, we need to produce a continuous function $H: \alpha X \rightarrow \beta X$ such that $H \circ \alpha=\beta$.

Let $C(X,I)$ be the set of all continuous functions from $X$ into $I$. For each $f \in C(X,I)$, let $I_f=I$. Recall that $\beta X$ is embedded in the cube $\prod \limits_{f \in C(X,I)} I_f$ by the mapping $\beta$. For each $f \in C(X,I)$, let $\pi_f$ be the projection map from this cube into $I_f$.

Each $f \in C(X,I)$ can be expressed as $f=\pi_f \circ \beta$. Thus by assumption, each $f$ can be extended by $\hat{f}: \alpha X \rightarrow I$. Now define $H: \alpha X \rightarrow \prod \limits_{f \in C(X,I)} I_f$ by the following:

For each $t \in \alpha X$, $H(t)=a=< a_f >_{f \in C(X,I)}$ such that $a_f=\hat{f}(t)$

For each $x \in \alpha(X)$, we have $H(\alpha(x))=\beta(x)$. Note that $\hat{f}$ agrees with $f$ on $\alpha(X)$ since $\hat{f}$ extends $f$. So we have $H(\alpha(x))=a$ where $a_f=\hat{f}(\alpha(x))=f(x)$ for each $f \in C(X,I)$. On the other hand, by definition of $\beta$, we have $\beta(x)=a$ where $a_f=f(x)$ for each $f \in C(X,I)$. Thus we have $H \circ \alpha=\beta$ and the following:

$H(\alpha(X)) \subset \beta(X) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (1)$

It is straightforward to verify that $H$ is continuous. Note that $\alpha(X)$ is dense in $\alpha X$. Since $H$ is continuous, $H(\alpha(X))$ is dense in $H(\alpha X)$. Thus we have:

$H(\alpha X)=\overline{H(\alpha(X))} \ \ \ \ \ \ \ \ \ \ \ \ \ (2)$

Putting $(1)$ and $(2)$ together, we have the following:

$H(\alpha X)=\overline{H(\alpha(X))} \subset \overline{\beta(X)}=\beta X$

Thus we can describe the map $H$ as $H: \alpha X \rightarrow \beta X$. As noted before, we have $H \circ \alpha=\beta$. Thus $\beta X \le \alpha X$. $\blacksquare$

Proof of Theorem U3.2
Suppose $\alpha X$ is a compactification of $X$ such that $X$ is $C^*$-embedded in $\alpha X$. Then every bounded continuous $f:X \rightarrow I_f$ can be extended to $\hat{f}:\alpha X \rightarrow I_f$ where $I_f$ is some closed and bounded interval containing the range. In particular, this means every continuous $f:X \rightarrow I$ can be extended. By Theorem U3.1, we have $\alpha X \approx \beta X$. $\blacksquare$

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Example

This is one example where we can use $C^*$-embedding to describe $\beta X$ explicitly.

Let $\omega_1$ be the first uncountable ordinal. Let $\omega_1+1$ be the successor ordinal of $\omega_1$ (i.e. $\omega_1$ with one additional point at the end). Consider $X=\omega_1$ and $Y=\omega_1+1$ as topological spaces with the order topology derived from the well ordering of the ordinals. The space $Y$ is a compactification of $X$. In fact $Y$ is the one-point compactification of $X$.

It is well known that every continuous real-valued function on $X$ is bounded (note that $X$ here is countably compact and hence pseudocompact). Furthermore, every continuous real-valued function on $X$ is eventually constant. This means that if $f:X \rightarrow \mathbb{R}$ is continuous, for some $\alpha < \omega_1$, $f$ is constant on the final segment $X_\alpha=\left\{\rho < \omega_1: \rho>\alpha \right\}$ (see result B in The First Uncountable Ordinal). As a result, every continuous bounded real-valued function $f:X \rightarrow \mathbb{R}$ can be extended to a continuous $\hat{f}:Y \rightarrow \mathbb{R}$. Then according to Theorem U3.2, $\beta X=\beta \omega_1=Y=\beta \omega_1+1$.

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Blog Posts on Stone-Cech Compactification

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Reference

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

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$\copyright \ \ 2012$

# Completely Regular Spaces and Pseudocompact Spaces

In proving theorems about properties in abstract topological spaces, it makes sense that the spaces in questions satisfy some axioms in addition to the ones stipulated in the definition of topological spaces. For example, authors typically assume certain separation axioms. Doing so will help authors know in advance what basic properties the spaces will have. For example, it is desirable to know that singleton sets (and finite sets) are closed (assuming the $T_1$ axiom or a $T_1$ space). In some circumstances, it may be desirable to be able to separate a single point from a closed set not containing it (assuming $T_3$ axiom or regularity). In some situation, it may be advantageous (and even necessary) to know in advance that there is a sufficient quantity in continuous real-valued functions that can be defined on the spaces in question. We give several reasons of needing enough continuous functions (this list is not meant to be exhaustive).

1. Certain notions involve continuous real-valued function defined on the space. Pseudocompactness is one such notion (this view point is discussed in this post).
2. Spaces of continuous functions are the objects being studied (this view point is discussed in this post).
3. Completely regular spaces are precisely the spaces that can be embedded in a cube (the product space of copies of the unit interval). Discussed in this post.

In this post, we discuss the importance of complete regularity from the first view point and use pseudocompactness as an illustration. See [1] and [2] and [3] for any notions not defined here. Steen and Seebach (section 2 of [2] starting on p.11) has an excellent discussion of separation axioms.

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Completely Regular Spaces and Pseudocompact Spaces

A space $X$ is said to be completely regular if $X$ is a $T_0$ space and for each $x \in X$ and for each closed subset $A$ of $X$ with $x \notin A$, there is a continuous function $f:X \rightarrow [0,1]$ such that $f(A) \subset \left\{0 \right\}$ and $f(x)=1$. Note that the $T_0$ axiom and the existence of the continuous function imply the $T_1$ axiom, which is equivalent to the property that single points are closed sets. Completely regular spaces are also called Tychonoff spaces.

As defined above, in a completely regular space, for any closed set and a point not in the closed set, we can always find a continuous function mapping the closed set to 0 and the point to 1. Essentially, to be a completely regular space, it suffices to provide a continuous function that maps a given closed set and a point (not in the closed set) to two different real numbers $a$ and $b$. So in a space that is not completely regular, there exist a closed set $H$ and a point $x \notin H$ such that every real-valued continuous function $g$ that can be defined on the space maps $H$ and the point $x$ to the same real number. Thus outside of completely regular spaces, notions that are based on continuous real-valued functions may be difficult to work with.

A space $X$ is said to be pseudocompact if every real-valued continuous function $f$ defined on $X$ is a bounded function (i.e. $f(X)$ is a bounded set in the real line $\mathbb{R}$). Even though the definition does not include complete regularity, an effective discussion of pseudocompactness typically make use of complete regularity. We illustrate this point using a proof of a theorem that gives a characterization of pseudocompactness.

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
This theorem is discussed in this discussion of pseudocompactness. We repeat the proof of $1 \Rightarrow 2$ to illustrate an application of complete regularity.

$1 \Rightarrow 2$
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. $\blacksquare$

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Remark

Complete regularity is an integral part of the above proof. It simplifies the proof and clarifies the argument. As a direct corollary to the Theorem 1, any pseudocompact paracompact space is compact. A less direct corollary is that any pseudocompact metacompact space is compact (see this discussion of pseudocompactness). These results are made possible by assuming that the pseudocompact spaces are also completely regular. They are restated below.

Corollary 2
Let $X$ be a completely regular space. If $X$ is pseudocompact and paracompact, then $X$ is compact.

Corollary 3
Let $X$ be a completely regular space. If $X$ is pseudocompact and metacompact, then $X$ is compact.

It could be a valid math question to ask whether the above two results are valid outside of the class of completely regular space. We do not know the answer. We also feel that it is also a valid approach to just assume complete regularity and focus our attention on exploring the main concepts involved (in this case pseudocompactness, paracompactness and metacompactness).

We would like to remark that in working with pseudocompactness, we also need to take care that we do not assume too much. For example, we do not want to assume normality since any normal pseudocompact space is countably compact (see Theorem 2 in this post). Then working with pseudocompactness is turned into the problem of working with the stronger concept of countably compactness.

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Reference

1. Engelking, R., General Topology, Revised and Completed edition, Heldermann Verlag, Berlin, 1989.
2. Steen, L. A., Seebach, J. A., Counterexamples in Topology, Dover Publications, Inc., New York, 1995.
3. Willard, S., General Topology, Addison-Wesley Publishing Company, 1970.

# Pseudocompact + Metacompact implies Compact

It is a well known result that any countably compact and metacompact space is compact (see Theorem 5.3.2 in [1]). A discussion of this result is also found in this blog (countably compact + metacompact). Since countably compactness implies pseudocompactness, a natural question arises: can this result be generalized to “any pseudocompact and metacompact space is compact?” The answer is yes and was established in [2] and [3]. In this post, we put together a proof of this result by using building blocks already worked out in this blog.

All spaces considered here are Tychonoff (completely regular). Refer to [1] and [4] for any terms and notions not defined here (or refer to elsewhere in this blog).

A space $X$ is said to be almost compact if for every open cover $\mathcal{U}$ of $X$, there is a finite $\mathcal{V} \subset \mathcal{U}$ such that $\bigcup \mathcal{V}$ is dense in $X$. It can be shown that for any regular space, almost compactness implies compactness. We have the following lemma.

Lemma 1
Let $X$ be a regular space. Then $X$ is compact if and only if $X$ is almost compact.

Theorem 2, Theorem 3 and Theorem 4 are building blocks proved in previous posts. Theorem 5 below is the main theorem. Corollary 6, the intended result, is obtained from applying Theorem 5 and Lemma 1.

Theorem 2 (see Theorem 4B in this post)
Every regular pseudocompact is a Baire space.

Theorem 3 (see Main Theorem in this post)
Let $X$ be a space. The following conditions are equivalent.

1. $X$ is a Baire space.
2. For any point-finite open cover $\mathcal{V}$ of $X$, the set $D=\left\{x \in X: \mathcal{V} \text{ is locally finite at } x \right\}$ is a dense set in $X$.

Theorem 4 (see Theorem 1 in this post)
Let $X$ be a space. The following conditions are equivalent.

1. $X$ is a pseudocompact space.
2. If $\mathcal{W}$ is a locally finite family of non-empty open subsets of $X$, then $\mathcal{W}$ is finite.

Theorem 5
Let $X$ be a pseudocompact and metacompact space. Then $X$ is almost compact.

Proof
Let $\mathcal{U}$ be an open cover of $X$. By metacompactness, there is a $\mathcal{V}$ which is a point-finite open refinement of $\mathcal{U}$. It suffices to find a finite $\mathcal{W} \subset \mathcal{V}$ such that $\mathcal{W}$ covers a dense set.

By Theorem 2, $X$ is a Baire space. By Theorem 3, the set $D$ is dense in $X$ where $D=\left\{x \in X: \mathcal{V} \text{ is locally finite at } x \right\}$. Let $\mathcal{W}$ be the collection of all $V \in \mathcal{V}$ such that $V \cap D \ne \varnothing$. Note that $\bigcup \mathcal{W}$ is open and dense in $X$. Furthermore, it is straightforward to show that $\mathcal{W}$ is locally finite at each point $x \in D$. By Theorem 4, $\mathcal{W}$ is finite. $\blacksquare$

Corollary 6
Let $X$ be a pseudocompact and metacompact space. Then $X$ is compact.

Reference

1. Engelking, R., General Topology, Revised and Completed edition, Heldermann Verlag, Berlin, 1989.
2. Scott, B., M., Pseudocompact Metacompact Spaces are Compact, Topology Proc., 4, 577-587, 1979.
3. Watson, W. S., Pseudocompact Metacompact Spaces are Compact, Proc. Amer. Math. Soc., 81, 151-152, 1981.
4. Willard, S., General Topology, Addison-Wesley Publishing Company, 1970.

# Baire Category Theorem and the Finite Intersection Property

A Baire space is a topological space in which the intersection of any countable family of open and dense sets is dense (equivalently every non-empty open subset is of second category). One version of the Baire category theorem states that every complete metric space is a Baire space. Another common version states that every compact Hausdorff space is a Baire space. Another version states that every locally compact Hausdorff space is a Baire space. The commonality among these versions is the finite intersection property (whenever a collection of a certain type of sets satisfies the property that any finite subcollection has non-empty intersection, the whole collection has non-empty intersection). For each of these classes of spaces, in addition to countably compact spaces and pseudocompact spaces, Baire category theorem is derived from having one specific form of the finite intersection property. In this post, we explore this relationship.

In each of the following theorem pairs, the B Theorem follows from the A theorem. The A theorem is a form of the finite intersection property and the B theorem is a version of Baire category theorem.

Another interesting observation is that the finite intersection properties discussed here can give a stronger property than being a Baire space. This stronger property is defined by the Banach-Mazur game.

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Theorem 1A
Let $(X, \rho)$ be a metric space. Then the following conditions are equivalent.

1. $(X, \rho)$ is a complete metric space.
2. For each decreasing sequence $C_1 \supset C_2 \supset C_3 \supset \cdots$ of non-empty closed subsets of $X$ such that the diameters of the sets $C_n$ converge to zero, we have $\bigcap \limits_{n=1}^\infty C_n \ne \varnothing$.

Theorem 1B
Every complete metric space is a Baire space.

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Theorem 2A
Let $X$ be a Hausdorff space. Then the following conditions are equivalent.

1. $X$ is a compact space.
2. For every family $\mathcal{F}$ consisting of non-empty closed subsets of $X$, if $\mathcal{F}$ has the finite intersection property, then $\mathcal{F}$ has non-empty intersection.

Theorem 2B

• Every compact Hausdorff space is a Baire space.
• Every locally compact Hausdorff space is a Baire space.

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

1. $X$ is a countably compact space.
2. For every countable family $\mathcal{F}$ consisting of non-empty closed subsets of $X$, if $\mathcal{F}$ has the finite intersection property, then $\mathcal{F}$ has non-empty intersection.
3. For each decreasing sequence $C_1 \supset C_2 \supset C_3 \supset \cdots$ of non-empty closed subsets of $X$, we have $\bigcap \limits_{n=1}^\infty C_n \ne \varnothing$.

Theorem 3B
Every countably compact Hausdorff space is a Baire space.

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Theorem 4A
Let $X$ be a regular 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 $\bigcap \limits_{n=1}^\infty \overline{V_n} \ne \varnothing$.

Theorem 4B
Every regular pseudocompact space is a Baire space.

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Remark
Theorem 1A (the Cantor Theorem) can be found in Engelking (page 269 in [1]). Theorem 2A and Theorem 3A can also be found in Engelking (they are also proved in this post). Theorem 4B is also found in Engelking (Theorem 3.10.23 in page 207 of [1]) and is proved this post.

We would like to explicitly point out that between Thoerem 1A and Theorem 2A, none of the two theorems implies the other. For example, even though both complete metric spaces and compact Hausdorff spaces are Baire spaces, complete metric spaces are not necessarily compact and there are compact spaces that are not even metrizable. However, the finite intersection property of Theorem 2A implies that of Theorem 3A, which in turn implies the finite intersection property of Theorem 4A.
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Baire Category Theorem

The proofs of all four B theorems are amazingly similar. It is a matter of exploiting the fact that whenever a decreasing sequence of open sets satisfying the condition that each closure is a subset of the previous open set (and satisfying some other condition), the sequence of open sets has non-empty intersection. For example, for complete metric space, make sure that the closures of the open sets have diameters going to zero. For any reader who is new to this material, it will be very instructive to walk through the arguments of these Baire category theorems. The proof of Theorem 1A can be found this post. We prove Theorem 4B.

Recall that $X$ is a Baire space if $\left\{U_1,U_2,U_3,\cdots \right\}$ is a countable family of open and dense sets in $X$, $\bigcap \limits_{i=1}^\infty U_i$ is dense in $X$, or equivalently every non-empty open subset of $X$ is of second category in $X$. For more background about the concepts of Baire space and category (see [1] or this post).

Proof of Theorem 4B
Let $X$ be a regular pseudocompact space. Let $\left\{U_1,U_2,U_3,\cdots \right\}$ be a countable family of open and dense sets in $X$. Let $O$ be a non-empty open subset of $X$. We show that $O$ has to contain points of $\bigcap \limits_{n=1}^\infty U_n$. We let $O_1=O \cap U_1$. We find open $O_2$ such that $O_2 \subset U_2$ and $\overline{O_2} \subset O_1$ (using regularity). Continue this inductive process, we have for each $n$, an open $O_n$ such that $O_n \subset U_n$ and $\overline{O_n} \subset O_{n-1}$. Then we have a decreasing sequence of open sets $O_n$ as in condition 2 of Theorem 4A. Then we have $\bigcap \limits_{n=1}^\infty \overline{O_n} \ne \varnothing$. Since $\overline{O_{n+1}} \subset O_n$ for each $n$, we also have $\bigcap \limits_{n=1}^\infty O_n \ne \varnothing$. It is clear that $\bigcap \limits_{n=1}^\infty O_n \subset \bigcap \limits_{n=1}^\infty U_n$. $\blacksquare$

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Banach-Mazur Game

In proving the above versions of Baire category theorem, we can exploit the appropriate version of the finite intersection property – the situation that any nested decreasing sequence of open sets (under some specified conditions) has non-empty intersection. In fact, the finite intersection property offers more than just Baire category theorem; it can endow the space in question a type of completeness property stronger than Baire space. This completeness property is defined using the Banach-Mazur game.

The Banach-Mazur game is a two-person game played on a topological space. Let $X$ be a space. There are two players, $\alpha$ and $\beta$. They take turn choosing nested decreasing nonempty open subsets of $X$ as follows. The player $\beta$ goes first by choosing a nonempty open subset $U_0$ of $X$. The player $\alpha$ then chooses a nonempty open subset $V_0 \subset U_0$. At the nth play where $n \ge 1$, $\beta$ chooses an open set $U_n \subset V_{n-1}$ and $\alpha$ chooses an open set $V_n \subset U_n$. The player $\alpha$ wins if $\bigcap \limits_{n=0}^\infty V_n \ne \varnothing$. Otherwise the player $\beta$ wins. For more detailed discussion of the game, see this post.

One interesting point that we like to make about the finite intersection property ranging from Theorem 1A to Theorem 4A is that the player $\alpha$ can always win the Banach-Mazur game as long as he/she plays the game according to each specific version of the finite intersection. For example, playing the game in a complete metric space, player $\alpha$ always wins as long as he/she makes the diameters of the closures of the open sets going to zero. In a regular pseudocompact space, player $\alpha$ can always win by making the closure of each of his/her open sets a subset of the previous move of other player.

A topological space in which the player $\alpha$ has a winning strategy is said to be a weakly $\alpha$-favorable space. Thus complete metric spaces, compact Hausdorff spaces, locally compact Hausdorff spaces, countably compact Hausdorff spaces, regular pseudocompact spaces are all weakly $\alpha$-favorable.

There is characterization of Baire spaces in terms of the Banach-Mazur game. A space $X$ is a Baire space if and only if the player $\beta$ has no winning strategy in the Banach-Mazur game played on the space $X$ (see theorem 1 in this post). If the player $\alpha$ can always win, then player $\beta$ can never win. In terms of game terminology, if player $\alpha$ has a winning strategy, then the other player (player $\beta$) has no winning strategy. Thus a space is weakly $\alpha$-favorable implies that it is a Baire space. But the implication is not reversible (see example in this post).

So all the spaces discussed from Theorem 1A to Theorem 4A are all weakly $\alpha$-favorable, a property stronger than Baire spaces. These observations are summarized in the following theorems.

Theorem 1C
Every complete metric space is a weakly $\alpha$-favorable space.

Theorem 2C

• Every compact Hausdorff space is a weakly $\alpha$-favorable space.
• Every locally compact Hausdorff space is a weakly $\alpha$-favorable space.

Theorem 3C
Every countably compact Hausdorff space is a weakly $\alpha$-favorable space.

Theorem 4C
Every regular pseudocompact space is a weakly $\alpha$-favorable space.

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Reference

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

# A Space with G-delta Diagonal that is not Submetrizable

The property of being submetrizable implies having a $G_\delta$-diagonal. There are several other properties lying between these two properties (see [1]). Before diving into these other properties, it may be helpful to investigate a classic example of a space with a $G_\delta$-diagonal that is not submetrizable.

The diagonal of a space $X$ is the set $\Delta=\left\{(x,x): x \in X \right\}$, a subset of the square $X \times X$. An interesting property is when the diagonal of a space is a $G_\delta$-set in $X \times X$ (the space is said to have a $G_\delta$-diagonal). Any compact space or a countably compact space with this property must be metrizable (see compact and countably compact space). A space $(X,\tau)$ is said to be submetrizable if there is a topology $\tau^*$ that can be defined on $X$ such that $(X,\tau^*)$ is a metrizable space and $\tau^* \subset \tau$. In other words, a submetrizable space is a space that has a coarser (weaker) metrizable topology. Every submetrizable space has a $G_\delta$-diagonal. Note that when $X$ has a weaker metric topology, the diagonal $\Delta$ is always a $G_\delta$-set in the metric square $X \times X$ (hence in the square in the original topology). The property of having a $G_\delta$-diagonal is strictly weaker than the property of having a weaker metric topology. In this post, we discuss the Mrowka space, which is a classic example of a space with a $G_\delta$-diagonal that is not submetrizable.

The Mrowka space (also called Psi space) was discussed previously in this blog (see this post). For the sake of completeness, the example is defined here.

First, we define some basic notions. Let $\omega$ be the first infinite ordinal (or more conveniently the set of all nonnegative integers). 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. An 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 $A \in \mathcal{A}$. In other words, if you put one more set into a maximal almost disjoint family, it ceases to be almost disjoint.

A natural question is whether there is an uncountable almost disjoint family of subsets of $\omega$. In fact, there is one whose cardinality is continuum (the cardinality of the real line). To see this, identify $\omega$ with $\mathbb{Q}=\lbrace{r_0,r_1,r_2,...}\rbrace$ (the set of all rational numbers). Let $\mathbb{P}=\mathbb{R}-\mathbb{Q}$ be the set of all irrational numbers. For each $x \in \mathbb{P}$, choose a subsequence of $\mathbb{Q}$ consisting of distinct elements that converges to $x$ (in the Euclidean topology). 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 this post).

Let $\mathcal{A}$ be an infinite almost disjoint family of subsets of $\omega$. We now define a Mrowka space (or $\Psi$-space), denoted by $\Psi(\mathcal{A})$. 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 is 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 would like to point out that the definition of a Mrowka space $\Psi(\mathcal{A})$ only requires that the family $\mathcal{A}$ is an almost disjoint family and does not necessarily have to be maximal. For the example discribed in the title, $\mathcal{A}$ needs to be a maximal almost disjoint family of subsets of $\omega$.

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Example
Let $\mathcal{A}$ be a maximal almost disjoint family of subsets of $\omega$. Then $\Psi(\mathcal{A})$ as defined above is a space in which there is a $G_\delta$-diagonal that is not submetrizable.

Note that $\Psi(\mathcal{A})$ is pseudocompact (proved in this post). Because there is no countable maximal almost disjoint family of subsets of $\omega$, $\mathcal{A}$ must be an uncountable in addition to being a closed and discrete subspace of $\Psi(\mathcal{A})$ (thus the space is not Lindelof). Since $\Psi(\mathcal{A})$ is separable and is not Lindelof, $\Psi(\mathcal{A})$ is not metrizable. Any psuedocompact submetrizable space is metrizable (see Theorem 4 in this post). Thus $\Psi(\mathcal{A})$ must not be submetrizable.

On the other hand, any $\Psi$-space $\Psi(\mathcal{A})$ (even if $\mathcal{A}$ is not maximal) is a Moore space. It is well known that any Moore space has a $G_\delta$-diagonal. The remainder of this post has a brief discussion of Moore space.

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Moore Space

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.

Suppose that $X$ is a Moore space. We show that $X$ has a $G_\delta$-diagonal. That is, we wish to show that $\Delta=\left\{(x,x): x \in X \right\}$ is a $G_\delta$-set in $X \times X$.

Let $\lbrace{\mathcal{D}_n}\rbrace_{n<\omega}$ be a development. For each $n$, let $U_n=\bigcup \lbrace{V \times V:V \in \mathcal{D}_n}\rbrace$. Clearly $\Delta \subset \bigcap_{n<\omega} U_n$. Let $(x,y) \in \bigcap_{n<\omega} U_n$. For each $n$, $(x,y) \in V_n \times V_n$ for some $V_n \in \mathcal{D}_n$. We claim that $x=y$. Suppose that $x \ne y$. By the definition of development, there exists some $m$ such that every open set in $\mathcal{D}_m$ containing the point $x$ has to be a subset of $X-\left\{y \right\}$. Then $V_m \subset X-\left\{y \right\}$, which contradicts $y \in V_m$. Thus we have $\Delta = \bigcap_{n<\omega} U_n$.

The remaining thing to show is that $\Psi(\mathcal{A})$ is a Moore space. For each positive integer $n$, let $F_n=\left\{0,1,\cdots,n-1 \right\}$ and let $F_0=\varnothing$. The development is defined by $\lbrace{\mathcal{E}_n}\rbrace_{n<\omega}$, where for each $n$, $\mathcal{E}_n$ consists of open sets of the form $\lbrace{A}\rbrace \cup (A-F_n)$ where $A \in \mathcal{A}$ plus any singleton $\left\{j \right\}$ ($j \in \omega$) that has not been covered by the sets $\lbrace{A}\rbrace \cup (A-F_n)$.

Reference

1. Arhangel’skii, A. V., Buzyakova, R. Z., The rank of the diagonal and submetrizability, Commentationes Mathematicae Universitatis Carolinae, Vol. 47 (2006), No. 4, 585-597.
2. Engelking, R., General Topology, Revised and Completed edition, Heldermann Verlag, Berlin, 1989.
3. Willard, S., General Topology, Addison-Wesley Publishing Company, 1970.

# 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.
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