# 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.