# Completely Regular Spaces and Function Spaces

This is a continuation of a discussion on completely regular spaces (continuing from this previous post). These spaces are an integral part of many discussions involving topological spaces and/or properties. Some notions are contingent on the existence of certain real-valued continuous functions. Discussion of such notions can be greatly facilitated by working in the class of completely regular spaces. One example given in a previous post is on a discussion of pseudocompact spaces. Another example for requiring complete regularity is when working with function spaces. When the object being studied is the space of real-valued continuous functions defined on a topological space $X$, it is desirable to have enough continuous functions in the function space being studied. In this post, we illustrate this point by giving the proofs of two simple results in $C_p(X)$, the space of real-valued continuous functions endowed with the pointwise convergence topology.

Basic references are [2] and [4]. Refer to [3] for a discussion of where complete regularity is placed among the separation axioms. An in-depth treatment for $C_p$ theory is [1].
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Completely Regular 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.

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Function Spaces

Let $X$ be a space. Let $C(X)$ be the set of all real-valued continuous functions defined on the space $X$. The set $C(X)$ is naturally a subspace of the product space $\prod \limits_{x \in X} Y_x$ where each $Y_x = \mathbb{R}$. We can also write $\prod \limits_{x \in X} Y_x = \mathbb{R}^X$. Thus $C(X)$ can be endowed with the subspace topology inherited from the product space $\mathbb{R}^X$. When this is the case, the resulting function space is denoted by $C_p(X)$.

Now we need a good handle on the open sets in the function space $C_p(X)$. A basic open set in the product space $\mathbb{R}^X$ is of the form $\prod \limits_{x \in X} U_x$ where each $U_x$ is open in $\mathbb{R}$ and $U_x = \mathbb{R}$ for all but finitely many $x \in X$ (equivalently $U_x \ne \mathbb{R}$ for only finitely many $x \in X$). Thus a basic open set in $C_p(X)$ is of the form:

$(1) \ \ \ \ \ \ \ \ C(X) \cap \prod \limits_{x \in X} U_x$

where each $U_x$ is open in $\mathbb{R}$ and $U_x = \mathbb{R}$ for all but finitely many $x \in X$. In addition, when $U_x \ne \mathbb{R}$, we can take $U_x$ to be an open interval of the form $(a,b)$. To make the basic open sets of $C_p(X)$ more explicit, $(1)$ is translated as follows:

$(2) \ \ \ \ \ \ \ \ \bigcap \limits_{x \in F} [x, O_x]$

where $F \subset X$ is a finite set, for each $x \in F$, $O_x$ is an open interval of $\mathbb{R}$, and $[x,O_x]$ is the set of all $f \in C(X)$ such that $f(x) \in O_x$. In proving results about $C_p(X)$, we can use basic open sets that are described in $(2)$.

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Completely Regular Spaces and Function Spaces

We now give two simple results in $C_p(X)$ that are good illustrations that complete regularity is the ideal setting for the domain space of $C_p$ function spaces. In Theorem 1, complete regularity is used to generate a non-separable subspace of the function space given a non-Lindelof subspace of the domain space. In Theorem 2, complete regularity is used to generate a non-Lindelof subspace of the function space given a non-separable subspace of the domain space.

Theorem 1
Let $X$ be a completely regular space. Then if $C_p(X)$ is hereditarily separable, then $X$ is hereditarily Lindelof.

Proof
We show that if $X$ has a subspace that is not Lindelof, then there is a subspace of $C_p(X)$ that is not separable. We use complete regularity to generate the continuous functions that form a non-separable subspace.

Let $Y \subset X$ be a non-Lindelof subspace. There exists an open cover $\mathcal{U}$ of $Y$ such that $\mathcal{U}$ has no countable subcover. Open sets in $\mathcal{U}$ are open sets in $Y$. We wish to expand these to open sets in $X$. Let $\mathcal{U}^*$ be the collection of open subsets $U$ of $X$ such that $U \cap Y \in \mathcal{U}$. It is clear that no countable subcollection of $\mathcal{U}^*$ can cover $Y$.

For each $y \in Y$, choose $U(y) \in \mathcal{U}^*$ such that $y \in U(y)$. Here’s where we use complete regularity. For each $y \in Y$, there is a continuous function $f_y:X \rightarrow [0,1]$ such that $f_y(X-U(y)) \subset \left\{0 \right\}$ and $f_y(y) =1$. Let $T=\left\{f_y:y \in Y \right\}$, which is a subspace of $C_p(X)$.

We claim that $T$ is not separable. To see this, let $A=\left\{g_1,g_2,g_3,\cdots \right\}$ be a countable subset of $T$ such that for each $i$, $g_i$ is obtained from the point $y(i) \in Y$ and the open set $U_i=U(y(i))$, i.e., $g_i=f_{y(i)}$. Note that $\left\{U_1,U_2,U_3,\cdots \right\}$ cannot be a cover of $Y$. Let $a \in Y$ be a point that is not in all $U_i$.

Consider the function $f_a$ chosen above using complete regularity. Note that $f_a(a)=1$ and $f_a(X-U(a)) \subset \left\{0 \right\}$. On the other hand, $g_i(a)=f_{y(i)}(a)=0$ for all $i$ since $a \notin U_i$ for all $i$. This means that $f_a$ is not in the closure of $A$. For example, $[a,(0.9,1.1)]$ is a basic open set containing $f_a$ such that $g_i \notin [a,(0.9,1.1)]$ for all $i$. Thus no countable subset of $T$ can be dense in $T$, completing the proof. $\blacksquare$

Theorem 2
Let $X$ be a completely regular space. Then if $C_p(X)$ is hereditarily Lindelof, then $X$ is hereditarily separable.

Proof
Let $Y$ be a non-separable subspace of $X$. For each countable $A \subset Y$, there must be some point $y(A) \in Y$ such that $y(A)$ is not a member of the closure of $A$ (relative to $Y$). For each countable $A \subset Y$, let $\overline{A}$ be the closure of $A$ in the entire space $X$. Clearly $y(A) \notin \overline{A}$.

Now apply the complete regularity of the space $X$. For each countable $A \subset Y$, let $f_A: X \rightarrow [0,1]$ be continuous such that $f_A(\overline{A}) \subset \left\{0 \right\}$ and $f_A(y(A))=1$. Let $W$ be the set of all $f_A$ where $A \subset Y$ is countable.

We now show that $W$ is a non-Lindelof subspace of $C_p(X)$. For each $f_A \in W$, let $U_A=[y(A),(0.9, 1.1)] \cap W$, which is open in $C_p(X)$ and contains $f_A$. Let $\mathcal{U}$ be the collection of all such open sets $U_A$.

Then $\mathcal{U}$ is an open cover of $W$ that has no countable subcover. To see this, suppose we have $\left\{U_1,U_2,U_3,\cdots \right\}$ such that for each $i$, $U_i=U_{A(i)}$ where $A(i) \subset Y$ is countable. Then let $B$ be the set of all $A(i)$ and all points $y(A(i))$. Note that the set $B$ is still a countable subset of $Y$. Consider the point $y(B)$ and the continuous function $f_B$, which is a member of $W$. We have $f_B(y(B))=1$ and $f_B(t)=0$ for all $t \in \overline{B}$. Note that each $y(A(i)) \in B$ and thus $f_B(y(A(i))=0$ for all $i$. This shows that $f_B \notin U_i=U_{A(i)}$ for all $i$. Then $\mathcal{U}$ is an open cover of $W$ that has no countable subcover, leading to the conclusion that $W$ is a non-Lindelof subspace of $C_p(X)$. $\blacksquare$

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Remark

The above proofs of Theorem 1 and Theorem 2 are only meant as demonstration of the role played by complete regularity in working with function spaces. They are much weaker versions of a deeper result. These two results can serve to motivate a deeper result that explores the relationship between the hereditary separability (respectively hereditary Lindelof property) of the domain space and the hereditary Lindelof property (respectively the hereditary separability) of the function space. The following theorem is the countable version of a theorem found in [5].

Theorem 3
Let $X$ be a completely regular space. The following conditions are equivalent.

1. $C_p(X)$ is hereditarily separable (respectively hereditarily Lindelof).
2. $X^\omega$ is hereditarily Lindelof (respectively hereditarily separable).
3. For each positive integer $n$, $X^n$ is hereditarily Lindelof (respectively hereditarily separable).

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Reference

1. Arhangel’skii, A., Topological Function Spaces, Kluwer Academic Publishers, Boston, 1992.
2. Engelking, R., General Topology, Revised and Completed edition, Heldermann Verlag, Berlin, 1989.
3. Steen, L. A., Seebach, J. A., Counterexamples in Topology, Dover Publications, Inc., New York, 1995.
4. Willard, S., General Topology, Addison-Wesley Publishing Company, 1970.
5. Zenor, P., Hereditarily m-separability and the hereditarily m-lindelof property in product spaces and function spaces, Fund. Math. 106, 175-180, 1980.