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 , 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 , 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 theory is [1].

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**Completely Regular Spaces**

A space is said to be completely regular if is a space and for each and for each closed subset of with , there is a continuous function such that and . Note that the axiom and the existence of the continuous function imply the 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 be a space. Let be the set of all real-valued continuous functions defined on the space . The set is naturally a subspace of the product space where each . We can also write . Thus can be endowed with the subspace topology inherited from the product space . When this is the case, the resulting function space is denoted by .

Now we need a good handle on the open sets in the function space . A basic open set in the product space is of the form where each is open in and for all but finitely many (equivalently for only finitely many ). Thus a basic open set in is of the form:

where each is open in and for all but finitely many . In addition, when , we can take to be an open interval of the form . To make the basic open sets of more explicit, is translated as follows:

where is a finite set, for each , is an open interval of , and is the set of all such that . In proving results about , we can use basic open sets that are described in .

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**Completely Regular Spaces and Function Spaces**

We now give two simple results in that are good illustrations that complete regularity is the ideal setting for the domain space of 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 be a completely regular space. Then if is hereditarily separable, then is hereditarily Lindelof.

**Proof**

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

Let be a non-Lindelof subspace. There exists an open cover of such that has no countable subcover. Open sets in are open sets in . We wish to expand these to open sets in . Let be the collection of open subsets of such that . It is clear that no countable subcollection of can cover .

For each , choose such that . Here’s where we use complete regularity. For each , there is a continuous function such that and . Let , which is a subspace of .

We claim that is not separable. To see this, let be a countable subset of such that for each , is obtained from the point and the open set , i.e., . Note that cannot be a cover of . Let be a point that is not in all .

Consider the function chosen above using complete regularity. Note that and . On the other hand, for all since for all . This means that is not in the closure of . For example, is a basic open set containing such that for all . Thus no countable subset of can be dense in , completing the proof.

**Theorem 2**

Let be a completely regular space. Then if is hereditarily Lindelof, then is hereditarily separable.

**Proof**

Let be a non-separable subspace of . For each countable , there must be some point such that is not a member of the closure of (relative to ). For each countable , let be the closure of in the entire space . Clearly .

Now apply the complete regularity of the space . For each countable , let be continuous such that and . Let be the set of all where is countable.

We now show that is a non-Lindelof subspace of . For each , let , which is open in and contains . Let be the collection of all such open sets .

Then is an open cover of that has no countable subcover. To see this, suppose we have such that for each , where is countable. Then let be the set of all and all points . Note that the set is still a countable subset of . Consider the point and the continuous function , which is a member of . We have and for all . Note that each and thus for all . This shows that for all . Then is an open cover of that has no countable subcover, leading to the conclusion that is a non-Lindelof subspace of .

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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 be a completely regular space. The following conditions are equivalent.

- is hereditarily separable (respectively hereditarily Lindelof).
- is hereditarily Lindelof (respectively hereditarily separable).
- For each positive integer , is hereditarily Lindelof (respectively hereditarily separable).

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*Reference*

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