Complete regularity is a separation axiom that is in between regularity and normality. In the previous posts we discuss some of the dominant roles played by complete regularity in the study of topology (see the three links below). In this post, we present an example of a regular space that is not completely regular. In such a space, all the construction techniques involving real-valued continuous maps discussed in these previous posts are not possible or are difficult to do. This example is another demonstration of the importance of completely regular spaces. Here’s the links to the previous posts.

- Completely Regular Spaces and Pseudocompact Spaces
- Completely Regular Spaces and Function Spaces
- Embedding Completely Regular Spaces into a Cube

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.

Essentially, to show a space to be a completely regular space, it suffices to provide for a given closed set and a point (not in the closed set), a bounded real-valued continuous function that maps the given closed set and the point to two different real numbers and . So in a space that is not completely regular, there exist a closed set and a point such that every real-valued continuous function that can be defined on the space maps and the point to the same real number.

The example of a regular but not completely regular space we define here is based on an elementary construction found in the literature (see [2] or Example 1.5.9 on page 40 of [1]). Other examples can also be found in the literature, e.g. the example called Deleted Tychonoff Corkscrew found in [3] (Example 91).

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**Defining the Example**

We define a completely regular space such that by adding a new point the resulting space is no longer completely regular. Let be the set of all points such that . Let be a point not in . For convenience we take . The underlying set is . The following figure illustrates the underlying set.

**Figure 1 – Underlying Set**

For each real number , define to be the set , define to be the set , and define . The topology on is defined by the following:

- Each point where is isolated.
- For each point , a basic open set is of the form where and is a finite subset of .
- A basic open set at the point is of the form where is a positive integer and .

The following figures illustrate the basic open sets at points and at .

**Figure 2 – Basic Open Sets at the x-Axis**

**Figure 3 – Basic Open Sets at p**

It is straightforward to verify that the basic open sets defined above form a base for a topology on the set and that the resulting topology is Hausdorff. The space is an interesting one that merits some attention. So we briefly discuss before we show that the space is regular but not completely regular.

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**Discussing the Space **

One observation we like to make is that the basic open sets for points in are both open and closed in . Whenever a space has a base consisting of closed and open sets, it is said to be a zero-dimensional space. Because the members of one particular base are both closed and open, it follows that any zero-dimensional space is completely regular. As a zero-dimensional space, is completely regular. In addition, is locally compact. Note that the basic open sets defined above are compact.

However, the space is not normal (shown in the last section below). Another point we would like to make is that is metacompact (i.e. every open cover has a point-finite open refinement).

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**Main Result**

We discuss a result that provides a lot of insight about the space and the space . For each integer , let . Let be the x-axis.

. Let be an integer. If be a continuous function such that for infinitely many points , then for each integer where , for all but countably many points .**Main Result**

We consider the following two claims.

* Claim 1*.

Let . If is continuous and , then for all but countably many points .

* Claim 2*.

Let . If is continuous and for countably infinitely many points , then .

Claim 1 follows from the way basic open sets are defined at the point and the fact that is a continuous function. Claim 2 also follows from the definition of open sets at .

We now prove the main result. Let be a subset of such that for all . For any point , let (the projection into the x-axis).

We first show that for all but countably many . For each , let be a countably infinite subset of such that for all . The sets are possible by Claim 1. For each , let be the set of all where . Consider . Note that is the complement of a countable set. Let which is a co-countable subset of . For each , for countably infinitely many (these points are in ). Thus by Claim 2, for each , .

Now that the for infinitely many , we can continue the same argument to prove the same for the next interval . Continue the same inductive process, we can show that for each integer , for all but countably many .

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**Discussing the Space **

We now show that is not completely regular. Let , which is a closed set in . Let be continuous such that . Then we show that . This follows from the main result. By the main result derived above, for each integer , for all by countably many . Then has no choice by to be zero as well.

It remains to be shown that is regular. Since the subspace is completely regular, we only need to verify the regularity at the point . Note that for the open set , the closure is . Furthermore, . For each closed set with , choose some integer such that . Then we have . This establishes the regularity of .

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**The Non-Normality of the Space **

The main result derived above also shows that the space is not normal. Let be the set of all where and is a rational number. Let be the set of all where and is an irrational number. Note that and are disjoint closed sets in . If is normal, then by the Urysohn lemma there is a continuous function such that and .

Such function is not possible. To see this, suppose exists. Note that for infinitely many , namely all where is rational. By the main result, for all but countably many in the x-axis to the right of . But for all irrational .

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*Reference*

- Engelking, R.,
*General Topology, Revised and Completed edition*, Heldermann Verlag, Berlin, 1989. - Mysior, A.,
*A Regular Space Which Is Not Completely Regular*, Proc. Amer. Math. Soc., 81852-853, 1981. - Steen, L. A., Seebach, J. A.,
*Counterexamples in Topology*, Dover Publications, Inc., New York, 1995. - Willard, S.,
*General Topology*, Addison-Wesley Publishing Company, 1970.