# Regular but not Completely Regular

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.

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.

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 $a$ 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.

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 $S$ such that by adding a new point $p$ the resulting space $T=S \cup \left\{p \right\}$ is no longer completely regular. Let $S$ be the set of all points $(x,y) \in \mathbb{R} \times \mathbb{R}$ such that $y \ge 0$. Let $p$ be a point not in $S$. For convenience we take $p=(0,-1)$. The underlying set is $T=S \cup \left\{p \right\}$. The following figure illustrates the underlying set.

Figure 1 – Underlying Set

For each real number $x$, define $V_x$ to be the set $V_x=\left\{(x,y) \in S: 0 \le y \le 2 \right\}$, define $D_x$ to be the set $D_x=\left\{(s,s-x) \in S: x \le s \le x+2 \right\}$, and define $O_x=V_x \cup D_x$. The topology on $T$ is defined by the following:

• Each point $(x,y) \in S$ where $y>0$ is isolated.
• For each point $(x,0) \in S$, a basic open set is of the form $O_x - F$ where $(x,0) \notin F$ and $F$ is a finite subset of $O_x$.
• A basic open set at the point $p$ is of the form $\left\{p \right\} \cup U_n$ where $n$ is a positive integer and $U_n=\left\{(x,y) \in S: x \ge n \right\}$.

The following figures illustrate the basic open sets at points $(x,0)$ and at $p$.

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 $T$ and that the resulting topology is Hausdorff. The space $S$ is an interesting one that merits some attention. So we briefly discuss $S$ before we show that the space $T$ is regular but not completely regular.

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Discussing the Space $\bold S$

One observation we like to make is that the basic open sets for points in $S$ are both open and closed in $S$. 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, $S$ is completely regular. In addition, $S$ is locally compact. Note that the basic open sets defined above are compact.

However, the space $S$ is not normal (shown in the last section below). Another point we would like to make is that $S$ 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 $S$ and the space $T$. For each integer $n$, let $H_n=\left\{(x,0): n \le x \le n+1 \right\}=[n,n+1] \times \left\{0 \right\}$. Let $H=\left\{(x,0): x \in \mathbb{R} \right\}$ be the x-axis.

• Main Result. Let $n$ be an integer. If $f:S \rightarrow \mathbb{R}$ be a continuous function such that $f((x,0))=0$ for infinitely many points $(x,0) \in H_n$, then for each integer $j$ where $j \ge n$, $f((x,0))=0$ for all but countably many points $(x,0) \in H_j$.

We consider the following two claims.

Claim 1.
Let $(a,0) \in H$. If $g:S \rightarrow \mathbb{R}$ is continuous and $g((a,0))=0$, then $g(w)=0$ for all but countably many points $w \in O_a=V_a \cup D_a$.

Claim 2.
Let $(a,0) \in H$. If $g:S \rightarrow \mathbb{R}$ is continuous and $g(w)=0$ for countably infinitely many points $w \in V_a -\left\{(a,0) \right\}$, then $g((a,0))=0$.

Claim 1 follows from the way basic open sets are defined at the point $(a,0)$ and the fact that $g$ is a continuous function. Claim 2 also follows from the definition of open sets at $(a,0)$.

We now prove the main result. Let $A=\left\{(x(1),0),(x(2),0),(x(3),0),\cdots \right\}$ be a subset of $H_n$ such that $f((x(j),0))=0$ for all $j$. For any point $(a,b) \in S$, let $\pi((a,b))=a$ (the projection into the x-axis).

We first show that $f(z)=0$ for all but countably many $z \in H_{n+1}$. For each $j \ge 1$, let $A_j$ be a countably infinite subset of $D_{x(j)}$ such that $f(w)=0$ for all $w \in D_{x(j)}-A_j$. The sets $A_j$ are possible by Claim 1. For each $j$, let $B_j$ be the set of all $\pi(w)$ where $w \in A_j$. Consider $J=[n+1,n+2] - \bigcup_{j \ge 1} B_j$. Note that $J$ is the complement of a countable set. Let $K=J \times \left\{0 \right\}$ which is a co-countable subset of $H_{n+1}$. For each $(x,0) \in K$, $f(w)=0$ for countably infinitely many $w \in V_x$ (these points $w$ are in $V_x \cap (D_{x(j)}-A_j)$). Thus by Claim 2, for each $(x,0) \in K$, $f((x,0))=0$.

Now that the $f(b)=0$ for infinitely many $b \in H_{n+1}$, we can continue the same argument to prove the same for the next interval $H_{n+2}$. Continue the same inductive process, we can show that for each integer $k>1$, $f(z)=0$ for all but countably many $z \in H_{n+k}$.

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Discussing the Space $\bold T$

We now show that $T$ is not completely regular. Let $H_0=[0,1] \times \left\{0 \right\}$, which is a closed set in $T$. Let $f:T \rightarrow \mathbb{R}$ be continuous such that $f(H_0) \subset \left\{0 \right\}$. Then we show that $f(p)=0$. This follows from the main result. By the main result derived above, for each integer $j \ge 1$, $f(w)=0$ for all by countably many $w \in H_j =[j,j+1] \times \left\{0 \right\}$. Then $f(p)$ has no choice by to be zero as well.

It remains to be shown that $T$ is regular. Since the subspace $S$ is completely regular, we only need to verify the regularity at the point $p$. Note that for the open set $U_j$, the closure is $\overline{U_j}=U_j \cup H_{j-1} \cup H_{j-2}$. Furthermore, $\overline{U_j} \subset U_{j-2}$. For each closed set $C \subset T$ with $x \notin C$, choose some integer $n$ such that $p \in U_n \subset T-C$. Then we have $p \in U_{n+2} \subset \overline{U_{n+2}} \subset U_n \subset T-C$. This establishes the regularity of $T$.

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The Non-Normality of the Space $\bold S$

The main result derived above also shows that the space $S$ is not normal. Let $L$ be the set of all $(x,0)$ where $x \ge 0$ and $x$ is a rational number. Let $M$ be the set of all $(x,0)$ where $x \ge 0$ and $x$ is an irrational number. Note that $L$ and $M$ are disjoint closed sets in $S$. If $S$ is normal, then by the Urysohn lemma there is a continuous function $\rho: S \rightarrow [0,1]$ such that $\rho(L) \subset \left\{0 \right\}$ and $\rho(M) \subset \left\{1 \right\}$.

Such function $\rho$ is not possible. To see this, suppose $\rho$ exists. Note that $\rho(w)=0$ for infinitely many $w \in H_0$, namely all $w=(x,0)$ where $x \ge 0$ is rational. By the main result, $\rho(w)=0$ for all but countably many $w$ in the x-axis to the right of $H_0$. But $\rho((x,0))=1$ for all irrational $x \ge 0$.

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Reference

1. Engelking, R., General Topology, Revised and Completed edition, Heldermann Verlag, Berlin, 1989.
2. Mysior, A., A Regular Space Which Is Not Completely Regular, Proc. Amer. Math. Soc., 81852-853, 1981.
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.