# An Example of a Completely Regular Space that is not Normal

We present an example of a space $S$ that is Hausdorff, locally compact, zero-dimensional (having a base consisting of closed and open sets), metacompact, completely regular but not normal. This space $S$ is a building block for defining other spaces. In a previous post, we show that the space $S$ can be turned into a space that is regular but not completely regular by adding one point (see Regular but not Completely Regular). Even though $S$ is discussed in the previous post, the space $S$ is an interesting example by itself and deserves a separate post.

Let $S$ be the set of all points $(x,y) \in \mathbb{R} \times \mathbb{R}$ such that $y \ge 0$. 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 $S$ 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 $O_x$.

It is straightforward to verify that the basic open sets defined above form a base for a topology on the set $S$ and that the resulting topology is Hausdorff. One important observation to make is that this base consists of sets that are both closed and open. Whenever a space has a base consisting of closed and open sets, it is said to be a zero-dimensional space. It is straightforward to show that any zero-dimensional space is completely regular. Another interesting point about the space $S$ is that it is metacompact. Recall that a space $X$ is metacompact if every open cover of $X$ has a point-finite open refinement.

Now we discuss why $S$ is not normal. Note that the x-axis in $S$ is a closed and discrete set of cardinality continuum. Jones’ lemma states that in a normal and separable space, the cardinality of any closed and discrete set must not equal to or exceed continuum. But Jones’ lemma is of no use here since $S$ is not separable. However, the two disjoint closed sets that destroyed normality are from the x-axis, namely $H=\left\{(x,0): x \in \mathbb{Q} \right\}$ and $K=\left\{(x,0): x \in \mathbb{R}-\mathbb{Q} \right\}$.

Note that $H$ and $K$ are disjoint closed sets in $S$. If $S$ were normal, there would be a continuous $f:S \rightarrow [0,1]$ such that $f(a)=0$ for all $a \in H$ and $f(a)=1$ for all $a \in K$ (using Urysohn lemma). But this function is not possible. It can be shown that any continuous function $g:S \rightarrow [0,1]$ that maps $H$ to zero would have to map the entire x-axis to zero except for a countable subset of x-axis. This fact follows from the Main Result presented in Regular but not Completely Regular.

________________________________________________________________________

Reference

1. Engelking, R., General Topology, Revised and Completed edition, Heldermann Verlag, Berlin, 1989.
2. Willard, S., General Topology, Addison-Wesley Publishing Company, 1970.

________________________________________________________________________

$\copyright \ \ 2012$