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.



  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


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