An Example of a Completely Regular Space that is not Normal

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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.

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Reference

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

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\copyright \ \ 2012

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