As the title suggests, this post discusses counterexample 106 in Steen and Seebach [2]. We extend the discussion by adding two facts not found in [2].

The counterexample 106 is the space , which is the product of with the interval topology and the product space where is of course the unit interval . The notation of , the first uncountable ordinal, in Steen and Seebach is .

Another way to notate the example is the product space where is and is the unit interval for all . Thus in this product space, all factors except for one factor is the unit interval and the lone non-compact factor is the first uncountable ordinal. The factor of makes this product space an interesting example.

The following lists out the basic topological properties of the space that are covered in [2].

- The space is Hausdorff and completely regular.
- The space is countably compact.
- The space is neither compact nor sequentially compact.
- The space is neither separable, Lindelof nor -compact.
- The space is not first countable.
- The space is locally compact.

All the above bullet points are discussed in Steen and Seebach. In this post we add the following two facts.

- The space is not normal.
- The space has a dense subspace that is normal.

It follows from these bullet points that the space is an example of a completely regular space that is not normal. Not being a normal space, is then not metrizable. Of course there are other ways to show that is not metrizable. One is that neither of the two factors or is metrizable. Another is that is not first countable.

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**The space is not normal**

Now we are ready to discuss the non-normality of the example. It is a natural question to ask whether the example is normal. The fact that it was not discussed in [2] could be that the tool for answering the normality question was not yet available at the time [2] was originally published, though we do not know for sure. It turns out that the tool became available in the paper [1] published a few years after the publication of [2]. The key to showing the normality (or the lack of) in the example is to show whether the second factor is a countably tight space.

The main result in [1] is discussed in this previous post. Theorem 1 in the previous post states that for any compact space , the product is normal if and only if is countably tight. Thus the normality of the space (or the lack of) hinges on whether the compact factor is countably tight.

A space is countably tight (or has countable tightness) if for each and for each , there exists some countable such that . The definitions of tightness in general and countable tightness in particular are discussed here.

To show that the product space is not countably tight, we let be the subspace of consisting of points, each of which is non-zero on at most countably many coordinates. Specifically is defined as follows:

The set just defined is also called the -product of copies of unit interval . Let be defined by for all . It follows that . It can also be verified that for any countable . This shows that the product space is not countably tight.

By Theorem 1 found in this link, the space is not normal.

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**The space has a dense subspace that is normal**

Now that we know is not normal, a natural question is whether it has a dense subspace that is normal. Consider the subspace where is the -product defined in the preceding section. The subspace is dense in the product space . Thus is dense in . The space is normal since the -product of separable metric spaces is normal. Furthermore, can be embedded as a closed subspace of . Then is homeomorphic to a closed subspace of . Since , the space is normal.

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**Reference**

- Nogura, T.,
*Tightness of compact Hausdorff space and normality of product spaces*, J. Math. Soc. Japan, 28, 360-362, 1976 - Steen, L. A., Seebach, J. A.,
*Counterexamples in Topology*, Dover Publications, Inc., New York, 1995.

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