Counterexample 106 from Steen and Seebach

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 $X=\omega_1 \times I^I$, which is the product of $\omega_1$ with the interval topology and the product space $I^I=\prod_{t \in I} I$ where $I$ is of course the unit interval $[0,1]$. The notation of $\omega_1$, the first uncountable ordinal, in Steen and Seebach is $[0,\Omega)$.

Another way to notate the example $X$ is the product space $\prod_{t \in I} X_t$ where $X_0$ is $\omega_1$ and $X_t$ is the unit interval $I$ for all $t>0$. 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 $\omega_1$ makes this product space an interesting example.

The following lists out the basic topological properties of the space that $X=\omega_1 \times I^I$ are covered in [2].

• The space $X$ is Hausdorff and completely regular.
• The space $X$ is countably compact.
• The space $X$ is neither compact nor sequentially compact.
• The space $X$ is neither separable, Lindelof nor $\sigma$-compact.
• The space $X$ is not first countable.
• The space $X$ 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 $X$ is not normal.
• The space $X$ has a dense subspace that is normal.

It follows from these bullet points that the space $X$ is an example of a completely regular space that is not normal. Not being a normal space, $X$ is then not metrizable. Of course there are other ways to show that $X$ is not metrizable. One is that neither of the two factors $\omega_1$ or $I^I$ is metrizable. Another is that $X$ is not first countable.

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The space $X$ 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 $X=\omega_1 \times I^I$ 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 $X=\omega_1 \times I^I$ is to show whether the second factor $I^I$ 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 $Y$, the product $\omega_1 \times Y$ is normal if and only if $Y$ is countably tight. Thus the normality of the space $X$ (or the lack of) hinges on whether the compact factor $I^I=\prod_{t \in I} I$ is countably tight.

A space $Y$ is countably tight (or has countable tightness) if for each $S \subset Y$ and for each $x \in \overline{S}$, there exists some countable $B \subset S$ such that $x \in \overline{B}$. The definitions of tightness in general and countable tightness in particular are discussed here.

To show that the product space $I^I=\prod_{t \in I} I$ is not countably tight, we let $S$ be the subspace of $I^I$ consisting of points, each of which is non-zero on at most countably many coordinates. Specifically $S$ is defined as follows:

$S=\Sigma_{t \in I} I=\left\{y \in I^I: y(t) \ne 0 \text{ for at most countably many } t \in I \right\}$

The set $S$ just defined is also called the $\Sigma$-product of copies of unit interval $I$. Let $g \in I^I$ be defined by $g(t)=1$ for all $t \in I$. It follows that $g \in \overline{S}$. It can also be verified that $g \notin \overline{B}$ for any countable $B \subset S$. This shows that the product space $I^I=\prod_{t \in I} I$ is not countably tight.

By Theorem 1 found in this link, the space $X=\omega_1 \times I^I$ is not normal.

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The space $X$ has a dense subspace that is normal

Now that we know $X=\omega_1 \times I^I$ is not normal, a natural question is whether it has a dense subspace that is normal. Consider the subspace $\omega_1 \times S$ where $S$ is the $\Sigma$-product $S=\Sigma_{t \in I} I$ defined in the preceding section. The subspace $S$ is dense in the product space $I^I$. Thus $\omega_1 \times S$ is dense in $X=\omega_1 \times I^I$. The space $S$ is normal since the $\Sigma$-product of separable metric spaces is normal. Furthermore, $\omega_1$ can be embedded as a closed subspace of $S=\Sigma_{t \in I} I$. Then $\omega_1 \times S$ is homeomorphic to a closed subspace of $S \times S$. Since $S \times S \cong S$, the space $\omega_1 \times S$ is normal.

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Reference

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

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$\copyright \ 2015 \text{ by Dan Ma}$