# Normal x Compact Needs Not be Normal

This is to present a non-normal cross product where one factor is a normal space and the other factor is a compact space. The normal space is $X=[0,\omega_1)$ and the compact space is $Y=[0,\omega_1]$. I would like to show that $X \times Y$ is not normal. For a basic discussion on these ordinal spaces with the order topology, see this previous post.

In addition to demonstrating an example meeting the criteria set forth in the title, I would like to look at this through two angles. Because these spaces have the ordered topology, they are hereditarily normal (completely normal). Note that $X \times Y$ is a subspace of $Y \times Y$. Thus $Y \times Y$ fails to be hereditarily normal, demonstrating that hereditarily normal x hereditarily normal needs not be hereditarily normal. Another example of this is the double arrow space. In this previous post, I showed that the square of the double arrow space is not hereditarily normal. The double arrow space is perfectly normal and is thus a stronger example in this regard since it shows the square of a perfectly normal space needs not be hereditarily normal. But the uncountable ordinal spaces of $[0,\omega_1)$ and $[0,\omega_1]$ are important counterexamples as well as building blocks for other counterexamples. So for the record, I present them here.

For the second angle, these two spaces show that hereditarily normality is not “nice” enough to prevent non-normal product.

In proving $X \times Y$ is not normal, the Pressing Down Lemma is used. A subset $S$ of $X=[0,\omega_1)$ is a stationary set if $S$ meets every closed and unbounded set in $X$. The following is one version of the Pressing Down Lemma (see Lemma 6.15 on p. 80 of  for a more general version).

Pressing Down Lemma

Let $S$ be a stationary subset of $[0,\omega_1)$. Let $f:S \rightarrow \omega_1$ such that for each $\gamma \in S$, $f(\gamma)<\gamma$, then for some $\alpha<\omega_1$, $f^{-1}\lbrace{\alpha}\rbrace$ is a stationary subset of $[0,\omega_1)$.

To show that $X \times Y$ is not normal, let $H$ and $K$ be the following sets. $H=\lbrace{(\alpha,\alpha):\alpha<\omega_1}\rbrace$ $K=\lbrace{(\alpha,\omega_1):\alpha<\omega_1}\rbrace$

These two sets are disjoint and closed in $X \times Y$. Let $U$ and $V$ be open such that $H \subset U$ and $K \subset V$. It follows that $U \cap V \neq \phi$.

For each $\alpha<\omega_1$, let $g(\alpha)<\omega_1$ such that $[g(\alpha),\alpha] \times [g(\alpha),\alpha] \subset U$. Since $g$ is a pressing down function, there is some $\delta<\omega_1$ and there is a stationary set $S$ such that all points in $S$ are mapped to $\delta$ by $g$. Specifically, for each $\alpha \in S$, we have: $[\delta,\alpha] \times [\delta,\alpha] \subset U$

Choose $\beta_0>\delta$ and let $\beta=\beta_0+1$. We have $(\beta,\omega_1) \in K \subset V$. Choose $\gamma>\delta$ such that $\lbrace{\beta}\rbrace \times [\gamma,\omega_1] \subset V$

Choose $\alpha \in S$ such that $\delta < \beta<\alpha$ and $\delta<\gamma<\alpha<\omega_1$. We have the following set inclusions: $\lbrace{\beta}\rbrace \times [\gamma,\alpha] \subset [\delta,\alpha] \times [\delta,\alpha] \subset U$ $\lbrace{\beta}\rbrace \times [\gamma,\alpha] \subset \lbrace{\beta}\rbrace \times [\gamma,\omega_1] \subset V$

Thus $U \cap V \neq \phi$. This completes the proof that the product $X \times Y$ is not normal.

____________________________________________________________________

Reference

1. Kunen, K., Set Theory, An Introduction to Independence Proofs, North-Holland, Amsterdam, 1980.

____________________________________________________________________

Revised February 3, 2014. $\copyright \ 2014 \text{ by Dan Ma}$