# The product of a normal countably compact space and a metric space is normal

It is well known that normality is not preserved by taking products. When nothing is known about the spaces $X$ and $Y$ other than the facts that they are normal spaces, there is not enough to go on for determining whether $X \times Y$ is normal. In fact even when one factor is a metric space and the other factor is a hereditarily paracompact space, the product can be non-normal (discussed here). This post discusses a productive scenario – the first factor is a normal space and second factor is a metric space with the first factor having the additional property that it is countably compact. In this scenario the product is always normal. This is a well known result in general topology. The goal here is to nail down a proof for use as future reference.

Main Theorem
Let $X$ be a normal and countably compact space. Then $X \times Y$ is a normal space for every metric space $Y$.

The proof of the main theorem uses the notion of shrinkable open covers.

Remarks
The main theorem is a classic result and is often used as motivation for more advanced results for products of normal spaces. Thus we would like to present a clear and complete proof of this classic result for anyone who would like to study the topics of normality (or the lack of) in product spaces. We found that some proofs of this result in the literature are hard to follow. In A. H. Stone’s paper [2], the result is stated in a footnote, stating that “it can be shown that the topological product of a metric space and a normal countably compact space is normal, though not necessarily paracompact”. We had seen several other papers citing [2] as a reference for the result. The Handbook [1] also has a proof (Corollary 4.10 in page 805), which we feel may not be the best proof to learn from. We found a good proof in [3] using the idea of shrinking of open covers.

____________________________________________________________________

The Notion of Shrinking

The key to the proof is the notion of shrinkable open covers and shrinking spaces. Let $X$ be a space. Let $\mathcal{U}$ be an open cover of $X$. The open cover of $\mathcal{U}$ is said to be shrinkable if there is an open cover $\mathcal{V}=\left\{V(U): U \in \mathcal{U} \right\}$ of $X$ such that $\overline{V(U)} \subset U$ for each $U \in \mathcal{U}$. When this is the case, the open cover $\mathcal{V}$ is said to be a shrinking of $\mathcal{U}$. If an open cover is shrinkable, we also say that the open cover can be shrunk (or has a shrinking). Whenever an open cover has a shrinking, the shrinking is indexed by the open cover that is being shrunk. Thus if the original cover is indexed, e.g. $\left\{U_\alpha: \alpha<\kappa \right\}$, then a shrinking has the same indexing, e.g. $\left\{V_\alpha: \alpha<\kappa \right\}$.

A space $X$ is a shrinking space if every open cover of $X$ is shrinkable. Every open cover of a paracompact space has a locally finite open refinement. With a little bit of rearranging, the locally finite open refinement can be made to be a shrinking (see Theorem 2 here). Thus every paracompact space is a shrinking space. For other spaces, the shrinking phenomenon is limited to certain types of open covers. In a normal space, every finite open cover has a shrinking, as stated in the following theorem.

Theorem 1
The following conditions are equivalent.

1. The space $X$ is normal.
2. Every point-finite open cover of $X$ is shrinkable.
3. Every locally finite open cover of $X$ is shrinkable.
4. Every finite open cover of $X$ is shrinkable.
5. Every two-element open cover of $X$ is shrinkable.

The hardest direction in the proof is $1 \Longrightarrow 2$, which is established in this previous post. The directions $2 \Longrightarrow 3 \Longrightarrow 4 \Longrightarrow 5$ are immediate. To see $5 \Longrightarrow 1$, let $H$ and $K$ be two disjoint closed subsets of $X$. By condition 5, the two-element open cover $\left\{X-H,X-K \right\}$ has a shrinking $\left\{U,V \right\}$. Then $\overline{U} \subset X-H$ and $\overline{V} \subset X-K$. As a result, $H \subset X-\overline{U}$ and $K \subset X-\overline{V}$. Since the open sets $U$ and $V$ cover the whole space, $X-\overline{U}$ and $X-\overline{V}$ are disjoint open sets. Thus $X$ is normal.

In a normal space, all finite open covers are shrinkable. In general, an infinite open cover of a normal space may or may not be shrinkable. It turns out that finding a normal space with an infinite open cover that is not shrinkable is no trivial matter (see Dowker’s theorem in this previous post). However, if an open cover in a normal space point-finite or locally finite, then it is shrinkable.

____________________________________________________________________

Key Idea

We now discuss the key idea to the proof of the main theorem. Consider the produce space $X \times Y$. Let $\mathcal{U}$ be an open cover of $X \times Y$. Let $M \subset X \times Y$. The set $M$ is stable with respect to the open cover $\mathcal{U}$ if for each $x \in X$, there is an open set $O_x$ containing $x$ such that $O_x \times M \subset U$ for some $U \in \mathcal{U}$.

Let $\kappa$ be a cardinal number (either finite or infinite). A space $X$ is a $\kappa$-shrinking space if for each open cover $\mathcal{W}$ of $X$ such that the cardinality of $\mathcal{W}$ is $\le \kappa$, then $\mathcal{W}$ is shrinkable. According to Theorem 1, any normal space is 2-shrinkable.

Theorem 2
Let $\kappa$ be a cardinal number (either finite or infinite). Let $X$ be a $\kappa$-shrinking space. Let $Y$ be a paracompact space. Suppose that $\mathcal{U}$ is an open cover of $X \times Y$ such that the following two conditions are satisfied:

• Each point $y \in Y$ has an open set $V_y$ containing $y$ such that $V_y$ is stable with respect to $\mathcal{U}$.
• $\lvert \mathcal{U} \lvert = \kappa$.

Then $\mathcal{U}$ is shrinkable.

Proof of Theorem 2
Let $\mathcal{U}$ be any open cover of $X \times Y$ satisfying the hypothesis. We show that $\mathcal{U}$ has a shrinking.

For each $y \in Y$, obtain the open covers $\left\{G(U,y): U \in \mathcal{U} \right\}$ and $\left\{H(U,y): U \in \mathcal{U} \right\}$ of $X$ as follows. For each $U \in \mathcal{U}$, define the following:

$G(U,y)=\cup \left\{O: O \text{ is open in } X \text{ such that } O \times V_y \subset U \right\}$

Then $\left\{G(U,y): U \in \mathcal{U} \right\}$ is an open cover of $X$. Since $X$ is $\kappa$-shrinkable, there is an open cover $\left\{H(U,y): U \in \mathcal{U} \right\}$ of $X$ such that $\overline{H(U,y)} \subset G(U,y)$ for each $U \in \mathcal{U}$.

Now $\left\{V_y: y \in Y \right\}$ is an open cover of $Y$. By the paracompactness of $Y$, let $\left\{W_y: y \in Y \right\}$ be a locally finite open cover of $Y$ such that $\overline{W_y} \subset V_y$ for each $y \in Y$. For each $U \in \mathcal{U}$, define the following:

$W_U=\cup \left\{H(U,y) \times W_y: y \in Y \text{ such that } \overline{H(U,y) \times W_y} \subset U \right\}$

We claim that $\mathcal{W}=\left\{ W_U: U \in \mathcal{U} \right\}$ is a shrinking of $\mathcal{U}$. First it is a cover of $X \times Y$. Let $(x,t) \in X \times Y$. Then $t \in W_y$ for some $y \in Y$. There exists $U \in \mathcal{U}$ such that $x \in H(U,y)$. Note the following.

$\overline{H(U,y) \times W_y} \subset \overline{H(U,y)} \times \overline{W_y} \subset G(U,y) \times V_y \subset U$

This means that $H(U,y) \times W_y \subset W_U$. Since $(x,t) \in H(U,y) \times W_y$, $(x,t) \in W_U$. Thus $\mathcal{W}$ is an open cover of $X \times Y$.

Now we show that $\mathcal{W}$ is a shrinking of $\mathcal{U}$. Let $U \in \mathcal{U}$. To show that $\overline{W_U} \subset U$, let $(x,t) \in \overline{W_U}$. Let $L$ be open in $Y$ such that $t \in L$ and that $L$ meets only finitely many $W_y$, say for $y=y_1,y_2,\cdots,y_n$. Immediately we have the following relations.

$\forall \ i=1,\cdots,n, \ \overline{W_{y_i}} \subset V_{y_i}$

$\forall \ i=1,\cdots,n, \ \overline{H(U,y_i)} \subset G(U,y_i)$

$\forall \ i=1,\cdots,n, \ \overline{H(U,y_i) \times W_{y_i}} \subset \overline{H(U,y_i)} \times \overline{W_{y_i}} \subset G(U,y_i) \times V_{y_i} \subset U$

Then it follows that

$\displaystyle (x,t) \in \overline{\bigcup \limits_{j=1}^n H(U,y_j) \times W_{y_j}}=\bigcup \limits_{j=1}^n \overline{H(U,y_j) \times W_{y_j}} \subset U$

Thus $U \in \mathcal{U}$. This shows that $\mathcal{W}$ is a shrinking of $\mathcal{U}$. $\square$

Remark
Theorem 2 is the Theorem 3.2 in [3]. Theorem 2 is a formulation of Theorem 3.2 [3] for the purpose of proving Theorem 3 below.

____________________________________________________________________

Main Theorem

Theorem 3 (Main Theorem)
Let $X$ be a normal and countably compact space. Let $Y$ be a metric space. Then $X \times Y$ is a normal space.

Proof of Theorem 3
Let $\mathcal{U}$ be a 2-element open cover of $X \times Y$. We show that $\mathcal{U}$ is shrinkable. This would mean that $X \times Y$ is normal (according to Theorem 1). To show that $\mathcal{U}$ is shrinkable, we show that the open cover $\mathcal{U}$ satisfies the two bullet points in Theorem 2.

Fix $y \in Y$. Let $\left\{B_n: n=1,2,3,\cdots \right\}$ be a base at the point $y$. Define $G_n$ as follows:

$G_n=\cup \left\{O \subset X: O \text{ is open such that } O \times B_n \subset U \text{ for some } U \in \mathcal{U} \right\}$

It is clear that $\mathcal{G}=\left\{G_n: n=1,2,3,\cdots \right\}$ is an open cover of $X$. Since $X$ is countably compact, choose $m$ such that $\left\{G_1,G_2,\cdots,G_m \right\}$ is a cover of $X$. Let $E_y=\bigcap_{j=1}^m B_j$. We claim that $E_y$ is stable with respect to $\mathcal{U}$. To see this, let $x \in X$. Then $x \in G_j$ for some $j \le m$. By the definition of $G_j$, there is some open set $O_x \subset X$ such that $x \in O_x$ and $O_x \times B_j \subset U$ for some $U \in \mathcal{U}$. Furthermore, $O_x \times E_y \subset O_x \times B_j \subset U$.

To summarize: for each $y \in Y$, there is an open set $E_y$ such that $y \in E_y$ and $E_y$ is stable with respect to the open cover $\mathcal{U}$. Thus the first bullet point of Theorem 2 is satisfied. The open cover $\mathcal{U}$ is a 2-element open cover. Thus the second bullet point of Theorem 2 is satisfied. By Theorem 2, the open cover $\mathcal{U}$ is shrinkable. Thus $X \times Y$ is normal. $\square$

Corollary 4
Let $X$ be a normal and pseudocompact space. Let $Y$ be a metric space. Then $X \times Y$ is a normal space.

The corollary follows from the fact that any normal and pseudocompact space is countably compact (see here).

Remarks
The proof of Theorem 3 actually gives a more general result. Note that the second factor only needs to be paracompact and that every point has a countable base (i.e. first countable). The first factor $X$ has to be countably compact. The shrinking requirement for $X$ is flexible – if open covers of a certain size for $X$ are shrinkable, then open covers of that size for the product are shrinkable. We have the following corollaries.

Corollary 5
Let $X$ be a $\kappa$-shrinking and countably compact space and let $Y$ be a paracompact first countable space. Then $X \times Y$ is a $\kappa$-shrinking space.

Corollary 6
Let $X$ be a shrinking and countably compact space and let $Y$ be a paracompact first countable space. Then $X \times Y$ is a shrinking space.

____________________________________________________________________

Remarks

The main theorem (Theorem 3) says that any normal and countably compact space is productively normal with one class of spaces, namely the metric spaces. Thus if one wishes to find a non-normal product space with one factor being countably compact, the other factor must not be a metric space. For example, if $W=\omega_1$, the first uncountable ordinal with the ordered topology, then $W \times X$ is always normal for every metric $X$. For non-normal example, $W \times C$ is not normal for any compact space $C$ with uncountable tightness (see Theorem 1 in this previous post). Another example, $W \times L_{\omega_1}$ is not normal where $L_{\omega_1}$ is the one-point Lindelofication of a discrete space of cardinality $\omega_1$ (follows from Example 1 and Theorem 7 in this previous post).

Another comment is that normal countably paracompact spaces are examples of Normal P-spaces. K. Morita defined the notion of P-space and he proved that a space $Y$ is a Normal P-space if and only if $X \times Y$ is normal for every metric space $X$.

____________________________________________________________________

Reference

1. Przymusinski T. C., Products of Normal Spaces, Handbook of Set-Theoretic Topology (K. Kunen and J. E. Vaughan, eds), Elsevier Science Publishers B. V., Amsterdam, 781-826, 1984.
2. Stone A. H., Paracompactness and Product Spaces, Bull. Amer. Math. Soc., Vol. 54, 977-982, 1948. (paper)
3. Yang L., The Normality in Products with a Countably Compact Factor, Canad. Math. Bull., Vol. 41 (2), 245-251, 1998. (abstract, paper)

____________________________________________________________________
$\copyright \ 2017 \text{ by Dan Ma}$