# The product of locally compact paracompact spaces

It is well known that when $X$ and $Y$ are paracompact spaces, the product space $X \times Y$ is not necessarily normal. Classic examples include the product of the Sorgenfrey line with itself (discussed here) and the product of the Michael line and the space of irrational numbers (discussed here). However, if one of the paracompact factors is “compact”, the product can be normal or even paracompact. This post discusses several classic results along this line. All spaces are Hausdorff and regular.

Suppose that $X$ and $Y$ are paracompact spaces. We have the following results:

1. If $Y$ is a compact space, then $X \times Y$ is paracompact.
2. If $Y$ is a $\sigma$-compact space, then $X \times Y$ is paracompact.
3. If $Y$ is a locally compact space, then $X \times Y$ is paracompact.
4. If $Y$ is a $\sigma$-locally compact space, then $X \times Y$ is paracompact.

The proof of the first result makes uses the tube lemma. The second result is a corollary of the first. The proofs of both results are given here. The third result is a corollary of the fourth result. We give a proof of the fourth result.

____________________________________________________________________

Proof of the Fourth Result

The fourth result indicated above is restated as Theorem 2 below. It is a theorem of K. Morita [1]. This is one classic result on product of paracompact spaces. After proving the theorem, comments are made about interesting facts and properties that follow from this result. Theorem 2 is also Theorem 3.22 in chapter 18 in the Handbook of Set-Theoretic Topology [2].

A space $W$ is a locally compact space if for each $w \in W$, there is an open subset $O$ of $W$ such that $w \in O$ and $\overline{O}$ is compact. When we say $Y$ is a $\sigma$-locally compact space, we mean that $Y=\bigcup_{j=1}^\infty Y_j$ where each $Y_j$ is a locally compact space. In proving the result discussed here, we also assume that each $Y_j$ is a closed subspace of $Y$. The following lemma will be helpful.

Lemma 1
Let $Y$ be a paracompact space. Suppose that $Y$ is $\sigma$-locally compact. Then there exists a cover $\mathcal{C}=\bigcup_{j=1}^\infty \mathcal{C}_j$ of $Y$ such that each $\mathcal{C}_j$ is a locally finite family consisting of compact sets.

Proof of Lemma 1
Let $Y=\bigcup_{n=1}^\infty Y_n$ such that each $Y_n$ is closed and is locally compact. Fix an integer $n$. For each $y \in Y_n$, let $O_{n,y}$ be an open subset of $Y_n$ such that $y \in O_{n,y}$ and $\overline{O_{n,y}}$ is compact (the closure is taken in $Y_n$). Consider the open cover $\mathcal{O}=\left\{ O_{n,y}: y \in Y_j \right\}$ of $Y_n$. Since $Y_n$ is a closed subspace of $Y$, $Y_n$ is also paracompact. Let $\mathcal{V}=\left\{ V_{n,y}: y \in Y_j \right\}$ be a locally finite open cover of $Y_n$ such that $\overline{V_{n,y}} \subset O_{n,y}$ for each $y \in Y_n$ (again the closure is taken in $Y_n$). Each $\overline{V_{n,y}}$ is compact since $\overline{V_{n,y}} \subset O_{n,y} \subset \overline{O_{n,y}}$. Let $\mathcal{C}_n=\left\{ \overline{V_{n,y}}: y \in Y_n \right\}$.

We claim that $\mathcal{C}_n$ is a locally finite family with respect to the space $Y$. For each $y \in Y-Y_n$, $Y-Y_n$ is an open set containing $y$ that intersects no set in $\mathcal{C}_n$. For each $y \in Y_n$, there is an open set $O \subset Y_n$ that meets only finitely many sets in $\mathcal{C}_n$. Extend $O$ to an open subset $O_1$ of $Y$. That is, $O_1$ is an open subset of $Y$ such that $O=O_1 \cap Y_n$. It is clear that $O_1$ can only meets finitely many sets in $\mathcal{C}_n$.

Then $\mathcal{C}=\bigcup_{j=1}^\infty \mathcal{C}_j$ is the desired $\sigma$-locally finite cover of $Y$. $\square$

Theorem 2
Let $X$ be any paracompact space and let $Y$ be any $\sigma$-locally compact paracompact space. Then $X \times Y$ is paracompact.

Proof of Theorem 2
By Lemma 1, let $\mathcal{C}=\bigcup_{n=1}^\infty \mathcal{C}_n$ be a $\sigma$-locally finite cover of $Y$ such that each $\mathcal{C}_n$ consists of compact sets. To show that $X \times Y$ is paracompact, let $\mathcal{U}$ be an open cover of $X \times Y$. For each $C \in \mathcal{C}$ and for each $x \in X$, the set $\left\{ x \right\} \times C$ is obviously compact.

Fix $C \in \mathcal{C}$ and fix $x \in X$. For each $y \in C$, the point $(x,y) \in U_{y}$ for some $U_{y} \in \mathcal{U}$. Choose open $H_y \subset X$ and open $K_y \subset Y$ such that $(x,y) \in H_y \times K_y \subset U_{x,y}$. Letting $y$ vary, the open sets $H_y \times K_y$ cover the compact set $\left\{ x \right\} \times C$. Choose finitely many open sets $H_y \times K_y$ that also cover $\left\{ x \right\} \times C$. Let $H(C,x)$ be the intersection of these finitely many $H_y$. Let $\mathcal{K}(C,x)$ be the set of these finitely many $K_y$.

To summarize what we have obtained in the previous paragraph, for each $C \in \mathcal{C}$ and for each $x \in X$, there exists an open subset $H(C,x)$ containing $x$, and there exists a finite set $\mathcal{K}(C,x)$ of open subsets of $Y$ such that

• $C \subset \bigcup \mathcal{K}(C,x)$,
• for each $K \in \mathcal{K}(C,x)$, $H(C,x) \times K \subset U$ for some $U \in \mathcal{U}$.

For each $C \in \mathcal{C}$, the set of all $H(C,x)$ is an open cover of $X$. Since $X$ is paracompact, for each $C \in \mathcal{C}$, there exists a locally finite open cover $\mathcal{L}_C=\left\{L(C,x): x \in X \right\}$ such that $L(C,x) \subset H(C,x)$ for all $x$. Consider the following families of open sets.

$\mathcal{E}_n=\left\{L(C,x) \times K: C \in \mathcal{C}_n \text{ and } x \in X \text{ and } K \in \mathcal{K}(C,x) \right\}$

$\mathcal{E}=\bigcup_{n=1}^\infty \mathcal{E}_n$

We claim that $\mathcal{E}$ is a $\sigma$-locally finite open refinement of $\mathcal{U}$. First, show that $\mathcal{E}$ is an open cover of $X \times Y$. Let $(a,b) \in X \times Y$. Then for some $n$, $b \in C$ for some $C \in \mathcal{C}_n$. Furthermore, $a \in L(C,x)$ for some $x \in X$. The information about $C$ and $x$ are detailed above. For example, $C \subset \bigcup \mathcal{K}(C,x)$. Thus there exists some $K \in \mathcal{K}(C,x)$ such that $b \in K$. We now have $(a,b) \in L(C,x) \times K \in \mathcal{E}_n$.

Next we show that $\mathcal{E}$ is a refinement of $\mathcal{U}$. Fix $L(C,x) \times K \in \mathcal{E}_n$. Immediately we see that $L(C,x) \subset H(C,x)$. Since $K \in \mathcal{K}(C,x)$, $H(C,x) \times K \subset U$ for some $U \in \mathcal{U}$. Then $L(C,x) \times K \subset U$.

The remaining point to make is that each $\mathcal{E}_n$ is a locally finite family of open subsets of $X \times Y$. Let $(a,b) \in X \times Y$. Since $\mathcal{C}_n$ is locally finite in $Y$, there exists some open $Q \subset Y$ such that $b \in Q$ and $Q$ meets only finitely many sets in $\mathcal{C}_n$, say $C_1,C_2,\cdots,C_m$. Recall that $\mathcal{L}_{C_j}$ is the set of all $L(C_j,x)$ and is locally finite. Thus there exists an open $O \subset X$ such that $a \in O$ and $O$ meets only finitely many sets in each $\mathcal{L}_{C_j}$ where $j=1,2,\cdots,m$. Thus the open set $O$ meets only finitely many sets $L(C,x)$ for finitely many $C \in \mathcal{C}_n$ and finitely many $x \in X$. These finitely many $C$ and $x$ lead to finitely many $K$. Thus it follows that $O \times Q$ meets only finitely many sets $L(C,x) \times K$ in $\mathcal{E}_n$. Thus $\mathcal{E}_n$ is locally finite.

What has been established is that every open cover of $X \times Y$ has a $\sigma$-locally finite open refinement. This fact is equivalent to paracompactness (according to Theorem 1 in this previous post). This concludes the proof of the theorem. $\square$

____________________________________________________________________

Productively Paracompact Spaces

Consider this property for a space $X$.

(*) The space $X$ satisfies the property that $X \times Y$ is a paracompact space for every paracompact space $Y$.

Such a space can be called a productively paracompact space (for some reason, this term is not used in the literature).

According to the four results stated at the beginning, any space in any one of the following four classes

1. Compact spaces.
2. $\sigma$-compact spaces.
3. Locally compact paracompact spaces.
4. $\sigma$-locally compact paracompact spaces.

satisfies this property. Both the Michael line and the space of the irrational numbers are examples of paracompact spaces that do not have this productively paracompact property. According to comments made on page 799 [2], the theorem of Morita (Theorem 2 here) triggered extensive research to investigate this class of spaces. The class of spaces is broader than the four classes listed here. For example, the productively paracompact spaces also include the closed images of locally compact paracompact spaces. The handbook [2] has more references.

____________________________________________________________________

Normal P-Spaces

Consider this property.

(**) The space $X$ satisfies the property that $X \times Y$ is a normal space for every metric space $Y$.

These spaces can be called productively normal spaces with respect to metric spaces. They go by another name. Morita defined the notion of P-spaces and proved that a space $X$ is a normal P-space if and only if the product of $X$ with any metric space is normal.

Since the class of metric spaces contain the paracompact spaces, any space has property (*) would have property (**), i.e. a normal P-space.Thus any locally compact paracompact space is a normal P-space. Any $\sigma$-locally compact paracompact space is a normal P-space. If a paracompact space has any one of the four “compact” properties discussed here, it is a normal P-space.

Other examples of normal P-spaces are countably compact normal spaces (see here) and perfectly normal spaces (see here).

____________________________________________________________________

Looking at Diagrams

Let’s compare these classes of spaces: productively paracompact spaces (the spaces satisfying property (*)), normal P-spaces and paracompact spaces. We have the following diagram.

Diagram 1

$\displaystyle \begin{array}{ccccc} \text{ } &\text{ } & \text{Productively Paracompact} & \text{ } & \text{ } \\ \text{ } & \swarrow & \text{ } & \searrow & \text{ } \\ \text{Paracompact} &\text{ } & \text{ } & \text{ } & \text{Normal P-space} \\ \text{ } & \text{ } & \text{ } & \text{ } & \text{ } \\ \end{array}$

Clearly productively paracompact implies paracompact. As discussed in the previous section, productively paracompact implies normal P. If a space $X$ is such that the product of $X$ with every paracompact space is paracompact, then the product of $X$ with every metric space is paracompact and hence normal.

However, the arrows in Diagram 1 are not reversible. The Michael line mentioned at the beginning will shed some light on this point. Here’s the previous post on Michael line. Let $\mathbb{M}$ be the Michael line. Let $\mathbb{P}$ be the space of the irrational numbers. The space $\mathbb{M}$ would be a paracompact space that is not productively paracompact since its product with $\mathbb{P}$ is not normal, hence not paracompact.

On the other hand, the space of irrational numbers $\mathbb{P}$ is a normal P-space since it is a metric space. But it is not productively paracompact since its product with the Michael line $\mathbb{M}$ is not normal, hence not paracompact.

The two classes of spaces at the bottom of Diagram 1 do not relate. The Michael line $\mathbb{M}$ is a paracompact space that is not a normal P-space since its product with $\mathbb{P}$ is not normal. Normal P-space does not imply paracompact. Any space that is normal and countably compact is a normal P-space. For example, the space $\omega_1$, the first uncountable ordinal, with the ordered topology is normal and countably compact and is not paracompact.

There are other normal P-spaces that are not paracompact. For example, Bing’s Example H is perfectly normal and not paracompact. As mentioned in the previous section, any perfectly normal space is a normal P-space.

The class of spaces whose product with every paracompact space is paracompact is stronger than both classes of paracompact spaces and normal P-spaces. It is a strong property and an interesting class of spaces. It is also an excellent topics for any student who wants to dig deeper into paracompact spaces.

Let’s add one more property to Diagram 1.

Diagram 2

$\displaystyle \begin{array}{ccccc} \text{ } &\text{ } & \text{Productively Paracompact} & \text{ } & \text{ } \\ \text{ } & \swarrow & \text{ } & \searrow & \text{ } \\ \text{Paracompact} &\text{ } & \text{ } & \text{ } & \text{Normal P-space} \\ \text{ } & \searrow & \text{ } & \swarrow & \text{ } \\ \text{ } &\text{ } & \text{Normal Countably Paracompact} & \text{ } & \text{ } \\ \text{ } & \text{ } & \text{ } & \text{ } & \text{ } \\ \end{array}$

All properties in Diagram 2 except for paracompact are productive. Normal countably paracompact spaces are productive. According to Dowker’s theorem, the product of any normal countably paracompact space with any compact metric space is normal (see Theorem 1 in this previous post). The last two arrows in Diagram 2 are also not reversible.

____________________________________________________________________

Reference

1. Morita K., On the Product of Paracompact Spaces, Proc. Japan Acad., Vol. 39, 559-563, 1963.
2. 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.

____________________________________________________________________
$\copyright$ 2017 – Dan Ma

# The product of a perfectly normal space and a metric space is perfectly normal

The previous post gives a positive result for normality in product space. It shows that the product of a normal countably compact space and a metric space is always normal. In this post, we discuss another positive result, which is the following theorem.

Main Theorem
If $X$ is a perfectly normal space and $Y$ is a metric space, then $X \times Y$ is a perfectly normal space.

As a result of this theorem, perfectly normal spaces belong to a special class of spaces called 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 $X$ (see the section below on P-spaces). Thus any perfectly normal space is a Normal P-space.

All spaces under consideration are Hausdorff. A subset $A$ of the space $X$ is a $G_\delta$-subset of the space $X$ if $A$ is the intersection of countably many open subsets of $X$. A subset $B$ of the space $X$ is an $F_\sigma$-subset of the space $X$ if $B$ is the union of countably many closed subsets of $X$. Clearly, a set $A$ is a $G_\delta$-subset of the space $X$ if and only if $X-A$ is an $F_\sigma$-subset of the space $X$.

A space $X$ is said to be a perfectly normal space if $X$ is normal with the additional property that every closed subset of $X$ is a $G_\delta$-subset of $X$ (or equivalently every open subset of $X$ is an $F_\sigma$-subset of $X$).

The perfect normality has a characterization in terms of zero-sets and cozero-sets. A subset $A$ of the space $X$ is said to be a zero-set if there exists a continuous function $f: X \rightarrow [0,1]$ such that $A=f^{-1}(0)$, where $f^{-1}(0)=\left\{x \in X: f(x)=0 \right\}$. A subset $B$ of the space $X$ is a cozero-set if $X-B$ is a zero-set, or more explicitly if there is a continuous function $f: X \rightarrow [0,1]$ such that $B=\left\{x \in X: f(x)>0 \right\}$.

It is well known that the space $X$ is perfectly normal if and only if every closed subset of $X$ is a zero-set, equivalently every open subset of $X$ is a cozero-set. See here for a proof of this result. We use this result to show that $X \times Y$ is perfectly normal.

____________________________________________________________________

The Proof

Let $X$ be a perfectly normal space and $Y$ be a metric space. Since $Y$ is a metric space, let $\mathcal{B}=\bigcup_{j=1}^\infty \mathcal{B}_j$ be a base for $Y$ such that each $\mathcal{B}_j$ is locally finite. We show that $X \times Y$ is perfectly normal. To that end, we show that every open subset of $X \times Y$ is a cozero-set. Let $U$ be an open subset of $X \times Y$.

For each $(x,y) \in X \times Y$, there exists open $O_{x,y} \subset X$ and there exists $B_{x,y} \in \mathcal{B}$ such that $(x,y) \in O_{x,y} \times B_{x,y} \subset U$. Then $U$ is the union of all sets $O_{x,y} \times B_{x,y}$. Observe that $B_{x,y} \in \mathcal{B}_{j}$ for some integer $j$. For each $B \in \mathcal{B}$ such that $B=B_{x,y}$ for some $(x,y) \in X \times Y$, let $O(B)$ be the union of all corresponding open sets $O_{x,y}$ for all applicable $(x,y)$.

For each positive integer $j$, let $\mathcal{W}_j$ be the collection of all open sets $O(B) \times B$ such that $B \in \mathcal{B}_j$ and $B=B_{x,y}$ for some $(x,y) \in X \times Y$. Let $\mathcal{V}_j=\cup \mathcal{W}_j$. As a result, $U=\bigcup_{j=1}^\infty \mathcal{V}_j$.

Since both $X$ and $Y$ are perfectly normal, for each $O(B) \times B \in \mathcal{W}_j$, there exist continuous functions

$F_{O(B),j}: X \rightarrow [0,1]$

$G_{B,j}: Y \rightarrow [0,1]$

such that

$O(B)=\left\{x \in X: F_{O(B),j}(x) >0 \right\}$

$B=\left\{y \in Y: G_{B,j}(y) >0 \right\}$

Now define $H_j: X \times Y \rightarrow [0,1]$ by the following:

$\displaystyle H_j(x,y)=\sum \limits_{O(B) \times B \in \mathcal{W}_j} F_{O(B),j}(x) \ G_{B,j}(y)$

for all $(x,y) \in X \times Y$. Note that the function $H_j$ is well defined. Since $\mathcal{B}_j$ is locally finite in $Y$, $\mathcal{W}_j$ is locally finite in $X \times Y$. Thus $H_j(x,y)$ is obtained by summing a finite number of values of $F_{O(B),j}(x) \ G_{B,j}(y)$. On the other hand, it can be shown that $H_j$ is continuous for each $j$. Based on the definition of $H_j$, it can be readily verified that $H_j(x,y)>0$ for all $(x,y) \in \cup \mathcal{W}_j$ and $H_j(x,y)=0$ for all $(x,y) \notin \cup \mathcal{W}_j$.

Define $H: X \times Y \rightarrow [0,1]$ by the following:

$\displaystyle H(x,y)=\sum \limits_{j=1}^\infty \biggl[ \frac{1}{2^j} \ \frac{H_j(x,y)}{1+H_j(x,y)} \biggr]$

It is clear that $H$ is continuous. We claim that $U=\left\{(x,y) \in X \times Y: H(x,y) >0 \right\}$. Recall that the open set $U$ is the union of all $O(B) \times B \in \mathcal{W}_j$ for all $j$. Thus if $(x,y) \in \cup \mathcal{W}_j$ for some $j$, then $H(x,y)>0$ since $H_j(x,y)>0$. If $(x,y) \notin \cup \mathcal{W}_j$ for all $j$, $H(x,y)=0$ since $H_j(x,y)=0$ for all $j$. Thus the open set $U$ is an $F_\sigma$-subset of $X \times Y$. This concludes the proof that $X \times Y$ is perfectly normal. $\square$

____________________________________________________________________

Remarks

The main theorem here is a classic result in general topology. An alternative proof is to show that any perfectly normal space is a P-space (definition given below). Then by Morita’s theorem, the product of any perfectly normal space and any metric space is normal (Theorem 1 below). For another proof that is elementary, see Lemma 7 in this previous post.

The notions of perfectly normal spaces and paracompact spaces are quite different. By the theorem discussed here, perfectly normal spaces are normally productive with metric spaces. It is possible for a paracompact space to have a non-normal product with a metric space. The classic example is the Michael line (discussed here).

On the other hand, there are perfectly normal spaces that are not paracompact. One example is Bing’s Example H, which is perfectly normal and not paracompact (see here).

Even though a perfectly normal space is normally productive with metric spaces, it cannot be normally productive in general. For each non-discrete perfectly normal space $X$, there exists a normal space $Y$ such that $X \times Y$ is not normal. This follows from Morita’s first conjecture (now a true statement). Morita’s first conjecture is discussed here.

____________________________________________________________________

P-Space in the Sense of Morita

Morita defined the notion of P-spaces [1] and [2]. Let $\kappa$ be a cardinal number such that $\kappa \ge 1$. Let $\Gamma$ be the set of all finite ordered sequences $(\alpha_1,\alpha_2,\cdots,\alpha_n)$ where $n=1,2,\cdots$ and all $\alpha_i < \kappa$. Let $X$ be a space. The collection $\left\{F_\sigma \subset X: \sigma \in \Gamma \right\}$ is said to be decreasing if this condition holds: $\sigma =(\alpha_1,\alpha_2,\cdots,\alpha_n)$ and $\delta =(\alpha_1,\alpha_2,\cdots,\alpha_n, \cdots, \alpha_m)$ with $n imply that $F_{\delta} \subset F_{\sigma}$. The space $X$ is a P-space if for any cardinal $\kappa \ge 1$ and for any decreasing collection $\left\{F_\sigma \subset X: \sigma \in \Gamma \right\}$ of closed subsets of $X$, there exists open set $U_\sigma$ for each $\sigma \in \Gamma$ such that the following conditions hold:

• for all $\sigma \in \Gamma$, $F_\sigma \subset U_\sigma$,
• for any infinite sequence $(\alpha_1,\alpha_2,\cdots,\alpha_n,\cdots)$ where each each finite subsequence $\sigma_n=(\alpha_1,\alpha_2,\cdots,\alpha_n)$ is an element of $\Gamma$, if $\bigcap_{n=1}^\infty F_{\sigma_n}=\varnothing$, then $\bigcap_{n=1}^\infty U_{\sigma_n}=\varnothing$.

If $\kappa=1$ where $1=\left\{0 \right\}$. Then the index set $\Gamma$ defined above can be viewed as the set of all positive integers. As a result, the definition of P-space with $\kappa=1$ implies the a condition in Dowker’s theorem (see condition 6 in Theorem 1 here). Thus any space $X$ that is normal and a P-space is countably paracompact (or countably shrinking or that $X \times Y$ is normal for every compact metric space or any other equivalent condition in Dowker’s theorem). The following is a theorem of Morita.

Theorem 1 (Morita)
Let $X$ be a space. Then $X$ is a normal P-space if and only if $X \times Y$ is normal for every metric space $Y$.

In light of Theorem 1, both perfectly normal spaces and normal countably compact spaces are P-spaces (see here). According to Theorem 1 and Dowker’s theorem, it follows that any normal P-space is countably paracompact.

____________________________________________________________________

Reference

1. Morita K., On the Product of a Normal Space with a Metric Space, Proc. Japan Acad., Vol. 39, 148-150, 1963. (article information; paper)
2. Morita K., Products of Normal Spaces with Metric Spaces, Math. Ann., Vol. 154, 365-382, 1964.

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

# 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)

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