# Normality in Cp(X)

Any collectionwise normal space is a normal space. Any perfectly normal space is a hereditarily normal space. In general these two implications are not reversible. In function spaces $C_p(X)$, the two implications are reversible. There is a normal space that is not countably paracompact (such a space is called a Dowker space). If a function space $C_p(X)$ is normal, it is countably paracompact. Thus normality in $C_p(X)$ is a strong property. This post draws on Dowker’s theorem and other results, some of them are previously discussed in this blog, to discuss this remarkable aspect of the function spaces $C_p(X)$.

Since we are discussing function spaces, the domain space $X$ has to have sufficient quantity of real-valued continuous functions, e.g. there should be enough continuous functions to separate the points from closed sets. The ideal setting is the class of completely regular spaces (also called Tychonoff spaces). See here for a discussion on completely regular spaces in relation to function spaces.

Let $X$ be a completely regular space. Let $C(X)$ be the set of all continuous functions from $X$ into the real line $\mathbb{R}$. When $C(X)$ is endowed with the pointwise convergence topology, the space is denoted by $C_p(X)$ (see here for further comments on the definition of the pointwise convergence topology).

When Function Spaces are Normal

Let $X$ be a completely regular space. We discuss these four facts of $C_p(X)$:

1. If the function space $C_p(X)$ is normal, then $C_p(X)$ is countably paracompact.
2. If the function space $C_p(X)$ is hereditarily normal, then $C_p(X)$ is perfectly normal.
3. If the function space $C_p(X)$ is normal, then $C_p(X)$ is collectionwise normal.
4. Let $X$ be a normal space. If $C_p(X)$ is normal, then $X$ has countable extent, i.e. every closed and discrete subset of $X$ is countable, implying that $X$ is collectionwise normal.

Fact #1 and Fact #2 rely on a representation of $C_p(X)$ as a product space with one of the factors being the real line. For $x \in X$, let $Y_x=\left\{f \in C_p(X): f(x)=0 \right\}$. Then $C_p(X) \cong Y_x \times \mathbb{R}$. This representation is discussed here.

Another useful tool is Dowker’s theorem, which essentially states that for any normal space $W$, the space $W$ is countably paracompact if and only if $W \times C$ is normal for all compact metric space $C$ if and only if $W \times [0,1]$ is normal. For the full statement of the theorem, see Theorem 1 in this previous post, which has links to the proofs and other discussion.

To show Fact #1, suppose that $C_p(X)$ is normal. Immediately we make use of the representation $C_p(X) \cong Y_x \times \mathbb{R}$ where $x \in X$. Since $Y_x \times \mathbb{R}$ is normal, $Y_x \times [0,1]$ is also normal. By Dowker’s theorem, $Y_x$ is countably paracompact. Note that $Y_x$ is a closed subspace of the normal $C_p(X)$. Thus $Y_x$ is also normal.

One more helpful tool is Theorem 5 in in this previous post, which is like an extension of Dowker’s theorem, which states that a normal space $W$ is countably paracompact if and only if $W \times T$ is normal for any $\sigma$-compact metric space $T$. This means that $Y_x \times \mathbb{R} \times \mathbb{R}$ is normal.

We want to show $C_p(X) \cong Y_x \times \mathbb{R}$ is countably paracompact. Since $Y_x \times \mathbb{R} \times \mathbb{R}$ is normal (based on the argument in the preceding paragraph), $(Y_x \times \mathbb{R}) \times [0,1]$ is normal. Thus according to Dowker’s theorem, $C_p(X) \cong Y_x \times \mathbb{R}$ is countably paracompact.

For Fact #2, a helpful tool is Katetov’s theorem (stated and proved here), which states that for any hereditarily normal $X \times Y$, one of the factors is perfectly normal or every countable subset of the other factor is closed (in that factor).

To show Fact #2, suppose that $C_p(X)$ is hereditarily normal. With $C_p(X) \cong Y_x \times \mathbb{R}$ and according to Katetov’s theorem, $Y_x$ must be perfectly normal. The product of a perfectly normal space and any metric space is perfectly normal (a proof is found here). Thus $C_p(X) \cong Y_x \times \mathbb{R}$ is perfectly normal.

The proof of Fact #3 is found in Problems 294 and 295 of [2]. The key to the proof is a theorem by Reznichenko, which states that any dense convex normal subspace of $[0,1]^X$ has countable extent, hence is collectionwise normal (problem 294). See here for a proof that any normal space with countable extent is collectionwise normal (see Theorem 2). The function space $C_p(X)$ is a dense convex subspace of $[0,1]^X$ (problem 295). Thus if $C_p(X)$ is normal, then it has countable extent and hence collectionwise normal.

Fact #4 says that normality of the function space imposes countable extent on the domain. This result is discussed in this previous post (see Corollary 3 and Corollary 5).

Remarks

The facts discussed here give a flavor of what function spaces are like when they are normal spaces. For further and deeper results, see [1] and [2].

Fact #1 is essentially driven by Dowker’s theorem. It follows from the theorem that whenever the product space $X \times Y$ is normal, one of the factor must be countably paracompact if the other factor has a non-trivial convergent sequence (see Theorem 2 in this previous post). As a result, there is no Dowker space that is a $C_p(X)$. No pathology can be found in $C_p(X)$ with respect to finding a Dowker space. In fact, not only $C_p(X) \times C$ is normal for any compact metric space $C$, it is also true that $C_p(X) \times T$ is normal for any $\sigma$-compact metric space $T$ when $C_p(X)$ is normal.

The driving force behind Fact #2 is Katetov’s theorem, which basically says that the hereditarily normality of $X \times Y$ is a strong statement. Coupled with the fact that $C_p(X)$ is of the form $Y_x \times \mathbb{R}$, Katetov’s theorem implies that $Y_x \times \mathbb{R}$ is perfectly normal. The argument also uses the basic fact that perfectly normality is preserved when taking product with metric spaces.

There are examples of normal but not collectionwise normal spaces (e.g. Bing’s Example G). Resolution of the question of whether normal but not collectionwise normal Moore space exists took extensive research that spanned decades in the 20th century (the normal Moore space conjecture). The function $C_p(X)$ is outside of the scope of the normal Moore space conjecture. The function space $C_p(X)$ is usually not a Moore space. It can be a Moore space only if the domain $X$ is countable but then $C_p(X)$ would be a metric space. However, it is still a powerful fact that if $C_p(X)$ is normal, then it is collectionwise normal.

On the other hand, a more interesting point is on the normality of $X$. Suppose that $X$ is a normal Moore space. If $C_p(X)$ happens to be normal, then Fact #4 says that $X$ would have to be collectionwise normal, which means $X$ is metrizable. If the goal is to find a normal Moore space $X$ that is not collectionwise normal, the normality of $C_p(X)$ would kill the possibility of $X$ being the example.

Reference

1. Arkhangelskii, A. V., Topological Function Spaces, Mathematics and Its Applications Series, Kluwer Academic Publishers, Dordrecht, 1992.
2. Tkachuk V. V., A $C_p$-Theory Problem Book, Topological and Function Spaces, Springer, New York, 2011.

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$\copyright$ 2017 – Dan Ma

Countably paracompact spaces are discussed in a previous post. The discussion of countably paracompactness in the previous post is through discussing Dowker’s theorem. In this post, we discuss a few more facts that can be derived from Dowker’s theorem.

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Dowker’s Theorem

Essentially, Dowker’s theorem is the statement that for a normal space $X$, the space $X$ is countably paracompact if any only if $X \times Y$ is normal for any infinite compact metric space. The following is the full statement of Dowker’s theorem. The long list of equivalent conditions is important for applications in various scenarios.

Theorem 1 (Dowker’s Theorem)
Let $X$ be a normal space. The following conditions are equivalent.

1. The space $X$ is countably paracompact.
2. Every countable open cover of $X$ has a point-finite open refinement.
3. If $\left\{U_n: n=1,2,3,\cdots \right\}$ is an open cover of $X$, there exists an open refinement $\left\{V_n: n=1,2,3,\cdots \right\}$ such that $\overline{V_n} \subset U_n$ for each $n$.
4. The product space $X \times Y$ is normal for any infinite compact metric space $Y$.
5. The product space $X \times [0,1]$ is normal where $[0,1]$ is the closed unit interval with the usual Euclidean topology.
6. The product space $X \times S$ is normal where $S$ is a non-trivial convergent sequence with the limit point. Note that $S$ can be taken as a space homeomorphic to $\left\{1,\frac{1}{2},\frac{1}{3},\cdots \right\} \cup \left\{0 \right\}$ with the Euclidean topology.
7. For each sequence $\left\{A_n \subset X: n=1,2,3,\cdots \right\}$ of closed subsets of $X$ such that $A_1 \supset A_2 \supset A_3 \supset \cdots$ and $\cap_n A_n=\varnothing$, there exist open sets $B_1,B_2,B_3,\cdots$ such that $A_n \subset B_n$ for each $n$ such that $\cap_n B_n=\varnothing$.

A Dowker space is any normal space that is not countably paracompact. The notion of Dowker space was motivated by Dowker’s theorem since such a space would be a normal space $X$ for which $X \times [0,1]$ is not normal. The search for such a space took about 20 years from 1951 when C. H. Dowker proved the theorem to 1971 when M. E. Rudin constructed a ZFC example of a Dowker space.

Theorem 1 (Dowker’s theorem) is proved here and is further discussed in this previous post on countably paracompact space. The statement appears in Condition 6 here is not found in the previous version of the theorem. However, no extra effort is required to support it. Condition 5 trivially implies condition 6. The proof of condition 5 implying condition 7 (the proof of 4 implies 5 shown here) only requires that the product of $X$ and a convergent sequence is normal. So inserting condition 6 does not require extra proof.

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Getting More from Dowker’s Theorem

As a result of Theorem 1, normal countably paracompact spaces are productive in normality with respect to compact metric spaces (condition 4 in Dowker’s theorem as stated above). Another way to look at condition 4 is that the normality in the product $X \times Y$ is a strong property. Whenever the product $X \times Y$ is normal, we know that each factor is normal. Dowker’s theorem tells us that whenever $X \times Y$ is normal and one of the factor is a compact metric space such as the unit interval $[0,1]$, the other factor is countably paracompact. The fact can be extended. Even if the factors are not metric spaces, as long as one of the factors has a non-discrete point with “countable” tightness, normality of the product confers countably paracompactness on one of the factors. The following two theorems make this clear.

Theorem 2
Suppose that the product $X \times Y$ is normal. If one of the factor contains a non-trivial convergent sequence, then the other factor is countably paracompact.

Proof of Theorem 2
Suppose $Y$ contains a non-trivial convergent sequence. Let this sequence be denoted by $S =\left\{ x_n:n=1,2,3,\cdots \right\} \cup \left\{x \right\}$ such that the point $x$ is the limit point. Since $X \times Y$ is normal, both $X$ and $Y$ are normal and that $X \times S$ is normal. By Theorem 1, $X$ is countably paracompact. $\square$

Theorem 3
Suppose that the product $X \times Y$ is normal. If one of the factor contains a countable subset that is non-discrete, then the other factor is countably paracompact.

Proof of Theorem 3
To discuss this fact, we need to turn to the generalized Dowker’s theorem, which is Theorem 2 in this previous post. We will not re-state the theorem. The crucial direction is $7 \longrightarrow 4$ in that theorem. To avoid confusion, we call these two conditions A7 and A4. The following are the conditions.

A7

The product $X \times Y$ is a normal space for some space $Y$ containing a non-discrete subspace of cardinality $\kappa$.

A4

For each decreasing family $\left\{F_\alpha: \alpha<\kappa \right\}$ of closed subsets of $X$ such that $\bigcap_{\alpha<\kappa} F_\alpha=\varnothing$, there exists a family $\left\{G_\alpha: \alpha<\kappa \right\}$ of open subsets of $X$ satisfying $\bigcap_{\alpha<\kappa} G_\alpha=\varnothing$ and $F_\alpha \subset G_\alpha$ for all $\alpha<\kappa$.

Actually the proof in the previous post shows that A7 implies another condition that is equivalent to A4 for any infinite cardinal $\kappa$. In particular, A7 $\longrightarrow$ A4 would hold for the countably infinite $\kappa=\omega$. Note that under $\kappa=\omega$, A4 would be the same as condition 7 in Theorem 1 above.

Thus by Theorem 2 in this previous post for the countably infinite case and by Theorem 1 in this post, the theorem is established. $\square$

Remarks
In Theorem 2, the second factor $Y$ does not have to be a metric space. As long as it has a non-trivial convergent sequence, the normality of the product (a big if in some situation) implies countably paracompactness in the other factor.

Theorem 3 is essentially a corollary of the proof of Theorem 2 in the previous post. One way to look at Theorem 3 is that the normality of the product $X \times Y$ is a strong statement. If the product is normal and if one factor has a countable non-discrete subspace, then the other factor is countably paracompact. Another way to look at it is through the angle of Dowker spaces. By Dowker’s theorem (Theorem 1), the product of any Dowker space with any infinite compact metric space is not normal. The pathology is actually more severe. A Dowker space is severely lacking in ability to form normal product, as the following corollary makes clear.

Corollary 4
If $X$ is a Dowker space, then $X \times Y$ is not normal for any space $Y$ containing a non-discrete countable subspace.

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More Results

Two more results are discussed. According to Dowker’s theorem, the product of a countably paracompact space $X$ and any compact metric space is normal. In particular, $X \times [0,1]$ is normal. Theorem 5 is saying that with a little extra work, it can be shown that $X \times \mathbb{R}$ is normal. What makes this works is that the metric factor is $\sigma$-compact.

Theorem 5
Let $X$ be a normal space. The following conditions are equivalent.

1. The space $X$ is countably paracompact.
2. The product space $X \times Y$ is normal for any non-discrete $\sigma$-compact metric space $Y$.
3. The product space $X \times \mathbb{R}$ is normal where $\mathbb{R}$ is the real number line with the usual Euclidean topology.

Proof of Theorem 5
$1 \rightarrow 2$
Suppose that $X$ is countably paracompact. Let $Y=\bigcup_{j=1}^\infty Y_j$ where each $Y_j$ is compact. Since $Y$ is a $\sigma$-compact metric space, it is Lindelof. The Lindelof number and the weight agree in a metric space. Thus $Y$ has a countable base. According to Urysohn’s metrization theorem (discussed here), $Y$ can be embedded into the compact metric space $\prod_{j=1}^\infty W_j$ where each $W_j=[0,1]$. For convenience, we consider $Y$ as a subspace of $\prod_{j=1}^\infty W_j$. Furthermore, $X \times Y=\bigcup_{j=1}^\infty (X \times Y_j) \subset X \times\prod_{j=1}^\infty W_j$.

By Theorem 1, each $X \times Y_j$ is normal and that $X \times\prod_{j=1}^\infty W_j$ is normal. Note that $X \times Y$ is an $F_\sigma$-subset of the normal space $X \times\prod_{j=1}^\infty W_j$. Since normality is passed to $F_\sigma$-subsets, $X \times Y$ is normal.

Note. For a proof that $F_\sigma$-subsets of normal spaces are normal, see 2.7.2(b) on p. 112 of Englelking [1].

$2 \rightarrow 3$ is immediate.

$3 \rightarrow 1$
Suppose that $X \times \mathbb{R}$ is normal. Then $X \times [0,1]$ is normal since it is a closed subspace of $X \times \mathbb{R}$. By Theorem 1, $X$ is countably paracompact. $\square$

Theorem 6
Let $X$ be a normal space. Let $Y$ be a non-discrete $\sigma$-compact metric space. Then $X \times Y$ is a normal space if and only if $X \times Y$ is countably paracompact.

Proof of Theorem 6
Let $Y=\bigcup_{j=1}^\infty Y_j$ where each $Y_j$ is compact. As in the proof of Theorem 5, we use the compact metric space $\prod_{j=1}^\infty W_j$ where each $W_j=[0,1]$.

Suppose that $X \times Y$ is normal. Since $Y$ is a non-discrete metric space, $Y$ contains a countable non-discrete subspace. Then by either Theorem 2 or Theorem 3, $X$ is countably paracompact.

By Theorem 1, $X \times\prod_{j=1}^\infty W_j$ is normal. Note that $X \times \prod_{j=1}^\infty W_j \times [0,1]$ is normal since $(\prod_{j=1}^\infty W_j) \times [0,1]$ is a compact metric space. By Theorem 1 again, $X \times\prod_{j=1}^\infty W_j$ is countably paracompact.

As in the proof of Theorem 5, we can consider $Y$ as a subspace of $\prod_{j=1}^\infty W_j$. Furthermore, $X \times Y=\bigcup_{j=1}^\infty X \times Y_j \subset X \times\prod_{j=1}^\infty W_j$.

Note that $X \times Y$ is $F_\sigma$-subset of the countably paracompact space $X \times\prod_{j=1}^\infty W_j$. Since countably paracompactness is passed to $F_\sigma$-subsets, we conclude that $X \times Y$ is countably paracompact.

Note. For a proof that countably paracompactness is passed to $F_\sigma$-subsets, see the proof that paracompactness is passed to $F_\sigma$-subsets in this previous post. Just apply the same proof but start with a countable open cover.

For the other direction, suppose that $X \times Y$ is countably paracompact. Since $X \times \left\{y \right\}$ is a closed subspace of $Y$ with $y \in Y$ and is a copy of $X$, $X$ is countably paracompact. Then by Theorem 5, $X \times Y$ is a normal space. $\square$

Remarks
Theorem 5 seems like an extension of Theorem 1. But the amount of extra work is very little. So normal countably paracompact spaces are productive with not just compact metric spaces but also with $\sigma$-compact metric spaces. The $\sigma$-compactness is absolutely crucial. The product of a normal countably paracompact space with a metric space does not have to be normal. For example, the Michael line $\mathbb{M}$ is paracompact and thus countably paracompact. The product of $\mathbb{M}$ and metric space is not necessarily normal (discussed here). However, the product of $\mathbb{M}$ and $\mathbb{R}$ or other $\sigma$-compact metric space is normal.

Recall that a space is called a Dowker space if it is normal and not countably paracompact. For the type of product $X \times Y$ discussed in Theorem 6, it cannot be Dowker (if it is normal, it is countably paracompact). The two notions are the same with such product $X \times Y$. Theorem 6 actually holds for a wider class than indicated. The following is Corollary 4.3 in [2].

Theorem 7
Let $X$ be a normal space. Let $Y$ be a non-discrete metric space. Then $X \times Y$ is a normal space if and only if $X \times Y$ is countably paracompact.

So $\sigma$-compactness is not necessary for Theorem 6. However, when the metric factor is $\sigma$-compact, the proof is simplified considerably. For the full proof, see Corollary 4.3 in [2].

Among the products $X \times Y$, the two notions of normality and countably paracompactness are the same as long as one factor is normal and the other factor is a non-discrete metric space. For such product, determining normality is equivalent to determining countably paracompactness, a covering property. In showing countably paracompactness, a shrinking property as well as a condition about decreasing sequence of closed sets being expanded by open sets (see Theorem 4 and Theorem 5 in this previous post) can be used.

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Reference

1. Engelking R., General Topology, Revised and Completed edition, Elsevier Science Publishers B. V., Heldermann Verlag, Berlin, 1989.
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.

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$\copyright$ 2017 – Dan Ma

# 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.

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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. $\square$

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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.

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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).

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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.

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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.

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$\copyright$ 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.

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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. $\square$

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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.

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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.

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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.

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

# 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.

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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 is point-finite or locally finite, then it is shrinkable.

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Key Idea

We now discuss the key idea to the proof of the main theorem. Consider the product space $X \times Y$. Let $\mathcal{U}$ be an open cover of $X \times Y$. Let $M \subset Y$. 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.

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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.

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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 $X$.

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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)

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

# kappa-Dowker space and the first conjecture of Morita

Recall the product space of the Michael line and the space of the irrational numbers. Even though the first factor is a normal space (in fact a paracompact space) and the second factor is a metric space, their product space is not normal. This is one of the classic examples demonstrating that normality is not well behaved with respect to product space. This post presents an even more striking result, i.e., for any non-discrete normal space $Y$, there exists another normal space $X$ such that $X \times Y$ is not normal. The example of the non-normal product of the Michael line and the irrationals is not some isolated example. Rather it is part of a wide spread phenomenon. This result guarantees that no matter how nice a space $Y$ is, a counter part $X$ can always be found that the product of the two spaces is not normal. This result is known as Morita’s first conjecture and was proved by Atsuji and Rudin. The solution is based on a generalization of Dowker’s theorem and a construction done by Rudin. This post demonstrates how the solution is put together.

All spaces under consideration are Hausdorff.

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Morita’s First Conjecture

In 1976, K. Morita posed the following conjecture.

Morita’s First Conjecture
If $Y$ is a normal space such that $X \times Y$ is a normal space for every normal space $X$, then $Y$ is a discrete space.

The proof given in this post is for proving the contrapositive of the above statement.

Morita’s First Conjecture
If $Y$ is a non-discrete normal space, then there exists some normal space $X$ such that $X \times Y$ is not a normal space.

Though the two forms are logically equivalent, the contrapositive form seems to have a bigger punch. The contrapositive form gives an association. Each non-discrete normal space is paired with a normal space to form a non-normal product. Examples of such pairings are readily available. Michael line is paired with the space of the irrational numbers (as discussed above). The Sogenfrey line is paired with itself. The first uncountable ordinal $\omega_1$ is paired with $\omega_1+1$ (see here) or paired with the cube $I^I$ where $I=[0,1]$ with the usual topology (see here). There are plenty of other individual examples that can be cited. In this post, we focus on a constructive proof of finding such a pairing.

Since the conjecture had been affirmed positively, it should no longer be called a conjecture. Calling it Morita’s first theorem is not appropriate since there are other results that are identified with Morita. In this discussion, we continue to call it a conjecture. Just know that it had been proven.

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Dowker’s Theorem

Next, we examine Dowker’s theorem, which characterizes normal countably paracompact spaces. The following is the statement.

Theorem 1 (Dowker’s Theorem)
Let $X$ be a normal space. The following conditions are equivalent.

1. The space $X$ is countably paracompact.
2. Every countable open cover of $X$ has a point-finite open refinement.
3. If $\left\{U_n: n=1,2,3,\cdots \right\}$ is an open cover of $X$, there exists an open refinement $\left\{V_n: n=1,2,3,\cdots \right\}$ such that $\overline{V_n} \subset U_n$ for each $n$.
4. The product space $X \times Y$ is normal for any compact metric space $Y$.
5. The product space $X \times [0,1]$ is normal where $[0,1]$ is the closed unit interval with the usual Euclidean topology.
6. For each sequence $\left\{A_n \subset X: n=1,2,3,\cdots \right\}$ of closed subsets of $X$ such that $A_1 \supset A_2 \supset A_3 \supset \cdots$ and $\cap_n A_n=\varnothing$, there exist open sets $B_1,B_2,B_3,\cdots$ such that $A_n \subset B_n$ for each $n$ such that $\cap_n B_n=\varnothing$.

The theorem is discussed here and proved here. Any normal space that violates any one of the conditions in the theorem is said to be a Dowker space. One such space was constructed by Rudin in 1971 [2]. Any Dowker space would be one factor in a non-normal product space with the other factor being a compact metric space. Actually much more can be said.

The Dowker space constructed by Rudin is the solution of Morita’s conjecture for a large number of spaces. At minimum, the product of any infinite compact metric space and the Dowker space is not normal as indicated by Dowker’s theorem. Any nontrivial convergent sequence plus the limit point is a compact metric space since it is homeomorphic to $S=\left\{0 \right\} \cup \left\{\frac{1}{n}: n=1,2,3,\cdots \right\}$ (as a subspace of the real line). Thus Rudin’s Dowker space has non-normal product with $S$. Furthermore, the product of Rudin’s Dowker space and any space containing a copy of $S$ is not normal.

Spaces that contain a copy of $S$ extend far beyond the compact metric spaces. Spaces that have lots of convergent sequences include first countable spaces, Frechet spaces and many sequential spaces (see here for an introduction for these spaces). Thus any Dowker space is an answer to Morita’s first conjecture for the non-discrete members of these classes of spaces. Actually, the range for the solution is wider than these spaces. It turns out that any space that has a countable non-discrete subspace would have a non-normal product with a Dowker space. These would include all the classes mentioned above (first countable, Frechet, sequential) as well as countably tight spaces and more.

Therefore, any Dowker space, a normal space that is not countably paracompact, is severely lacking in ability in forming normal product with another space. In order to obtain a complete solution to Morita’s first conjecture, we would need a generalized Dowker’s theorem.

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Shrinking Properties

The key is to come up with a generalized Dowker’s theorem, a theorem like Theorem 1 above, except that it is for arbitrary infinite cardinality. Then a $\kappa$-Dowker space is a space that violates one condition in the theorem. That space would be a candidate for the solution of Morita’s first conjecture. Note that Theorem 1 is for the infinite countable cardinal $\omega$ only. Before stating the theorem, let’s gather all the notions that will go into the theorem.

Let $X$ be a space. Let $\mathcal{U}$ be an open cover of the space $X$. The open cover $\mathcal{U}$ is said to be shrinkable if there is an open cover $\mathcal{V}=\left\{V(U): U \in \mathcal{U} \right\}$ 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).

Let $\kappa$ be a cardinal. The space $X$ is said to be a $\kappa$-shrinking space if every open cover of cardinality $\le \kappa$ of the space $X$ is shinkable. The space $X$ is a shrinking space if it is a $\kappa$-shrinking space for every cardinal $\kappa$.

When a family of sets are indexed by ordinals, the notion of an increasing or decreasing family of sets is possible. For example, the family $\left\{A_\alpha \subset X: \alpha<\kappa \right\}$ of subsets of the space $X$ is said to be increasing if $A_\beta \subset A_\gamma$ whenever $\beta<\gamma$. In other words, for an increasing family, the sets are getting larger whenever the index becomes larger. A decreasing family of sets is defined in the reverse way. These two notions are important for some shrinking properties discussed here – e.g. using an open cover that is increasing or using a family of closed sets that is decreasing.

In the previous discussion on shrinking spaces, two other shrinking properties are discussed – property $\mathcal{D}(\kappa)$ and property $\mathcal{B}(\kappa)$. A space $X$ is said to have property $\mathcal{D}(\kappa)$ if every increasing open cover of cardinality $\le \kappa$ for the space $X$ is shrinkable. A space $X$ is said to have property $\mathcal{B}(\kappa)$ if every increasing open cover of cardinality $\le \kappa$ for the space $X$ has a shrinking that is increasing. See this previous post for a discussion on property $\mathcal{D}(\kappa)$ and property $\mathcal{B}(\kappa)$.

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An Attempt for a Generalized Dowker’s Theorem

Let $\kappa$ be an infinite cardinal. The space $X$ is said to be a $\kappa$-paracompact space if every open cover $\mathcal{U}$ of $X$ with $\lvert \mathcal{U} \lvert \le \kappa$ has a locally finite open refinement. Thus a space is paracompact if it is $\kappa$-paracompact for every infinite cardinal $\kappa$. Of course, an $\omega$-paracompact space is a countably paracompact space.

For any infinite $\kappa$, let $D_\kappa$ be a discrete space of size $\kappa$. Let $p$ be a point not in $D_\kappa$. Define the space $Y_\kappa=D_\kappa \cup \left\{p \right\}$ as follows. The subspace $D_\kappa$ is discrete as before. The open neighborhoods at $p$ are of the form $\left\{ p \right\} \cup B$ where $B \subset D_\kappa$ and $\lvert D_\kappa-B \lvert<\kappa$. In other words, any open set containing $p$ contains all but less than $\kappa$ many discrete points.

Another concept that is needed is the cardinal function called minimal tightness. Let $Y$ be any space. Define the minimal tightness $mt(Y)$ as the least infinite cardinal $\kappa$ such that there is a non-discrete subspace of $Y$ of cardinality $\kappa$. If $Y$ is a discrete space, then let $mt(Y)=0$. For any non-discrete space $Y$, $mt(Y)=\kappa$ for some infinite $\kappa$. Note that for the space $Y_\kappa$ defined above would have $mt(Y_\kappa)=\kappa$. For any space $Y$, $mt(Y)=\omega$ if and only if $Y$ has a countable non-discrete subspace.

The following theorem can be called a $\kappa$-Dowker’s Theorem.

Theorem 2
Let $X$ be a normal space. Let $\kappa$ be an infinite cardinal. Consider the following conditions.

1. The space $X$ is a $\kappa$-paracompact space.
2. The space $X$ is a $\kappa$-shrinking space.
• For each open cover $\left\{U_\alpha: \alpha<\kappa \right\}$ of $X$, there exists an open cover $\left\{V_\alpha: \alpha<\kappa \right\}$ such that $\overline{V_\alpha} \subset U_\alpha$ for each $\alpha<\kappa$.
3. The space $X$ has property $\mathcal{D}(\kappa)$.
• For each increasing open cover $\left\{U_\alpha: \alpha<\kappa \right\}$ of $X$, there exists an open cover $\left\{V_\alpha: \alpha<\kappa \right\}$ such that $\overline{V_\alpha} \subset U_\alpha$ for each $\alpha<\kappa$.
4. For each decreasing family $\left\{F_\alpha: \alpha<\kappa \right\}$ of closed subsets of $X$ such that $\bigcap_{\alpha<\kappa} F_\alpha=\varnothing$, there exists a family $\left\{G_\alpha: \alpha<\kappa \right\}$ of open subsets of $X$ such that $\bigcap_{\alpha<\kappa} G_\alpha=\varnothing$ and $F_\alpha \subset G_\alpha$ for each $\alpha<\kappa$.
5. The space $X$ has property $\mathcal{B}(\kappa)$.
• For each increasing open cover $\left\{U_\alpha: \alpha<\kappa \right\}$ of $X$, there exists an increasing open cover $\left\{V_\alpha: \alpha<\kappa \right\}$ such that $\overline{V_\alpha} \subset U_\alpha$ for each $\alpha<\kappa$.
6. The product space $X \times Y_\kappa$ is a normal space.
7. The product space $X \times Y$ is a normal space for some space $Y$ with $mt(Y)=\kappa$.

The following diagram shows how these conditions are related.

Diagram 1
$\displaystyle \begin{array}{ccccc} 1 &\text{ } & \Longrightarrow & \text{ } & 5 \\ \text{ } & \text{ } & \text{ } & \text{ } & \text{ } \\ \Downarrow & \text{ } & \text{ } & \text{ } & \Updownarrow \\ \text{ } & \text{ } & \text{ } & \text{ } & \text{ } \\ 2 &\text{ } & \text{ } & \text{ } & 6 \\ \text{ } & \text{ } & \text{ } & \text{ } & \text{ } \\ \Downarrow & \text{ } & \text{ } & \text{ } & \Downarrow \\ \text{ } & \text{ } & \text{ } & \text{ } & \text{ } \\ 3 & \text{ } & \Longleftarrow & \text{ } & 7 \\ \text{ } & \text{ } & \text{ } & \text{ } & \text{ } \\ \Updownarrow & \text{ } & \text{ } & \text{ } & \text{ } \\ \text{ } & \text{ } & \text{ } & \text{ } & \text{ } \\ 4 & \text{ } & \text{ } & \text{ } & \text{ } \end{array}$

In addition to Diagram 1, we have the relations $5 \Longrightarrow 3$ and $2 \not \Longrightarrow 5$.

Remarks
At first glance, Diagram 1 might give the impression that the conditions in the theorem form a loop. It turns out the strongest property is $\kappa$-paracompactness (condition 1). Since condition 2 does not imply condition 5, condition 2 does not imply condition 1. Thus the conditions do not form a loop.

The implications $1 \Longrightarrow 2 \Longrightarrow 3 \Longleftarrow 5$ and $6 \Longrightarrow 7$ are immediate. The following implications are established in this previous post.

$3 \Longleftrightarrow 4$ (Theorem 4)

$5 \Longleftrightarrow 6$ (Theorem 7)

$2 \not \Longrightarrow 5$ (Example 1)

The remaining implications to be shown are $1 \Longrightarrow 5$ and $7 \Longrightarrow 3$.

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Proof of Theorem 2

$1 \Longrightarrow 5$
Let $\mathcal{U}=\left\{U_\alpha: \alpha<\kappa \right\}$ be an increasing open cover of $X$. By $\kappa$-paracompactness, let $\mathcal{G}$ be a locally finite open refinement of $\mathcal{U}$. For each $\alpha<\kappa$, define $W_\alpha$ as follows:

$W_\alpha=\cup \left\{G \in \mathcal{G}: G \subset U_\alpha \right\}$

Then $\mathcal{W}=\left\{W_\alpha: \alpha<\kappa \right\}$ is still a locally finite refinement of $\mathcal{U}$. Since the space $X$ is normal, any locally finite open cover is shrinkable. Let $\mathcal{E}=\left\{E_\alpha: \alpha<\kappa \right\}$ be a shrinking of $\mathcal{W}$. The open cover $\mathcal{E}$ is also locally finite. For each $\alpha$, let $V_\alpha=\bigcup_{\beta<\alpha} E_\beta$. Then $\mathcal{V}=\left\{V_\alpha: \alpha<\kappa \right\}$ is an increasing open cover of $X$. Note that

$\overline{V_\alpha}=\overline{\bigcup_{\beta<\alpha} E_\beta}=\bigcup_{\beta<\alpha} \overline{E_\beta}$

since $\mathcal{E}$ is locally finite and thus closure preserving. Since $\mathcal{U}$ is increasing, $\overline{E_\beta} \subset W_\beta \subset U_\beta \subset U_\alpha$ for all $\beta<\alpha$. This means that $\overline{V_\alpha} \subset U_\alpha$ for all $\alpha$.

$7 \Longrightarrow 3$
Since condition 3 is equivalent to condition 4, we show $7 \Longrightarrow 4$. Suppose that $X \times Y$ is normal where $Y$ is a space such that $mt(Y)=\kappa$. Let $D=\left\{d_\alpha: \alpha<\kappa \right\}$ be a non-discrete subset of $Y$. Let $p$ be a point such that $p \ne d_\alpha$ for all $\alpha$ and such that $p$ is a limit point of $D$ (this means that every open set containing $p$ contains some $d_\alpha$). Let $\mathcal{F}=\left\{F_\alpha: \alpha<\kappa \right\}$ be a decreasing family of closed subsets of $X$ such that $\bigcap_{\alpha<\kappa} F_\alpha=\varnothing$. Define $H$ and $K$ as follows:

$H=\cup \left\{F_\alpha \times \left\{d_\alpha \right\}: \alpha<\kappa \right\}$

$K=X \times \left\{p \right\}$

The sets $H$ and $K$ are clearly disjoint. The set $K$ is clearly a closed subset of $X \times Y$. To show that $H$ is closed, let $(x,y) \in (X \times Y)-H$. Two cases to consider: $x \in F_0$ or $x \notin F_0$ where $F_0$ is the first closed set in the family $\mathcal{F}$.

The first case $x \in F_0$. Let $\beta<\kappa$ be least such that $x \notin F_\beta$. Then $y \ne d_\gamma$ for all $\gamma<\beta$ since $(x,y) \in (X \times Y)-H$. In the space $Y$, any subset of cardinality $<\kappa$ is a closed set. Let $E=Y-\left\{d_\gamma: \gamma<\beta \right\}$, which is open containing $y$. Let $O \subset X$ be open such that $x \in O$ and $O \cap F_\beta=\varnothing$. Then $(x,y) \in O \times E$ and $O \times E$ misses points of $H$.

Now consider the second case $x \notin F_0$. Let $O \subset X$ be open such that $x \in O$ and $O$ misses $F_0$. Then $O \times Y$ is an open set containing $(x,y)$ such that $O \times Y$ misses $H$. Thus $H$ is a closed subset of $X \times Y$.

Since $X \times Y$ is normal, choose open $V \subset X \times Y$ such that $H \subset V$ and $\overline{V} \cap K=\varnothing$. For each $\alpha<\kappa$, define $G_\alpha$ as follows:

$G_\alpha=\left\{x \in X: (x,d_\alpha) \in V \right\}$

Note that each $G_\alpha$ is open in $X$ and that $F_\alpha \subset G_\alpha$ for each $\alpha<\kappa$. We claim that $\bigcap_{\alpha<\kappa} G_\alpha=\varnothing$. Let $x \in X$. The point $(x,p)$ is in $K$. Thus $(x,p) \notin \overline{V}$. Choose an open set $L \times M$ such that $(x,p) \in L \times M$ and $(L \times M) \cap \overline{V}=\varnothing$. Since $p \in M$, there is some $\gamma<\kappa$ such that $d_\gamma \in M$. Since $(x,d_\gamma) \notin \overline{V}$, $(x,d_\gamma) \notin V$. Thus $x \notin G_\gamma$. This establishes the claim that $\bigcap_{\alpha<\kappa} G_\alpha=\varnothing$.

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$\kappa$-Dowker Space

Analogous to the Dowker space, a $\kappa$-Dowker space is a normal space that violates one condition in Theorem 2. Since the seven conditions listed in Theorem 7 are not all equivalent, which condition to use? Condition 1 is the strongest condition since it implies all the other condition. At the lower left corner of Diagram 1 is condition 3, which follows from every other condition. Thus condition 3 (or 4) is the weakest property. An appropriate definition of a $\kappa$-Dowker space is through negating condition 3 or condition 4. Thus, given an infinite cardinal $\kappa$, a $\kappa$-Dowker space is a normal space $X$ that satisfies the following condition:

There exists a decreasing family $\left\{F_\alpha: \alpha<\kappa \right\}$ of closed subsets of $X$ with $\bigcap_{\alpha<\kappa} F_\alpha=\varnothing$ such that for every family $\left\{G_\alpha: \alpha<\kappa \right\}$ of open subsets of $X$ with $F_\alpha \subset G_\alpha$ for each $\alpha$, $\bigcap_{\alpha<\kappa} G_\alpha \ne \varnothing$.

The definition of $\kappa$-Dowker space is through negating condition 4. Of course, negating condition 3 would give an equivalent definition.

When $\kappa$ is the countably infinite cardinal $\omega$, a $\kappa$-Dowker space is simply the ordinary Dowker space constructed by M. E. Rudin [2]. Rudin generalized the construction of the ordinary Dowker space to obtain a $\kappa$-Dowker space for every infinite cardinal $\kappa$ [4]. The space that Rudin constructed in [4] would be a normal space $X$ such that condition 4 of Theorem 2 is violated. This means that the space $X$ would violate condition 7 in Theorem 2. Thus $X \times Y$ is not normal for every space $Y$ with $mt(Y)=\kappa$.

Here’s the solution of Morita’s first conjecture. Let $Y$ be a normal and non-discrete space. Determine the least cardinality $\kappa$ of a non-discrete subspace of $Y$. Obtain the $\kappa$-Dowker space $X$ as in [4]. Then $X \times Y$ is not normal according to the preceding paragraph.

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Remarks

Answering Morita’s first conjecture is a two-step approach. First, figure out what a generalized Dowker’s theorem should be. Then a $\kappa$-Dowker space is one that violates an appropriate condition in the generalized Dowker’s theorem. By violating the right condition in the theorem, we have a way to obtain non-normal product space needed in the answer. The second step is of course the proof of the existence of a space that violates the condition in the generalized Dowker’s theorem.

Figuring out the form of the generalized Dowker’s theorem took some work. It is more than just changing the countable infinite cardinal in Dowker’s theorem (Theorem 1 above) to an arbitrary infinite cardinal. This is because the conditions in Theorem 1 are unequal when the cardinality is changed to an uncountable one.

We take the cue from Rudin’s chapter on Dowker spaces [3]. In the last page of that chapter, Rudin pointed out the conditions that should go into a generalized Dowker’s theorem. However, the explanation of the relationship among the conditions is not clear. The previous post and this post are an attempt to sort out the conditions and fill in as much details as possible.

Rudin’s chapter did have the right condition for defining $\kappa$-Dowker space. It seems that prior to the writing of that chapter, there was some confusion on how to define a $\kappa$-Dowker space, i.e. a condition in the theorem the violation of which would give a $\kappa$-Dowker space. If the condition used is a stronger property, the violation may not yield enough information to get non-normal products. According to Diagram 1, condition 3 in Theorem 2 is the right one to use since it is the weakest condition and is down streamed from the conditions about normal product space. So the violation of condition 3 would answer Morita’s first conjecture.

We do not discuss the other step in the solution in any details. Any interested reader can review Rudin’s construction in [2] and [4]. The $\kappa$-Dowker space is an appropriate subspace of a product space with the box topology.

One interesting observation about the ordinary Dowker space (the one that violates a condition in Theorem 1) is that the product of any Dowker space and any space with a countable non-discrete subspace is not normal. This shows that Dowker space is badly non-productive with respect to normality. This fact is actually not obvious in the usual formulation of Dowker’s theorem (Theorem 1 above). What makes this more obvious in the direction $7 \Longrightarrow 3$ in Theorem 2. For the countably infinite case, $7 \Longrightarrow 3$ is essentially this: If $X \times Y$ is normal where $Y$ has a countable non-discrete subspace, then $X$ is not a Dowker space. Thus if the goal is to find a non-normal product space, a Dowker space should be one space to check.

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Loose Ends

In the course of working on the contents in this post and the previous post, there are some questions that we do not know how to answer and have not spent time to verify one way or the other. Possibly there are some loose ends to tie. They for the most parts are not open questions, but they should be interesting questions to consider.

For the $\kappa$-Dowker’s theorem (Theorem 2), one natural question is on the relative strengths of the conditions. It will be interesting to find out the implications not shown in Diagram 1. For example, for the three shrinking properties (conditions 2, 3 and 5), it is straightforward from definition that $2 \Longrightarrow 3$ and $5 \Longrightarrow 3$. The example of $X=\omega_1$ (the first uncountable ordinal) shows that $2 \not \Longrightarrow 5$ and hence $2 \not \Longrightarrow 1$. What about $3 \Longrightarrow 2$? In [5], Beslagic and Rudin showed that $3 \not \Longrightarrow 2$ using $\Diamond ^{++}$. A natural question would be: can there be ZFC example? Perhaps searching on more recent papers can yield some answers.

Another question is $5 \Longrightarrow 1$? The answer is no with the example being a Navy space – Example 7.6 in p. 194 [1]. The other two directions that have not been accounted for are: $7 \Longrightarrow 6$ and $3 \Longrightarrow 7$? We do not know the answer.

Another small question that we come across is about $X=\omega_1$ (the first uncountable ordinal). This is an example for showing $2 \not \Longrightarrow 5$. Thus condition 6 is false. Thus $X \times Y_{\omega_1}$ is not normal. Here $Y_{\omega_1}$ is simply the one-point Lindelofication of a discrete space of cardinality $\omega_1$. The question is: is condition 7 true for $X=\omega_1$? The product of $X=\omega_1$ and $Y_{\omega_1}$ (a space with minimal tightness $\omega_1$) is not normal. Is there a normal $X \times Y$ where $Y$ is another space with minimal tightness $\omega_1$?

Dowker’s theorem and $\kappa$-Dowker’s theorem show that finding a normal space that is not shrinking is not a simple matter. To find a normal space that is not countably shrinking took 20 years (1951 to 1971). For any uncountable $\kappa$, the $\kappa$-Dowker space that is based on the same construction of an ordinary Dowker space is also a space that is not $\kappa$-shrinking. With an uncountable $\kappa$, is the $\kappa$-Dowker space countably shrinking? This is not obvious one way or the other just from the definition of $\kappa$-Dowker space. Perhaps there is something obvious and we have not connected the dots. Perhaps we need to go into the definition of the $\kappa$-Dowker space in [4] to show that it is countably shrinking. The motivation is that we tried to find a normal space that is countably shrinking but not $\kappa$-shrinking for some uncountable $\kappa$. It seems that the $\kappa$-Dowker space in [4] is the natural candidate.

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Reference

1. Morita K., Nagata J.,Topics in General Topology, Elsevier Science Publishers, B. V., The Netherlands, 1989.
2. Rudin M. E., A Normal Space $X$ for which $X \times I$ is not Normal, Fund. Math., 73, 179-486, 1971. (link)
3. Rudin M. E., Dowker Spaces, Handbook of Set-Theoretic Topology (K. Kunen and J. E. Vaughan, eds), Elsevier Science Publishers B. V., Amsterdam, (1984) 761-780.
4. Rudin M. E., $\kappa$-Dowker Spaces, Czechoslovak Mathematical Journal, 28, No.2, 324-326, 1978. (link)
5. Rudin M. E., Beslagic A.,Set-Theoretic Constructions of Non-Shrinking Open Covers, Topology Appl., 20, 167-177, 1985. (link)
6. Yasui Y., On the Characterization of the $\mathcal{B}$-Property by the Normality of Product Spaces, Topology and its Applications, 15, 323-326, 1983. (abstract and paper)
7. Yasui Y., Some Characterization of a $\mathcal{B}$-Property, TSUKUBA J. MATH., 10, No. 2, 243-247, 1986.

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

# Spaces with shrinking properties

Certain covering properties and separation properties allow open covers to shrink, e.g. paracompact spaces, normal spaces, and countably paracompact spaces. The shrinking property is also interesting on its own. This post gives a more in-depth discussion than the one in the previous post on countably paracompact spaces. After discussing shrinking spaces, we introduce three shrinking related properties. These properties show that there is a deep and delicate connection among shrinking properties and normality in products. This post is also a preparation for the next post on $\kappa$-Dowker space and Morita’s first conjecture.

All spaces under consideration are Hausdorff and normal or Hausdorff and regular (if not normal).

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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 in a certain way, 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. The property can also be broken up according to the cardinality of the open cover. Let $\kappa$ be a cardinal. A space $X$ is $\kappa$-shrinking if every open cover of cardinality $\le \kappa$ for $X$ is shrinkable. A space $X$ is countably shrinking if it is $\omega$-shrinking.

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Examples of Shrinking

Let’s look at a few situations where open covers can be shrunk either all the time or on a limited basis. For a normal space, certain covers can be shrunk as indicated by 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 does not have to be shrinkable unless it is a point-finite or locally finite open cover.

The theorem of C. H. Dowker states that a normal space $X$ is countably paracompact if and only every countable open cover of $X$ is shrinkable if and only if the product space $X \times Y$ is normal for every compact metric space $Y$ if and only if the product space $X \times [0,1]$ is normal. The theorem is discussed here. A Dowker space is a normal space that violates the theorem. Thus any Dowker space has a countably infinite open cover that cannot be shrunk, or equivalently a normal space that forms a non-normal product with a compact metric space. Thus the notion of shrinking has a connection with normality in the product spaces. A Dowker space space was constructed by M. E. Rudin in ZFC [2]. So far Rudin’s example is essentially the only ZFC Dowker space. This goes to show that finding a normal space that is not countably shrinking is not a trivial matter.

Several facts can be derived easily from Theorem 1 and Dowker’s theorem. For clarity, they are called out as corollaries.

Corollary 2

• All shrinking spaces are normal.
• All shrinking spaces are normal and countably paracompact.
• Any normal and metacompact space is a shrinking space.

For the first corollary, if every open cover of a space can be shrunk, then all finite open covers can be shrunk and thus the space must be normal. As indicated above, Dowker’s theorem states that in a normal space, countably paracompactness is equivalent to countably shrinking. Thus any shrinking space is normal and countably paracompact.

Though an infinite open cover of a normal space may not be shrinkable, adding an appropriate covering property to any normal space will make it into a shrinking space. An easy way is through point-finite open covers. If every open cover has a point-finite open refinement (i.e. a metacompact space), then the point-finite open refinement can be shrunk (if the space is also normal). Thus the third corollary is established. Note that the metacompact is not the best possible result. For example, it is known that any normal and submetacompact space is a shrinking space – see Theorem 6.2 of [1].

In paracompact spaces, all open covers can be shrunk. One way to see this is through Corollary 2. Any paracompact space is normal and metacompact. It is also informative to look at the following characterization of paracompact spaces.

Theorem 3
A space $X$ is paracompact if and only if every open cover $\left\{U_\alpha: \alpha<\kappa \right\}$ of $X$ has a locally finite open refinement $\left\{V_\alpha: \alpha<\kappa \right\}$ such that $\overline{V_\alpha} \subset U_\alpha$ for each $\alpha$.

A proof can be found here. Thus every open cover of a paracompact space can be shrunk by a locally finite shrinking. To summarize, we have discussed the following implications.

Diagram 1

\displaystyle \begin{aligned} \text{Paracompact} \Longrightarrow & \text{ Normal + Metacompact} \\&\ \ \ \ \ \ \Big \Downarrow \\&\text{ Shrinking} \\&\ \ \ \ \ \ \Big \Downarrow \\& \text{ Normal + Countably Paracompact} \\&\ \ \ \ \ \ \Big \Downarrow \\& \text{ Normal} \end{aligned}

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Three Shrinking Related Properties

None of the implications in Diagram 1 can be reversed. The last implication in the diagram cannot be reversed due to Rudin’es Dowker space. One natural example to look for would be spaces that are normal and countably paracompact but fail in shrinking at some uncountable cardinal. As indicated by the the theorem of C. H, Dowker, the notion of shrinking is intimately connected to normality in product spaces $X \times Y$. To further investigate, consider the following three properties.

Let $X$ be a space. Let $\kappa$ be an infinite cardinal. Consider the following three properties.

The space $X$ is $\kappa$-shrinking if and only if any open cover of cardinality $\le \kappa$ for the space $X$ is shrinkable, i.e. the following condition holds.

For each open cover $\left\{U_\alpha: \alpha<\kappa \right\}$ of $X$, there exists an open cover $\left\{V_\alpha: \alpha<\kappa \right\}$ such that $\overline{V_\alpha} \subset U_\alpha$ for each $\alpha<\kappa$.

The space $X$ has Property $\mathcal{D}(\kappa)$ if and only if every increasing open cover of cardinality $\le \kappa$ for the space $X$ is shrinkable, i.e. the following holds.

For each increasing open cover $\left\{U_\alpha: \alpha<\kappa \right\}$ of $X$, there exists an open cover $\left\{V_\alpha: \alpha<\kappa \right\}$ such that $\overline{V_\alpha} \subset U_\alpha$ for each $\alpha<\kappa$.

The space $X$ has Property $\mathcal{B}(\kappa)$ if and only if the following holds.

For each increasing open cover $\left\{U_\alpha: \alpha<\kappa \right\}$ of $X$, there exists an increasing open cover $\left\{V_\alpha: \alpha<\kappa \right\}$ such that $\overline{V_\alpha} \subset U_\alpha$ for each $\alpha<\kappa$.

A family $\left\{A_\alpha: \alpha<\kappa \right\}$ is increasing if $A_\alpha \subset A_\beta$ for any $\alpha<\beta<\kappa$. It is decreasing if $A_\beta \subset A_\alpha$ for any $\alpha<\beta<\kappa$.

In general, any space that is $\kappa$-shrinking for all cardinals $\kappa$ is a shrinking space as defined earlier. Any space that has property $\mathcal{D}(\kappa)$ for all cardinals $\kappa$ is said to have property $\mathcal{D}$. Any space that has property $\mathcal{B}(\kappa)$ for all cardinals $\kappa$ is said to have property $\mathcal{B}$.

The first property $\kappa$-shrinking is simply the shrinking property for open covers of cardinality $\le \kappa$. The property $\mathcal{D}(\kappa)$ is $\kappa$-shrinking with the additional requirement that the open covers to be shrunk must be increasing. It is clear that $\kappa$-shrinking implies property $\mathcal{D}(\kappa)$. The property $\mathcal{B}(\kappa)$ appears to be similar to $\mathcal{D}(\kappa)$ except that $\mathcal{B}(\kappa)$ has the additional requirement that the shrinking is also increasing. As a result $\mathcal{B}(\kappa)$ implies $\mathcal{D}(\kappa)$. The following diagram shows the implications.

Diagram 2

$\displaystyle \begin{array}{ccccc} \kappa \text{-Shrinking} &\text{ } & \not \longrightarrow & \text{ } & \text{Property } \mathcal{B}(\kappa) \\ \text{ } & \searrow & \text{ } & \swarrow & \text{ } \\ \text{ } &\text{ } & \text{Property } \mathcal{D}(\kappa) & \text{ } & \text{ } \\ \text{ } & \text{ } & \text{ } & \text{ } & \text{ } \\ \end{array}$

The implications in Diagram 2 are immediate. An example is given below showing that $\omega_1$-shrinking does not imply property $\mathcal{B}(\omega_1)$. If $\kappa=\omega$, then all three properties are equivalent in normal spaces, as displayed in the following diagram. The proof is in Theorem 5.

Diagram 3

$\displaystyle \begin{array}{ccccc} \omega \text{-Shrinking} &\text{ } & \longrightarrow & \text{ } & \text{Property } \mathcal{B}(\omega) \\ \text{ } & \nwarrow & \text{ } & \swarrow & \text{ } \\ \text{ } &\text{ } & \text{Property } \mathcal{D}(\omega) & \text{ } & \text{ } \\ \text{ } & \text{ } & \text{ } & \text{ } & \text{ } \\ \end{array}$

The property $\mathcal{D}(\kappa)$ has a dual statement in terms of decreasing closed sets. The following theorem gives the dual statement.

Theorem 4
Let $X$ be a normal space. Let $\kappa$ be an infinite cardinal. The following two properties are equivalent.

• The space $X$ has property $\mathcal{D}(\kappa)$.
• For each decreasing family $\left\{F_\alpha: \alpha<\kappa \right\}$ of closed subsets of $X$ such that $\bigcap_{\alpha<\kappa} F_\alpha=\varnothing$, there exists a family $\left\{G_\alpha: \alpha<\kappa \right\}$ of open subsets of $X$ such that $\bigcap_{\alpha<\kappa} G_\alpha=\varnothing$ and $F_\alpha \subset G_\alpha$ for each $\alpha<\kappa$.

First bullet implies second bullet
Let $\left\{F_\alpha: \alpha<\kappa \right\}$ be a decreasing family of closed subsets of $X$ with empty intersection. Then $\left\{U_\alpha: \alpha<\kappa \right\}$ is an increasing family of open subsets of $X$ where $U_\alpha=X-F_\alpha$. Let $\left\{V_\alpha: \alpha<\kappa \right\}$ be an open cover of $X$ such that $\overline{V_\alpha} \subset U_\alpha$ for each $\alpha$. Then $\left\{G_\alpha: \alpha<\kappa \right\}$ where $G_\alpha=X-\overline{V_\alpha}$ is the needed open expansion.

Second bullet implies first bullet
Let $\left\{U_\alpha: \alpha<\kappa \right\}$ be an increasing open cover of $X$. Then $\left\{F_\alpha: \alpha<\kappa \right\}$ is a decreasing family of closed subsets of $X$ where $F_\alpha=X-U_\alpha$. Note that $\bigcap_{\alpha<\kappa} F_\alpha=\varnothing$. Let $\left\{G_\alpha: \alpha<\kappa \right\}$ be a family of open subsets of $X$ such that $\bigcap_{\alpha<\kappa} G_\alpha=\varnothing$ and $F_\alpha \subset G_\alpha$ for each $\alpha$. For each $\alpha$, there is open set $W_\alpha$ such that $F_\alpha \subset W_\alpha \subset \overline{W_\alpha} \subset G_\alpha$ since $X$ is normal. For each $\alpha$, let $V_\alpha=X-\overline{W_\alpha}$. Then $\left\{V_\alpha: \alpha<\kappa \right\}$ is a family of open subsets of $X$ required by the first bullet. It is a cover because $\bigcap_{\alpha<\kappa} \overline{W_\alpha}=\varnothing$. To show $\overline{V_\alpha} \subset U_\alpha$, let $x \in \overline{V_\alpha}$ such that $x \notin U_\alpha$. Then $x \in W_\alpha$. Since $x \in \overline{V_\alpha}$ and $W_\alpha$ is open, $W_\alpha \cap V_\alpha \ne \varnothing$. Let $y \in W_\alpha \cap V_\alpha$. Since $y \in V_\alpha$, $y \notin \overline{W_\alpha}$, which means $y \notin W_\alpha$, a contradiction. Thus $\overline{V_\alpha} \subset U_\alpha$.

Now we show that the three properties in Diagram 3 are equivalent.

Theorem 5
Let $X$ be a normal space. Then the following implications hold.
$\omega$-shrinking $\Longrightarrow$ Property $\mathcal{B}(\omega)$ $\Longrightarrow$ Property $\mathcal{D}(\omega)$ $\Longrightarrow$ $\omega$-shrinking

Proof of Theorem 5
$\omega$-shrinking $\Longrightarrow$ Property $\mathcal{B}(\omega)$
Suppose that $X$ is $\omega$-shrinking. By Dowker’s theorem, $X \times (\omega+1)$ is a normal space. We can think of $\omega+1$ as a convergent sequence with $\omega$ as the limit point. Let $\left\{U_n:n=0,1,2,\cdots \right\}$ be an increasing open cover of $X$. Define $H$ and $K$ as follows:

$H=\cup \left\{(X-U_n) \times \left\{n \right\}: n=0,1,2,\cdots \right\}$

$K=X \times \left\{\omega \right\}$

It is straightforward to verify that $H$ and $K$ are disjoint closed subsets of $X \times (\omega+1)$. By normality, let $V$ and $W$ be disjoint open subsets of $X \times (\omega+1)$ such that $H \subset W$ and $K \subset V$. For each integer $n=0,1,2,\cdots$, define $V_n$ as follows:

$V_n=\left\{x \in X: \exists \ \text{open } O \subset X \text{ such that } x \in O \text{ and } O \times [n, \omega] \subset V \right\}$

The set $[n, \omega]$ consists of all integers $\ge n$ and the limit point $\omega$. From the way the sets $V_n$ are defined, $\left\{V_n:n=0,1,2,\cdots \right\}$ is an increasing open cover of $X$. The remaining thing to show is that $\overline{V_n} \subset U_n$ for each $n$. Suppose that $x \in \overline{V_n}$ and $x \notin U_n$. Then $(x,n) \in H$ by definition of $H$. There exists an open set $E \times \left\{n \right\}$ such that $(x,n) \in E \times \left\{n \right\}$ and $(E \times \left\{n \right\}) \cap V=\varnothing$. Since $E$ is an open set containing $x$, $E \cap V_n \ne \varnothing$. Let $y \in E \cap V_n$. By definition of $V_n$, there is some open set $O$ such that $y \in O$ and $O \times [n, \omega] \subset V$, a contradiction since $(E \cap O) \times \left\{n \right\}$ is supposed to miss $V$. Thus $\overline{V_n} \subset U_n$ for all integers $n$.

The direction Property $\mathcal{B}(\omega)$ $\Longrightarrow$ Property $\mathcal{D}(\omega)$ is immediate.

Property $\mathcal{D}(\omega)$ $\Longrightarrow$ $\omega$-shrinking
Consider the dual condition of $\mathcal{D}(\omega)$ in Theorem 4, which is equivalent to $\omega$-shrinking according to Dowker’s theorem. $\square$

Remarks
The direction $\omega$-shrinking $\Longrightarrow$ Property $\mathcal{B}(\omega)$ is true because $\omega$-shrinking is equivalent to the normality in the product $X \times (\omega+1)$. The same is not true when $\kappa$ becomes an uncountable cardinal. We now show that $\kappa$-shrinking does not imply $\mathcal{B}(\kappa)$ in general.

Example 1
The space $X=\omega_1$ is the set of all ordinals less than $\omega_1$ with the ordered topology. Since it is a linearly ordered space, it is a shrinking space. Thus in particular it is $\omega_1$-shrinking. To show that $X$ does not have property $\mathcal{B}(\omega_1)$, consider the increasing open cover $\left\{U_\alpha: \alpha<\omega_1 \right\}$ where $U_\alpha=[0,\alpha)$ for each $\alpha<\omega_1$. Here $[0,\alpha)$ consists of all ordinals less than $\alpha$. Suppose $X$ has property $\mathcal{B}(\omega_1)$. Then let $\left\{V_\alpha: \alpha<\omega_1 \right\}$ be an increasing open cover of $X$ such that $\overline{V_\alpha} \subset U_\alpha$ for each $\alpha$.

Let $L$ be the set of all limit ordinals in $X$. For each $\alpha \in L$, $\alpha \notin U_\alpha$ and thus $\alpha \notin \overline{V_\alpha}$. Thus there exists a countable ordinal $f(\alpha)<\alpha$ such that $(f(\alpha),\alpha]$ misses points in $\overline{V_\alpha}$. Thus the map $f: L \rightarrow \omega_1$ is a pressing down map. By the pressing down lemma, there exists some $\alpha<\omega_1$ such that $S=f^{-1}(\alpha)$ is a stationary set in $\omega_1$, which means that $S$ intersects with every closed and unbounded subset of $X=\omega_1$. This means that for each $\gamma>\alpha$, $(\alpha, \gamma]$ would miss $\overline{V_\gamma}$. This means that for each $\gamma>\alpha$, $\overline{V_\gamma} \subset [0,\alpha]$. As a result $\left\{V_\alpha: \alpha<\omega_1 \right\}$ would not be a cover of $X$, a contradiction. So $X$ does not have property $\mathcal{B}(\omega_1)$. $\square$

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Property $\mathcal{B}(\kappa)$

Of the three properties discussed in the above section, we would like to single out property $\mathcal{B}(\kappa)$. This property has a connection with normality in the product $X \times Y$ (see Theorem 7). First, we prove a lemma that is used in proving Theorem 7.

Lemma 6
Show that the property $\mathcal{B}(\kappa)$ is hereditary with respect to closed subsets.

Proof of Lemma 6
Let $X$ be a space with property $\mathcal{B}(\kappa)$. Let $A$ be a closed subspace of $X$. Let $\left\{U_\alpha \subset A: \alpha<\kappa \right\}$ be an increasing open cover of $A$. For each $\alpha$, let $W_\alpha$ be an open subset of $X$ such that $U_\alpha=W_\alpha \cap A$. Since the open sets $U_\alpha$ are increasing, the open sets $W_\alpha$ can be chosen inductively such that $W_\alpha \supset W_\gamma$ for all $\gamma<\alpha$. This will ensure that $W_\alpha$ will form an increasing cover.

Then $\left\{W_\alpha^* \subset X: \alpha<\kappa \right\}$ is an increasing open cover of $X$ where $W_\alpha^*=W_\alpha \cup (X-A)$. By property $\mathcal{B}(\kappa)$, let $\left\{E_\alpha \subset X: \alpha<\kappa \right\}$ be an increasing open cover of $X$ such that $\overline{E_\alpha} \subset W_\alpha^*$. For each $\alpha$, let $V_\alpha=E_\alpha \cap A$. It can be readily verified that $\left\{V_\alpha \subset A: \alpha<\kappa \right\}$ is an increasing open cover of $A$. Furthermore, $\overline{V_\alpha} \subset U_\alpha$ for each $\alpha$ (closure taken in $A$). $\square$

Let $\kappa$ be an infinite cardinal. Let $D_\kappa=\left\{d_\alpha: \alpha<\kappa \right\}$ be a discrete space of cardinality $\kappa$. Let $p$ be a point not in $D_\kappa$. Let $Y_\kappa=D_\kappa \cup \left\{p \right\}$. Define a topology on $Y_\kappa$ by letting $D_\kappa$ be discrete and by letting open neighborhood of $p$ be of the form $\left\{p \right\} \cup E$ where $E \subset D_\kappa$ and $D_\kappa-E$ has cardinality less than $\kappa$. Note the similarity between $Y_\kappa$ and the convergent sequence $\omega+1$ in the proof of Theorem 5.

Theorem 7
Let $X$ be a normal space. Then the product space $X \times Y_\kappa$ is normal if and only if $X$ has property $\mathcal{B}(\kappa)$.

Remarks
The property $\mathcal{B}(\kappa)$ involves the shrinking of any increasing open cover with the added property that the shrinking is also increasing. The increasing shrinking is just what is needed to show that disjoint closed subsets of the product space can be separated.

Notations
Let’s set some notations that are useful in proving Theorem 7.

• The set $[d_\alpha,p]$ is an open set in $Y_\kappa$ containing the point $p$ and is defined as follows.
• $[d_\alpha,p]=\left\{d_\beta: \alpha \le \beta<\kappa \right\} \cup \left\{p \right\}$.
• For any two disjoint closed subsets $H$ and $K$ of the product space $X \times Y_\kappa$, define the following sets.
• For each $\alpha<\kappa$, let $H_\alpha=H \cap (X \times \left\{d_\alpha \right\})$ and $K_\alpha=K \cap (X \times \left\{d_\alpha \right\})$.
• Let $H_p=H \cap (X \times \left\{p \right\})$ and $K_p=K \cap (X \times \left\{p \right\})$.
• For each $\alpha<\kappa$, choose open $O_\alpha \subset X$ such that $G_\alpha=O_\alpha \times \left\{d_\alpha \right\}$, $H_\alpha \subset G_\alpha$ and $\overline{G_\alpha} \cap K_\alpha=\varnothing$ (due to normality of $X$).
• Choose open $O_p \subset X$ such that $G_p=O_p \times \left\{p \right\}$, $H_p \subset G_p$ and $\overline{G_p} \cap K_p=\varnothing$ (due to normality of $X$).

Proof of Theorem 7
Suppose that $X$ has property $\mathcal{B}(\kappa)$. Let $H$ and $K$ be two disjoint closed sets of $X \times Y_\kappa$. Consider the following cases based on the locations of the closed sets $H$ and $K$.

Case 1. $H \subset X \times D_\kappa$ and $K \subset X \times D_\kappa$.
Case 2a. $H=X \times \left\{p\right\}$
Case 2b. Exactly one of $H$ and $K$ intersect the set $X \times \left\{p\right\}$.
Case 3. Both $H$ and $K$ intersect the set $X \times \left\{p\right\}$.

Remarks
Case 1 is easy. Case 2a is the pivotal case. Case 2b and Case 3 use a similar idea. The result in Theorem 7 is found in [1] (Theorem 6.9 in p. 189) and [4]. The authors in these two sources claimed that Case 2a is the only case that matters, citing a lemma in another source. The lemma was not stated in these two sources and the source for the lemma is a PhD dissertation that is not readily available. Case 3 essentially uses the same idea but it has enough differences. For the sake of completeness, we work out all the cases. Case 3 applies property $\mathcal{B}(\kappa)$ twice. Despite the complicated notations, the essential idea is quite simple. If any reader finds the proof too long, just understand Case 2a and then get the gist of how the idea is applied in Case 2b and Case 3.

Case 1.
$H \subset X \times D_\kappa$ and $K \subset X \times D_\kappa$.

Let $M =\bigcup_{\alpha<\kappa} G_\alpha$. It is clear that $H \subset M$ and $\overline{M} \cap K=\varnothing$.

Case 2a.
Assume that $H=X \times \left\{p\right\}$. We now proceed to separate $H$ and $K$ with disjoint open sets. For each $\alpha<\kappa$, define $U_\alpha$ as follows:

$U_\alpha=\cup \left\{O \subset X: O \text{ is open such that } (O \times [d_\alpha,p]) \cap K =\varnothing \right\}$

Then $\left\{U_\alpha: \alpha<\kappa \right\}$ is an increasing open cover of $X$. By property $\mathcal{B}(\kappa)$, there is an increasing open cover $\mathcal{V}=\left\{V_\alpha: \alpha<\kappa \right\}$ of $X$ such that $\overline{V_\alpha} \subset U_\alpha$ for each $\alpha$. The shrinking $\mathcal{V}$ allows us to define an open set $G$ such that $H \subset G$ and $\overline{G} \cap K=\varnothing$.

Let $G=\cup \left\{V_\alpha \times [d_\alpha,p]: \alpha<\kappa \right\}$. It is clear that $H \subset G$. Next, we show that $\overline{G} \cap K=\varnothing$. Suppose that $(x,d_\alpha) \in K$. Then $(x,d_\alpha) \notin U_\alpha \times [d_\alpha,p]$. As a result, $(x,d_\alpha) \notin \overline{V_\alpha} \times [d_\alpha,p]$. Let $O \subset X$ be open such that $x \in O$ and $(O \times \left\{d_\alpha \right\}) \cap (\overline{V_\alpha} \times [d_\alpha,p])=\varnothing$. Since $V_\beta \subset V_\alpha$ for all $\beta<\alpha$, it follows that $(O \times \left\{d_\alpha \right\}) \cap (V_\beta \times [d_\beta,p])=\varnothing$ for all $\beta < \alpha$. It is clear that $(O \times \left\{d_\alpha \right\}) \cap (V_\gamma \times [d_\gamma,p])=\varnothing$ for all $\gamma>\alpha$. What has been shown is that there is an open set containing the point $(x,d_\alpha)$ that contains no point of $G$. This means that $(x,d_\alpha) \notin \overline{G}$. We have established that $\overline{G} \cap K=\varnothing$.

Case 2b.
Exactly one of $H$ and $K$ intersect the set $X \times \left\{p\right\}$. We assume that $H$ is the set that intersects the set $X \times \left\{p\right\}$. The only difference between Case 2b and Case 2a is that there can be points of $H$ outside of $X \times \left\{p\right\}$ in Case 2b.

Now proceed as in Case 2a. Obtain the open cover $\left\{U_\alpha: \alpha<\kappa \right\}$, the open cover $\left\{V_\alpha: \alpha<\kappa \right\}$ and the open set $G$ as in Case 2a. Let $M=G \cup (\bigcup_{\alpha<\kappa} G_\alpha)$. It is clear that $H \subset M$. We claim that $\overline{M} \cap K=\varnothing$. Suppose that $(x,d_\gamma) \in K$. Since $\overline{G} \cap K=\varnothing$ (as in Case 2a), there exists open set $W=O \times \left\{ d_\gamma \right\}$ such that $(x,d_\gamma) \in W$ and $W \cap \overline{G}=\varnothing$. There also exists open $W_1 \subset W$ such that $(x,d_\gamma) \in W_1$ and $W_1 \cap \overline{G_\gamma}=\varnothing$. It is clear that $W_1 \cap G_\beta=\varnothing$ for all $\beta \ne \gamma$. This means that $W_1$ is an open set containing the point $(x,d_\gamma)$ such that $W_1$ misses the open set $M$. Thus $\overline{M} \cap K=\varnothing$.

Case 3.
Both $H$ and $K$ intersect the set $X \times \left\{p\right\}$.

Now project $H_p$ and $K_p$ onto the space $X$.

$H_p^*=\left\{x \in X: (x,p) \in H_p \right\}$

$K_p^*=\left\{x \in X: (x,p) \in K_p \right\}$

Note that $H_p^*$ is simply the copy of $H_p$ and $K_p^*$ is the copy of $K_p$ in $X$. Since $X$ is normal, choose disjoint open sets $E_1$ and $E_1$ such that $H_p^* \subset E_1$ and $K_p^* \subset E_2$.

Let $A_1=\overline{E_1}$ and $B_1=X-K_p^*$. Let $A_2=\overline{E_2}$ and $B_2=X-H_p^*$. Note that $A_1$ is closed in $X$, $B_1$ is open in $X$ and $A_1 \subset B_1$. Similarly $A_2$ is closed in $X$, $B_2$ is open in $X$ and $A_2 \subset B_2$.

We now define two increasing open covers using property $\mathcal{B}(\kappa)$. Define $U_{\alpha,1}$ and $T_{\alpha,1}$ and $U_{\alpha,2}$ and $T_{\alpha,2}$ as follows:

$U_{\alpha,1}=\cup \left\{O \subset B_1: O \text{ is open such that } (O \times [d_\alpha,p]) \cap K =\varnothing \right\}$

$T_{\alpha,1}=U_{\alpha,1} \cap A_1$

$U_{\alpha,2}=\cup \left\{O \subset B_2: O \text{ is open such that } (O \times [d_\alpha,p]) \cap H =\varnothing \right\}$

$T_{\alpha,2}=U_{\alpha,2} \cap A_2$

The open cover $\mathcal{T}_1=\left\{T_{\alpha,1}: \alpha<\kappa \right\}$ is an increasing open cover of $A_1$. The open cover $\mathcal{T}_2=\left\{T_{\alpha,2}: \alpha<\kappa \right\}$ is an increasing open cover of $A_2$.By property $\mathcal{B}(\kappa)$ of $A_1$ and $A_2$, both covers have the following as shrinking (by Lemma 6). The two shrinkings are:

$\mathcal{V}_1=\left\{V_{\alpha,1} \subset A_1: \alpha<\kappa \right\}$

$\mathcal{V}_2=\left\{V_{\alpha,2} \subset A_2: \alpha<\kappa \right\}$

such that

$\overline{V_{\alpha,1}} \subset T_{\alpha,1}$

$\overline{V_{\alpha,2}} \subset T_{\alpha,2}$

for each $\alpha<\kappa$ and such that both $\mathcal{V}_1$ and $\mathcal{V}_2$ are increasing open covers. Note that the closure $\overline{V_{\alpha,1}}$ is taken in $A_1$ and the closure $\overline{V_{\alpha,2}}$ is taken in $A_2$.

For each $\alpha$, let $W_{\alpha,1}$ be the interior of $V_{\alpha,1}$ and $W_{\alpha,2}$ be the interior of $V_{\alpha,2}$ (with respect to $X$). Note that $W_{\alpha,1}$ is meaningful since $V_{\alpha,1}$ is a subset of the closure of the open set $E_1$. Similar observation for $W_{\alpha,2}$. To make the rest of the argument easier to see, note the following fact about $W_{\alpha,1}$ and $W_{\alpha,2}$.

$\overline{W_{\alpha,1}} \subset \overline{V_{\alpha,1}} \subset T_{\alpha,1} \subset U_{\alpha,1}$ (closure with respect to $X$)

$\overline{W_{\alpha,2}} \subset \overline{V_{\alpha,2}} \subset T_{\alpha,2} \subset U_{\alpha,2}$ (closure with respect to $X$)

For each $\alpha<\kappa$, choose open set $O_\alpha \subset X$ such that

$L_\alpha=O_\alpha \times \left\{d_\alpha \right\}$

$H_\alpha \subset L_\alpha$

$\overline{L_\alpha} \cap K_\alpha=\varnothing$

$L_\alpha \cap (\overline{W_{\alpha,2}} \times [d_\alpha,p])=\varnothing$

The last point is possible because $U_{\alpha,2} \times [d_\alpha,p]$ misses $H$ and $\overline{W_{\alpha,2}} \subset U_{\alpha,2}$. Define the open sets $G$ and $M$ as follows:

$G=\cup \left\{W_{\alpha,1} \times [d_\alpha,p]: \alpha<\kappa \right\}$

$M=G \cup (\bigcup_{\alpha<\kappa} L_\alpha)$

It is clear that $H \subset M$. We claim that $\overline{M} \cap K=\varnothing$. To this end, we show that if $(x,y) \in K$, then $(x,y) \notin \overline{M}$. If $(x,y) \in K$, then either $(x,y)=(x,d_\gamma)$ for some $\gamma$ or $(x,y)=(x,p)$.

Let $(x,d_\gamma) \in K$. Note that $(x,d_\gamma) \notin U_{\gamma,1} \times [d_\gamma,p]$. Since $\overline{W_{\gamma,1}} \subset \overline{V_{\gamma,1}} \subset T_{\gamma,1} \subset U_{\gamma,1}$, $(x,d_\gamma) \notin \overline{W_{\gamma,1}} \times [d_\gamma,p]$. Choose an open set $O \subset X$ such that $x \in O$ and $C=O \times \left\{d_\gamma \right\}$ misses $\overline{W_{\gamma,1}} \times [d_\gamma,p]$. Note that $C$ misses $W_{\beta,1} \times [d_\beta,p]$ for all $\beta<\gamma$ since $W_{\beta,1} \subset W_{\gamma,1}$ for all $\beta<\gamma$. It is clear that $C$ misses $W_{\beta,1} \times [d_\beta,p]$ for all $\beta>\gamma$.

We can also choose open $C_1 \subset C$ such that $(x,d_\gamma) \in C_1$ and $C_1$ misses $\overline{L_\gamma}$. It is clear that $C_1$ misses $L_\beta$ for all $\beta \ne \gamma$. Thus there is an open set $C_1$ containing the point $(x,d_\gamma)$ such that $C_1$ contains no point of $M$.

Let $(x,p) \in K$. First we find an open set $Q$ containing $(x,p)$ such that $Q$ misses $G$. From the way the open sets $U_{\alpha,1}$ are defined, it follows that $(x,p) \notin \overline{W_{\alpha,1}} \times [d_\alpha,p]$ for all $\alpha$. Furthermore $W_{\alpha,1} \subset \overline{A_1}$. Thus $Q=(X-\overline{A_1}) \times Y_\kappa$ is the desired open set. On the other hand, there exists $\alpha<\kappa$ such that $x \in W_{\alpha,2}$. Note that $L_\gamma$ are chosen so that $(W_{\gamma,2} \times [d_\gamma,p]) \cap L_\gamma=\varnothing$ for all $\gamma$. Since $W_{\alpha,2} \subset W_{\beta,2}$ for all $\beta \ge \alpha$, $(W_{\alpha,2} \times [d_\alpha,p]) \cap L_\beta=\varnothing$ for all $\beta \ge \alpha$. Thus the open set $W_{\alpha,2} \times [d_\alpha,p]$ contains no points of $L_\gamma$ for any $\gamma$. Then the open set $Q \cap (W_{\alpha,2} \times [d_\alpha,p])$ contains no point of $M$. This means that $(x,p) \notin \overline{M}$. Thus $\overline{M} \cap K=\varnothing$.

In each of the four cases (1, 2a, 2b and 3), there exists an open set $M \subset X \times Y_\kappa$ such that $H \subset M$ and $\overline{M} \cap K=\varnothing$. This completes the proof that $X \times Y_\kappa$ is normal assuming that $X$ has property $\mathcal{B}(\kappa)$.

Now the other direction. Suppose that $X \times Y_\kappa$ is normal. Then it can be shown that $X$ has property $\mathcal{B}(\kappa)$. The proof is similar to the proof for $\omega$-shrinking $\Longrightarrow$ Property $\mathcal{B}(\omega)$ in Theorem 5. $\square$

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Reference

1. Morita K., Nagata J.,Topics in General Topology, Elsevier Science Publishers, B. V., The Netherlands, 1989.
2. Rudin M. E., A Normal Space $X$ for which $X \times I$ is not Normal, Fund. Math., 73, 179-486, 1971. (link)
3. Rudin M. E., Dowker Spaces, Handbook of Set-Theoretic Topology (K. Kunen and J. E. Vaughan, eds), Elsevier Science Publishers B. V., Amsterdam, (1984) 761-780.
4. Yasui Y., On the Characterization of the $\mathcal{B}$-Property by the Normality of Product Spaces, Topology and its Applications, 15, 323-326, 1983. (abstract and paper)
5. Yasui Y., Some Characterization of a $\mathcal{B}$-Property, TSUKUBA J. MATH., 10, No. 2, 243-247, 1986.

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