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 Urysohnâ€™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 $X \times Y$ 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

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

# Countably paracompact spaces

This post is a basic discussion on countably paracompact space. A space is a paracompact space if every open cover has a locally finite open refinement. The definition can be tweaked by saying that only open covers of size not more than a certain cardinal number $\tau$ can have a locally finite open refinement (any space with this property is called a $\tau$-paracompact space). The focus here is that the open covers of interest are countable in size. Specifically, a space is a countably paracompact space if every countable open cover has a locally finite open refinement. Even though the property appears to be weaker than paracompact spaces, the notion of countably paracompactness is important in general topology. This post discusses basic properties of such spaces. All spaces under consideration are Hausdorff.

Basic discussion of paracompact spaces and their Cartesian products are discussed in these two posts (here and here).

A related notion is that of metacompactness. A space is a metacompact space if every open cover has a point-finite open refinement. For a given open cover, any locally finite refinement is a point-finite refinement. Thus paracompactness implies metacompactness. The countable version of metacompactness is also interesting. A space is countably metacompact if every countable open cover has a point-finite open refinement. In fact, for any normal space, the space is countably paracompact if and only of it is countably metacompact (see Corollary 2 below).

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Normal Countably Paracompact Spaces

A good place to begin is to look at countably paracompactness along with normality. In 1951, C. H. Dowker characterized countably paracompactness in the class of normal spaces.

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 $\cap_n B_n=\varnothing$.

Dowker’s Theorem is proved in this previous post. Condition 2 in the above formulation of the Dowker’s theorem is not in the Dowker’s theorem in the previous post. In the proof for $1 \rightarrow 2$ in the previous post is essentially $1 \rightarrow 2 \rightarrow 3$ for Theorem 1 above. As a result, we have the following.

Corollary 2
Let $X$ be a normal space. Then $X$ is countably paracompact if and only of $X$ is countably metacompact.

Theorem 1 indicates that normal countably paracompact spaces are important for the discussion of normality in product spaces. As a result of this theorem, we know that normal countably paracompact spaces are productively normal with compact metric spaces. The Cartesian product of normal spaces with compact spaces can be non-normal (an example is found here). When the normal factor is countably paracompact and the compact factor is upgraded to a metric space, the product is always normal. The connection with normality in products is further demonstrated by the following corollary of Theorem 1.

Corollary 3
Let $X$ be a normal space. Let $Y$ be a non-discrete metric space. If $X \times Y$ is normal, then $X$ is countably paracompact.

Since $Y$ is non-discrete, there is a non-trivial convergent sequence (i.e. the sequence represents infinitely many points). Then the sequence along with the limit point is a compact metric subspace of $Y$. Let’s call this subspace $S$. Then $X \times S$ is a closed subspace of the normal $X \times Y$. As a result, $X \times S$ is normal. By Theorem 1, $X$ is countably paracompact.

C. H. Dowker in 1951 raised the question: is every normal space countably paracompact? Put it in another way, is the product of a normal space and the unit interval always a normal space? As a result of Theorem 1, any normal space that is not countably paracompact is called a Dowker space. The search for a Dowker space took about 20 years. In 1955, M. E. Rudin showed that a Dowker space can be constructed from assuming a Souslin line. In the mid 1960s, the existence of a Souslin line was shown to be independent of the usual axioms of set theorey (ZFC). Thus the existence of a Dowker space was known to be consistent with ZFC. In 1971, Rudin constructed a Dowker space in ZFC. Rudin’s Dowker space has large cardinality and is pathological in many ways. Zoltan Balogh constructed a small Dowker space (cardinality continuum) in 1996. Various Dowker space with nicer properties have also been constructed using extra set theory axioms. The first ZFC Dowker space constructed by Rudin is found in [2]. An in-depth discussion of Dowker spaces is found in [3]. Other references on Dowker spaces is found in [4].

Since Dowker spaces are rare and are difficult to come by, we can employ a “probabilistic” argument. For example, any “concrete” normal space (i.e. normality can be shown without using extra set theory axioms) is likely to be countably paracompact. Thus any space that is normal and not paracompact is likely countably paracompact (if the fact of being normal and not paracompact is established in ZFC). Indeed, any well known ZFC example of normal and not paracompact must be countably paracompact. In the long search for Dowker spaces, researchers must have checked all the well known examples! This probability thinking is not meant to be a proof that a given normal space is countably paracompact. It is just a way to suggest a possible answer. In fact, a good exercise is to pick a normal and non-paracompact space and show that it is countably paracompact.

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Some Examples

The following lists out a few classes of spaces that are always countably paracompact.

• Metric spaces are countably paracompact.
• Paracompact spaces are countably paracompact.
• Compact spaces are countably paracompact.
• Countably compact spaces are countably paracompact.
• Perfectly normal spaces are countably paracompact.
• Normal Moore spaces are countably paracompact.
• Linearly ordered spaces are countably paracompact.
• Shrinking spaces are countably paracompact.

The first four bullet points are clear. Metric spaces are paracompact. It is clear from definition that paracompact spaces, compact and countably compact spaces are countably paracompact. One way to show perfect normal spaces are countably paracompact is to show that they satisfy condition 6 in Theorem 1 (shown here). Any Moore space is perfect (closed sets are $G_\delta$). Thus normal Moore space are perfectly normal and hence countably paracompact. The proof of the countably paracompactness of linearly ordered spaces can be found in [1]. See Theorem 5 and Corollary 6 below for the proof of the last bullet point.

As suggested by the probability thinking in the last section, we now look at examples of countably paracompact spaces among spaces that are “normal and not paracompact”. The first uncountable ordinal $\omega_1$ is normal and not paracompact. But it is countably compact and is thus countably paracompact.

Example 1
Any $\Sigma$-product of uncountably many metric spaces is normal and countably paracompact.

For each $\alpha<\omega_1$, let $X_\alpha$ be a metric space that has at least two points. Assume that each $X_\alpha$ has a point that is labeled 0. Consider the following subspace of the product space $\prod_{\alpha<\omega_1} X_\alpha$.

$\displaystyle \Sigma_{\alpha<\omega_1} X_\alpha =\left\{f \in \prod_{\alpha<\omega_1} X_\alpha: \ f(\alpha) \ne 0 \text{ for at most countably many } \alpha \right\}$

The space $\Sigma_{\alpha<\omega_1} X_\alpha$ is said to be the $\Sigma$-product of the spaces $X_\alpha$. It is well known that the $\Sigma$-product of metric spaces is normal, in fact collectionwise normal (this previous post has a proof that $\Sigma$-product of separable metric spaces is collectionwise normal). On the other hand, any $\Sigma$-product always contains $\omega_1$ as a closed subset as long as there are uncountably many factors and each factor has at least two points (see the lemma in this previous post). Thus any such $\Sigma$-product, including the one being discussed, cannot be paracompact.

Next we show that $T=(\Sigma_{\alpha<\omega_1} X_\alpha) \times [0,1]$ is normal. The space $T$ can be reformulated as a $\Sigma$-product of metric spaces and is thus normal. Note that $T=\Sigma_{\alpha<\omega_1} Y_\alpha$ where $Y_0=[0,1]$, for any $n$ with $1 \le n<\omega$, $Y_n=X_{n-1}$ and for any $\alpha$ with $\alpha>\omega$, $Y_\alpha=X_\alpha$. Thus $T$ is normal since it is the $\Sigma$-product of metric spaces. By Theorem 1, the space $\Sigma_{\alpha<\omega_1} X_\alpha$ is countably paracompact. $\square$

Example 2
Let $\tau$ be any uncountable cardinal number. Let $D_\tau$ be the discrete space of cardinality $\tau$. Let $L_\tau$ be the one-point Lindelofication of $D_\tau$. This means that $L_\tau=D_\tau \cup \left\{\infty \right\}$ where $\infty$ is a point not in $D_\tau$. In the topology for $L_\tau$, points in $D_\tau$ are isolated as before and open neighborhoods at $\infty$ are of the form $L_\tau - C$ where $C$ is any countable subset of $D_\tau$. Now consider $C_p(L_\tau)$, the space of real-valued continuous functions defined on $L_\tau$ endowed with the pointwise convergence topology. The space $C_p(L_\tau)$ is normal and not Lindelof, hence not paracompact (discussed here). The space $C_p(L_\tau)$ is also homeomorphic to a $\Sigma$-product of $\tau$ many copies of the real lines. By the same discussion in Example 1, $C_p(L_\tau)$ is countably paracompact. For the purpose at hand, Example 2 is similar to Example 1. $\square$

Example 3
Consider R. H. Bing’s example G, which is a classic example of a normal and not collectionwise normal space. It is also countably paracompact. This previous post shows that Bing’s Example G is countably metacompact. By Corollary 2, it is countably paracompact. $\square$

Based on the “probabilistic” reasoning discussed at the end of the last section (based on the idea that Dowker spaces are rare), “normal countably paracompact and not paracompact” should be in plentiful supply. The above three examples are a small demonstration of this phenomenon.

Existence of Dowker spaces shows that normality by itself does not imply countably paracompactness. On the other hand, paracompact implies countably paracompact. Is there some intermediate property that always implies countably paracompactness? We point that even though collectionwise normality is intermediate between paracompactness and normality, it is not sufficiently strong to imply countably paracompactness. In fact, the Dowker space constructed by Rudin in 1971 is collectionwise normal.

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More on Countably Paracompactness

Without assuming normality, the following is a characterization of countably paracompact spaces.

Theorem 4
Let $X$ be a topological space. Then the space $X$ is countably paracompact if and only of the following condition holds.

• For any decreasing sequence $\left\{A_n: n=1,2,3,\cdots \right\}$ of closed subsets of $X$ such that $\cap_n A_n=\varnothing$, there exists a decreasing sequence $\left\{B_n: n=1,2,3,\cdots \right\}$ of open subsets of $X$ such that $A_n \subset B_n$ for each $n$ and $\cap_n \overline{B_n}=\varnothing$.

Proof of Theorem 4
Suppose that $X$ is countably paracompact. Suppose that $\left\{A_n: n=1,2,3,\cdots \right\}$ is a decreasing sequence of closed subsets of $X$ as in the condition in the theorem. Then $\mathcal{U}=\left\{X-A_n: n=1,2,3,\cdots \right\}$ is an open cover of $X$. Let $\mathcal{V}$ be a locally finite open refinement of $\mathcal{U}$. For each $n=1,2,3,\cdots$, define the following:

$B_n=\cup \left\{V \in \mathcal{V}: V \cap A_n \ne \varnothing \right\}$

It is clear that $A_n \subset B_n$ for each $n$. The open sets $B_n$ are decreasing, i.e. $B_1 \supset B_2 \supset \cdots$ since the closed sets $A_n$ are decreasing. To show that $\cap_n \overline{B_n}=\varnothing$, let $x \in X$. The goal is to find $B_j$ such that $x \notin \overline{B_j}$. Once $B_j$ is found, we will obtain an open set $V$ such that $x \in V$ and $V$ contains no points of $B_j$.

Since $\mathcal{V}$ is locally finite, there exists an open set $V$ such that $x \in V$ and $V$ meets only finitely many sets in $\mathcal{V}$. Suppose that these finitely many open sets in $\mathcal{V}$ are $V_1,V_2,\cdots,V_m$. Observe that for each $i=1,2,\cdots,m$, there is some $j(i)$ such that $V_i \cap A_{j(i)}=\varnothing$ (i.e. $V_i \subset X-A_{j(i)}$). This follows from the fact that $\mathcal{V}$ is a refinement $\mathcal{U}$. Let $j$ be the maximum of all $j(i)$ where $i=1,2,\cdots,m$. Then $V_i \cap A_{j}=\varnothing$ for all $i=1,2,\cdots,m$. It follows that the open set $V$ contains no points of $B_j$. Thus $x \notin \overline{B_j}$.

For the other direction, suppose that the space $X$ satisfies the condition given in the theorem. Let $\mathcal{U}=\left\{U_n: n=1,2,3,\cdots \right\}$ be an open cover of $X$. For each $n$, define $A_n$ as follows:

$A_n=X-U_1 \cup U_2 \cup \cdots \cup U_n$

Then the closed sets $A_n$ form a decreasing sequence of closed sets with empty intersection. Let $B_n$ be decreasing open sets such that $\bigcap_{i=1}^\infty \overline{B_i}=\varnothing$ and $A_n \subset B_n$ for each $n$. Let $C_n=X-B_n$ for each $n$. Then $C_n \subset \cup_{j=1}^n U_j$. Define $V_1=U_1$. For each $n \ge 2$, define $V_n=U_n-\bigcup_{j=1}^{n-1}C_{j}$. Clearly each $V_n$ is open and $V_n \subset U_n$. It is straightforward to verify that $\mathcal{V}=\left\{V_n: n=1,2,3,\cdots \right\}$ is a cover of $X$.

We claim that $\mathcal{V}$ is locally finite in $X$. Let $x \in X$. Choose the least $n$ such that $x \notin \overline{B_n}$. Choose an open set $O$ such that $x \in O$ and $O \cap \overline{B_n}=\varnothing$. Then $O \cap B_n=\varnothing$ and $O \subset C_n$. This means that $O \cap V_k=\varnothing$ for all $k \ge n+1$. Thus the open cover $\mathcal{V}$ is a locally finite refinement of $\mathcal{U}$. $\square$

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We present another characterization of countably paracompact spaces that involves the notion of shrinkable open covers. An open cover $\mathcal{U}$ of a space $X$ is said to be shrinkable if there exists an open cover $\mathcal{V}=\left\{V(U): U \in \mathcal{U} \right\}$ of the space $X$ such that for each $U \in \mathcal{U}$, $\overline{V(U)} \subset U$. If $\mathcal{U}$ is shrinkable by $\mathcal{V}$, then we also say that $\mathcal{V}$ is a shrinking of $\mathcal{U}$. Note that Theorem 1 involves a shrinking. Condition 3 in Theorem 1 (Dowker’s Theorem) can rephrased as: every countable open cover of $X$ has a shrinking. This for any normal countably paracompact space, every countable open cover has a shrinking (or is shrinkable).

A space $X$ is a shrinking space if every open cover of $X$ is shrinkable. Every shrinking space is a normal space. This follows from this lemma: A space $X$ is normal if and only if every point-finite open cover of $X$ is shrinkable (see here for a proof). With this lemma, it follows that every shrinking space is normal. The converse is not true. To see this we first show that any shrinking space is countably paracompact. Since any Dowker space is a normal space that is not countably paracompact, any Dowker space is an example of a normal space that is not a shrinking space. To show that any shrinking space is countably paracompact, we first prove the following characterization of countably paracompactness.

Theorem 5
Let $X$ be a space. Then $X$ is countably paracompact if and only of every countable increasing open cover of $X$ is shrinkable.

Proof of Theorem 5
Suppose that $X$ is countably paracompact. Let $\mathcal{U}=\left\{U_1,U_2,U_3,\cdots \right\}$ be an increasing open cover of $X$. Then there exists a locally open refinement $\mathcal{V}_0$ of $\mathcal{U}$. For each $n$, define $V_n=\cup \left\{O \in \mathcal{V}_0: O \subset U_n \right\}$. Then $\mathcal{V}=\left\{V_1,V_2,V_3,\cdots \right\}$ is also a locally finite refinement of $\mathcal{U}$. For each $n$, define

$G_n=\cup \left\{O \subset X: O \text{ is open and } \forall \ m > n, O \cap V_m=\varnothing \right\}$

Let $\mathcal{G}=\left\{G_n: n=1,2,3,\cdots \right\}$. It follows that $G_n \subset G_m$ if $n. Then $\mathcal{G}$ is an increasing open cover of $X$. Observe that for each $n$, $\overline{G_n} \cap V_m=\varnothing$ for all $m > n$. Then we have the following:

\displaystyle \begin{aligned} \overline{G_n}&\subset X-\cup \left\{V_m: m > n \right\} \\&\subset \cup \left\{V_k: k=1,2,\cdots,n \right\} \\&\subset \cup \left\{U_k: k=1,2,\cdots,n \right\}=U_n \end{aligned}

We have just established that $\mathcal{G}$ is a shrinking of $\mathcal{U}$, or that $\mathcal{U}$ is shrinkable.

For the other direction, to show that $X$ is countably paracompact, we show that the condition in Theorem 4 is satisfied. Let $\left\{A_1,A_2,A_3,\cdots \right\}$ be a decreasing sequence of closed subsets of $X$ with empty intersection. Then $\mathcal{U}=\left\{U_1,U_2,U_3,\cdots \right\}$ be an open cover of $X$ where $U_n=X-A_n$ for each $n$. By assumption, $\mathcal{U}$ is shrinkable. Let $\mathcal{V}=\left\{V_1,V_2,V_3,\cdots \right\}$ be a shrinking. We can assume that $\mathcal{V}$ is an increasing sequence of open sets.

For each $n$, let $B_n=X-\overline{V_n}$. We claim that $\left\{B_1,B_2,B_3,\cdots \right\}$ is a decreasing sequence of open sets that expand the closed sets $A_n$ and that $\bigcap_{n=1}^\infty \overline{B_n}=\varnothing$. The expansion part follows from the following:

$A_n=X-U_n \subset X-\overline{V_n}=B_n$

The part about decreasing follows from:

$B_{n+1}=X-\overline{V_{n+1}} \subset X-\overline{V_n}=B_n$

We show that $\bigcap_{n=1}^\infty \overline{B_n}=\varnothing$. To this end, let $x \in X$. Then $x \in V_n$ for some $n$. We claim that $x \notin \overline{B_n}$. Suppose $x \in \overline{B_n}$. Since $V_n$ is an open set containing $x$, $V_n$ must contain a point of $B_n$, say $y$. Since $y \in B_n$, $y \notin \overline{V_n}$. This in turns means that $y \notin V_n$, a contradiction. Thus we have $x \notin \overline{B_n}$ as claimed. We have established that every point of $X$ is not in $\overline{B_n}$ for some $n$. Thus the intersection of all the $\overline{B_n}$ must be empty. We have established the condition in Theorem 4 is satisfied. Thus $X$ is countably paracompact. $\square$

Corollary 6
If $X$ is a shrinking space, then $X$ is countably paracompact.

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Reference

1. Ball, B. J., Countable Paracompactness in Linearly Ordered Spaces, Proc. Amer. Math. Soc., 5, 190-192, 1954. (link)
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. Wikipedia Entry on Dowker Spaces (link)

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$\copyright \ 2016 \text{ by Dan Ma}$

# On Spaces That Can Never Be Dowker

A Dowker space is a normal space $X$ for which the product with the closed unit interval $[0,1]$ is not normal. In 1951, Dowker characterized Dowker’s spaces as those spaces that are normal but not countably paracompact ([1]). Soon after, spaces that are normal but not countably paracompactÂ became known as Dowker spaces. In 1971, M. E. Rudin ([2]) constructed a ZFC example of a Dowker’s space. But this Dowker’s space is large. It has cardinality $(\omega_\omega)^\omega$ and is pathological in many ways. Thus the search for “nice” Dowker’s spaces continued. The Dowker’s spaces being sought were those with additional properties such as having various cardinal functions (e.g. density, character and weight) countable. Many “nice” Dowker’s spaces had been constructed using various additional set-theoretic assumptions.Â In 1996, Balogh constructed a first “small” Dowker’s space (cardinaltiy continuum) without additional set-theoretic axioms beyond ZFC ([4]). Rudin’s survey article is an excellent reference for Dowker’s spaces ([3]).

In this note, I make several additional observations on Dowker’s spaces. In this previous post, I presented a proof of the Dowker’s theorem characterizing the normal spaces for which the product with the unit interval is normal (see the statement of the Dowker’s theorem below). In another post, I showed that perfectly normal spaces can never be Dowker’s spaces. Based on the Dowker’s theorem, several other classes of spaces are easily seen as not Dowker.

Dowker’s Theorem. For a normal space $X$, the following conditions are equivalent.

1. The space $X$ is countably paracompact.
2. The product $X \times Y$ is normal for any infinite compact metric space $Y$.
3. The product $X \times [0,1]$ is normal.
4. For each sequence of closed subsets $\lbrace{A_0,A_1,A_2,...}\rbrace$ of $X$Â such that $A_0 \supset A_1 \supset A_2 \supset ...$ and $\bigcap_{n<\omega} A_n=\phi$, there is open sets $U_n \supset A_n$ for each $n$ such that $\bigcap_{n<\omega} U_n=\phi$.

Observations. If $X$ is perfectly compact, then it can be shown that it is countably paracompact by showing that it satisfies condition 4 in the Dowker’s theorem (there is a proof in this blog). Thus thereÂ are no perfectly normal Dowker’s spaces. There are no countably compact Dowker’s spaces since any countably compact space is countably paracompact. This can also be seen using condition 4 above. In a countably compact space, any decreasing nested sequence of closed sets has non-empty intersection and thus condition 4 is satisfied vacuously. Furthermore, all metric spaces, compact spaces, regular Lindelof spaces cannot be Dowker since these spaces are paracomapct.

Normal Moore spaces are perfectly normal. ThusÂ there are no Dowker’s spaces that are Moore spaces. Note thatÂ a space is perfectly normal if it is normal and if every closed set is $G_\delta$. We show that inÂ a Moore space, every closed set is $G_\delta$.Â Let $\lbrace{\mathcal{O}_n:n \in \omega}\rbrace$ be a development for the regular space $X$. Let $A$ be a closed set in $X$. We show that $A$ is a $G_\delta-$ set in $X$. For each $n$, let $U_n=\lbrace{O \in \mathcal{O}_n:O \bigcap A \neq \phi}\rbrace$. Obviously, $A \subset \bigcap_n U_n$. Let $x \in \bigcap_n U_n$. If $x \notin A$, there is some $n$ such that for each $O \in \mathcal{O}_n$ with $x \in O$, we have $O \subset X-A$. Since $x \in \bigcap_n U_n$, $x \in O$ for someÂ $O \in \mathcal{O}_n$ and $O \cap A \neq \phi$, a contradiction. Thus we have $A=\bigcap_n U_n$.

There are other classes of spaces that can never be Dowker. We point these out without proof. For example, there are no linearly ordered Dowker’s spaces and there are no monotonically normal Dowker’s spaces (see Rudin’s survey article [3]).

Reference

1. Dowker, C. H., On Countably Paracompact Spaces, Canad. J. Math. 3, (1951) 219-224.
2. Rudin, M. E., A normal space $X$ for which $X \times I$ is not normal, Fund. Math., 73 (1971), 179-186.
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. Balogh, Z., A small Dowker space in ZFC, Proc. Amer. Math. Soc., 124 (1996), 2555-2560.

# Perfectly Normal Spaces Can Never Be Dowker Spaces

The Dowker’s theorem states that for a normal space $X$, $X \times [0,1]$ is normal if and only if $X$ is countably paracompact. Since this theorem was published, any normal space that is not countably paracompact became known as Dowker space. There are classes of spaces that can never be Dowker spaces (e.g. metrizable spaces, paracompact spaces, compact spaces and Lindelof spaces). In [Katetov], it was shown that there are no perfectly normal Dowker spaces. My blog has a proof of the Dowker’s theorem (see the proof here). For more background on Dowker’s spaces, see the survey article [Rudin]. Dowker’s theorem was published in [Dowker].

Theorem. If $X$ is perfectly normal, then $X$ is countably paracompact.

To prove this theorem, we use the following characterization of countably paracompactness (you can find a proof here).

Lemma. Let $X$ be a normal space. Then $X$ is countably paracompact if and only if for each sequence $\lbrace{A_n:n \in \omega}\rbrace$ of closed subsets of $X$ such that $A_0 \supset A_1 \supset ...$ and $\bigcap_n A_n=\phi$, there exist open sets $B_n \supset A_n$ such that $\bigcap_n B_n=\phi$.

Proof of Theorem. Suppose $X$ is perfectly normal. Let $A_0 \supset A_1 \supset ...$ be a sequence of closed sets such that $\bigcap_n A_n=\phi$. For each $n$, let $A_n=\bigcap_{i<\omega} U_{n,i}$ where each $U_{n,i}$ is open in $X$. For each $n$, define $B_n=\bigcap_{i, j \leq n}U_{i,j}$. Clearly, $B_n \supset A_n$. It is easy to see that $\bigcap_n B_n=\phi$. Note that all the open sets $U_{n,j}$ are used in defining the sequence $B_0,B_1,B_2,\cdots$. Thus $\bigcap_n B_n \neq \phi$ would imply $\bigcap_n A_n \neq \phi$.

Comment. As a consequence of this theorem and the Dowker’s theorem, if $X$ is perfectly normal, then $X \times Y$ is normal for any compact metric space $Y$.

Reference
[Dowker]
Dowker, C. H. [1951], On Countably Paracompact Spaces, Canad. J. Math. 3, 219-224.

[Katetov]
Katetov, M., On real-valued functions in topological spaces, Fund. Math. 38 (1951), pp. 85-91.

[Rudin]
Rudin, M. E., [1984], Dowker Spaces, Handbook of Set-Theoretic Topology (K. Kunen and J. E. Vaughan, eds), Elsevier Science Publishers B. V., Amsterdam, 761-780.

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$\copyright \ 2009-2016 \text{ by Dan Ma}$
Revised November 23, 2016

# A Theorem About Hereditarily Normality

Let $Y$ be either the closed unit interval $\mathbb{I}=[0,1]$ or $[0,\omega]=\omega+1$. Let $X$ be any one of the following spaces:

$X=[0,\omega_1]=\omega_1+1$,
$X=[0,\omega_1)=\omega_1$,
$X=$ Michael Line,
$X=$ Sorgenfrey Line.

The cross product $X \times Y$ is normal in all four cases. The $X$ factor, in the order listed, is compact, countably compact, paracompact and Lindelof. Thus they are all countably paracompact. According to the Dowker’s theorem, the product of a normal space $X$ and a compact metric space $Y$ is normal if and only if $X$ is countably paracompact. My goal here is to show that the four cases of $X \times Y$ here cannot be hereditarily normal. This is from a theorem due to Katetov. We prove the following theorem.

Theorem. If $X \times Y$ is hereditarily normal, then either $X$ is perfectly normal or every countably infinite subspace of $Y$ is closed and discrete.

Proof. Suppose that $X$ is not perfectly normal and $Y$ has a countably infinite subset that is not closed and discrete. Let $H \subset X$ be a closed set that is not a $G_\delta-$set. Let $C=\lbrace{y_n:n \in \omega}\rbrace \subset Y$ be an infinite set with an accumulation point $y$. We assume that $y \notin C$.

We show that the open subspace $U=(X \times Y)-(H \times \lbrace{y}\rbrace)$ is not normal. To this end, let $A=H \times (Y-\lbrace{y}\rbrace)$ and $B=(X-H) \times \lbrace{y}\rbrace$. The sets $A$ and $B$ are disjoint closed subspaces of the open subspace $U$. Suppose we have disjoint open sets $S$ and $T$ such that $A \subset S$ and $B \subset T$.

For each $n \in \omega$, let $O_n=\lbrace{x \in X:(x,y_n) \in S}\rbrace$. Each $O_n$ is open and $H \subset O_n$. Thus $H \subset \bigcap_n O_n$. Let $x \in \bigcap_n O_n$. Then $(x,y) \in \overline{S}$. This means $x \in H$. If $x \notin H$, then $(x,y) \in B \subset T$ (which is impossible). So we have $H=\bigcap_n O_n$, indicating that $H$ is a $G_\delta-$set, and leading to a contradiction. So the subspace $U=(X \times Y)-(H \times \lbrace{y}\rbrace)$ is not normal.

Corollary. For countably compact spaces (in particular compact spaces) $X$ and $Y$, if the Cartesian product $X \times Y$ is hereditarily normal, then both $X$ and $Y$ are prefectly normal.

Proof. Note that both factors, being countably compact, cannot have closed and discrete infinite subsets.

Comment. Note that the converse of the corollary is not true. Let both factors be the double arrow space, which is perfectly normal. But the square of the double arrow space contains a copy of the Sorgenfrey Plane, which is not normal.

Reference
[Katetov] Katetov, M., [1948] Complete normality of Cartesian products, Fund. Math., 36, 271-274.

# Dowker’s Theorem

Let $\mathbb{I}$ be the closed unit interval $[0,1]$. In 1951, Dowker proved that for normal spaces $X$, $X \times \mathbb{I}$ is normal if and only if $X$ is countably paracompact. A space $X$ is countably paracompact if every countable open cover of $X$ has a locally finite open refinement. Dowker’s theorem is a fundamental result on products of normal spaces. With this theorem, a question was raised about the existence of a normal but not countably paracompact space (such a space became known as a Dowker space). For a detailed discussion on Dowker spaces, see the survey paper [Rudin]. The focus here is on presenting a proof for the Dowker’s theorem, laying the groundwork for future discussion.

An open cover $\mathcal{V}=\lbrace{V_\alpha:\alpha \in S}\rbrace$ of a space $X$ is shrinkable if there exists an open cover$\mathcal{W}=\lbrace{W_\alpha:\alpha \in S}\rbrace$ such that $\overline{W_\alpha} \subset V_\alpha$ for each $\alpha \in S$. The open cover $\mathcal{W}$ is called a shrinking of $\mathcal{V}$. We use the following lemma in proving Dowker’s theorem. Go here to see a proof of this lemma.

Lemma
A space $X$ is normal if and only if every point-finite open cover of $X$ is shrinkable.

Dowker’s Theorem
For a normal space $X$, the following conditions are equivalent:

1. $X$ is a countably paracompact space.
2. If $\lbrace{U_n:n<\omega}\rbrace$ is an open cover of $X$, then there is a locally finite open refinement $\lbrace{V_n:n<\omega}\rbrace$ such that $\overline{V_n} \subset U_n$ for each $n$.
3. $X \times Y$ is normal for any compact metrizable space $Y$.
4. $X \times \mathbb{I}$ is normal.
5. For each sequence of closed sets such that $A_0 \supset A_1 \supset ...$ and $\cap_n A_n=\phi$, there exists open sets $B_n \supset A_n$ such that $\cap_n B_n=\phi$.

Proof
$1 \Longrightarrow 2$
Suppose $X$ is countably paracompact. Let $\mathcal{U}=\lbrace{U_n:n<\omega}\rbrace$ be an open cover of $X$. Then $\mathcal{U}$ has a locally open refinement $\mathcal{V}$. For each $n<\omega$, let $W_n=\bigcup \lbrace{V \in \mathcal{V}:V \subset U_n}\rbrace$. Note that $\lbrace{W_n:n<\omega}\rbrace$ is still locally finite (thus point-finite). By the lemma, it has a shrinking $\lbrace{V_n:n<\omega}\rbrace$.

$2 \Longrightarrow 3$
Let $Y$ be a compact metrizable space. Let $\lbrace{E_0,E_1,E_2,...}\rbrace$ be a base of $Y$ such that it is closed under finite unions. Let $H,K$ be two disjoint closed subsets of $X \times Y$. The goal here is to find an open set $\mathcal{V}$ such that $H \subset \mathcal{V}$ and $\overline{\mathcal{V}}$ misses $K$.

For each $x \in X$, consider the following:
$S_x=\lbrace{y \in Y: (x,y) \in H}\rbrace$
$T_x=\lbrace{y \in Y: (x,y) \in K}\rbrace$.

For each $n \in \omega$, let $O_n$ be defined by:
$O_n=\lbrace{x \in X:S_x \subset E_n}$ and $T_x \subset Y-\overline{E_n}\rbrace$.

Claim 1.
Each $x \in X$ is an element of some $O_n$.

Both $S_x$ and $T_x$ are compact. Thus we can find some $E_n$ such that $S_x \subset E_n$ and $T_x \subset Y-\overline{E_n}$.

Claim 2.
Each $O_n$ is open in $X$.

Fix $z \in O_n$. For each $y \in Y-E_n$, $(z,y) \notin H$. Choose open set $A_y \times B_y$ such that $(z,y) \in A_y \times B_y$ and $A_y \times B_y$ misses $H$. The set of all $B_y$ is a cover of $Y-E_n$. We can find a finite subcover covering $Y-E_n$, say $B_{y(0)},B_{y(1)},...,B_{y(k)}$. Let $A=\bigcap_i A_{y(i)}$.

For each $a \in \overline{E_n}$, $(z,a) \notin K$. Choose open set $C_a \times D_a$ such that $(z,a) \in C_a \times D_a$ and $C_a \times D_a$ misses $K$. The set of all $D_a$ is a cover of $\overline{E_n}$. We can find a finite subcover covering $\overline{E_n}$, say $D_{a(0)},D_{a(1)},...,D_{a(j)}$. Let $C=\bigcap_i C_{a(i)}$.

Now, let $O=A \cap C$. Clearly $z \in O$. To show that $O \subset O_n$, pick $x \in O$. We want to show $S_x \subset E_n$ and $T_x \subset Y-\overline{E_n}$. Suppose we have $y \in S_x$ and $y \in Y-E_n$. Then $y \in B_{y(i)}$ for some $i$. As a result, $(x,y) \in A_{y(i)} \times B_{y(i)}$. This would mean that $(x,y) \notin H$. But this contradicts with $y \in S_x$. So we have $S_x \subset E_n$. On the other hand, suppose we have $y \in T_x$ and $y \in \overline{E_n}$. Then $y \in D_{a(i)}$ for some $i$. As a result, $(x,y) \in C_{a(i)} \times D_{a(i)}$. This means that $(x,y) \notin K$. This contradicts with $y \in T_x$. So we have $T_x \subset Y-\overline{E_n}$. It follows that $O \subset O_n$ and $O_n$ is open in $X$.

Claim 3.
We can find an open set $\mathcal{V}$ such that $H \subset \mathcal{V}$ and $\overline{\mathcal{V}}$ misses $K$.

Let $\mathcal{O}=\bigcup_n O_n \times \overline{E_n}$. Note that $H \subset \mathcal{O}$ and $\mathcal{O}$ misses $K$. The open set $\mathcal{V}$ being constructed will satisfiy $H \subset \mathcal{V} \subset \overline{\mathcal{V}} \subset \mathcal{O}$. The open cover $\lbrace{O_n}\rbrace$ has a locally finite open refinement $\lbrace{V_n}\rbrace$ such that $\overline{V_n} \subset O_n$ for each $n$. Now define $\mathcal{V}=\bigcup_n V_n \times E_n$. Note that $H \subset \mathcal{V}$.

We also have $\overline{\mathcal{V}}=\overline{\bigcup_n V_n \times E_n}=\bigcup_n \overline{V_n \times E_n}$. We have a closure preserving situation because the open sets $V_n$ are from a locally finite collection. Continue the derivation and we have:

$=\bigcup_n \overline{V_n} \times \overline{E_n} \subset \bigcup_n O_n \times \overline{E_n}=\mathcal{O}$.

$3 \Longrightarrow 4$ is obvious.

$4 \Longrightarrow 5$
Let $A_0 \supset A_1 \supset ...$ be closed sets with empty intersection. We show that $A_n$ can be expanded by open sets $B_n$ such that $A_n \subset B_n$ and $\bigcap_n B_n=\phi$.

Choose $p \in \mathbb{I}$ and a sequence of distinct points $p_n \in \mathbb{I}$ converging to $p$. Let $H=\cup \lbrace{A_n \times \lbrace{p_n}\rbrace:n \in \omega}\rbrace$ and $K=X \times \lbrace{p}\rbrace$. These are disjoint closed sets. Since $X \times \mathbb{I}$ is normal, we can find open set $V \subset X \times \mathbb{I}$ such that $H \subset V$ and $\overline{V}$ misses $K$.

Let $B_n=\lbrace{x \in X:(x,p_n) \in V}\rbrace$. Note that each $B_n$ is open in $X$. Also, $A_n \subset B_n$. We want to show that $\bigcap_n B_n=\phi$. Let $x \in X$. The point $(x,p) \in K$ and $(x,p) \notin \overline{V}$. There exist open $E \subset X$ and open $F \subset Y$ such that $(x,p) \in E \times F$ and $(E \times F) \cap \overline{V}=\phi$. Then $p_n \in F$ for some $n$. Note that $(x,p_n) \notin V$. Thus $x \notin B_n$. It follows that $\bigcap_n B_n=\phi$.

$5 \Longrightarrow 1$
Let $\lbrace{T_n:n \in \omega}\rbrace$ be an open cover of $X$. Let $E_n=\bigcup_{i \leq n} T_i$. Let $A_n=X-E_n$. Each $A_n$ is closed and $\bigcap_n A_n=\phi$. So there exist open sets $B_n \supset A_n$ such that $\bigcap_n B_n=\phi$.

Since $X$ is normal, there exists open $W_n$ such that $A_n \subset W_n \subset \overline{W_n} \subset B_n$. It follows that $\bigcap_n \overline{W_n}=\phi$. Let $O_n=X-\bigcap_{i \leq n} \overline{W_n}$.

Claim. $\lbrace{O_n:n \in \omega}\rbrace$ is an open cover of $X$. Furthermore, $\overline{O_n} \subset \bigcup_{i \leq n}T_i$.
Let $x \in X$. Since $\bigcap_n B_n=\phi$, $x \notin B_n$ for some $n$. Then $x \notin \overline{W_n}$ and $x \in O_n$. To show the second half of the claim, let $y \notin \bigcup_{i \leq n}T_i$. Then $y \notin E_i$ for each $i \leq n$. This means $y \in A_i$ and $y \in W_i$ for each $i \leq n$. Then $\bigcap_{i \leq n}W_i$ is an open set containing $y$ that contains no point of $O_n$. Thus we have $\overline{O_n} \subset \bigcup_{i \leq n}T_i$.

Let $S_n=T_n-\bigcup_{i < n} \overline{O_i}$. We show that $\lbrace{S_n:n \in \omega}\rbrace$ is a locally finite open refinement of $\lbrace{T_n:n \in \omega}\rbrace$. Clearly $S_n \subset T_n$. To show that the open sets $S_n$ form a cover, let $x \in X$. Choose least $n$ such that $x \in T_n$. Then for each $i, $x \notin E_i$. This means $x \in A_i$ and $x \in \overline{W_i}$ for each $i. It follows that $x \notin O_i$ for each $i. This implies $x \in S_n$. To see that the open sets $S_n$ form a locally finite collection, note that each $x \in X$ belong to some $O_n$. The open set $O_n$ misses $S_m$ for all $m>n$. Thus $O_n$ can only meet $S_i$ for $i \leq n$. We just prove that $X$ is countably paracompact.

Note. In $4 \Longrightarrow 5$, all we need is that the factor $Y$ has a non-trivial convergent sequence. That is, if $X \times Y$ is normal and $Y$ has a non-trivial convergent sequence, then $X$ satisfies condition $5$.

Reference
[Dowker] Dowker, C. H. [1951], On Countably Paracompact Spaces, Canad. J. Math. 3, 219-224.

[Rudin] Rudin, M. E., [1984], Dowker Spaces, Handbook of Set-Theoretic Topology (K. Kunen and J. E. Vaughan, eds), Elsevier Science Publishers B. V., Amsterdam, 761-780.

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$\copyright \ 2009-2016 \text{ by Dan Ma}$
Revised November 27, 2016