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