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).

____________________________________________________________________

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.

____________________________________________________________________

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}

____________________________________________________________________

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

____________________________________________________________________

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

____________________________________________________________________

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.

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

Advertisements

One thought on “Spaces with shrinking properties

  1. Pingback: The product of a normal countably compact space and a metric space is normal | Dan Ma's Topology Blog

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s