# kappa-Dowker space and the first conjecture of Morita

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

All spaces under consideration are Hausdorff.

____________________________________________________________________

Morita’s First Conjecture

In 1976, K. Morita posed the following conjecture.

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

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

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

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

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

____________________________________________________________________

Dowker’s Theorem

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

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

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

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

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

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

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

____________________________________________________________________

Shrinking Properties

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

Let $X$ be a space. Let $\mathcal{U}$ be an open cover of the space $X$. The open cover $\mathcal{U}$ is said to be shrinkable if there is an open cover $\mathcal{V}=\left\{V(U): U \in \mathcal{U} \right\}$ such that $\overline{V(U)} \subset U$ for each $U \in \mathcal{U}$. When this is the case, the open cover $\mathcal{V}$ is said to be a shrinking of $\mathcal{U}$. If an open cover is shrinkable, we also say that the open cover can be shrunk (or has a shrinking).

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

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

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

____________________________________________________________________

An Attempt for a Generalized Dowker’s Theorem

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

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

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

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

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

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

The following diagram shows how these conditions are related.

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

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

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

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

$3 \Longleftrightarrow 4$ (Theorem 4)

$5 \Longleftrightarrow 6$ (Theorem 7)

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

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

____________________________________________________________________

Proof of Theorem 2

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

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

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

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

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

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

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

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

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

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

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

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

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

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

____________________________________________________________________

$\kappa$-Dowker Space

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

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

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

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

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

____________________________________________________________________

Remarks

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

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

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

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

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

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

____________________________________________________________________

Loose Ends

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

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

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

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

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

____________________________________________________________________

Reference

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

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

# Spaces with shrinking properties

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

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

____________________________________________________________________

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.

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

# 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).

____________________________________________________________________

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.

____________________________________________________________________

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.

____________________________________________________________________

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 $\mathcal{U}$. $\square$

___________________________________

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.

____________________________________________________________________

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)

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

# A stroll in Bing’s Example G

In this post we take a leisurely walk in Bing’s Example G, which is a classic example of a normal but not collectionwise normal space. Hopefully anyone who is new to this topological space can come away with an intuitive feel and further learn about it. Indeed this is a famous space that had been extensively studied. This example has been written about in several posts in this topology blog. In this post, we explain how Example G is defined, focusing on intuitive idea as much as possible. Of course, the intuitive idea is solely the perspective of the author. Any reader who is interested in building his/her own intuition on this example can skip this post and go straight to the previous introduction. Other blog posts on various subspaces of Example G are here, here and here. Bing’s Example H is discussed here.

At the end of the post, we will demonstrate that the product of Bing’s Example G with the closed unit interval, $F \times [0,1]$, is a normal space.

____________________________________________________________________

The Product Space Angle

The topology in Example G is tweaked from the product space topology. It is thus a good idea to first examine the relevant product space. Let $P$ be any uncountable set. Let $Q$ be the set of all subsets of $P$. In other words, $Q$ is the power set of $P$. Consider the product of $\lvert Q \lvert$ many copies of the two element set $\left\{0,1 \right\}$. The usual notation of this product space is $2^Q$. The elements of $2^Q$ are simply the functions from $Q$ into $\left\{0,1 \right\}$. An arbitrary element of $2^Q$ is a function $f$ that maps every subset of $P$ to either 0 or 1.

Though the base set $P$ can be any uncountable set, it is a good idea to visualize clearly what $P$ is. In the remainder of this section, think of $P$ as the real line $\mathbb{R}$. Then $Q$ is simply the collection of all subsets of the real line. The elements of the product space are simply functions that map each set of real numbers to either 0 or 1. Or think of each function as a 2-color labeling of the subsets of the real line, where each subset is either red or green for example. There are $2^c$ many subsets of the real line where $c$ is the cardinality of the continuum.

To further visualize the product space, let’s look at a particular subspace of $2^Q$. For each real number $p$, define the function $f_p$ such that $f_p$ always maps any set of real numbers that contains $p$ to 1 and maps any set of real numbers that does not contain $p$ to 0. For example, the following are several values of the function $f_0$.

$f_0([0,1])=1$

$f_0([1,2])=0$

$f_0(\left\{0 \right\})=1$

$f_0(\mathbb{R}-\left\{0 \right\})=0$

$f_0(\mathbb{R})=1$

$f_0(\varnothing)=0$

$f_0(\mathbb{P})=0$

where $\mathbb{P}$ is the set of all irrational numbers. Consider the subspace $F_P=\left\{f_p: p \in P \right\}$. Members of $F_P$ are easy to describe. Each function in $F_P$ maps a subset of the real line to 0 or 1 depending on whether the subscript belongs to the given subset. Another reason that $F_P$ is important is that Bing’s Example is defined by declaring all points not in $F_P$ isolated points and by allowing all points in $F_P$ retaining the open sets in the product topology.

Any point $f$ in $F_P$ determines $f(q)=0 \text{ or } 1$ based on membership (whether the reference point belongs to the set $q$). Points not in $F_P$ have no easy characterization. It seems that any set can be mapped to 0 or 1. Note that any $f$ in $F_P$ maps equally to 0 or 1. So the constant functions $f(q)=0$ and $f(q)=1$ are not in $F_P$. Furthermore, any $f$ such that $f(q)=1$ for at most countably many $q$ would not be in $F_P$.

Let’s continue focusing on the product space for the time being. When $F_P$ is considered as a subspace of the product space $2^Q$, $F_P$ is a discrete space. For each $p \in P$, there is an open set $W_p$ containing $f_p$ such that $W_p$ contains no other points of $F_P$. So $F_P$ is relatively discrete in the product space $2^Q$. Of course $F_P$ cannot be closed in $2^Q$ since $2^Q$ is a compact space. The open set $W_p$ is defined as follows:

$W_p=\left\{f \in 2^Q: f(\left\{p \right\})=1 \text{ and } f(P-\left\{p \right\})=0 \right\}$

It is clear that $f_p \in W_p$ and that $f_t \notin W_p$ for any real number $t \ne p$.

Two properties of the product space $2^Q$ would be very relevant for the discussion. By the well known Tychonoff theorem, the product space $2^Q$ is compact. Since $P$ is uncountable, $2^Q$ always has the countable chain condition (CCC) since it is the product of separable spaces. A space having CCC means that there can only be at most countably many pairwise disjoint open sets. As a result, the uncountably many open sets $W_p$ cannot be all pairwise disjoint. So there exist at least a pair of $W_p$, say $W_{a}$ and $W_{b}$, with nonempty intersection.

The last observation can be generalized. For each $p \in P$, let $V_p$ be any open set containing $f_p$ (open in the product topology). We observe that there are at least two $a$ and $b$ from $P$ such that $V_a \cap V_b \ne \varnothing$. If there are only countably many distinct sets $V_p$, then there are uncountably many $V_p$ that are identical and the observation is valid. So assume that there are uncountably many distinct $V_p$. By the CCC in the product space, there are at least two $a$ and $b$ with $V_a \cap V_b \ne \varnothing$. This observation shows that the discrete points in $F_P$ cannot be separated by disjoint open sets. This means that Bing’s Example G is not collectionwise Hausdorff and hence not collectionwise normal.

Another observation is that any disjoint $A_1, A_2 \subset F_P$ can be separated by disjoint open sets. To see this, define the following two open sets $E_1$ and $E_2$ in the product topology.

$q_1=\left\{p \in P: f_p \in A_1 \right\}$

$q_2=\left\{p \in P: f_p \in A_2 \right\}$

$E_1=\left\{f \in 2^Q: f(q_1)=1 \text{ and } f(q_2)=0 \right\}$

$E_2=\left\{f \in 2^Q: f(q_1)=0 \text{ and } f(q_2)=1 \right\}$

It is clear that $A_1 \subset E_1$ and $A_2 \subset E_2$. Furthermore, $E_1 \cap E_2=\varnothing$. This observation will be the basis for showing that Bing’s Example G is normal.

____________________________________________________________________

The Topology of Bing’s Example G

The topology for Bing’s Example G is obtained by tweaking the product topology on $2^Q$. Let $P$ be any uncountable set. Let $Q$ be the set of all subsets of $P$. The set $F_P$ is defined as above. Bing’s Example G is $F=2^Q$ with points in $F_P$ retaining the open sets in the product topology and with points not in $F_P$ declared isolated. For some reason, in Bing’s original paper, the notation $F$ is used even though the example is identified by G. We will follow Bing’s notation.

The subspace $F_P$ is discrete but not closed in the product topology. However, $F_P$ is both discrete and closed in Bing’s Example G. Based on the discussion in the previous section, one immediate conclusion we can made is that the space $F$ is not collectionwise Hausdorff. This follows from the fact that points in the uncountable closed and discrete set $F_P$ cannot be separated by disjoint open sets. By declaring points not in $F_P$ isolated, the countable chain condition in the original product space $2^Q$ is destroyed. However, there is still a strong trace of CCC around the points in the set $F_P$, which is sufficient to prevent collectionwise Hausdorffness, and consequently collectionwise normality.

To show that $F$ is normal, let $H$ and $K$ be disjoint closed subsets of $F$. To make it easy to follow, let $H=A_1 \cup B_1$ and $K=A_2 \cup B_2$ where

$A_1=H \cap F_P \ \ \ \ B_1=H \cap (F-F_P)$

$A_2=K \cap F_P \ \ \ \ B_2=K \cap (F-F_P)$

In other words, $A$ is the non-isolated part and $B$ is the isolated part of the respective closed set. Based on the observation made in the previous section, obtain the disjoint open sets $E_1$ and $E_2$ where $A_1 \subset E_1$ and $A_2 \subset E_2$. Set the following open sets.

$O_1=(E_1 \cup B_1) - K$

$O_2=(E_2 \cup B_2) - H$

It follows that $O_1$ and $O_2$ are disjoint open sets and that $A_1 \subset O_1$ and $A_2 \subset O_2$. Thus Bing’s Example G is a normal space.

____________________________________________________________________

Bing’s Example G is Countably Paracompact

We discuss one more property of Bing’s Example G. A space $X$ is countably paracompact if every countable countable open cover of $X$ has a locally finite open refinement. In other words, such a space satisfies the property of being a paracompact space but just for countable open covers. A space is countably metacompact if every countable open cover has a point-finite open refinement (i.e. replacing locally finite in the paracompact definition with point-finite). It is well known that in the class of normal spaces, the two notions are equivalent (see Corollary 2 here). Since Bing’s Example G is normal, we only need to show that it is countably metacompact. Note that Bing’s Example G is not metacompact (see here).

Let $\mathcal{U}$ be a countable open cover of $F$. Let $\mathcal{U}^*=\left\{U_1,U_2,U_3,\cdots \right\}$ be the set of all open sets in $\mathcal{U}$ that contain points in $F_P$. For each $i$, let $A_i=U_i \cap F_P$. From the perspective of Bing’s Example G, the sets $A_i$ are discrete closed sets. In any normal space, countably many discrete closed sets can be separated by disjoint open sets (see Lemma 1 here). Let $O_1,O_2,O_3,\cdots$ be disjoint open sets such that $A_i \subset O_i$ for each $i$.

We now build a point-finite open refinement of $\mathcal{U}$. For each $i$, let $V_i=U_i \cap O_i$. Let $V=\cup_{i=1}^\infty V_i$. Consider the following.

$\mathcal{V}=\left\{V_i: i=1,2,3,\cdots \right\} \cup \left\{\left\{ x \right\}: x \in F-V \right\}$

It follows that $\mathcal{V}$ is an open cover of $F$. All points of $F_P$ belong to the open sets $V_i$. Any point that is not in one of the $V_i$ belongs to a singleton open set. It is also clear that $\mathcal{V}$ is a refinement of $\mathcal{U}$. For each $i$, $V_i \subset U_i$ and each singleton set is contained in some member of $\mathcal{U}$. It follows that each point in $F$ belongs to at most finitely many sets in $\mathcal{V}$. In fact, each point belongs to exactly one set in $\mathcal{V}$. Each point in $F_P$ belongs to exactly one $V_i$ since the open sets $O_i$ are disjoint. Any point in $V$ belongs to exactly one singleton open set. What we just show is slightly stronger than countably metacompact. The technical term would be countably 1-bounded metacompact.

Since among normal spaces, countably paracompactness is equivalent to countably metacompact, we can now say that Bing’s Example G is a topological space that is normal and countably paracompact. By Dowker’s Theorem, we can conclude that the product of Bing’s Example G with the closed unit interval, $F \times [0,1]$, is a normal space.

____________________________________________________________________

Previous Posts

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

# A strategy for finding CCC and non-separable spaces

In this post we present a general strategy for finding CCC spaces that are not separable. As illustration, we give four implementations of this strategy.

In searching for counterexamples in topology, one good place to look is of course the book by Steen and Seebach [2]. There are four examples of spaces that are CCC but not separable found in [2] – counterexamples 20, 21, 24 and 63. Counterexamples 20 and 21 are not Hausdorff. Counterexample 24 is the uncountable Fort space (it is completely normal but not perfectly normal). Counterexample 63 (Countable Complement Extension Topology) is Hausdorff but is not regular. These are valuable examples especially the last two (24 and 63). The examples discussed below expand the offerings in Steen and Seebach.

The discussion of CCC but not separable in this post does not use axioms beyond the usual axioms of set theory (i.e. ZFC). The discussion here does not touch on Suslin lines or other examples that require extra set theory. The existence of Suslin lines is independent of ZFC. A Suslin line would produce an example of a perfectly normal first countable CCC non-separable space. In models of set theory where Suslin lines do not exist, a perfectly normal first countable CCC non-separable space can also be produced using other set-theoretic assumptions. The examples discussed below are not as nice as the set-theoretic examples since they usually are not first countable and perfectly normal.

____________________________________________________________________

The countable chain conditon

A topological space $X$ is said to have the countable chain condition (to have the CCC for short) if $\mathcal{U}$ is a disjoint collection of non-empty open subsets of $X$ (meaning that for any $A,B \in \mathcal{U}$ with $A \ne B$, we have $A \cap B=\varnothing$), then $\mathcal{U}$ is countable. In other words, in a space with the CCC, there cannot be uncountably many pairwise disjoint non-empty open sets. For ease of discussion, if $X$ has the CCC, we also say that $X$ is a CCC space or X is CCC. A space $X$ is separable if there exists a countable subset $A$ of $X$ such that $A$ is dense in $X$ (meaning that if $U$ is a nonempty open subset of $X$, then $U \cap A \ne \varnothing$).

It is clear that any separable space has the CCC. In metric spaces, these two properties are equivalent. Among topological spaces in general, the two properties are not identical. Thus “CCC but not separable” is one way to distinguish between metrizable spaces and non-metrizable spaces. Even in non-metrizable spaces, “CCC but not separable” is also a way to obtain more information about the spaces being investigated.

____________________________________________________________________

The strategy

Here’s the strategy for finding CCC and not separable.

The strategy is to narrow the focus to spaces where “CCC and not separable” is likely to exist. Specifically, look for a space or a class of spaces such that each space in the class has the countable chain condition but is not hereditarily separable. If the non-separable subspace is also a dense subspace of the starting space, it would be “CCC and not separable.”

Any dense subspace of a CCC space always has the CCC. Thus the search focuses on the subspaces in a CCC space that are reliably CCC. The strategy is to find non-separable spaces among these dense subspaces. The search is given an assist if the space or class of spaces in question has a characteristic that delineate the “separable” from the CCC (see Example 3 and Example 4 below).

In the following sections, we illustrate four different ways to apply the strategy.

____________________________________________________________________

Example 1

The first way is a standard example found in the literature. The space to start from is the product space of separable spaces, which is always CCC. By a theorem of Ross and Stone, the product of more than continuum many separable spaces is not separable. Thus one way to get an example of CCC but not separable space is to take the product of more than continuum many separable spaces. For example, if $c$ is the cardinality of continuum, then consider $\left\{0,1 \right\}^{2^c}$, the product of $2^c$ many copies of $\left\{0,1 \right\}$, or consider $X^{2^c}$ where $X$ is your favorite separable space.

____________________________________________________________________

Example 2

The second implementation of the strategy is also from taking the product of separable spaces. This time the number of factors does not have to be more than continuum. Here, we focus on one particular dense subspace of the product space, the $\Sigma$-products. To make this clear, let’s focus on a specific example. Consider $X=\left\{0,1 \right\}^{c}$ where $c$ is the cardinality of continuum. Consider the following subspace.

$\Sigma(\left\{0,1 \right\}^{c})= \left\{x \in X: x(\alpha) \ne 0 \text{ for at most countably many } \alpha < c \right\}$

The subspace $\Sigma(\left\{0,1 \right\}^{c})$ is dense in $X$, thus has CCC. It is straightforward to verify that $\Sigma(\left\{0,1 \right\}^{c})$ is not separable.

To implement this example, find any space $X$ which is a product space of separable spaces, each of which has at least two point (one of the points is labeled 0). The dense subspace is the $\Sigma$-product, which is the subspace consisting of all points that are non-zero at only countably many coordinates. The $\Sigma$-product has the countable chain condition since it is a dense subspace of the CCC space $X$. The $\Sigma$-product is not separable since there are uncountably many factors in the product space $X$ and that each factor has at least two points. This idea had been implemented in this previous post.

____________________________________________________________________

Example 3

The third class of spaces is the class of Pixley-Roy spaces, which are hyperspaces. Given a space $X$, let $\mathcal{F}[X]$ be the set of all non-empty finite subsets of $X$. For $F \in \mathcal{F}[X]$ and for any open subset $U$ of $X$, let $[F,U]=\left\{B \in \mathcal{F}[X]: F \subset B \subset U \right\}$. The sets $[F,U]$ over all $F$ and $U$ form a base for a topology on $\mathcal{F}[X]$. This topology is called the Pixley-Roy topology (or Pixley-Roy hyperspace topology). The set $\mathcal{F}[X]$ with this topology is called a Pixley-Roy space.

The Pixley-Roy hyperspaces are discussed in this previous post. Whenever the ground space $X$ is uncountable, $\mathcal{F}[X]$ is not a separable space. We need to identify the $\mathcal{F}[X]$ that are CCC. According to the previous post on Pixley-Roy hyperspaces, for any space $X$ with a countable network, $\mathcal{F}[X]$ is CCC. Thus for any uncountable space $X$ with a countable network, the Pixley-Roy space $\mathcal{F}[X]$ is a CCC space that is not separable. The following gives a few such examples.

$\mathcal{F}[\mathbb{R}]$

$\mathcal{F}[X]$ where $X$ is any uncountable, separable and metrizable space.

$\mathcal{F}[X]$ where $X$ is uncountable and is the continuous image of a separable metrizable space.

Spaces with countable networks are discussed in this previous post. An example of a space $X$ that is the continuous image of a separable metrizable space is the bow-tie space found this previous post. Another example is any quote space of a separable metrizable space.

____________________________________________________________________

Example 4

For the fourth implementation of the strategy, we go back to the product space of separable spaces in Example 2, with the exception that the focus is on the product of the real line $\mathbb{R}$. Let $X$ be any uncountable completely regular space. The product space $\mathbb{R}^X$ always has the CCC since it is a product of separable space. Now we single out a dense subspace of $\mathbb{R}^X$ for which there is a characterization for separability, namely the subspace $C(X)$, which is the set of all continuous functions from $X$ into $\mathbb{R}$. The subspace $C(X)$ as a topological space is usually denoted by $C_p(X)$. For a basic discussion of $C_p(X)$, see this previous post.

We know precisely when $C_p(X)$ is separable. The following theorem captures the idea.

Theorem 1 – Theorem I.1.3 [1]
The function space $C_p(X)$ is separable if and only if the domain space $X$ has a weaker (or coarser) separable metric topology (in other words, $X$ is submetrizable with a separable metric topology).

Based on the theorem, $C_p(X)$ is separable for any separable metric space $X$. Other examples of separable $C_p(X)$ include spaces $X$ that are created by tweaking the usual Euclidean topology on the real line and at the same time retaining the usual real line topology as a weaker topology, e.g. the Sorgenfrey line and the Michael line. Thus $C_p(X)$ would be separable if $X$ is a space such as the Sorgenfrey line or the Michael line. For our purpose at hand, we need to look for spaces that are not like the Sorgenfrey line or the Michael line. Here’s some examples of spaces $X$ that have no weaker separable metric topology.

• Any compact space $X$ that is not metrizable.
• The space $X=\omega_1$, the first uncountable ordinal with the order topology.
• Any space $X=C_p(Y)$ where $Y$ is not separable.

The function space $C_p(X)$ for any one of the above three spaces has the CCC but is not separable. It is well known that any compact space with a weaker metrizable topology is metrizable. Some examples for compact $X$ are: the first uncountable successor ordinal $\omega_1+1$, the double arrow space, and the product space $\left\{0,1 \right\}^{\omega_1}$.

It can be shown that $C_p(\omega_1)$ is not separable (see this previous post). The last example is due to the following theorem.

Theorem 2 – Theorem I.1.4 [1]
The function space $C_p(Y)$ has a weaker (or coarser) separable metric topology if and only if the domain space $Y$ is separable.

Thus picking a non-separable space $Y$ would guarantee that $C_p(Y)$ has a weaker separable metric topology. As a result, $C_p(C_p(Y))$ is a CCC and not separable space.

Interestingly, Theorem 1 and Theorem 2 show a duality existing between the property of having a weaker separable metric topology and the property of being separable. The two theorems allow us to switch the two properties between the domain space and the function space.

____________________________________________________________________

Remarks

Another interesting point to make is that Theorem 1 and Theorem 2 together allow us to “buy one get one free.” Once we obtain a space $X$ that is CCC and not separable from any one of the avenues discussed here, the function space $C_p(X)$ has no weaker separable metric topology (by Theorem 2) and the function space $C_p(C_p(X))$ is another example of CCC and not separable.

The strategy discussed above unifies all four examples. Undoubtedly there will be other examples that can come from the strategy. To find more examples, find a space or a class of spaces that are reliably CCC and then look for potential non-separable spaces among the dense subspaces of the starting space.

____________________________________________________________________

Exercises

1. Show that in metrizable spaces, CCC and separable are equivalent. The only part that needs to be shown is that if $X$ is metrizable and CCC, then $X$ is separable.
2. Show that any dense subspace of a CCC space is also CCC.
3. Verify that the space $\Sigma(\left\{0,1 \right\}^{c})$ defined in Example 2 is dense in $X$ and is not separable.
4. Verify that the Pixley-Roy space $\mathcal{F}[\mathbb{R}]$ defined in Example 3 is CCC and not separable.
5. Verify that function space $C_p(\omega_1)$ mentioned in Example 4 is not separable. Hint: use the pressing down lemma.

____________________________________________________________________

Reference

1. Arkhangelskii, A. V., Topological Function Spaces, Mathematics and Its Applications Series, Kluwer Academic Publishers, Dordrecht, 1992.
2. Steen, L. A., Seebach, J. A., Counterexamples in Topology, Dover Publications, Inc., New York, 1995.

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

# Product Space – Exercise Set 1

This post presents several exercises concerning product spaces. All the concepts involved in the exercises have been discussed in the blog.

____________________________________________________________________

Exercise 1

Exercise 1a
Prove or disprove:
If $X$ and $Y$ are both hereditarily separable, then $X \times Y$ is hereditarily separable.

Exercise 1b
Show that if each $X_\alpha$ is separable, then the product space $\prod_{\alpha < \omega} \ X_\alpha$ is separable.

Exercise 1c
Prove or disprove:
If each $X_\alpha$ is separable, then the product space $\prod_{\alpha < \omega_1} \ X_\alpha$ is not separable.

____________________________________________________________________

Exercise 2

Exercise 2a
Show that if the space $X$ is normal, then every closed subspace of $X$ is a normal space.

Exercise 2b
Prove or disprove:
If the space $X$ is normal, then every dense open subspace of $X$ is a normal space.

____________________________________________________________________

Exercise 3

Consider the product space $\prod_{\alpha \in W} \ X_\alpha$.

Exercise 3a
Suppose that $X_\alpha$ is compact for all but one $\alpha \in W$ such that the non-compact factor is a Lindelof space. Show that the product space $\prod_{\alpha \in W} \ X_\alpha$ is a normal space.

Exercise 3b
Prove or disprove:
Suppose that $X_\alpha$ is compact for all but one $\alpha \in W$ such that the non-compact factor is a normal space. Then the product space $\prod_{\alpha \in W} \ X_\alpha$ is a normal space.

____________________________________________________________________

Exercise 4

Exercise 4a
Let $X$ be a compact space.
Show that if $X^n$ is hereditarily Lindelof for all positive integer $n$, then $X$ is metrizable.

Exercise 4b
Prove or disprove:
If $X^n$ is hereditarily Lindelof for all positive integer $n$, then $X$ is metrizable.

____________________________________________________________________

Exercise 5

Let $Y$ the product of uncountably many copies of the real line $\mathbb{R}$. If a specific example is desired, try $Y=\mathbb{R}^{\omega_1}$ ($\omega_1$ many copies of $\mathbb{R}$) or $Y=\mathbb{R}^{\mathbb{R}}$ (continuum many copies of $\mathbb{R}$). It is also OK to use a larger number of copies of the real line.

Note that the space $Y$ is not normal (see here).

Exercise 5a
Since the product space $Y$ is not normal, it is not Lindelof. As an exercise, find an open cover of $Y$ that proves that $Y$ is not Lindelof, i.e. an open cover $\mathcal{U}$ of $Y$ such that no countable subcollection of $\mathcal{U}$ can cover $Y$.

Exercise 5b
Show that for every open cover $\mathcal{U}$ of the space $Y$, there is a countable $\mathcal{V} \subset \mathcal{U}$ of $Y$ such that $\overline{\mathcal{V}}=Y$, i.e. $\cup \mathcal{V}$ is dense in $Y$. Note that with this property, the space $Y$ is said to be weakly Lindelof.

____________________________________________________________________

Exercise 6

This exercise is about the product $Y=\mathbb{R}^{\mathbb{R}}$ (continuum many copies of $\mathbb{R}$). Show the following.

1. Show that $Y$ is separable by exhibiting a countable dense set.
2. Show that $Y$ is not hereditarily separable by exhibiting a non-separable subspace.
3. Show that the space $Y$ has a closed and discrete subspace of cardinality continuum.
4. Show that $Y$ is not first countable.
5. Show that $Y$ is not a Frechet space.
6. Show that $Y$ is not a countably tight space.

See here for the definition of Frechet space.

See here for the definition of countably tight space.

____________________________________________________________________

Exercise 7

Consider the product space $Y=\mathbb{\omega}^{\omega_1}$. It is not normal (see here).

Exercise 7a
Construct a dense normal subspace of $Y$.

Exercise 7b
Construct a dense Lindelof subspace of $Y$.

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

# Counterexample 106 from Steen and Seebach

As the title suggests, this post discusses counterexample 106 in Steen and Seebach [2]. We extend the discussion by adding two facts not found in [2].

The counterexample 106 is the space $X=\omega_1 \times I^I$, which is the product of $\omega_1$ with the interval topology and the product space $I^I=\prod_{t \in I} I$ where $I$ is of course the unit interval $[0,1]$. The notation of $\omega_1$, the first uncountable ordinal, in Steen and Seebach is $[0,\Omega)$.

Another way to notate the example $X$ is the product space $\prod_{t \in I} X_t$ where $X_0$ is $\omega_1$ and $X_t$ is the unit interval $I$ for all $t>0$. Thus in this product space, all factors except for one factor is the unit interval and the lone non-compact factor is the first uncountable ordinal. The factor of $\omega_1$ makes this product space an interesting example.

The following lists out the basic topological properties of the space that $X=\omega_1 \times I^I$ are covered in [2].

• The space $X$ is Hausdorff and completely regular.
• The space $X$ is countably compact.
• The space $X$ is neither compact nor sequentially compact.
• The space $X$ is neither separable, Lindelof nor $\sigma$-compact.
• The space $X$ is not first countable.
• The space $X$ is locally compact.

All the above bullet points are discussed in Steen and Seebach. In this post we add the following two facts.

• The space $X$ is not normal.
• The space $X$ has a dense subspace that is normal.

It follows from these bullet points that the space $X$ is an example of a completely regular space that is not normal. Not being a normal space, $X$ is then not metrizable. Of course there are other ways to show that $X$ is not metrizable. One is that neither of the two factors $\omega_1$ or $I^I$ is metrizable. Another is that $X$ is not first countable.

____________________________________________________________________

The space $X$ is not normal

Now we are ready to discuss the non-normality of the example. It is a natural question to ask whether the example $X=\omega_1 \times I^I$ is normal. The fact that it was not discussed in [2] could be that the tool for answering the normality question was not yet available at the time [2] was originally published, though we do not know for sure. It turns out that the tool became available in the paper [1] published a few years after the publication of [2]. The key to showing the normality (or the lack of) in the example $X=\omega_1 \times I^I$ is to show whether the second factor $I^I$ is a countably tight space.

The main result in [1] is discussed in this previous post. Theorem 1 in the previous post states that for any compact space $Y$, the product $\omega_1 \times Y$ is normal if and only if $Y$ is countably tight. Thus the normality of the space $X$ (or the lack of) hinges on whether the compact factor $I^I=\prod_{t \in I} I$ is countably tight.

A space $Y$ is countably tight (or has countable tightness) if for each $S \subset Y$ and for each $x \in \overline{S}$, there exists some countable $B \subset S$ such that $x \in \overline{B}$. The definitions of tightness in general and countable tightness in particular are discussed here.

To show that the product space $I^I=\prod_{t \in I} I$ is not countably tight, we let $S$ be the subspace of $I^I$ consisting of points, each of which is non-zero on at most countably many coordinates. Specifically $S$ is defined as follows:

$S=\Sigma_{t \in I} I=\left\{y \in I^I: y(t) \ne 0 \text{ for at most countably many } t \in I \right\}$

The set $S$ just defined is also called the $\Sigma$-product of copies of unit interval $I$. Let $g \in I^I$ be defined by $g(t)=1$ for all $t \in I$. It follows that $g \in \overline{S}$. It can also be verified that $g \notin \overline{B}$ for any countable $B \subset S$. This shows that the product space $I^I=\prod_{t \in I} I$ is not countably tight.

By Theorem 1 found in this link, the space $X=\omega_1 \times I^I$ is not normal.

____________________________________________________________________

The space $X$ has a dense subspace that is normal

Now that we know $X=\omega_1 \times I^I$ is not normal, a natural question is whether it has a dense subspace that is normal. Consider the subspace $\omega_1 \times S$ where $S$ is the $\Sigma$-product $S=\Sigma_{t \in I} I$ defined in the preceding section. The subspace $S$ is dense in the product space $I^I$. Thus $\omega_1 \times S$ is dense in $X=\omega_1 \times I^I$. The space $S$ is normal since the $\Sigma$-product of separable metric spaces is normal. Furthermore, $\omega_1$ can be embedded as a closed subspace of $S=\Sigma_{t \in I} I$. Then $\omega_1 \times S$ is homeomorphic to a closed subspace of $S \times S$. Since $S \times S \cong S$, the space $\omega_1 \times S$ is normal.

____________________________________________________________________

Reference

1. Nogura, T., Tightness of compact Hausdorff space and normality of product spaces, J. Math. Soc. Japan, 28, 360-362, 1976
2. Steen, L. A., Seebach, J. A., Counterexamples in Topology, Dover Publications, Inc., New York, 1995.

____________________________________________________________________
$\copyright \ 2015 \text{ by Dan Ma}$