# Countably paracompact spaces

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

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

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

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Normal Countably Paracompact Spaces

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

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

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

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

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

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

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

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

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

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

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Some Examples

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

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

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

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

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

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

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

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

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

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

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

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

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

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More on Countably Paracompactness

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

The part about decreasing follows from:

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

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

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

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Reference

1. Ball, B. J., Countable Paracompactness in Linearly Ordered Spaces, Proc. Amer. Math. Soc., 5, 190-192, 1954. (link)
2. Rudin, M. E., A Normal Space $X$ for which $X \times I$ is not Normal, Fund. Math., 73, 179-486, 1971. (link)
3. Rudin, M. E., Dowker Spaces, Handbook of Set-Theoretic Topology (K. Kunen and J. E. Vaughan, eds), Elsevier Science Publishers B. V., Amsterdam, (1984) 761-780.
4. Wikipedia Entry on Dowker Spaces (link)

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

# 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.

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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.

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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.

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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.

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