# 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 compact? 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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$\copyright \ 2016 \text{ by Dan Ma}$

# Normal dense subspaces of a product of “continuum” many separable metric factors

Is every normal dense subspace of a product of separable metric spaces collectionwise normal? This question was posed by Arkhangelskii in [1] (see Problem I.5.25). A partial positive answer is provided by a theorem that is usually attributed to Corson: If $Y$ is a normal dense subspace of a product of separable metric spaces and if $Y \times Y$ is also normal, then $Y$ is collectionwise normal. In this post, using a simple combinatorial argument, we show that any normal dense subspace of a product of continuum many separable metric space is collectionwise normal (see Corollary 4 below), which is a corollary of the following theorem.

Theorem 1
Let $X$ be a normal space with character $\le 2^\omega$. If $2^\omega<2^{\omega_1}$, then the following holds:

• If $Y$ is a closed and discrete subspace of $X$ with $\lvert Y \lvert=\omega_1$, then $Y$ contains a separated subset of cardinality $\omega_1$.

Theorem 1 gives the corollary indicated at the beginning and several other interesting results. The statement $2^\omega<2^{\omega_1}$ means that the cardinality of the power set (the set of all subsets) of $\omega$ is strictly less than the cardinality of the power set of $\omega_1$. Note that the statement $2^\omega<2^{\omega_1}$ follows from the continuum hypothesis (CH), the statement that $2^\omega=\omega_1$. With the assumption $2^\omega<2^{\omega_1}$, Theorem 1 is a theorem that goes beyond ZFC. We also present an alternative to Theorem 1 that removes the assumption $2^\omega<2^{\omega_1}$ (see Theorem 6 below).

A subset $T$ of a space $S$ is a separated set (in $S$) if for each $t \in T$, there is an open subset $O_t$ of $S$ with $t \in O_t$ such that $\left\{O_t: t \in T \right\}$ is a pairwise disjoint collection. First we prove Theorem 1 and then discuss the corollaries.

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Proof of Theorem 1

Suppose $Y$ is a closed and discrete subset of $X$ with $\lvert Y \lvert=\omega_1$ such that no subset of $Y$ of cardinality $\omega_1$ can be separated. We then show that $2^{\omega_1} \le 2^{\omega}$.

For each $y \in Y$, let $\mathcal{B}_y$ be a local base at the point $y$ such that $\lvert \mathcal{B}_y \lvert \le 2^\omega$. Let $\mathcal{B}=\bigcup_{y \in Y} \mathcal{B}_y$. Thus $\lvert \mathcal{B} \lvert \le 2^\omega$. By normality, for each $W \subset Y$, let $U_W$ be an open subset of $X$ such that $W \subset U_W$ and $\overline{U_W} \cap (Y-W)=\varnothing$. For each $W \subset Y$, consider the following collection of open sets:

$\mathcal{G}_W=\left\{V \in \mathcal{B}_y: y \in W \text{ and } V \subset U_W \right\}$

For each $W \subset Y$, choose a maximal disjoint collection $\mathcal{M}_W$ of open sets in $\mathcal{G}_W$. Because no subset of $Y$ of cardinality $\omega_1$ can be separated, each $\mathcal{M}_W$ is countable. If $W_1 \ne W_2$, then $\mathcal{M}_{W_1} \ne \mathcal{M}_{W_2}$.

Let $\mathcal{P}(Y)$ be the power set (i.e. the set of all subsets) of $Y$. Let $\mathcal{P}_\omega(\mathcal{B})$ be the set of all countable subsets of $\mathcal{B}$. Then the mapping $W \mapsto \mathcal{M}_W$ is a one-to-one map from $\mathcal{P}(Y)$ into $\mathcal{P}_\omega(\mathcal{B})$. Note that $\lvert \mathcal{P}(Y) \lvert=2^{\omega_1}$. Also note that since $\lvert \mathcal{B} \lvert \le 2^\omega$, $\lvert \mathcal{P}_\omega(\mathcal{B}) \lvert \le 2^\omega$. Thus $2^{\omega_1} \le 2^{\omega}$. $\blacksquare$

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Some Corollaries of Theorem 1

Here’s some corollaries that follow easily from Theorem 1. A space $X$ has the countable chain condition (CCC) if every pairwise disjoint collection of non-empty open subset of $X$ is countable. For convenience, if $X$ has the CCC, we say $X$ is CCC. The following corollaries make use of the fact that any normal space with countable extent is collectionwise normal (see Theorem 2 in this previous post).

Corollary 2
Let $X$ be a CCC space with character $\le 2^\omega$. If $2^\omega<2^{\omega_1}$, then the following conditions hold:

• If $X$ is normal, then every closed and discrete subset of $X$ is countable, i.e., $X$ has countable extent.
• If $X$ is normal, then $X$ is collectionwise normal.

Corollary 3
Let $X$ be a CCC space with character $\le 2^\omega$. If CH holds, then the following conditions hold:

• If $X$ is normal, then every closed and discrete subset of $X$ is countable, i.e., $X$ has countable extent.
• If $X$ is normal, then $X$ is collectionwise normal.

Corollary 4
Let $X=\prod_{\alpha<2^\omega} X_\alpha$ be a product where each factor $X_\alpha$ is a separable metric space. If $2^\omega<2^{\omega_1}$, then the following conditions hold:

• If $Y$ is a normal dense subspace of $X$, then $Y$ has countable extent.
• If $Y$ is a normal dense subspace of $X$, then $Y$ is collectionwise normal.

Corollary 4 is the result indicated in the title of the post. The product of separable spaces has the CCC. Thus the product space $X$ and any dense subspace of $X$ have the CCC. Because $X$ is a product of continuum many separable metric spaces, $X$ and any subspace of $X$ have characters $\le 2^\omega$. Then Corollary 4 follows from Corollary 2.

When dealing with the topic of normal versus collectionwise normal, it is hard to avoid the connection with the normal Moore space conjecture. Theorem 1 gives the result of F. B. Jones from 1937 (see [3]). We have the following theorem.

Theorem 5
If $2^\omega<2^{\omega_1}$, then every separable normal Moore space is metrizable.

Though this was not how Jones proved it in [3], Theorem 5 is a corollary of Corollary 2. By Corollary 2, any separable normal Moore space is collectionwise normal. It is well known that collectionwise normal Moore space is metrizable (Bing’s metrization theorem, see Theorem 5.4.1 in [2]).

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A ZFC Theorem

We now prove a result that is similar to Corollary 2 but uses no set-theory beyond the Zermelo–Fraenkel set theory plus axiom of choice (abbreviated by ZFC). Of course the conclusion is not as strong. Even though the assumption $2^\omega<2^{\omega_1}$ is removed in Theorem 6, note the similarity between the proof of Theorem 1 and the proof of Theorem 6.

Theorem 6
Let $X$ be a CCC space with character $\le 2^\omega$. Then the following conditions hold:

• If $X$ is normal, then every closed and discrete subset of $X$ has cardinality less than continuum.

Proof of Theorem 6
Let $X$ be a normal CCC space with character $\le 2^\omega$. Let $Y$ be a closed and discrete subset of $X$. We show that $\lvert Y \lvert < 2^\omega$. Suppose that $\lvert Y \lvert = 2^\omega$.

For each $y \in Y$, let $\mathcal{B}_y$ be a local base at the point $y$ such that $\lvert \mathcal{B}_y \lvert \le 2^\omega$. Let $\mathcal{B}=\bigcup_{y \in Y} \mathcal{B}_y$. Thus $\lvert \mathcal{B} \lvert = 2^\omega$. By normality, for each $W \subset Y$, let $U_W$ be an open subset of $X$ such that $W \subset U_W$ and $\overline{U_W} \cap (Y-W)=\varnothing$. For each $W \subset Y$, consider the following collection of open sets:

$\mathcal{G}_W=\left\{V \in \mathcal{B}_y: y \in W \text{ and } V \subset U_W \right\}$

For each $W \subset Y$, choose $\mathcal{M}_W \subset \mathcal{G}_W$ such that $\mathcal{M}_W$ is a maximal disjoint collection. Since $X$ is CCC, $\mathcal{M}_W$ is countable. It is clear that if $W_1 \ne W_2$, then $\mathcal{M}_{W_1} \ne \mathcal{M}_{W_2}$.

Let $\mathcal{P}(Y)$ be the power set (i.e. the set of all subsets) of $Y$. Let $\mathcal{P}_\omega(\mathcal{B})$ be the set of all countable subsets of $\mathcal{B}$. Then the mapping $W \mapsto \mathcal{M}_W$ is a one-to-one map from $\mathcal{P}(Y)$ into $\mathcal{P}_\omega(\mathcal{B})$. Note that since $\lvert \mathcal{B} \lvert = 2^\omega$, $\lvert \mathcal{P}_\omega(\mathcal{B}) \lvert = 2^\omega$. Thus $\lvert \mathcal{P}(Y) \lvert \le 2^{\omega}$. However, $Y$ is assumed to be of cardinality continuum. Then $\lvert \mathcal{P}(Y) \lvert>2^{\omega_1}$, leading to a contradiction. Thus it must be the case that $\lvert Y \lvert < 2^\omega$. $\blacksquare$

With Theorem 6, Corollary 3 still holds. Theorem 6 removes the set-theoretic assumption of $2^\omega<2^{\omega_1}$. As a result, the upper bound for cardinalities of closed and discrete sets is (at least potentially) higher.

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Reference

1. Arkhangelskii, A. V., Topological Function Spaces, Mathematics and Its Applications Series, Kluwer Academic Publishers, Dordrecht, 1992.
2. Engelking, R., General Topology, Revised and Completed edition, Heldermann Verlag, Berlin, 1989.
3. Jones, F. B., Concerning normal and completely normal spaces, Bull. Amer. Math. Soc., 43, 671-677, 1937.

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$\copyright \ 2014 \text{ by Dan Ma}$

# One theorem about normality of Cp(X)

Assuming that the function space $C_p(X)$ is normal, what can be said about the domain space $X$? In this post, we prove a theorem that yields a corollary that for any normal space $X$, if $C_p(X)$ is normal, then $X$ has countable extent (i.e. every closed and discrete subset of $X$ is countable). Thus the normality of the function space limits the size of a closed and discrete subset of the domain space. It then follows that for any metric space $X$, if $C_p(X)$ is normal, $X$ has is second countable (i.e. having a countable base). Another immediate, but slightly less obvious, corollary is that for any $X$ that is a normal Moore space, if $C_p(X)$ is normal, then $X$ is metrizable.

For definitions of basic open sets and other background information on the function space $C_p(X)$, see this previous post.

Let $X$ be a space. Let $Y \subset X$. Let $\pi_Y$ be the natural projection from the product space $\mathbb{R}^X$ into the product space $\mathbb{R}^Y$. Specifically, if $f \in \mathbb{R}^X$, then $\pi_Y(f)=f \upharpoonright Y$, i.e., the function $f$ restricted to $Y$. In the discussion below, $\pi_Y$ is defined just on $C_p(X)$, i.e., $\pi_Y$ is the natural projection from $C_p(X)$ into $C_p(Y)$. It is always the case that $\pi_Y(C_p(X)) \subset C_p(Y)$. It is not necessarily the case that $\pi_Y(C_p(X))=C_p(Y)$. However, if $X$ is a normal space and $Y$ is closed in $X$, then $\pi_Y(C_p(X))=C_p(Y)$ and $\pi_Y$ is the natural projection from $C_p(X)$ onto $C_p(Y)$. We prove the following theorem.

Theorem 1

Suppose that $C_p(X)$ is a normal space. Let $Y$ be a closed subspace of $X$. Then $\pi_Y(C_p(X))$ is a normal space.

Theorem 1 is found in [1] (see Theorem I.6.2). In proving Theorem 1, we need the following lemma.

Lemma 2

Let $T=\prod_{\alpha \in A} T_\alpha$ be a product of separable metrizable spaces. Let $S$ be a dense subspace of $T$. Then the following conditions are equivalent.

1. $S$ is normal.
2. For any pair of disjoint closed subsets $H$ and $K$ of $S$, there exists a countable $B \subset A$ such that $\overline{\pi_B(H)} \cap \overline{\pi_B(K)}=\varnothing$.
3. For any pair of disjoint closed subsets $H$ and $K$ of $S$, there exists a countable $B \subset A$ such that $\pi_B(H)$ and $\pi_B(K)$ are separated in $\pi_B(S)$, meaning that $\overline{\pi_B(H)} \cap \pi_B(K)=\pi_B(H) \cap \overline{\pi_B(K)}=\varnothing$.

For a proof of Lemma 2, see Lemma 1 in this previous post.

Proof of Theorem 1
Note that $\pi_Y(C_p(X))$ is a dense subspace of $\mathbb{R}^Y$. Let $H$ and $K$ be disjoint closed subsets of $\pi_Y(C_p(X))$. To show $\pi_Y(C_p(X))$ is normal, by Lemma 2, we only need to produce a countable $B \subset Y$ such that $\overline{\pi_B(H)} \cap \overline{\pi_B(K)}=\varnothing$. The closure here is taken in $\pi_B(\pi_Y(C_p(X)))$.

Let $H_1=\pi_Y^{-1}(H)$ and $K_1=\pi_Y^{-1}(K)$. Both $H_1$ and $K_1$ are closed subsets of $C_p(X)$. By Lemma 2, there exists some countable $C \subset X$ such that $\overline{\pi_C(H_1)} \cap \overline{\pi_C(K_1)}=\varnothing$. The closure here is taken in $\pi_C(C_p(X))$. According to the remark at the end of this previous post, for any countable $D \subset X$ such that $C \subset D$, $\overline{\pi_D(H_1)} \cap \overline{\pi_D(K_1)}=\varnothing$. In other words, the countable set $C$ can be enlarged and the conclusion of the lemma still holds. With this observation in mind, we can assume that $C \cap Y \ne \varnothing$. If not, we can always throw countably many points of $Y$ into $C$ and still have $\overline{\pi_C(H_1)} \cap \overline{\pi_C(K_1)}=\varnothing$.

Let $B=C \cap Y$. We claim that $\overline{\pi_B(H_1)} \cap \overline{\pi_B(K_1)}=\varnothing$. The closure here is taken $\pi_B(C_p(X))$. Suppose that $\overline{\pi_B(H_1)} \cap \overline{\pi_B(K_1)} \ne \varnothing$. Choose $f \in C_p(X)$ such that $f \upharpoonright B \in \overline{\pi_B(H_1)} \cap \overline{\pi_B(K_1)}$. It follows that $f \upharpoonright C \in \overline{\pi_C(H_1)}$. To see this, let $f \upharpoonright C \in U=\prod_{x \in C} U_x$ where $U$ is a standard basic open set. Let $F$ be the support of $U$, i.e., the finite set of $x \in C$ such that $U_x \ne \mathbb{R}$. Let $F_1=F \cap Y$ and $F_2=F \cap (X-Y)$. Let $U^*=\prod_{x \in B} U_x$. Note that $f \upharpoonright B \in U^*$. Since $f \upharpoonright B \in \overline{\pi_B(H_1)}$, there is some $g \in H_1$ such that $\pi_B(g) \in U^*$. Note that $F_1$ is the support of $U^*$.

Because the space $X$ is completely regular, there is a $h \in C_p(X)$ such that $h(x)=0$ for all $x \in Y$ and $h(x)=f(x)-g(x)$ for all $x \in F_2$. Let $w=h+g$. Since $w \upharpoonright Y=g \upharpoonright Y$, $w \in H_1$. Note that $w=g$ on $Y$, hence on $F_1$ and that $w=f$ on $F_2$. Thus $w \upharpoonright C \in U$. Since $U$ is an arbitrary open set containing $f \upharpoonright C$, it follows that $f \upharpoonright C \in \overline{\pi_C(H_1)}$. By a similar argument, it can be shown that $f \upharpoonright C \in \overline{\pi_C(K_1)}$. This is a contradiction since $\overline{\pi_C(H_1)} \cap \overline{\pi_C(K_1)}=\varnothing$. Therefore the claim that $\overline{\pi_B(H_1)} \cap \overline{\pi_B(K_1)}=\varnothing$ is true, with the closure being taken in $\pi_B(C_p(X))$.

Because $B \subset Y$, observe that $\pi_B(H_1)=\pi_B(H)$ and $\pi_B(K_1)=\pi_B(K)$. Furthermore, $\pi_B(\pi_Y(C_p(X)))=\pi_B(C_p(X))$. Thus we can claim that $\overline{\pi_B(H)} \cap \overline{\pi_B(K)}=\varnothing$, with the closure being taken in $\pi_B(\pi_Y(C_p(X)))$. By Lemma 2, $\pi_Y(C_p(X))$ is normal. $\blacksquare$

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Some Corollaries

Corollary 3

Let $X$ be a normal space. If $C_p(X)$ is normal, then $X$ has countable extent, i.e., every closed and discrete subset of $X$ is countable.

Proof of Corollary 3
Let $Y$ be a closed and discrete subset of $X$. We show that $Y$ must be countable. Since $Y$ is closed and $X$ is normal, $\pi_Y(C_p(X))=C_p(Y)$. By Theorem 1, $C_p(Y)$ is normal. Since $Y$ is discrete, $C_p(Y)=\mathbb{R}^Y$. If $Y$ is uncountable, $\mathbb{R}^Y$ is not normal. Thus $Y$ must be countable. $\blacksquare$

Corollary 4

Let $X$ be a metrizable space. If $C_p(X)$ is normal, then $X$ has a countable base.

Proof of Corollary 4
Note that in any metrizable space, the weight equals the extent. By Corollary 3, $X$ has countable extent and thus has countable base. $\blacksquare$

Corollary 5

Let $X$ be a normal space. If $C_p(X)$ is normal, then $X$ is collectionwise normal.

Proof of Corollary 5
Any normal space with countable extent is collectionwise normal. See Theorem 2 in this previous post. $\blacksquare$

Corollary 6

Let $X$ be a normal Moore space. If $C_p(X)$ is normal, then $X$ is metrizable.

Proof of Corollary 6
Suppose $C_p(X)$ is normal. By Theorem 1, $X$ has countable extent. By Corollary 5, $X$ is collectionwise normal. According to Bing’s metrization theorem, any collectionwise normal Moore space is metrizable (see [2] Theorem 5.4.1 in page 329). $\blacksquare$

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Reference

1. Arkhangelskii, A. V., Topological Function Spaces, Mathematics and Its Applications Series, Kluwer Academic Publishers, Dordrecht, 1992.
2. Engelking, R., General Topology, Revised and Completed edition, Heldermann Verlag, Berlin, 1989.

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$\copyright \ 2014 \text{ by Dan Ma}$

# Pixley-Roy hyperspaces

In this post, we introduce a class of hyperspaces called Pixley-Roy spaces. This is a well-known and well studied set of topological spaces. Our goal here is not to be comprehensive but rather to present some selected basic results to give a sense of what Pixley-Roy spaces are like.

A hyperspace refers to a space in which the points are subsets of a given “ground” space. There are more than one way to define a hyperspace. Pixley-Roy spaces were first described by Carl Pixley and Prabir Roy in 1969 (see [5]). In such a space, the points are the non-empty finite subsets of a given ground space. More precisely, let $X$ be a $T_1$ space (i.e. finite sets are closed). Let $\mathcal{F}[X]$ be the set of all non-empty finite subsets of $X$. For each $F \in \mathcal{F}[X]$ and for each open subset $U$ of $X$ with $F \subset U$, we define:

$[F,U]=\left\{B \in \mathcal{F}[X]: F \subset B \subset U \right\}$

The sets $[F,U]$ over all possible $F$ 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 hyperspace as defined above was first defined by Pixley and Roy on the real line (see [5]) and was later generalized by van Douwen (see [7]). These spaces are easy to define and is useful for constructing various kinds of counterexamples. Pixley-Roy played an important part in answering the normal Moore space conjecture. Pixley-Roy spaces have also been studied in their own right. Over the years, many authors have investigated when the Pixley-Roy spaces are metrizable, normal, collectionwise Hausdorff, CCC and homogeneous. For a small sample of such investigations, see the references listed at the end of the post. Our goal here is not to discuss the results in these references. Instead, we discuss some basic properties of Pixley-Roy to solidify the definition as well as to give a sense of what these spaces are like. Good survey articles of Pixley-Roy are [3] and [7].

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Basic Discussion

In this section, we focus on properties that are always possessed by a Pixley-Roy space given that the ground space is at least $T_1$. Let $X$ be a $T_1$ space. We discuss the following points:

1. The topology defined above is a legitimate one, i.e., the sets $[F,U]$ indeed form a base for a topology on $\mathcal{F}[X]$.
2. $\mathcal{F}[X]$ is a Hausdorff space.
3. $\mathcal{F}[X]$ is a zero-dimensional space.
4. $\mathcal{F}[X]$ is a completely regular space.
5. $\mathcal{F}[X]$ is a hereditarily metacompact space.

Let $\mathcal{B}=\left\{[F,U]: F \in \mathcal{F}[X] \text{ and } U \text{ is open in } X \right\}$. Note that every finite set $F$ belongs to at least one set in $\mathcal{B}$, namely $[F,X]$. So $\mathcal{B}$ is a cover of $\mathcal{F}[X]$. For $A \in [F_1,U_1] \cap [F_2,U_2]$, we have $A \in [A,U_1 \cap U_2] \subset [F_1,U_1] \cap [F_2,U_2]$. So $\mathcal{B}$ is indeed a base for a topology on $\mathcal{F}[X]$.

To show $\mathcal{F}[X]$ is Hausdorff, let $A$ and $B$ be finite subsets of $X$ where $A \ne B$. Then one of the two sets has a point that is not in the other one. Assume we have $x \in A-B$. Since $X$ is $T_1$, we can find open sets $U, V \subset X$ such that $x \in U$, $x \notin V$ and $A \cup B-\left\{ x \right\} \subset V$. Then $[A,U \cup V]$ and $[B,V]$ are disjoint open sets containing $A$ and $B$ respectively.

To see that $\mathcal{F}[X]$ is a zero-dimensional space, we show that $\mathcal{B}$ is a base consisting of closed and open sets. To see that $[F,U]$ is closed, let $C \notin [F,U]$. Either $F \not \subset C$ or $C \not \subset U$. In either case, we can choose open $V \subset X$ with $C \subset V$ such that $[C,V] \cap [F,U]=\varnothing$.

The fact that $\mathcal{F}[X]$ is completely regular follows from the fact that it is zero-dimensional.

To show that $\mathcal{F}[X]$ is metacompact, let $\mathcal{G}$ be an open cover of $\mathcal{F}[X]$. For each $F \in \mathcal{F}[X]$, choose $G_F \in \mathcal{G}$ such that $F \in G_F$ and let $V_F=[F,X] \cap G_F$. Then $\mathcal{V}=\left\{V_F: F \in \mathcal{F}[X] \right\}$ is a point-finite open refinement of $\mathcal{G}$. For each $A \in \mathcal{F}[X]$, $A$ can only possibly belong to $V_F$ for the finitely many $F \subset A$.

A similar argument show that $\mathcal{F}[X]$ is hereditarily metacompact. Let $Y \subset \mathcal{F}[X]$. Let $\mathcal{H}$ be an open cover of $Y$. For each $F \in Y$, choose $H_F \in \mathcal{H}$ such that $F \in H_F$ and let $W_F=([F,X] \cap Y) \cap H_F$. Then $\mathcal{W}=\left\{W_F: F \in Y \right\}$ is a point-finite open refinement of $\mathcal{H}$. For each $A \in Y$, $A$ can only possibly belong to $W_F$ for the finitely many $F \subset A$ such that $F \in Y$.

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More Basic Results

We now discuss various basic topological properties of $\mathcal{F}[X]$. We first note that $\mathcal{F}[X]$ is a discrete space if and only if the ground space $X$ is discrete. Though we do not need to make this explicit, it makes sense to focus on non-discrete spaces $X$ when we look at topological properties of $\mathcal{F}[X]$. We discuss the following points:

1. If $X$ is uncountable, then $\mathcal{F}[X]$ is not separable.
2. If $X$ is uncountable, then every uncountable subspace of $\mathcal{F}[X]$ is not separable.
3. If $\mathcal{F}[X]$ is Lindelof, then $X$ is countable.
4. If $\mathcal{F}[X]$ is Baire space, then $X$ is discrete.
5. If $\mathcal{F}[X]$ has the CCC, then $X$ has the CCC.
6. If $\mathcal{F}[X]$ has the CCC, then $X$ has no uncountable discrete subspaces,i.e., $X$ has countable spread, which of course implies CCC.
7. If $\mathcal{F}[X]$ has the CCC, then $X$ is hereditarily Lindelof.
8. If $\mathcal{F}[X]$ has the CCC, then $X$ is hereditarily separable.
9. If $X$ has a countable network, then $\mathcal{F}[X]$ has the CCC.
10. The Pixley-Roy space of the Sorgenfrey line does not have the CCC.
11. If $X$ is a first countable space, then $\mathcal{F}[X]$ is a Moore space.

Bullet points 6 to 9 refer to properties that are never possessed by Pixley-Roy spaces except in trivial cases. Bullet points 6 to 8 indicate that $\mathcal{F}[X]$ can never be separable and Lindelof as long as the ground space $X$ is uncountable. Note that $\mathcal{F}[X]$ is discrete if and only if $X$ is discrete. Bullet point 9 indicates that any non-discrete $\mathcal{F}[X]$ can never be a Baire space. Bullet points 10 to 13 give some necessary conditions for $\mathcal{F}[X]$ to be CCC. Bullet 14 gives a sufficient condition for $\mathcal{F}[X]$ to have the CCC. Bullet 15 indicates that the hereditary separability and the hereditary Lindelof property are not sufficient conditions for the CCC of Pixley-Roy space (though they are necessary conditions). Bullet 16 indicates that the first countability of the ground space is a strong condition, making $\mathcal{F}[X]$ a Moore space.

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To see bullet point 6, let $X$ be an uncountable space. Let $\left\{F_1,F_2,F_3,\cdots \right\}$ be any countable subset of $\mathcal{F}[X]$. Choose a point $x \in X$ that is not in any $F_n$. Then none of the sets $F_i$ belongs to the basic open set $[\left\{x \right\} ,X]$. Thus $\mathcal{F}[X]$ can never be separable if $X$ is uncountable.

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To see bullet point 7, let $Y \subset \mathcal{F}[X]$ be uncountable. Let $W=\cup \left\{F: F \in Y \right\}$. Let $\left\{F_1,F_2,F_3,\cdots \right\}$ be any countable subset of $Y$. We can choose a point $x \in W$ that is not in any $F_n$. Choose some $A \in Y$ such that $x \in A$. Then none of the sets $F_n$ belongs to the open set $[A ,X] \cap Y$. So not only $\mathcal{F}[X]$ is not separable, no uncountable subset of $\mathcal{F}[X]$ is separable if $X$ is uncountable.

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To see bullet point 8, note that $\mathcal{F}[X]$ has no countable open cover consisting of basic open sets, assuming that $X$ is uncountable. Consider the open collection $\left\{[F_1,U_1],[F_2,U_2],[F_3,U_3],\cdots \right\}$. Choose $x \in X$ that is not in any of the sets $F_n$. Then $\left\{ x \right\}$ cannot belong to $[F_n,U_n]$ for any $n$. Thus $\mathcal{F}[X]$ can never be Lindelof if $X$ is uncountable.

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For an elementary discussion on Baire spaces, see this previous post.

To see bullet point 9, let $X$ be a non-discrete space. To show $\mathcal{F}[X]$ is not Baire, we produce an open subset that is of first category (i.e. the union of countably many closed nowhere dense sets). Let $x \in X$ a limit point (i.e. an non-isolated point). We claim that the basic open set $V=[\left\{ x \right\},X]$ is a desired open set. Note that $V=\bigcup \limits_{n=1}^\infty H_n$ where

$H_n=\left\{F \in \mathcal{F}[X]: x \in F \text{ and } \lvert F \lvert \le n \right\}$

We show that each $H_n$ is closed and nowhere dense in the open subspace $V$. To see that it is closed, let $A \notin H_n$ with $x \in A$. We have $\lvert A \lvert>n$. Then $[A,X]$ is open and every point of $[A,X]$ has more than $n$ points of the space $X$. To see that $H_n$ is nowhere dense in $V$, let $[B,U]$ be open with $[B,U] \subset V$. It is clear that $x \in B \subset U$ where $U$ is open in the ground space $X$. Since the point $x$ is not an isolated point in the space $X$, $U$ contains infinitely many points of $X$. So choose an finite set $C$ with at least $2 \times n$ points such that $B \subset C \subset U$. For the the open set $[C,U]$, we have $[C,U] \subset [B,U]$ and $[C,U]$ contains no point of $H_n$. With the open set $V$ being a union of countably many closed and nowhere dense sets in $V$, the open set $V$ is not of second category. We complete the proof that $\mathcal{F}[X]$ is not a Baire space.

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To see bullet point 10, let $\mathcal{O}$ be an uncountable and pairwise disjoint collection of open subsets of $X$. For each $O \in \mathcal{O}$, choose a point $x_O \in O$. Then $\left\{[\left\{ x_O \right\},O]: O \in \mathcal{O} \right\}$ is an uncountable and pairwise disjoint collection of open subsets of $\mathcal{F}[X]$. Thus if $\mathcal{F}[X]$ is CCC then $X$ must have the CCC.

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To see bullet point 11, let $Y \subset X$ be uncountable such that $Y$ as a space is discrete. This means that for each $y \in Y$, there exists an open $O_y \subset X$ such that $y \in O_y$ and $O_y$ contains no point of $Y$ other than $y$. Then $\left\{[\left\{y \right\},O_y]: y \in Y \right\}$ is an uncountable and pairwise disjoint collection of open subsets of $\mathcal{F}[X]$. Thus if $\mathcal{F}[X]$ has the CCC, then the ground space $X$ has no uncountable discrete subspace (such a space is said to have countable spread).

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To see bullet point 12, let $Y \subset X$ be uncountable such that $Y$ is not Lindelof. Then there exists an open cover $\mathcal{U}$ of $Y$ such that no countable subcollection of $\mathcal{U}$ can cover $Y$. We can assume that sets in $\mathcal{U}$ are open subsets of $X$. Also by considering a subcollection of $\mathcal{U}$ if necessary, we can assume that cardinality of $\mathcal{U}$ is $\aleph_1$ or $\omega_1$. Now by doing a transfinite induction we can choose the following sequence of points and the following sequence of open sets:

$\left\{x_\alpha \in Y: \alpha < \omega_1 \right\}$

$\left\{U_\alpha \in \mathcal{U}: \alpha < \omega_1 \right\}$

such that $x_\beta \ne x_\gamma$ if $\beta \ne \gamma$, $x_\alpha \in U_\alpha$ and $x_\alpha \notin \bigcup \limits_{\beta < \alpha} U_\beta$ for each $\alpha < \omega_1$. At each step $\alpha$, all the previously chosen open sets cannot cover $Y$. So we can always choose another point $x_\alpha$ of $Y$ and then choose an open set in $\mathcal{U}$ that contains $x_\alpha$.

Then $\left\{[\left\{x_\alpha \right\},U_\alpha]: \alpha < \omega_1 \right\}$ is a pairwise disjoint collection of open subsets of $\mathcal{F}[X]$. Thus if $\mathcal{F}[X]$ has the CCC, then $X$ must be hereditarily Lindelof.

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To see bullet point 13, let $Y \subset X$. Consider open sets $[A,U]$ where $A$ ranges over all finite subsets of $Y$ and $U$ ranges over all open subsets of $X$ with $A \subset U$. Let $\mathcal{G}$ be a collection of such $[A,U]$ such that $\mathcal{G}$ is pairwise disjoint and $\mathcal{G}$ is maximal (i.e. by adding one more open set, the collection will no longer be pairwise disjoint). We can apply a Zorn lemma argument to obtain such a maximal collection. Let $D$ be the following subset of $Y$.

$D=\bigcup \left\{A: [A,U] \in \mathcal{G} \text{ for some open } U \right\}$

We claim that the set $D$ is dense in $Y$. Suppose that there is some open set $W \subset X$ such that $W \cap Y \ne \varnothing$ and $W \cap D=\varnothing$. Let $y \in W \cap Y$. Then $[\left\{y \right\},W] \cap [A,U]=\varnothing$ for all $[A,U] \in \mathcal{G}$. So adding $[\left\{y \right\},W]$ to $\mathcal{G}$, we still get a pairwise disjoint collection of open sets, contradicting that $\mathcal{G}$ is maximal. So $D$ is dense in $Y$.

If $\mathcal{F}[X]$ has the CCC, then $\mathcal{G}$ is countable and $D$ is a countable dense subset of $Y$. Thus if $\mathcal{F}[X]$ has the CCC, the ground space $X$ is hereditarily separable.

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A collection $\mathcal{N}$ of subsets of a space $Y$ is said to be a network for the space $Y$ if any non-empty open subset of $Y$ is the union of elements of $\mathcal{N}$, equivalently, for each $y \in Y$ and for each open $U \subset Y$ with $y \in U$, there is some $A \in \mathcal{N}$ with $x \in A \subset U$. Note that a network works like a base but the elements of a network do not have to be open. The concept of network and spaces with countable network are discussed in these previous posts Network Weight of Topological Spaces – I and Network Weight of Topological Spaces – II.

To see bullet point 14, let $\mathcal{N}$ be a network for the ground space $X$ such that $\mathcal{N}$ is also countable. Assume that $\mathcal{N}$ is closed under finite unions (for example, adding all the finite unions if necessary). Let $\left\{[A_\alpha,U_\alpha]: \alpha < \omega_1 \right\}$ be a collection of basic open sets in $\mathcal{F}[X]$. Then for each $\alpha$, find $B_\alpha \in \mathcal{N}$ such that $A_\alpha \subset B_\alpha \subset U_\alpha$. Since $\mathcal{N}$ is countable, there is some $B \in \mathcal{N}$ such that $M=\left\{\alpha< \omega_1: B=B_\alpha \right\}$ is uncountable. It follows that for any finite $E \subset M$, $\bigcap \limits_{\alpha \in E} [A_\alpha,U_\alpha] \ne \varnothing$.

Thus if the ground space $X$ has a countable network, then $\mathcal{F}[X]$ has the CCC.

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The implications in bullet points 12 and 13 cannot be reversed. Hereditarily Lindelof property and hereditarily separability are not sufficient conditions for $\mathcal{F}[X]$ to have the CCC. See [4] for a study of the CCC property of the Pixley-Roy spaces.

To see bullet point 15, let $S$ be the Sorgenfrey line, i.e. the real line $\mathbb{R}$ with the topology generated by the half closed intervals of the form $[a,b)$. For each $x \in S$, let $U_x=[x,x+1)$. Then $\left\{[ \left\{ x \right\},U_x]: x \in S \right\}$ is a collection of pairwise disjoint open sets in $\mathcal{F}[S]$.

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A Moore space is a space with a development. For the definition, see this previous post.

To see bullet point 16, for each $x \in X$, let $\left\{B_n(x): n=1,2,3,\cdots \right\}$ be a decreasing local base at $x$. We define a development for the space $\mathcal{F}[X]$.

For each finite $F \subset X$ and for each $n$, let $B_n(F)=\bigcup \limits_{x \in F} B_n(x)$. Clearly, the sets $B_n(F)$ form a decreasing local base at the finite set $F$. For each $n$, let $\mathcal{H}_n$ be the following collection:

$\mathcal{H}_n=\left\{[F,B_n(F)]: F \in \mathcal{F}[X] \right\}$

We claim that $\left\{\mathcal{H}_n: n=1,2,3,\cdots \right\}$ is a development for $\mathcal{F}[X]$. To this end, let $V$ be open in $\mathcal{F}[X]$ with $F \in V$. If we make $n$ large enough, we have $[F,B_n(F)] \subset V$.

For each non-empty proper $G \subset F$, choose an integer $f(G)$ such that $[F,B_{f(G)}(F)] \subset V$ and $F \not \subset B_{f(G)}(G)$. Let $m$ be defined by:

$m=\text{max} \left\{f(G): G \ne \varnothing \text{ and } G \subset F \text{ and } G \text{ is proper} \right\}$

We have $F \not \subset B_{m}(G)$ for all non-empty proper $G \subset F$. Thus $F \notin [G,B_m(G)]$ for all non-empty proper $G \subset F$. But in $\mathcal{H}_m$, the only sets that contain $F$ are $[F,B_m(F)]$ and $[G,B_m(G)]$ for all non-empty proper $G \subset F$. So $[F,B_m(F)]$ is the only set in $\mathcal{H}_m$ that contains $F$, and clearly $[F,B_m(F)] \subset V$.

We have shown that for each open $V$ in $\mathcal{F}[X]$ with $F \in V$, there exists an $m$ such that any open set in $\mathcal{H}_m$ that contains $F$ must be a subset of $V$. This shows that the $\mathcal{H}_n$ defined above form a development for $\mathcal{F}[X]$.

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Examples

In the original construction of Pixley and Roy, the example was $\mathcal{F}[\mathbb{R}]$. Based on the above discussion, $\mathcal{F}[\mathbb{R}]$ is a non-separable CCC Moore space. Because the density (greater than $\omega$ for not separable) and the cellularity ($=\omega$ for CCC) do not agree, $\mathcal{F}[\mathbb{R}]$ is not metrizable. In fact, it does not even have a dense metrizable subspace. Note that countable subspaces of $\mathcal{F}[\mathbb{R}]$ are metrizable but are not dense. Any uncountable dense subspace of $\mathcal{F}[\mathbb{R}]$ is not separable but has the CCC. Not only $\mathcal{F}[\mathbb{R}]$ is not metrizable, it is not normal. The problem of finding $X \subset \mathbb{R}$ for which $\mathcal{F}[X]$ is normal requires extra set-theoretic axioms beyond ZFC (see [6]). In fact, Pixley-Roy spaces played a large role in the normal Moore space conjecture. Assuming some extra set theory beyond ZFC, there is a subset $M \subset \mathbb{R}$ such that $\mathcal{F}[M]$ is a CCC metacompact normal Moore space that is not metrizable (see Example I in [8]).

On the other hand, Pixley-Roy space of the Sorgenfrey line and the Pixley-Roy space of $\omega_1$ (the first uncountable ordinal with the order topology) are metrizable (see [3]).

The Sorgenfrey line and the first uncountable ordinal are classic examples of topological spaces that demonstrate that topological spaces in general are not as well behaved like metrizable spaces. Yet their Pixley-Roy spaces are nice. The real line and other separable metric spaces are nice spaces that behave well. Yet their Pixley-Roy spaces are very much unlike the ground spaces. This inverse relation between the ground space and the Pixley-Roy space was noted by van Douwen (see [3] and [7]) and is one reason that Pixley-Roy hyperspaces are a good source of counterexamples.

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Reference

1. Bennett, H. R., Fleissner, W. G., Lutzer, D. J., Metrizability of certain Pixley-Roy spaces, Fund. Math. 110, 51-61, 1980.
2. Daniels, P, Pixley-Roy Spaces Over Subsets of the Reals, Topology Appl. 29, 93-106, 1988.
3. Lutzer, D. J., Pixley-Roy topology, Topology Proc. 3, 139-158, 1978.
4. Hajnal, A., Juahasz, I., When is a Pixley-Roy Hyperspace CCC?, Topology Appl. 13, 33-41, 1982.
5. Pixley, C., Roy, P., Uncompletable Moore spaces, Proc. Auburn Univ. Conf. Auburn, AL, 1969.
6. Przymusinski, T., Normality and paracompactness of Pixley-Roy hyperspaces, Fund. Math. 113, 291-297, 1981.
7. van Douwen, E. K., The Pixley-Roy topology on spaces of subsets, Set-theoretic Topology, Academic Press, New York, 111-134, 1977.
8. Tall, F. D., Normality versus Collectionwise Normality, Handbook of Set-Theoretic Topology (K. Kunen and J. E. Vaughan, eds), Elsevier Science Publishers B. V., Amsterdam, 685-732, 1984.
9. Tanaka, H, Normality and hereditary countable paracompactness of Pixley-Roy hyperspaces, Fund. Math. 126, 201-208, 1986.

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$\copyright \ 2014 \text{ by Dan Ma}$

# Bing’s Example G

Bing’s Example G is an example of a topological space that is normal but not collectionwise normal. It was introduced in an influential paper of R. H. Bing in 1951 (see [1]). This paper has a metrization theorem that is now called Bing’s metrization theorem (any regular space is metrizable if and only if it has a $\sigma$-discrete base). The paper also introduced the notion of collectionwise normality and discussed the roles it plays in metrization theory (e.g. a Moore space is metrizable if and only if it is collectionwise normal). Example G was an influential example from an influential paper. It became the basis of construction for many other counterexamples (see [5] for one example). Investigations were also conducted by looking at various covering properties among subspaces of Example G (see [2] and [4] are two examples).

In this post we prove some basic results about Bing’s Example G. Some of the results we prove are found in Bing’s 1951 paper. The other results shown here are usually mentioned without proof in various places in the literature.

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Bing’s Example G – Definition

Let $P$ be any uncountable set. Let $Q$ be the set of all subsets of $P$. Let $F=2^Q$ be the set of all functions $f: Q \rightarrow 2=\left\{0,1 \right\}$. Another notation for $2^Q$ is the Cartesian product $\prod \limits_{q \in Q} \left\{0,1 \right\}$. For each $p \in P$, define the function $f_p: Q \rightarrow 2$ by the following:

$\forall q \in Q$, $f_p(q)=1$ if $p \in q$ and $f_p(q)=0$ if $p \notin q$

Let $F_P=\left\{f_p: p \in P \right\}$. Let $\tau$ be the set of all open subsets of $2^Q$ in the product topology. We now consider another topology on $2^Q$ generated by the following base:

$\mathcal{B}=\tau \cup \left\{\left\{x \right\}: x \in F-F_P \right\}$

Bing’s Example G is the set $F=2^Q$ with the topology generated by the base $\mathcal{B}$. In other words, each $x \in F-F_P$ is made an isolated point and points in $F_P$ retain the usual product open sets.

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Bing’s Example G – Initial Discussion

Bing’s Example G, i.e. the space $F$ as defined above, is obtained by altering the topology of the product space of $2^{\lvert \mathcal{K} \lvert}$ many copies of the two-point discrete space where $\mathcal{K}$ is the cardinality of the power set of the uncountable index set $P$ we start with. Out of this product space, a set $F_P$ of points is carefully chosen such that $F_P$ has the same cardinality as $P$ and such that $F_P$ is relatively discrete in the product space. Points in $F_P$ are made to retain the product topology and all points outside of $F_P$ are declared as isolated points.

We now show that the set $F_P$ is a discrete set in the space $F$. For each $p \in P$, let $W_p$ be the open set defined by

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

It is clear that $f_p$ is the only point of $F_P$ belonging to $W_p$. Therefore, in the Example G topology, the set $F_P$ is discrete and closed . In the section “Bing’s Example G is not Collectionwise Hausdorff” below, we show below that $F_P$ cannot be separated by any pairwise disjoint collection of open sets.

The character at a point is the minimum cardinality of a local base at that point. The character at a point in $F_P$ in the Example G topology agrees with the product topology. Points in $F_P$ have character $\lvert Q \lvert=2^{\lvert P \lvert}$. Specifically if the starting $P$ has cardinality $\omega_1$, then points in $F_P$ have character $2^{\omega_1}$. Thus Example G has large character and cannot be a Moore space (any Moore space has a countable base at every point).

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Bing’s Example G is Normal

Let $H_1$ and $H_2$ be disjoint closed subsets of $F$. The easy case is that one of $H_1$ and $H_2$ is a subset of $F-F_P$, say $H_1 \subset F-F_P$. Then $H_1$ is a closed and open set in $F$. Then $H_1$ and $F-H_1$ are disjoint open sets containing $H_1$ and $H_2$, respectively. So we can assume that both $H_1 \cap F_P \ne \varnothing$ and $H_2 \cap F_P \ne \varnothing$.

Let $A_1=H_1 \cap F_P$ and $A_2=H_2 \cap F_P$. Let $q_1=\left\{p \in P: f_p \in A_1 \right\}$ and $q_2=\left\{p \in P: f_p \in A_2 \right\}$. Define the following open sets:

$U_1=\left\{f \in F: f(q_1)=1 \text{ and } f(q_2)=0 \right\}$
$U_2=\left\{f \in F: f(q_1)=0 \text{ and } f(q_2)=1 \right\}$

Because $H_1 \cap H_2=\varnothing$, we have $A_1 \subset U_1$ and $A_2 \subset U_2$. Furthermore, $U_1 \cap U_2=\varnothing$. Let $B_1=H_1 \cap (F-F_P)$ and $B_2=H_2 \cap (F-F_P)$, which are open since they consist of isolated points. Then $O_1=(U_1 \cup B_1)-H_2$ and $O_2=(U_2 \cup B_2)-H_1$ are disjoint open subsets of $F$ with $H_1 \subset O_1$ and $H_2 \subset O_2$.

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Collectionwise Normal Spaces

Let $X$ be a space. Let $\mathcal{A}$ be a collection of subsets of $X$. We say $\mathcal{A}$ is pairwise disjoint if $A \cap B=\varnothing$ whenever $A,B \in \mathcal{A}$ with $A \ne B$. We say $\mathcal{A}$ is discrete if for each $x \in X$, there is an open set $O$ containing $x$ such that $O$ intersects at most one set in $\mathcal{A}$.

The space $X$ is said to be collectionwise normal if for every discrete collection $\mathcal{D}$ of closed subsets fo $X$, there is a pairwise disjoint collection $\left\{U_D: D \in \mathcal{D} \right\}$ of open subsets of $X$ such that $D \subset U_D$ for each $D \in \mathcal{D}$. Every paracompact space is collectionwise normal (see Theorem 5.1.18, p.305 of [3]). Thus Bing’s Example G is not paracompact.

When discrete collection of closed sets in the definition of “collectionwise normal” is replaced by discrete collection of singleton sets, the space is said to be collectionwise Hausdorff. Clearly any collectionwise normal space is collectionwise Hausdorff. Bing’s Example is actually not collectionwise Hausdorff.

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Bing’s Example G is not Collectionwise Hausdorff

The discrete set $F_P$ cannot be separated by disjoint open sets. For each $p \in P$, let $O_p$ be an open subset of $F$ such that $p \in O_p$. We show that the open sets $O_p$ cannot be pairwise disjoint. For each $p \in P$, choose an open set $L_p$ in the product topology of $2^Q$ such that $p \in L_p \subset O_p$. The product space $2^Q$ is a product of separable spaces, hence has the countable chain condition (CCC). Thus the open sets $L_p$ cannot be pairwise disjoint. Thus $L_t \cap L_s \ne \varnothing$ and $O_t \cap O_s \ne \varnothing$ for at least two points $s,t \in P$.

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Bing’s Example G is Completely Normal

The proof for showing Bing’s Example G is normal can be modified to show that it is completely normal. First some definitions. Let $X$ be a space. Let $A \subset X$ and $B \subset X$. The sets $A$ and $B$ are separated sets if $A \cap \overline{B}=\varnothing=\overline{A} \cap B$. Essentially, any two disjoint sets are separated sets if and only if none of them contains limit points (i.e. accumulation points) of the other set. A space $X$ is said to be completely normal if for every two separated sets $A$ and $B$ in $X$, there exist disjoint open subsets $U$ and $V$ of $X$ such that $A \subset U$ and $B \subset V$. Any two disjoint closed sets are separated sets. Thus any completely normal space is normal. It is well known that for any regular space $X$, $X$ is completely normal if and only if $X$ is hereditarily normal. For more about completely normality, see [3] and [6].

Let $H_1 \subset F$ and $H_2 \subset F$ such that $H_1 \cap \overline{H_2}=\varnothing=\overline{H_1} \cap H_2$. We consider two cases. One is that one of $H_1$ and $H_2$ is a subset of $F-F_P$. The other is that both $H_1 \cap F_P \ne \varnothing$ and $H_2 \cap F_P \ne \varnothing$.

The first case. Suppose $H_1 \subset F-F_P$. Then $H_1$ consists of isolated points and is an open subset of $F$. For each $x \in H_2 \cap F_P$, choose an open subset $V_x$ of $F$ such that $x \in V_x$ and $V_x$ contains no points of $F_P-\left\{ x \right\}$ and $V_x \cap \overline{H_1}=\varnothing$. For each $x \in H_2 \cap (F-F_P)$, let $V_x=\left\{x \right\}$. Let $V$ be the union of all $V_x$ where $x \in H_2$. Let $U=H_1$. Then $U$ and $V$ are disjoint open sets with $H_1 \subset U$ and $H_2 \subset V$.

The second case. Suppose $A_1=H_1 \cap F_P \ne \varnothing$ and $A_2=H_2 \cap F_P \ne \varnothing$. Let $q_1=\left\{p \in P: f_p \in A_1 \right\}$ and $q_2=\left\{p \in P: f_p \in A_2 \right\}$. Define the following open sets:

$U_1=\left\{f \in F: f(q_1)=1 \text{ and } f(q_2)=0 \right\}$
$U_2=\left\{f \in F: f(q_1)=0 \text{ and } f(q_2)=1 \right\}$

Because $H_1 \cap H_2=\varnothing$, we have $A_1 \subset U_1$ and $A_2 \subset U_2$. Furthermore, $U_1 \cap U_2=\varnothing$. Let $B_1=H_1 \cap (F-F_P)$ and $B_2=H_2 \cap (F-F_P)$, which are open since they consist of isolated points. Then $O_1=(U_1 \cup B_1)-\overline{H_2}$ and $O_2=(U_2 \cup B_2)-\overline{H_1}$ are disjoint open subsets of $F$ with $H_1 \subset O_1$ and $H_2 \subset O_2$.

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Bing’s Example G is not Perfectly Normal

A space is perfectly normal if it is normal and that every closed subset is $G_\delta$ (i.e. the intersection of countably many open subsets). The set $F_P$ of non-isolated points is a closed set in $F$. We show that $F_P$ cannot be a $G_\delta$-set. Before we do so, we need to appeal to a fact about the product space $2^Q$.

According to the Tychonoff theorem, the product space $2^Q$ is a compact space since it is a product of compact spaces. On the other hand, $2^Q$ is a product of uncountably many factors and is thus not first countable. It is a well known fact that in a compact Hausdorff space, if a point is a $G_\delta$-point, then there is a countable local base at that point (i.e. the space is first countable at that point). Thus no point of the compact product space $2^Q$ can be a $G_\delta$-point. Since points of $F_P$ retain the open sets of the product topology, no point of $F_P$ can be a $G_\delta$-point in the Bing’s Example G topology.

For each $p \in P$, let $W_p$ be open in $F$ such that $f_p \in W_p$ and $W_p$ contains no points $F_P-\left\{f_p \right\}$. For example, we can define $W_p$ as in the above section “Bing’s Example G – Initial Discussion”.

Suppose that $F_P$ is a $G_\delta$-set. Then $F_P=\bigcap \limits_{i=1}^\infty O_i$ where each $O_i$ is an open subset of $F$. Now for each $p \in P$, we have $\left\{f_p \right\}=\bigcap \limits_{i=1}^\infty (O_i \cap W_p)$, contradicting the fact that the point $f_p$ cannot be a $G_\delta$-point in the space $F$ (and in the product space $2^Q$). Thus $F_P$ is not a $G_\delta$-set in the space $F$, leading to the conclusion that Bing’s Example G is not perfectly normal.

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Bing’s Example G is not Metacompact

A space $M$ is said to have caliber $\omega_1$ if for every uncountable collection $\left\{U_\alpha: \alpha < \omega_1 \right\}$ of non-empty open subsets of $M$, there is an uncountable $A \subset \omega_1$ such that $\bigcap \left\{U_\alpha: \alpha \in A \right\} \ne \varnothing$. Any product of separable spaces has this property (see Topological Spaces with Caliber Omega 1). Thus the product space $2^Q$ has caliber $\omega_1$. Thus in the product space $2^Q$, no collection of uncountably many non-empty open sets can be a point-finite collection (in fact cannot even be point-countable).

To see that the Example G is not metacompact, let $\mathcal{W}=\left\{W_p: p \in P \right\}$ be a collection of open sets such that for $p \in P$, $f_p \in W_p$, $W_p$ is open in the product topology of $2^Q$ and $W_p$ contains no points $F_P-\left\{f_p \right\}$. For example, we can define $W_p$ as in the above section “Bing’s Example G – Initial Discussion”.

Let $W=\bigcup \mathcal{W}$. Let $\mathcal{V}=\mathcal{W} \cup \left\{\left\{ x \right\}: x \in F-W \right\}$. Any open refinement of $\mathcal{V}$ would contain uncountably many open sets in the product topology and thus cannot be point-finite. Thus the space $F$ cannot be metacompact.

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Reference

1. Bing, R. H., Metrization of Topological Spaces, Canad. J. Math., 3, 175-186, 1951.
2. Burke, D. K., A note on R. H. Bing’s example G, Top. Conf. VPI, Lectures Notes in Mathematics, 375, Springer Verlag, New York, 47-52, 1974.
3. Engelking, R., General Topology, Revised and Completed edition, Heldermann Verlag, Berlin, 1989.
4. Lewis, I. W., On covering properties of subspaces of R. H. Bing’s Example G, Gen. Topology Appl., 7, 109-122, 1977.
5. Michael, E., Point-finite and locally finite coverings, Canad. J. Math., 7, 275-279, 1955.
6. Willard, S., General Topology, Addison-Wesley Publishing Company, 1970.

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$\copyright \ \ 2012$

# Michael Line Basics

Like the Sorgenfrey line, the Michael line is a classic counterexample that is covered in standard topology textbooks and in first year topology courses. This easily accessible example helps transition students from the familiar setting of the Euclidean topology on the real line to more abstract topological spaces. One of the most famous results regarding the Michael line is that the product of the Michael line with the space of the irrational numbers is not normal. Thus it is an important example in demonstrating the pathology in products of paracompact spaces. The product of two paracompact spaces does not even have be to be normal, even when one of the factors is a complete metric space. In this post, we discuss this classical result and various other basic results of the Michael line.

Let $\mathbb{R}$ be the real number line. Let $\mathbb{P}$ be the set of all irrational numbers. Let $\mathbb{Q}=\mathbb{R}-\mathbb{P}$, the set of all rational numbers. Let $\tau$ be the usual topology of the real line $\mathbb{R}$. The following is a base that defines a topology on $\mathbb{R}$.

$\mathcal{B}=\tau \cup \left\{\left\{ x \right\}: x \in \mathbb{P}\right\}$

The real line with the topology generated by $\mathcal{B}$ is called the Michael line and is denoted by $\mathbb{M}$. In essense, in $\mathbb{M}$, points in $\mathbb{P}$ are made isolated and points in $\mathbb{Q}$ retain the usual Euclidean open sets.

The Euclidean topology $\tau$ is coarser (weaker) than the Michael line topology (i.e. $\tau$ being a subset of the Michael line topology). Thus the Michael line is Hausdorff. Since the Michael line topology contains a metrizable topology, $\mathbb{M}$ is submetrizable (submetrized by the Euclidean topology). It is clear that $\mathbb{M}$ is first countable. Having uncountably many isolated points, the Michael line does not have the countable chain condition (thus is not separable). The following points are discussed in more details.

1. The space $\mathbb{M}$ is paracompact.
2. The space $\mathbb{M}$ is not Lindelof.
3. The extent of the space $\mathbb{M}$ is $c$ where $c$ is the cardinality of the real line.
4. The space $\mathbb{M}$ is not locally compact.
5. The space $\mathbb{M}$ is not perfectly normal, thus not metrizable.
6. The space $\mathbb{M}$ is not a Moore space, but has a $G_\delta$-diagonal.
7. The product $\mathbb{M} \times \mathbb{P}$ is not normal where $\mathbb{P}$ has the usual topology.
8. The product $\mathbb{M} \times \mathbb{P}$ is metacompact.
9. The space $\mathbb{M}$ has a point-countable base.
10. For each $n=1,2,3,\cdots$, the product $\mathbb{M}^n$ is paracompact.
11. The product $\mathbb{M}^\omega$ is not normal.
12. There exist a Lindelof space $L$ and a separable metric space $W$ such that $L \times W$ is not normal.

Results 10, 11 and 12 are shown in some subsequent posts.

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Baire Category Theorem

Before discussing the Michael line in greater details, we point out one connection between the Michael line topology and the Euclidean topology on the real line. The Michael line topology on $\mathbb{Q}$ coincides with the Euclidean topology on $\mathbb{Q}$. A set is said to be a $G_\delta$-set if it is the intersection of countably many open sets. By the Baire category theorem, the set $\mathbb{Q}$ is not a $G_\delta$-set in the Euclidean real line (see the section called “Discussion of the Above Question” in the post A Question About The Rational Numbers). Thus the set $\mathbb{Q}$ is not a $G_\delta$-set in the Michael line. This fact is used in Result 5.

The fact that $\mathbb{Q}$ is not a $G_\delta$-set in the Euclidean real line implies that $\mathbb{P}$ is not an $F_\sigma$-set in the Euclidean real line. This fact is used in Result 7.

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Result 1

Let $\mathcal{U}$ be an open cover of $\mathbb{M}$. We proceed to derive a locally finite open refinement $\mathcal{V}$ of $\mathcal{U}$. Recall that $\tau$ is the usual topology on $\mathbb{R}$. Assume that $\mathcal{U}$ consists of open sets in the base $\mathcal{B}$. Let $\mathcal{U}_\tau=\mathcal{U} \cap \tau$. Let $Y=\cup \mathcal{U}_\tau$. Note that $Y$ is a Euclidean open subspace of the real line (hence it is paracompact). Then there is $\mathcal{V}_\tau \subset \tau$ such that $\mathcal{V}_\tau$ is a locally finite open refinement $\mathcal{V}_\tau$ of $\mathcal{U}_\tau$ and such that $\mathcal{V}_\tau$ covers $Y$ (locally finite in the Euclidean sense). Then add to $\mathcal{V}_\tau$ all singleton sets $\left\{ x \right\}$ where $x \in \mathbb{M}-Y$ and let $\mathcal{V}$ denote the resulting open collection.

The resulting $\mathcal{V}$ is a locally finite open collection in the Michael line $\mathbb{M}$. Furthermore, $\mathcal{V}$ is also a refinement of the original open cover $\mathcal{U}$. $\blacksquare$

A similar argument shows that $\mathbb{M}$ is hereditarily paracompact.

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Result 2

To see that $\mathbb{M}$ is not Lindelof, observe that there exist Euclidean uncountable closed sets consisting entirely of irrational numbers (i.e. points in $\mathbb{P}$). For example, it is possible to construct a Cantor set entirely within $\mathbb{P}$.

Let $C$ be an uncountable Euclidean closed set consisting entirely of irrational numbers. Then this set $C$ is an uncountable closed and discrete set in $\mathbb{M}$. In any Lindelof space, there exists no uncountable closed and discrete subset. Thus the Michael line $\mathbb{M}$ cannot be Lindelof. $\blacksquare$

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Result 3

The argument in Result 2 indicates a more general result. First, a brief discussion of the cardinal function extent. The extent of a space $X$ is the smallest infinite cardinal number $\mathcal{K}$ such that every closed and discrete set in $X$ has cardinality $\le \mathcal{K}$. The extent of the space $X$ is denoted by $e(X)$. When the cardinal number $e(X)$ is $e(X)=\aleph_0$ (the first infinite cardinal number), the space $X$ is said to have countable extent, meaning that in this space any closed and discrete set must be countably infinite or finite. When $e(X)>\aleph_0$, there are uncountable closed and discrete subsets in the space.

It is straightforward to see that if a space $X$ is Lindelof, the extent is $e(X)=\aleph_0$. However, the converse is not true.

The argument in Result 2 exhibits a closed and discrete subset of $\mathbb{M}$ of cardinality $c$. Thus we have $e(\mathbb{M})=c$. $\blacksquare$

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Result 4

The Michael line $\mathbb{M}$ is not locally compact at all rational numbers. Observe that the Michael line closure of any Euclidean open interval is not compact in $\mathbb{M}$. $\blacksquare$

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Result 5

A set is said to be a $G_\delta$-set if it is the intersection of countably many open sets. A space is perfectly normal if it is a normal space with the additional property that every closed set is a $G_\delta$-set. In the Michael line $\mathbb{M}$, the set $\mathbb{Q}$ of rational numbers is a closed set. Yet, $\mathbb{Q}$ is not a $G_\delta$-set in the Michael line (see the discussion above on the Baire category theorem). Thus $\mathbb{M}$ is not perfectly normal and hence not a metrizable space. $\blacksquare$

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Result 6

The diagonal of a space $X$ is the subset of its square $X \times X$ that is defined by $\Delta=\left\{(x,x): x \in X \right\}$. If the space is Hausdorff, the diagonal is always a closed set in the square. If $\Delta$ is a $G_\delta$-set in $X \times X$, the space $X$ is said to have a $G_\delta$-diagonal. It is well known that any metric space has $G_\delta$-diagonal. Since $\mathbb{M}$ is submetrizable (submetrized by the usual topology of the real line), it has a $G_\delta$-diagonal too.

Any Moore space has a $G_\delta$-diagonal. However, the Michael line is an example of a space with $G_\delta$-diagonal but is not a Moore space. Paracompact Moore spaces are metrizable. Thus $\mathbb{M}$ is not a Moore space. For a more detailed discussion about Moore spaces, see Sorgenfrey Line is not a Moore Space. $\blacksquare$

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Result 7

We now show that $\mathbb{M} \times \mathbb{P}$ is not normal where $\mathbb{P}$ has the usual topology. In this proof, the following two facts are crucial:

• The set $\mathbb{P}$ is not an $F_\sigma$-set in the real line.
• The set $\mathbb{P}$ is dense in the real line.

Let $H$ and $K$ be defined by the following:

$H=\left\{(x,x): x \in \mathbb{P} \right\}$
$K=\mathbb{Q} \times \mathbb{P}$.

The sets $H$ and $K$ are disjoint closed sets in $\mathbb{M} \times \mathbb{P}$. We show that they cannot be separated by disjoint open sets. To this end, let $H \subset U$ and $K \subset V$ where $U$ and $V$ are open sets in $\mathbb{M} \times \mathbb{P}$.

To make the notation easier, for the remainder of the proof of Result 7, by an open interval $(a,b)$, we mean the set of all real numbers $t$ with $a. By $(a,b)^*$, we mean $(a,b) \cap \mathbb{P}$. For each $x \in \mathbb{P}$, choose an open interval $U_x=(a,b)^*$ such that $\left\{x \right\} \times U_x \subset U$. We also assume that $x$ is the midpoint of the open interval $U_x$. For each positive integer $k$, let $P_k$ be defined by:

$P_k=\left\{x \in \mathbb{P}: \text{ length of } U_x > \frac{1}{k} \right\}$

Note that $\mathbb{P}=\bigcup \limits_{k=1}^\infty P_k$. For each $k$, let $T_k=\overline{P_k}$ (Euclidean closure in the real line). It is clear that $\bigcup \limits_{k=1}^\infty P_k \subset \bigcup \limits_{k=1}^\infty T_k$. On the other hand, $\bigcup \limits_{k=1}^\infty T_k \not\subset \bigcup \limits_{k=1}^\infty P_k=\mathbb{P}$ (otherwise $\mathbb{P}$ would be an $F_\sigma$-set in the real line). So there exists $T_n=\overline{P_n}$ such that $\overline{P_n} \not\subset \mathbb{P}$. So choose a rational number $r$ such that $r \in \overline{P_n}$.

Choose a positive integer $j$ such that $\frac{2}{j}<\frac{1}{n}$. Since $\mathbb{P}$ is dense in the real line, choose $y \in \mathbb{P}$ such that $r-\frac{1}{j}. Now we have $(r,y) \in K \subset V$. Choose another integer $m$ such that $\frac{1}{m}<\frac{1}{j}$ and $(r-\frac{1}{m},r+\frac{1}{m}) \times (y-\frac{1}{m},y+\frac{1}{m})^* \subset V$.

Since $r \in \overline{P_n}$, choose $x \in \mathbb{P}$ such that $r-\frac{1}{m}. Now it is clear that $(x,y) \in V$. The following inequalities show that $(x,y) \in U$.

$\lvert x-y \lvert \le \lvert x-r \lvert + \lvert r-y \lvert < \frac{1}{m}+\frac{1}{j} \le \frac{2}{j} < \frac{1}{n}$

The open interval $U_x$ is chosen to have length $> \frac{1}{n}$. Since $\lvert x-y \lvert < \frac{1}{n}$, $y \in U_x$. Thus $(x,y) \in \left\{ x \right\} \times U_x \subset U$. We have shown that $U \cap V \ne \varnothing$. Thus $\mathbb{M} \times \mathbb{P}$ is not normal. $\blacksquare$

Remark
As indicated above, the proof of Result 7 hinges on two facts about $\mathbb{P}$, namely that it is not an $F_\sigma$-set in the real line and it is dense in the real line. We can modify the construction of the Michael line by using other partition of the real line (where one set is isolated and its complement retains the usual topology). As long as the set $D$ that is isolated is not an $F_\sigma$-set in the real line and is dense in the real line, the same proof will show that the product of the modified Michael line and the space $D$ (with the usual topology) is not normal. This will be how Result 12 is derived.

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Result 8

The product $\mathbb{M} \times \mathbb{P}$ is not paracompact since it is not normal. However, $\mathbb{M} \times \mathbb{P}$ is metacompact.

A collection of subsets of a space $X$ is said to be point-finite if every point of $X$ belongs to only finitely many sets in the collection. A space $X$ is said to be metacompact if each open cover of $X$ has an open refinement that is a point-finite collection.

Note that $\mathbb{M} \times \mathbb{P}=(\mathbb{P} \times \mathbb{P}) \cup (\mathbb{Q} \times \mathbb{P})$. The first $\mathbb{P}$ in $\mathbb{P} \times \mathbb{P}$ is discrete (a subspace of the Michael line) and the second $\mathbb{P}$ has the Euclidean topology.

Let $\mathcal{U}$ be an open cover of $\mathbb{M} \times \mathbb{P}$. For each $a=(x,y) \in \mathbb{Q} \times \mathbb{P}$, choose $U_a \in \mathcal{U}$ such that $a \in U_a$. We can assume that $U_a=A \times B$ where $A$ is a usual open interval in $\mathbb{R}$ and $B$ is a usual open interval in $\mathbb{P}$. Let $\mathcal{G}=\lbrace{U_a:a \in \mathbb{Q} \times \mathbb{P}}\rbrace$.

Fix $x \in \mathbb{P}$. For each $b=(x,y) \in \lbrace{x}\rbrace \times \mathbb{P}$, choose some $U_b \in \mathcal{U}$ such that $b \in U_b$. We can assume that $U_b=\lbrace{x}\rbrace \times B$ where $B$ is a usual open interval in $\mathbb{P}$. Let $\mathcal{H}_x=\lbrace{U_b:b \in \lbrace{x}\rbrace \times \mathbb{P}}\rbrace$.

As a subspace of the Euclidean plane, $\bigcup \mathcal{G}$ is metacompact. So there is a point-finite open refinement $\mathcal{W}$ of $\mathcal{G}$. For each $x \in \mathbb{P}$, $\mathcal{H}_x$ has a point-finite open refinement $\mathcal{I}_x$. Let $\mathcal{V}$ be the union of $\mathcal{W}$ and all the $\mathcal{I}_x$ where $x \in \mathbb{P}$. Then $\mathcal{V}$ is a point-finite open refinement of $\mathcal{U}$.

Note that the point-finite open refinement $\mathcal{V}$ may not be locally finite. The vertical open intervals in $\lbrace{x}\rbrace \times \mathbb{P}$, $x \in \mathbb{P}$ can “converge” to a point in $\mathbb{Q} \times \mathbb{P}$. Thus, metacompactness is the best we can hope for. $\blacksquare$

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Result 9

A collection of sets is said to be point-countable if every point in the space belongs to at most countably many sets in the collection. A base $\mathcal{G}$ for a space $X$ is said to be a point-countable base if $\mathcal{G}$, in addition to being a base for the space $X$, is also a point-countable collection of sets. The Michael line is an example of a space that has a point-countable base and that is not metrizable. The following is a point-countable base for $\mathbb{M}$:

$\mathcal{G}=\mathcal{H} \cup \left\{\left\{ x \right\}: x \in \mathbb{P}\right\}$

where $\mathcal{H}$ is the set of all Euclidean open intervals with rational endpoints. One reason for the interest in point-countable base is that any countable compact space (hence any compact space) with a point-countable base is metrizable (see Metrization Theorems for Compact Spaces).

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Reference

1. Engelking, R., General Topology, Revised and Completed edition, Heldermann Verlag, Berlin, 1989.
2. Willard, S., General Topology, Addison-Wesley Publishing Company, 1970.

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$\copyright \ \ 2012$

# Sorgenfrey Line is not a Moore Space

We found an incorrect statement about the Sorgenfrey line in an entry in Wikipedia about Moore space (link). This statement opens up a discussion on the question of whether the Sorgenfrey line is a Moore space as well as a discussion on Moore space. The following is the incorrect statement found in Wikipedia by the author.

The Sorgenfrey line is the space whose underlying set is the real line $S=\mathbb{R}$ where the topology is generated by a base consisting the half open intervals of the form $[a,b)$. The Sorgenfrey plane is the square $S \times S$.

Even though the Sorgenfrey line is normal, the Sorgenfrey plane is not normal. In fact, the Sorgenfrey line is the classic example of a normal space whose square is not normal. Both the Sorgenfrey line and the Sorgenfrey plane are not Moore space but not for the reason given. The statement seems to suggest that any normal Moore space is second countable. But this flies in the face of all the profound mathematics surrounding the normal Moore space conjecture, which is also discussed in the Wikipedia entry.

The statement indicated above is only a lead-in to a discussion of Moore space. We are certain that it will be corrected. We always appreciate readers who kindly alert us to errors found in this blog.

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Moore Spaces

Let $X$ be a regular space. A development for $X$ is a sequence $\mathcal{G}_1,\mathcal{G}_2,\mathcal{G}_3,\cdots$ of open covers of $X$ such that for each $x \in X$, and for each open subset $U$ of $X$ with $x \in U$, there exists one cover $\mathcal{G}_n$ satisfying the condition that for any open set $V \in \mathcal{G}_n$, $x \in V \Rightarrow V \subset U$. When $X$ has a development, $X$ is said to be a Moore space (also called developable space). A Note On The Sorgenfrey Line is an introductory note on the Sorgenfrey line.

Moore spaces can be viewed as a generalization of metrizable spaces. Moore spaces are first countable (having a countable base at each point). For a development $\mathcal{G}_1,\mathcal{G}_2,\mathcal{G}_3,\cdots$, the open sets in $\mathcal{G}_n$ are considered “smaller” as the index $n$ increases. In fact, this is how a development is defined for a metric space, where $\mathcal{G}_n$ consists of all open balls with diameters less than $\frac{1}{n}$. Thus metric spaces are developable. There are plenty of non-metrizable Moore space. One example is the Niemytzki’s Tangent Disc space.

In a Moore space, every closed set is a $G_\delta$-set. Thus if a Moore space is normal, it is perfectly normal. Any Moore space has a $G_\delta$-diagonal (the diagonal $\Delta=\left\{(x,x): x \in X \right\}$ is a $G_\delta$-set in $X \times X$). It is a well known theorem that every compact space with a $G_\delta$-diagonal is metrizable. Thus any compact Moore space is metrizable.

The last statement can be shown more directly. Suppose that $X$ is compact and has a development $\mathcal{G}_1,\mathcal{G}_2,\mathcal{G}_3,\cdots$. Then each $\mathcal{G}_n$ has a finite subcover $\mathcal{H}_n$. Then $\bigcup_{n=1}^\infty \mathcal{H}_n$ is a countable base for $X$. Thus any compact Moore space is second countable and hence metrizable.

What about paracompact Moore space? Suppose that $X$ is paracompact and has a development $\mathcal{G}_1,\mathcal{G}_2,\mathcal{G}_3,\cdots$. Then each $\mathcal{G}_n$ has a locally finite open refinement $\mathcal{H}_n$. Then $\bigcup_{n=1}^\infty \mathcal{H}_n$ is a $\sigma$-locally finite base for $X$. The Smirnov-Nagata metrization theorem states that a space is metrizable if and only if it has a $\sigma$-locally finite base (see Theorem 23.9 on page 170 of [2]). Thus any paracompact Moore space has a $\sigma$-locally finite base and is thus metrizable (after using the big gun of the Smirnov-Nagata metrization theorem).

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Sorgenfrey Line

The Sorgenfrey line is regular and Lindelof. Hence it is paracompact. Since the Sorgenfrey line is not metrizable, by the above discussion it cannot be a Moore space. The Sorgenfrey plane is also not a Moore space. Note that being a Moore space is a hereditary property. So if the Sorgenfrey plane is a Moore space, then every subspace of the Sorgenfrey plane (including the Sorgenfrey line) is a Moore space.

The following theorem is another way to show that the Sorgenfrey line is not a Moore space.

Bing’s Metrization Theorem
A topological space is metrizable if and only if it is a collectionwise normal Moore space.

Every paracompact space is collectionwise normal (see Theorem 5.1.18, p.305 of [1]). Thus the Sorgenfrey line is collectionwise normal and hence cannot be a Moore space. A space $X$ is said to be collectionwise normal if $X$ is a $T_1$-space and for every discrete collection $\left\{W_\alpha: \alpha \in A \right\}$ of closed sets in $X$, there exists a discrete collection $\left\{V_\alpha: \alpha \in A \right\}$ of open subsets of $X$ such that $W_\alpha \subset V_\alpha$. For a proof of Bing’s metrization theorem, see page 329 of [1].

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Remark

The normal Moore space conjecture is the statement that every normal Moore space is metrizable. This conjecture had been one of the key motivating questions for many set theorists and topologists during a large part of the twentieth century. The bottom line is that this statement cannot not be decided just on the basis of the set of generally accepted axioms called Zermelo–Fraenkel set theory with the axiom of choice, commonly abbreviated ZFC. But Bing’s metrization theorem states that if we strengthen normality to collectionwise normality, we have a definite answer.

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Reference

1. Engelking, R., General Topology, Revised and Completed edition, Heldermann Verlag, Berlin, 1989.
2. Willard, S., General Topology, Addison-Wesley Publishing Company, 1970.

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$\copyright \ \ 2012$