The product of a normal countably compact space and a metric space is normal

Posted on February 5, 2017 by Dan Ma

It is well known that normality is not preserved by taking products. When nothing is known about the spaces $X$ and $Y$ other than the facts that they are normal spaces, there is not enough to go on for determining whether $X \times Y$ is normal. In fact even when one factor is a metric space and the other factor is a hereditarily paracompact space, the product can be non-normal (discussed here). This post discusses a productive scenario – the first factor is a normal space and second factor is a metric space with the first factor having the additional property that it is countably compact. In this scenario the product is always normal. This is a well known result in general topology. The goal here is to nail down a proof for use as future reference.

Main Theorem
Let $X$ be a normal and countably compact space. Then $X \times Y$ is a normal space for every metric space $Y$ .

The proof of the main theorem uses the notion of shrinkable open covers.

Remarks
The main theorem is a classic result and is often used as motivation for more advanced results for products of normal spaces. Thus we would like to present a clear and complete proof of this classic result for anyone who would like to study the topics of normality (or the lack of) in product spaces. We found that some proofs of this result in the literature are hard to follow. In A. H. Stone’s paper [2], the result is stated in a footnote, stating that “it can be shown that the topological product of a metric space and a normal countably compact space is normal, though not necessarily paracompact”. We had seen several other papers citing [2] as a reference for the result. The Handbook [1] also has a proof (Corollary 4.10 in page 805), which we feel may not be the best proof to learn from. We found a good proof in [3] using the idea of shrinking of open covers.

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The Notion of Shrinking

The key to the proof is the notion of shrinkable open covers and shrinking spaces. Let $X$ be a space. Let $\mathcal{U}$ be an open cover of $X$ . The open cover of $\mathcal{U}$ is said to be shrinkable if there is an open cover $\mathcal{V}=\left\{V(U): U \in \mathcal{U} \right\}$ of $X$ such that $\overline{V(U)} \subset U$ for each $U \in \mathcal{U}$ . When this is the case, the open cover $\mathcal{V}$ is said to be a shrinking of $\mathcal{U}$ . If an open cover is shrinkable, we also say that the open cover can be shrunk (or has a shrinking). Whenever an open cover has a shrinking, the shrinking is indexed by the open cover that is being shrunk. Thus if the original cover is indexed, e.g. $\left\{U_\alpha: \alpha<\kappa \right\}$ , then a shrinking has the same indexing, e.g. $\left\{V_\alpha: \alpha<\kappa \right\}$ .

A space $X$ is a shrinking space if every open cover of $X$ is shrinkable. Every open cover of a paracompact space has a locally finite open refinement. With a little bit of rearranging, the locally finite open refinement can be made to be a shrinking (see Theorem 2 here). Thus every paracompact space is a shrinking space. For other spaces, the shrinking phenomenon is limited to certain types of open covers. In a normal space, every finite open cover has a shrinking, as stated in the following theorem.

Theorem 1
The following conditions are equivalent.

The space $X$ is normal.
Every point-finite open cover of $X$ is shrinkable.
Every locally finite open cover of $X$ is shrinkable.
Every finite open cover of $X$ is shrinkable.
Every two-element open cover of $X$ is shrinkable.

The hardest direction in the proof is $1 \Longrightarrow 2$ , which is established in this previous post. The directions $2 \Longrightarrow 3 \Longrightarrow 4 \Longrightarrow 5$ are immediate. To see $5 \Longrightarrow 1$ , let $H$ and $K$ be two disjoint closed subsets of $X$ . By condition 5, the two-element open cover $\left\{X-H,X-K \right\}$ has a shrinking $\left\{U,V \right\}$ . Then $\overline{U} \subset X-H$ and $\overline{V} \subset X-K$ . As a result, $H \subset X-\overline{U}$ and $K \subset X-\overline{V}$ . Since the open sets $U$ and $V$ cover the whole space, $X-\overline{U}$ and $X-\overline{V}$ are disjoint open sets. Thus $X$ is normal.

In a normal space, all finite open covers are shrinkable. In general, an infinite open cover of a normal space may or may not be shrinkable. It turns out that finding a normal space with an infinite open cover that is not shrinkable is no trivial matter (see Dowker’s theorem in this previous post). However, if an open cover in a normal space is point-finite or locally finite, then it is shrinkable.

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Key Idea

We now discuss the key idea to the proof of the main theorem. Consider the product space $X \times Y$ . Let $\mathcal{U}$ be an open cover of $X \times Y$ . Let $M \subset Y$ . The set $M$ is stable with respect to the open cover $\mathcal{U}$ if for each $x \in X$ , there is an open set $O_x$ containing $x$ such that $O_x \times M \subset U$ for some $U \in \mathcal{U}$ .

Let $\kappa$ be a cardinal number (either finite or infinite). A space $X$ is a $\kappa$ -shrinking space if for each open cover $\mathcal{W}$ of $X$ such that the cardinality of $\mathcal{W}$ is $\le \kappa$ , then $\mathcal{W}$ is shrinkable. According to Theorem 1, any normal space is 2-shrinkable.

Theorem 2
Let $\kappa$ be a cardinal number (either finite or infinite). Let $X$ be a $\kappa$ -shrinking space. Let $Y$ be a paracompact space. Suppose that $\mathcal{U}$ is an open cover of $X \times Y$ such that the following two conditions are satisfied:

Each point $y \in Y$ has an open set $V_y$ containing $y$ such that $V_y$ is stable with respect to $\mathcal{U}$ .
$\lvert \mathcal{U} \lvert = \kappa$ .

Then $\mathcal{U}$ is shrinkable.

Proof of Theorem 2
Let $\mathcal{U}$ be any open cover of $X \times Y$ satisfying the hypothesis. We show that $\mathcal{U}$ has a shrinking.

For each $y \in Y$ , obtain the open covers $\left\{G(U,y): U \in \mathcal{U} \right\}$ and $\left\{H(U,y): U \in \mathcal{U} \right\}$ of $X$ as follows. For each $U \in \mathcal{U}$ , define the following:

$G(U,y)=\cup \left\{O: O \text{ is open in } X \text{ such that } O \times V_y \subset U \right\}$

Then $\left\{G(U,y): U \in \mathcal{U} \right\}$ is an open cover of $X$ . Since $X$ is $\kappa$ -shrinkable, there is an open cover $\left\{H(U,y): U \in \mathcal{U} \right\}$ of $X$ such that $\overline{H(U,y)} \subset G(U,y)$ for each $U \in \mathcal{U}$ .

Now $\left\{V_y: y \in Y \right\}$ is an open cover of $Y$ . By the paracompactness of $Y$ , let $\left\{W_y: y \in Y \right\}$ be a locally finite open cover of $Y$ such that $\overline{W_y} \subset V_y$ for each $y \in Y$ . For each $U \in \mathcal{U}$ , define the following:

$W_U=\cup \left\{H(U,y) \times W_y: y \in Y \text{ such that } \overline{H(U,y) \times W_y} \subset U \right\}$

We claim that $\mathcal{W}=\left\{ W_U: U \in \mathcal{U} \right\}$ is a shrinking of $\mathcal{U}$ . First it is a cover of $X \times Y$ . Let $(x,t) \in X \times Y$ . Then $t \in W_y$ for some $y \in Y$ . There exists $U \in \mathcal{U}$ such that $x \in H(U,y)$ . Note the following.

$\overline{H(U,y) \times W_y} \subset \overline{H(U,y)} \times \overline{W_y} \subset G(U,y) \times V_y \subset U$

This means that $H(U,y) \times W_y \subset W_U$ . Since $(x,t) \in H(U,y) \times W_y$ , $(x,t) \in W_U$ . Thus $\mathcal{W}$ is an open cover of $X \times Y$ .

Now we show that $\mathcal{W}$ is a shrinking of $\mathcal{U}$ . Let $U \in \mathcal{U}$ . To show that $\overline{W_U} \subset U$ , let $(x,t) \in \overline{W_U}$ . Let $L$ be open in $Y$ such that $t \in L$ and that $L$ meets only finitely many $W_y$ , say for $y=y_1,y_2,\cdots,y_n$ . Immediately we have the following relations.

$\forall \ i=1,\cdots,n, \ \overline{W_{y_i}} \subset V_{y_i}$

$\forall \ i=1,\cdots,n, \ \overline{H(U,y_i)} \subset G(U,y_i)$

$\forall \ i=1,\cdots,n, \ \overline{H(U,y_i) \times W_{y_i}} \subset \overline{H(U,y_i)} \times \overline{W_{y_i}} \subset G(U,y_i) \times V_{y_i} \subset U$

Then it follows that

$\displaystyle (x,t) \in \overline{\bigcup \limits_{j=1}^n H(U,y_j) \times W_{y_j}}=\bigcup \limits_{j=1}^n \overline{H(U,y_j) \times W_{y_j}} \subset U$

Thus $U \in \mathcal{U}$ . This shows that $\mathcal{W}$ is a shrinking of $\mathcal{U}$ . $\square$

Remark
Theorem 2 is the Theorem 3.2 in [3]. Theorem 2 is a formulation of Theorem 3.2 [3] for the purpose of proving Theorem 3 below.

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Main Theorem

Theorem 3 (Main Theorem)
Let $X$ be a normal and countably compact space. Let $Y$ be a metric space. Then $X \times Y$ is a normal space.

Proof of Theorem 3
Let $\mathcal{U}$ be a 2-element open cover of $X \times Y$ . We show that $\mathcal{U}$ is shrinkable. This would mean that $X \times Y$ is normal (according to Theorem 1). To show that $\mathcal{U}$ is shrinkable, we show that the open cover $\mathcal{U}$ satisfies the two bullet points in Theorem 2.

Fix $y \in Y$ . Let $\left\{B_n: n=1,2,3,\cdots \right\}$ be a base at the point $y$ . Define $G_n$ as follows:

$G_n=\cup \left\{O \subset X: O \text{ is open such that } O \times B_n \subset U \text{ for some } U \in \mathcal{U} \right\}$

It is clear that $\mathcal{G}=\left\{G_n: n=1,2,3,\cdots \right\}$ is an open cover of $X$ . Since $X$ is countably compact, choose $m$ such that $\left\{G_1,G_2,\cdots,G_m \right\}$ is a cover of $X$ . Let $E_y=\bigcap_{j=1}^m B_j$ . We claim that $E_y$ is stable with respect to $\mathcal{U}$ . To see this, let $x \in X$ . Then $x \in G_j$ for some $j \le m$ . By the definition of $G_j$ , there is some open set $O_x \subset X$ such that $x \in O_x$ and $O_x \times B_j \subset U$ for some $U \in \mathcal{U}$ . Furthermore, $O_x \times E_y \subset O_x \times B_j \subset U$ .

To summarize: for each $y \in Y$ , there is an open set $E_y$ such that $y \in E_y$ and $E_y$ is stable with respect to the open cover $\mathcal{U}$ . Thus the first bullet point of Theorem 2 is satisfied. The open cover $\mathcal{U}$ is a 2-element open cover. Thus the second bullet point of Theorem 2 is satisfied. By Theorem 2, the open cover $\mathcal{U}$ is shrinkable. Thus $X \times Y$ is normal. $\square$

Corollary 4
Let $X$ be a normal and pseudocompact space. Let $Y$ be a metric space. Then $X \times Y$ is a normal space.

The corollary follows from the fact that any normal and pseudocompact space is countably compact (see here).

Remarks
The proof of Theorem 3 actually gives a more general result. Note that the second factor only needs to be paracompact and that every point has a countable base (i.e. first countable). The first factor $X$ has to be countably compact. The shrinking requirement for $X$ is flexible – if open covers of a certain size for $X$ are shrinkable, then open covers of that size for the product are shrinkable. We have the following corollaries.

Corollary 5
Let $X$ be a $\kappa$ -shrinking and countably compact space and let $Y$ be a paracompact first countable space. Then $X \times Y$ is a $\kappa$ -shrinking space.

Corollary 6
Let $X$ be a shrinking and countably compact space and let $Y$ be a paracompact first countable space. Then $X \times Y$ is a shrinking space.

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Remarks

The main theorem (Theorem 3) says that any normal and countably compact space is productively normal with one class of spaces, namely the metric spaces. Thus if one wishes to find a non-normal product space with one factor being countably compact, the other factor must not be a metric space. For example, if $W=\omega_1$ , the first uncountable ordinal with the ordered topology, then $W \times X$ is always normal for every metric $X$ . For non-normal example, $W \times C$ is not normal for any compact space $C$ with uncountable tightness (see Theorem 1 in this previous post). Another example, $W \times L_{\omega_1}$ is not normal where $L_{\omega_1}$ is the one-point Lindelofication of a discrete space of cardinality $\omega_1$ (follows from Example 1 and Theorem 7 in this previous post).

Another comment is that normal countably paracompact spaces are examples of Normal P-spaces. K. Morita defined the notion of P-space and he proved that a space $Y$ is a Normal P-space if and only if $X \times Y$ is normal for every metric space $X$ .

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Reference

Przymusinski T. C., Products of Normal Spaces, Handbook of Set-Theoretic Topology (K. Kunen and J. E. Vaughan, eds), Elsevier Science Publishers B. V., Amsterdam, 781-826, 1984.
Stone A. H., Paracompactness and Product Spaces, Bull. Amer. Math. Soc., Vol. 54, 977-982, 1948. (paper)
Yang L., The Normality in Products with a Countably Compact Factor, Canad. Math. Bull., Vol. 41 (2), 245-251, 1998. (abstract, paper)

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

Countably paracompact spaces

Posted on December 8, 2016 by Dan Ma

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.

The space $X$ is countably paracompact.
Every countable open cover of $X$ has a point-finite open refinement.
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 $n$ .
The product space $X \times Y$ is normal for any compact metric space $Y$ .
The product space $X \times [0,1]$ is normal where $[0,1]$ is the closed unit interval with the usual Euclidean topology.
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 $n<m$ . 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

Ball, B. J., Countable Paracompactness in Linearly Ordered Spaces, Proc. Amer. Math. Soc., 5, 190-192, 1954. (link)
Rudin, M. E., A Normal Space $X$ for which $X \times I$ is not Normal, Fund. Math., 73, 179-486, 1971. (link)
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.
Wikipedia Entry on Dowker Spaces (link)

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

Normality in the powers of countably compact spaces

Posted on May 17, 2015 by Dan Ma

Let $\omega_1$ be the first uncountable ordinal. The topology on $\omega_1$ we are interested in is the ordered topology, the topology induced by the well ordering. The space $\omega_1$ is also called the space of all countable ordinals since it consists of all ordinals that are countable in cardinality. It is a handy example of a countably compact space that is not compact. In this post, we consider normality in the powers of $\omega_1$ . We also make comments on normality in the powers of a countably compact non-compact space.

Let $\omega$ be the first infinite ordinal. It is well known that $\omega^{\omega_1}$ , the product space of $\omega_1$ many copies of $\omega$ , is not normal (a proof can be found in this earlier post). This means that any product space $\prod_{\alpha<\kappa} X_\alpha$ , with uncountably many factors, is not normal as long as each factor $X_\alpha$ contains a countable discrete space as a closed subspace. Thus in order to discuss normality in the product space $\prod_{\alpha<\kappa} X_\alpha$ , the interesting case is when each factor is infinite but contains no countable closed discrete subspace (i.e. no closed copies of $\omega$ ). In other words, the interesting case is that each factor $X_\alpha$ is a countably compact space that is not compact (see this earlier post for a discussion of countably compactness). In particular, we would like to discuss normality in $X^{\kappa}$ where $X$ is a countably non-compact space. In this post we start with the space $X=\omega_1$ of the countable ordinals. We examine $\omega_1$ power $\omega_1^{\omega_1}$ as well as the countable power $\omega_1^{\omega}$ . The former is not normal while the latter is normal. The proof that $\omega_1^{\omega}$ is normal is an application of the normality of $\Sigma$ -product of the real line.

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The uncountable product

Theorem 1
The product space $\prod_{\alpha<\omega_1} \omega_1=\omega_1^{\omega_1}$ is not normal.

Theorem 1 follows from Theorem 2 below. For any space $X$ , a collection $\mathcal{C}$ of subsets of $X$ is said to have the finite intersection property if for any finite $\mathcal{F} \subset \mathcal{C}$ , the intersection $\cap \mathcal{F} \ne \varnothing$ . Such a collection $\mathcal{C}$ is called an f.i.p collection for short. It is well known that a space $X$ is compact if and only collection $\mathcal{C}$ of closed subsets of $X$ satisfying the finite intersection property has non-empty intersection (see Theorem 1 in this earlier post). Thus any non-compact space has an f.i.p. collection of closed sets that have empty intersection.

In the space $X=\omega_1$ , there is an f.i.p. collection of cardinality $\omega_1$ using its linear order. For each $\alpha<\omega_1$ , let $C_\alpha=\left\{\beta<\omega_1: \alpha \le \beta \right\}$ . Let $\mathcal{C}=\left\{C_\alpha: \alpha < \omega_1 \right\}$ . It is a collection of closed subsets of $X=\omega_1$ . It is an f.i.p. collection and has empty intersection. It turns out that for any countably compact space $X$ with an f.i.p. collection of cardinality $\omega_1$ that has empty intersection, the product space $X^{\omega_1}$ is not normal.

Theorem 2
Let $X$ be a countably compact space. Suppose that there exists a collection $\mathcal{C}=\left\{C_\alpha: \alpha < \omega_1 \right\}$ of closed subsets of $X$ such that $\mathcal{C}$ has the finite intersection property and that $\mathcal{C}$ has empty intersection. Then the product space $X^{\omega_1}$ is not normal.

Proof of Theorem 2
Let’s set up some notations on product space that will make the argument easier to follow. By a standard basic open set in the product space $X^{\omega_1}=\prod_{\alpha<\omega_1} X$ , we mean a set of the form $O=\prod_{\alpha<\omega_1} O_\alpha$ such that each $O_\alpha$ is an open subset of $X$ and that $O_\alpha=X$ for all but finitely many $\alpha<\omega_1$ . Given a standard basic open set $O=\prod_{\alpha<\omega_1} O_\alpha$ , the notation $\text{Supp}(O)$ refers to the finite set of $\alpha$ for which $O_\alpha \ne X$ . For any set $M \subset \omega_1$ , the notation $\pi_M$ refers to the projection map from $\prod_{\alpha<\omega_1} X$ to the subproduct $\prod_{\alpha \in M} X$ . Each element $d \in X^{\omega_1}$ can be considered a function $d: \omega_1 \rightarrow X$ . By $(d)_\alpha$ , we mean $(d)_\alpha=d(\alpha)$ .

For each $t \in X$ , let $f_t: \omega_1 \rightarrow X$ be the constant function whose constant value is $t$ . Consider the following subspaces of $X^{\omega_1}$ .

$H=\prod_{\alpha<\omega_1} C_\alpha$

$\displaystyle K=\left\{f_t: t \in X \right\}$

Both $H$ and $K$ are closed subsets of the product space $X^{\omega_1}$ . Because the collection $\mathcal{C}$ has empty intersection, $H \cap K=\varnothing$ . We show that $H$ and $K$ cannot be separated by disjoint open sets. To this end, let $U$ and $V$ be open subsets of $X^{\omega_1}$ such that $H \subset U$ and $K \subset V$ .

Let $d_1 \in H$ . Choose a standard basic open set $O_1$ such that $d_1 \in O_1 \subset U$ . Let $S_1=\text{Supp}(O_1)$ . Since $S_1$ is the support of $O_1$ , it follows that $\pi_{S_1}^{-1}(\pi_{S_1}(d_1)) \subset O_1 \subset U$ . Since $\mathcal{C}$ has the finite intersection property, there exists $a_1 \in \bigcap_{\alpha \in S_1} C_\alpha$ .

Define $d_2 \in H$ such that $(d_2)_\alpha=a_1$ for all $\alpha \in S_1$ and $(d_2)_\alpha=(d_1)_\alpha$ for all $\alpha \in \omega_1-S_1$ . Choose a standard basic open set $O_2$ such that $d_2 \in O_2 \subset U$ . Let $S_2=\text{Supp}(O_2)$ . It is possible to ensure that $S_1 \subset S_2$ by making more factors of $O_2$ different from $X$ . We have $\pi_{S_2}^{-1}(\pi_{S_2}(d_2)) \subset O_2 \subset U$ . Since $\mathcal{C}$ has the finite intersection property, there exists $a_2 \in \bigcap_{\alpha \in S_2} C_\alpha$ .

Now choose a point $d_3 \in H$ such that $(d_3)_\alpha=a_2$ for all $\alpha \in S_2$ and $(d_3)_\alpha=(d_2)_\alpha$ for all $\alpha \in \omega_1-S_2$ . Continue on with this inductive process. When the inductive process is completed, we have the following sequences:

a sequence $d_1,d_2,d_3,\cdots$ of point of $H=\prod_{\alpha<\omega_1} C_\alpha$ ,
a sequence $S_1 \subset S_2 \subset S_3 \subset \cdots$ of finite subsets of $\omega_1$ ,
a sequence $a_1,a_2,a_3,\cdots$ of points of $X$

such that for all $n \ge 2$ , $(d_n)_\alpha=a_{n-1}$ for all $\alpha \in S_{n-1}$ and $\pi_{S_n}^{-1}(\pi_{S_n}(d_n)) \subset U$ . Let $A=\left\{a_1,a_2,a_3,\cdots \right\}$ . Either $A$ is finite or $A$ is infinite. Let’s examine the two cases.

Case 1
Suppose that $A$ is infinite. Since $X$ is countably compact, $A$ has a limit point $a$ . That means that every open set containing $a$ contains some $a_n \ne a$ . For each $n \ge 2$ , define $y_n \in \prod_{\alpha< \omega_1} X$ such that

$(y_n)_\alpha=(d_n)_\alpha=a_{n-1}$ for all $\alpha \in S_n$ ,
$(y_n)_\alpha=a$ for all $\alpha \in \omega_1-S_n$

From the induction step, we have $y_n \in \pi_{S_n}^{-1}(\pi_{S_n}(d_n)) \subset U$ for all $n$ . Let $t=f_a \in K$ , the constant function whose constant value is $a$ . It follows that $t$ is a limit of $\left\{y_1,y_2,y_3,\cdots \right\}$ . This means that $t \in \overline{U}$ . Since $t \in K \subset V$ , $U \cap V \ne \varnothing$ .

Case 2
Suppose that $A$ is finite. Then there is some $m$ such that $a_m=a_j$ for all $j \ge m$ . For each $n \ge 2$ , define $y_n \in \prod_{\alpha< \omega_1} X$ such that

$(y_n)_\alpha=(d_n)_\alpha=a_{n-1}$ for all $\alpha \in S_n$ ,
$(y_n)_\alpha=a_m$ for all $\alpha \in \omega_1-S_n$

As in Case 1, we have $y_n \in \pi_{S_n}^{-1}(\pi_{S_n}(d_n)) \subset U$ for all $n$ . Let $t=f_{a_m} \in K$ , the constant function whose constant value is $a_m$ . It follows that $t=y_n$ for all $n \ge m+1$ . Thus $U \cap V \ne \varnothing$ .

Both cases show that $U \cap V \ne \varnothing$ . This completes the proof the product space $X^{\omega_1}$ is not normal. $\blacksquare$

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The countable product

Theorem 3
The product space $\prod_{\alpha<\omega} \omega_1=\omega_1^{\omega}$ is normal.

Proof of Theorem 3
The proof here actually proves more than normality. It shows that $\prod_{\alpha<\omega} \omega_1=\omega_1^{\omega}$ is collectionwise normal, which is stronger than normality. The proof makes use of the $\Sigma$ -product of $\kappa$ many copies of $\mathbb{R}$ , which is the following subspace of the product space $\mathbb{R}^{\kappa}$ .

$\Sigma(\kappa)=\left\{x \in \mathbb{R}^{\kappa}: x(\alpha) \ne 0 \text{ for at most countably many } \alpha<\kappa \right\}$

It is well known that $\Sigma(\kappa)$ is collectionwise normal (see this earlier post). We show that $\prod_{\alpha<\omega} \omega_1=\omega_1^{\omega}$ is a closed subspace of $\Sigma(\kappa)$ where $\kappa=\omega_1$ . Thus $\omega_1^{\omega}$ is collectionwise normal. This is established in the following claims.

Claim 1
We show that the space $\omega_1$ is embedded as a closed subspace of $\Sigma(\omega_1)$ .

For each $\beta<\omega_1$ , define $f_\beta:\omega_1 \rightarrow \mathbb{R}$ such that $f_\beta(\gamma)=1$ for all $\gamma<\beta$ and $f_\beta(\gamma)=0$ for all $\beta \le \gamma <\omega_1$ . Let $W=\left\{f_\beta: \beta<\omega_1 \right\}$ . We show that $W$ is a closed subset of $\Sigma(\omega_1)$ and $W$ is homeomorphic to $\omega_1$ according to the mapping $f_\beta \rightarrow W$ .

First, we show $W$ is closed by showing that $\Sigma(\omega_1)-W$ is open. Let $y \in \Sigma(\omega_1)-W$ . We show that there is an open set containing $y$ that contains no points of $W$ .

Suppose that for some $\gamma<\omega_1$ , $y_\gamma \in O=\mathbb{R}-\left\{0,1 \right\}$ . Consider the open set $Q=(\prod_{\alpha<\omega_1} Q_\alpha) \cap \Sigma(\omega_1)$ where $Q_\alpha=\mathbb{R}$ except that $Q_\gamma=O$ . Then $y \in Q$ and $Q \cap W=\varnothing$ .

So we can assume that for all $\gamma<\omega_1$ , $y_\gamma \in \left\{0, 1 \right\}$ . There must be some $\theta$ such that $y_\theta=1$ . Otherwise, $y=f_0 \in W$ . Since $y \ne f_\theta$ , there must be some $\delta<\gamma$ such that $y_\delta=0$ . Now choose the open interval $T_\theta=(0.9,1.1)$ and the open interval $T_\delta=(-0.1,0.1)$ . Consider the open set $M=(\prod_{\alpha<\omega_1} M_\alpha) \cap \Sigma(\omega_1)$ such that $M_\alpha=\mathbb{R}$ except for $M_\theta=T_\theta$ and $M_\delta=T_\delta$ . Then $y \in M$ and $M \cap W=\varnothing$ . We have just established that $W$ is closed in $\Sigma(\omega_1)$ .

Consider the mapping $f_\beta \rightarrow W$ . Based on how it is defined, it is straightforward to show that it is a homeomorphism between $\omega_1$ and $W$ .

Claim 2
The $\Sigma$ -product $\Sigma(\omega_1)$ has the interesting property it is homeomorphic to its countable power, i.e.

$\Sigma(\omega_1) \cong \Sigma(\omega_1) \times \Sigma(\omega_1) \times \Sigma(\omega_1) \cdots \ \ \ \ \ \ \ \ \ \ \ \text{(countably many times)}$

Because each element of $\Sigma(\omega_1)$ is nonzero only at countably many coordinates, concatenating countably many elements of $\Sigma(\omega_1)$ produces an element of $\Sigma(\omega_1)$ . Thus Claim 2 can be easily verified. With above claims, we can see that

$\displaystyle \omega_1^{\omega}=\omega_1 \times \omega_1 \times \omega_1 \times \cdots \subset \Sigma(\omega_1) \times \Sigma(\omega_1) \times \Sigma(\omega_1) \cdots \cong \Sigma(\omega_1)$

Thus $\omega_1^{\omega}$ is a closed subspace of $\Sigma(\omega_1)$ . Any closed subspace of a collectionwise normal space is collectionwise normal. We have established that $\omega_1^{\omega}$ is normal. $\blacksquare$

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The normality in the powers of $X$

We have established that $\prod_{\alpha<\omega_1} \omega_1=\omega_1^{\omega_1}$ is not normal. Hence any higher uncountable power of $\omega_1$ is not normal. We have also established that $\prod_{\alpha<\omega} \omega_1=\omega_1^{\omega}$ , the countable power of $\omega_1$ is normal (in fact collectionwise normal). Hence any finite power of $\omega_1$ is normal. However $\omega_1^{\omega}$ is not hereditarily normal. One of the exercises below is to show that $\omega_1 \times \omega_1$ is not hereditarily normal.

Theorem 2 can be generalized as follows:

Theorem 4
Let $X$ be a countably compact space has an f.i.p. collection $\mathcal{C}$ of closed sets such that $\bigcap \mathcal{C}=\varnothing$ . Then $X^{\kappa}$ is not normal where $\kappa=\lvert \mathcal{C} \lvert$ .

The proof of Theorem 2 would go exactly like that of Theorem 2. Consider the following two theorems.

Theorem 5
Let $X$ be a countably compact space that is not compact. Then there exists a cardinal number $\kappa$ such that $X^{\kappa}$ is not normal and $X^{\tau}$ is normal for all cardinal number $\tau<\kappa$ .

By the non-compactness of $X$ , there exists an f.i.p. collection $\mathcal{C}$ of closed subsets of $X$ such that $\bigcap \mathcal{C}=\varnothing$ . Let $\kappa$ be the least cardinality of such an f.i.p. collection. By Theorem 4, that $X^{\kappa}$ is not normal. Because $\kappa$ is least, any smaller power of $X$ must be normal.

Theorem 6
Let $X$ be a space that is not countably compact. Then $X^{\kappa}$ is not normal for any cardinal number $\kappa \ge \omega_1$ .

Since the space $X$ in Theorem 6 is not countably compact, it would contain a closed and discrete subspace that is countable. By a theorem of A. H. Stone, $\omega^{\omega_1}$ is not normal. Then $\omega^{\omega_1}$ is a closed subspace of $X^{\omega_1}$ .

Thus between Theorem 5 and Theorem 6, we can say that for any non-compact space $X$ , $X^{\kappa}$ is not normal for some cardinal number $\kappa$ . The $\kappa$ from either Theorem 5 or Theorem 6 is at least $\omega_1$ . Interestingly for some spaces, the $\kappa$ can be much smaller. For example, for the Sorgenfrey line, $\kappa=2$ . For some spaces (e.g. the Michael line), $\kappa=\omega$ .

Theorems 4, 5 and 6 are related to a theorem that is due to Noble.

Theorem 7 (Noble)
If each power of a space $X$ is normal, then $X$ is compact.

A proof of Noble’s theorem is given in this earlier post, the proof of which is very similar to the proof of Theorem 2 given above. So the above discussion the normality of powers of $X$ is just another way of discussing Theorem 7. According to Theorem 7, if $X$ is not compact, some power of $X$ is not normal.

The material discussed in this post is excellent training ground for topology. Regarding powers of countably compact space and product of countably compact spaces, there are many topics for further discussion/investigation. One possibility is to examine normality in $X^{\kappa}$ for more examples of countably compact non-compact $X$ . One particular interesting example would be a countably compact non-compact $X$ such that the least power $\kappa$ for non-normality in $X^{\kappa}$ is more than $\omega_1$ . A possible candidate could be the second uncountable ordinal $\omega_2$ . By Theorem 2, $\omega_2^{\omega_2}$ is not normal. The issue is whether the $\omega_1$ power $\omega_2^{\omega_1}$ and countable power $\omega_2^{\omega}$ are normal.

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Exercises

Exercise 1
Show that $\omega_1 \times \omega_1$ is not hereditarily normal.

Exercise 2
Show that the mapping $f_\beta \rightarrow W$ in Claim 3 in the proof of Theorem 3 is a homeomorphism.

Exercise 3
The proof of Theorem 3 shows that the space $\omega_1$ is a closed subspace of the $\Sigma$ -product of the real line. Show that $\omega_1$ can be embedded in the $\Sigma$ -product of arbitrary spaces.

For each $\alpha<\omega_1$ , let $X_\alpha$ be a space with at least two points. Let $p \in \prod_{\alpha<\omega_1} X_\alpha$ . The $\Sigma$ -product of the spaces $X_\alpha$ is the following subspace of the product space $\prod_{\alpha<\omega_1} X_\alpha$ .

$\Sigma(X_\alpha)=\left\{x \in \prod_{\alpha<\omega_1} X_\alpha: x(\alpha) \ne p(\alpha) \ \text{for at most countably many } \alpha<\omega_1 \right\}$

The point $p$ is the center of the $\Sigma$ -product. Show that the space $\Sigma(X_\alpha)$ contains $\omega_1$ as a closed subspace.

Exercise 4
Find a direct proof of Theorem 3, that $\omega_1^{\omega}$ is normal.

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

One way to find collectionwise normal spaces

Posted on December 5, 2012 by Dan Ma

Collectionwise normality is a property that is weaker than paracompactness and stronger than normality (see the implications below). Normal spaces need not be collectionwise normal. Bing’s Example G is an example of a normal and not collectionwise normal space (see the blog post “Bing’s Example G”). We discuss one instance when normal spaces are collectionwise normal, giving a way to obtain collectionwise normal spaces that are not paracompact.

$\text{ }$

$\text{paracompact} \Longrightarrow \text{collectionwise normal} \Longrightarrow \text{normal}$

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Collectionwise Normal Spaces

A normal space is one in which any two disjoint closed sets can be separated by disjoint open sets. By induction, in a normal space any finite number of disjoint closed sets can be separated by disjoint open sets. Of course, the inductive reasoning cannot be carried over to the case of infinitely many disjoint closed sets. In the real line with the usual topology, the singleton sets $\left\{x \right\}$ , where $x$ is rational, are disjoint closed sets that cannot be simultaneously separated by disjoint open sets. In order to separate an infinite collection of disjoint closed sets, it makes sense to restrict on the type of collections of closed sets. A space $X$ is collectionwise normal if every discrete collection of closed subsets of $X$ can be separated by pairwise disjoint open subsets of $X$ . The following is a more specific definition.

Definition

$X$

$\mathcal{A}$

$X$

$\mathcal{U}=\left\{U_A: A \in \mathcal{A} \right\}$

$X$

$A \subset U_A$

$A \in \mathcal{A}$

For more details about the definitions of collectionwise normality, see “Definitions of Collectionwise Normal Spaces”. The implications displayed above are repeated below. None of the arrows is reversible.

$\text{ }$

$\text{paracompact} \Longrightarrow \text{collectionwise normal} \Longrightarrow \text{normal}$

As indicated above, Bing’s Example G is an example of a normal and not collectionwise normal space (see the blog post “Bing’s Example G”). The propositions in the next section can be used to obtain collectionwise normal spaces that are not paracompact.

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When Normal implies Collectionwise Normal

Being able to simultaneously separate any discrete collection of closed sets is stronger than the property of merely being able to separate finite collection of disjoint closed sets. It turns out that the stronger property of collectionwise normality is required only for separating uncountable discrete collections of closed sets. As the following lemma shows, normality is sufficient to separate any countable discrete collection of closed sets.

Lemma 1
Let $X$ be a normal space. Then for every discrete collection $\left\{C_1,C_2,C_3,\cdots \right\}$ of closed subsets of $X$ , there exists a pairwise disjoint collection $\left\{O_1,O_2,O_3,\cdots \right\}$ of open subsets of $X$ such that $C_i \subset O_i$ for each $i$ .

Proof of Lemma 1
Let $\left\{C_1,C_2,C_3,\cdots \right\}$ be a discrete collection of closed subsets of $X$ . For each $i$ , choose disjoint open sets $U_i$ and $V_i$ such that $C_i \subset U_i$ and $\cup \left\{C_j: j \ne i \right\} \subset V_i$ . Let $O_1=U_1$ . For each $i>1$ , let $O_i=U_i \cap V_1 \cap \cdots \cap V_{i-1}$ . It follows that $O_i \cap O_j = \varnothing$ for all $i \ne j$ . It is also clear that for each $i$ , $C_i \subset O_i$ . $\blacksquare$

We have the following propositions.

Proposition 2
Let $X$ be a normal space. If all discrete collections of closed subsets of $X$ are at most countable, then $X$ is collectionwise normal.

Proposition 3
Let $X$ be a normal space. If all closed and discrete subsets of $X$ are at most countable (such a space is said to have countable extent), then $X$ is collectionwise normal.

Proposition 4
Any normal and countably compact space is collectionwise normal.

Proposition 2 follows from Lemma 1. As noted in Proposition 3, any space in which all closed and discrete subsets are countable is said to have countable extent. It is easy to verify that $X$ has countable extent if and only if all discrete collections of closed subsets of $X$ are at most countable. If $X$ is a countably compact space, then every infinite subset of $X$ has a limit point. Thus Proposition 4 follows from the fact that any countably compact space has countable extent.

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Paracompact Spaces

One way to find a collectionwise normal space that is not paracompact is to find a non-paracompact space that satisfies Propositions 3, 4 or 5. For example, $\omega_1$ , the space of all countable ordinals with the order topology, is not paracompact. Since $\omega_1$ is normal and countably compact, it is collectionwise normal by Proposition 4. For a basic discussion of $\omega_1$ as a topological space, see “The First Uncountable Ordinal”.

As the following theorem shows, paracompact spaces are collectionwise normal. Thus the class of collectionwise normal spaces includes all metric spaces and paracompact spaces.

Theorem 5
If a space $X$ is paracompact, then $X$ is collectionwise normal.

Proof of Theorem 5
Let $X$ be a paracompact space. Let $\mathcal{A}$ be a discrete collection of closed subsets of $X$ . For each $x \in X$ , let $O_x$ be open such that $x \in O_x$ and $O_x$ meets at most one element of $\mathcal{A}$ . Let $\mathcal{O}=\left\{O_x: x \in X \right\}$ . By the paracompactness of $X$ , $\mathcal{O}$ has a locally finite open refinement $\mathcal{V}$ .

For each $A \in \mathcal{A}$ , let $W_A=\cup \left\{\overline{V}: V \in \mathcal{V} \text{ and } \overline{V} \cap A=\varnothing \right\}$ and let $U_A=X-W_A$ . Each $W_A$ is a closed set since $\mathcal{V}$ is locally finite. Thus each $U_A$ is open. Furthermore, for each $A \in \mathcal{A}$ , $A \subset U_A$ . It is easily checked that $\left\{U_A: A \in \mathcal{A} \right\}$ is pairwise disjoint. $\blacksquare$

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Reference

Bing, R. H., Metrization of Topological Spaces, Canad. J. Math., 3, 175-186, 1951.
Engelking, R., General Topology, Revised and Completed edition, Heldermann Verlag, Berlin, 1989.
Willard, S., General Topology, Addison-Wesley Publishing Company, 1970.

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$\copyright \ \ 2012$
Revised June 16, 2019

C*-Embedding Property and Stone-Cech Compactification

Posted on September 23, 2012 by Dan Ma

This is a continuation of an introduction of Stone-Cech compactification started in two previous posts (first post: A Beginning Look at Stone-Cech Compactification; second post: Two Characterizations of Stone-Cech Compactification). In this post, we present another characterization of the Stone-Cech compactification, that is, for any completely regular space $X$ , $X$ is $C^*$ -embedded in its Stone-Cech compactification $\beta X$ and that any compactification of $X$ in which $X$ is $C^*$ -embedded must be $\beta X$ . In other words, this property of $C^*$ -embedding is unique to Stone-Cech compactification. We prove the following two theorems (U3 has two versions).

The links for other posts on Stone-Cech compactification can be found toward the end of this post.

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Definition. Let $Y$ be a space. Let $A \subset Y$ . The subspace $A$ is $C^*$ -embedded in $Y$ if every bounded continuous function $f:A \rightarrow \mathbb{R}$ is extendable to a continuous $\hat{f}:Y \rightarrow \mathbb{R}$ .

Theorem C3
Let $X$ be a completely regular space. The space $X$ is $C^*$ -embedded in its Stone-Cech compactification $\beta X$ .

$\text{ }$

Theorem U3.1

$X$

$I=[0,1]$

$\alpha X$

$X$

$f:X \rightarrow I$

$\hat{f}:\alpha X \rightarrow I$

$\alpha X$

$\beta X$

$\text{ }$

Theorem U3.2

$\alpha X$

$X$

$C^*$

$\alpha X$

$\beta X$

$\text{ }$

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Other Characterizations

Two other characterizations of $\beta X$ are proved in the previous post (Two Characterizations of Stone-Cech Compactification).

Theorem C1

$X$

$f:X \rightarrow Y$

$X$

$Y$

$F: \beta X \rightarrow Y$

$F \circ \beta=f$

$\text{ }$

Theorem U1

$K$

$X$

$K$

$\beta X$

$\text{ }$

Theorem C2

$X$

$\beta X$

$X$

$\text{ }$

Theorem U2

$\beta X$

$\alpha X$

$X$

$\alpha X$

$\beta X$

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Remark

The C theorems and the U theorems are a great tool to determine whether a given compactification is $\beta X$ . Whenever a compactification $\alpha X$ of a space $X$ satisfies the property belonging to a C theorem, based on the corresponding U theorem, we know that this compactification $\alpha X$ must be $\beta X$ . For example, any compactification $\alpha X$ that satisfies the function extension property in Theorem C1 must be $\beta X$ . Th $C^*$ -embedding property in Theorem C3 and Theorem U3 (both versions) is also a function extension property much like that in Theorems C1 and U1, but is easier to use. The reason being that we only need to extend a smaller class of continuous functions (i.e., to check whether functions from $X$ into $I=[0,1]$ can be extended), rather than checking all continuous functions from $X$ to arbitrary compact spaces. As the following example below about $\beta \omega_1$ illustrates that the $C^*$ -embedding in Theorem C3 and U3.1 can be used to describe $\beta X$ explicitly.

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Proving Theorem U3.1 and Theorem U3.2

Let $Y$ be a space. Let $A$ be a subspace of $X$ . Recall that $A$ is $C^*$ -embedded in $Y$ if every bounded continuous function $f:A \rightarrow \mathbb{R}$ can be extended to a continuous $\hat{f}:Y \rightarrow \mathbb{R}$ .

Any bounded continuous function $f: X \rightarrow \mathbb{R}$ can be regarded as $f: X \rightarrow I_f$ where $I_f$ is some closed and bounded interval. The $C^*$ -embedding property in Theorem C3 is a function extension property like the one in Theorem C1, except that it deals with function from $X$ into a specific type of compact spaces $Y$ , namely the closed and bounded intervals in $\mathbb{R}$ . Theorem C3 is a corollary of Theorem C1 (see below). So we only need to prove Theorem U3.1 and Theorem U3.2. Theorem U3.2 is a corollary of Theorem U3.1.

Proof of Theorem U3.1
By Theorem C2, we have $\alpha X \le \beta X$ . So we only need to show $\beta X \le \alpha X$ . To this end, we need to produce a continuous function $H: \alpha X \rightarrow \beta X$ such that $H \circ \alpha=\beta$ .

Let $C(X,I)$ be the set of all continuous functions from $X$ into $I$ . For each $f \in C(X,I)$ , let $I_f=I$ . Recall that $\beta X$ is embedded in the cube $\prod \limits_{f \in C(X,I)} I_f$ by the mapping $\beta$ . For each $f \in C(X,I)$ , let $\pi_f$ be the projection map from this cube into $I_f$ .

Each $f \in C(X,I)$ can be expressed as $f=\pi_f \circ \beta$ . Thus by assumption, each $f$ can be extended by $\hat{f}: \alpha X \rightarrow I$ . Now define $H: \alpha X \rightarrow \prod \limits_{f \in C(X,I)} I_f$ by the following:

$t \in \alpha X$

$H(t)=a=< a_f >_{f \in C(X,I)}$

$a_f=\hat{f}(t)$

For each $x \in \alpha(X)$ , we have $H(\alpha(x))=\beta(x)$ . Note that $\hat{f}$ agrees with $f$ on $\alpha(X)$ since $\hat{f}$ extends $f$ . So we have $H(\alpha(x))=a$ where $a_f=\hat{f}(\alpha(x))=f(x)$ for each $f \in C(X,I)$ . On the other hand, by definition of $\beta$ , we have $\beta(x)=a$ where $a_f=f(x)$ for each $f \in C(X,I)$ . Thus we have $H \circ \alpha=\beta$ and the following:

$H(\alpha(X)) \subset \beta(X) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (1)$

It is straightforward to verify that $H$ is continuous. Note that $\alpha(X)$ is dense in $\alpha X$ . Since $H$ is continuous, $H(\alpha(X))$ is dense in $H(\alpha X)$ . Thus we have:

$H(\alpha X)=\overline{H(\alpha(X))} \ \ \ \ \ \ \ \ \ \ \ \ \ (2)$

Putting $(1)$ and $(2)$ together, we have the following:

$H(\alpha X)=\overline{H(\alpha(X))} \subset \overline{\beta(X)}=\beta X$

Thus we can describe the map $H$ as $H: \alpha X \rightarrow \beta X$ . As noted before, we have $H \circ \alpha=\beta$ . Thus $\beta X \le \alpha X$ . $\blacksquare$

Proof of Theorem U3.2
Suppose $\alpha X$ is a compactification of $X$ such that $X$ is $C^*$ -embedded in $\alpha X$ . Then every bounded continuous $f:X \rightarrow I_f$ can be extended to $\hat{f}:\alpha X \rightarrow I_f$ where $I_f$ is some closed and bounded interval containing the range. In particular, this means every continuous $f:X \rightarrow I$ can be extended. By Theorem U3.1, we have $\alpha X \approx \beta X$ . $\blacksquare$

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Example

This is one example where we can use $C^*$ -embedding to describe $\beta X$ explicitly.

Let $\omega_1$ be the first uncountable ordinal. Let $\omega_1+1$ be the successor ordinal of $\omega_1$ (i.e. $\omega_1$ with one additional point at the end). Consider $X=\omega_1$ and $Y=\omega_1+1$ as topological spaces with the order topology derived from the well ordering of the ordinals. The space $Y$ is a compactification of $X$ . In fact $Y$ is the one-point compactification of $X$ .

It is well known that every continuous real-valued function on $X$ is bounded (note that $X$ here is countably compact and hence pseudocompact). Furthermore, every continuous real-valued function on $X$ is eventually constant. This means that if $f:X \rightarrow \mathbb{R}$ is continuous, for some $\alpha < \omega_1$ , $f$ is constant on the final segment $X_\alpha=\left\{\rho < \omega_1: \rho>\alpha \right\}$ (see result B in The First Uncountable Ordinal). As a result, every continuous bounded real-valued function $f:X \rightarrow \mathbb{R}$ can be extended to a continuous $\hat{f}:Y \rightarrow \mathbb{R}$ . Then according to Theorem U3.2, $\beta X=\beta \omega_1=Y=\beta \omega_1+1$ .

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Blog Posts on Stone-Cech Compactification

Embedding Completely Regular Spaces into a Cube

Post #1: A Beginning Look at Stone-Cech Compactification

Post #2: Two Characterizations of Stone-Cech Compactification

Post #3 (this post): C*-Embedding Property and Stone-Cech Compactification

Post #4: Stone-Cech Compactification is Maximal

Post #5: Stone-Cech Compactification of the Integers – Basic Facts

Post #6: Stone-Cech Compactifications – Another Two Characterizations

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Reference

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

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$\copyright \ \ 2012$

Completely Regular Spaces and Pseudocompact Spaces

Posted on August 25, 2012 by Dan Ma

In proving theorems about properties in abstract topological spaces, it makes sense that the spaces in questions satisfy some axioms in addition to the ones stipulated in the definition of topological spaces. For example, authors typically assume certain separation axioms. Doing so will help authors know in advance what basic properties the spaces will have. For example, it is desirable to know that singleton sets (and finite sets) are closed (assuming the $T_1$ axiom or a $T_1$ space). In some circumstances, it may be desirable to be able to separate a single point from a closed set not containing it (assuming $T_3$ axiom or regularity). In some situation, it may be advantageous (and even necessary) to know in advance that there is a sufficient quantity in continuous real-valued functions that can be defined on the spaces in question. We give several reasons of needing enough continuous functions (this list is not meant to be exhaustive).

Certain notions involve continuous real-valued function defined on the space. Pseudocompactness is one such notion (this view point is discussed in this post).
Spaces of continuous functions are the objects being studied (this view point is discussed in this post).
Completely regular spaces are precisely the spaces that can be embedded in a cube (the product space of copies of the unit interval). Discussed in this post.

In this post, we discuss the importance of complete regularity from the first view point and use pseudocompactness as an illustration. See [1] and [2] and [3] for any notions not defined here. Steen and Seebach (section 2 of [2] starting on p.11) has an excellent discussion of separation axioms.

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Completely Regular Spaces and Pseudocompact Spaces

A space $X$ is said to be completely regular if $X$ is a $T_0$ space and for each $x \in X$ and for each closed subset $A$ of $X$ with $x \notin A$ , there is a continuous function $f:X \rightarrow [0,1]$ such that $f(A) \subset \left\{0 \right\}$ and $f(x)=1$ . Note that the $T_0$ axiom and the existence of the continuous function imply the $T_1$ axiom, which is equivalent to the property that single points are closed sets. Completely regular spaces are also called Tychonoff spaces.

As defined above, in a completely regular space, for any closed set and a point not in the closed set, we can always find a continuous function mapping the closed set to 0 and the point to 1. Essentially, to be a completely regular space, it suffices to provide a continuous function that maps a given closed set and a point (not in the closed set) to two different real numbers $a$ and $b$ . So in a space that is not completely regular, there exist a closed set $H$ and a point $x \notin H$ such that every real-valued continuous function $g$ that can be defined on the space maps $H$ and the point $x$ to the same real number. Thus outside of completely regular spaces, notions that are based on continuous real-valued functions may be difficult to work with.

A space $X$ is said to be pseudocompact if every real-valued continuous function $f$ defined on $X$ is a bounded function (i.e. $f(X)$ is a bounded set in the real line $\mathbb{R}$ ). Even though the definition does not include complete regularity, an effective discussion of pseudocompactness typically make use of complete regularity. We illustrate this point using a proof of a theorem that gives a characterization of pseudocompactness.

Theorem 1
Let $X$ be a space. The following conditions are equivalent:

The space $X$ is pseudocompact.
If $\mathcal{V}$ is a locally finite family of non-empty open subsets of $X$ , then $\mathcal{V}$ is finite.
If $\mathcal{V}$ is a locally finite open cover of $X$ , then $\mathcal{V}$ is finite.
If $\mathcal{V}$ is a locally finite open cover of $X$ , then $\mathcal{V}$ has a finite subcover.

Proof
This theorem is discussed in this discussion of pseudocompactness. We repeat the proof of $1 \Rightarrow 2$ to illustrate an application of complete regularity.

$1 \Rightarrow 2$
Suppose that condition $2$ does not hold. Then there is an infinite locally finite family of non-empty open sets $\mathcal{V}$ such that $\mathcal{V}=\left\{V_1,V_2,V_3,\cdots \right\}$ . We wish to define an unbounded continuous function using $\mathcal{V}$ .

This is where we need to invoke the assumption of complete regularity. For each $n$ choose a point $x_n \in V_n$ . Then for each $n$ , there is a continuous function $f_n:X \rightarrow [0,n]$ such that $f_n(x_n)=n$ and $f_n(X-V_n) \subset \left\{ 0 \right\}$ . Define $f:X \rightarrow [0,\infty)$ by $f(x)=f_1(x)+f_2(x)+f_3(x)+\cdots$ .

Because $\mathcal{V}$ is locally finite, the function $f$ is essentially pointwise the sum of finitely many $f_n$ . In other words, for each $x \in X$ , for some positive integer $N$ , $f_j(x)=0$ for all $j \ge N$ . Thus the function $f$ is well defined and is continuous at each $x \in X$ . Note that for each $x_n$ , $f(x_n) \ge n$ , showing that it is unbounded. $\blacksquare$

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Remark

Complete regularity is an integral part of the above proof. It simplifies the proof and clarifies the argument. As a direct corollary to the Theorem 1, any pseudocompact paracompact space is compact. A less direct corollary is that any pseudocompact metacompact space is compact (see this discussion of pseudocompactness). These results are made possible by assuming that the pseudocompact spaces are also completely regular. They are restated below.

Corollary 2
Let $X$ be a completely regular space. If $X$ is pseudocompact and paracompact, then $X$ is compact.

Corollary 3
Let $X$ be a completely regular space. If $X$ is pseudocompact and metacompact, then $X$ is compact.

It could be a valid math question to ask whether the above two results are valid outside of the class of completely regular space. We do not know the answer. We also feel that it is also a valid approach to just assume complete regularity and focus our attention on exploring the main concepts involved (in this case pseudocompactness, paracompactness and metacompactness).

We would like to remark that in working with pseudocompactness, we also need to take care that we do not assume too much. For example, we do not want to assume normality since any normal pseudocompact space is countably compact (see Theorem 2 in this post). Then working with pseudocompactness is turned into the problem of working with the stronger concept of countably compactness.

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Reference

Engelking, R., General Topology, Revised and Completed edition, Heldermann Verlag, Berlin, 1989.
Steen, L. A., Seebach, J. A., Counterexamples in Topology, Dover Publications, Inc., New York, 1995.
Willard, S., General Topology, Addison-Wesley Publishing Company, 1970.

Pseudocompact + Metacompact implies Compact

Posted on August 16, 2012 by Dan Ma

It is a well known result that any countably compact and metacompact space is compact (see Theorem 5.3.2 in [1]). A discussion of this result is also found in this blog (countably compact + metacompact). Since countably compactness implies pseudocompactness, a natural question arises: can this result be generalized to “any pseudocompact and metacompact space is compact?” The answer is yes and was established in [2] and [3]. In this post, we put together a proof of this result by using building blocks already worked out in this blog.

All spaces considered here are Tychonoff (completely regular). Refer to [1] and [4] for any terms and notions not defined here (or refer to elsewhere in this blog).

A space $X$ is said to be almost compact if for every open cover $\mathcal{U}$ of $X$ , there is a finite $\mathcal{V} \subset \mathcal{U}$ such that $\bigcup \mathcal{V}$ is dense in $X$ . It can be shown that for any regular space, almost compactness implies compactness. We have the following lemma.

Lemma 1
Let $X$ be a regular space. Then $X$ is compact if and only if $X$ is almost compact.

Theorem 2, Theorem 3 and Theorem 4 are building blocks proved in previous posts. Theorem 5 below is the main theorem. Corollary 6, the intended result, is obtained from applying Theorem 5 and Lemma 1.

Theorem 2 (see Theorem 4B in this post)
Every regular pseudocompact is a Baire space.

Theorem 3 (see Main Theorem in this post)
Let $X$ be a space. The following conditions are equivalent.

$X$ is a Baire space.
For any point-finite open cover $\mathcal{V}$ of $X$ , the set $D=\left\{x \in X: \mathcal{V} \text{ is locally finite at } x \right\}$ is a dense set in $X$ .

Theorem 4 (see Theorem 1 in this post)
Let $X$ be a space. The following conditions are equivalent.

$X$ is a pseudocompact space.
If $\mathcal{W}$ is a locally finite family of non-empty open subsets of $X$ , then $\mathcal{W}$ is finite.

Theorem 5
Let $X$ be a pseudocompact and metacompact space. Then $X$ is almost compact.

Proof
Let $\mathcal{U}$ be an open cover of $X$ . By metacompactness, there is a $\mathcal{V}$ which is a point-finite open refinement of $\mathcal{U}$ . It suffices to find a finite $\mathcal{W} \subset \mathcal{V}$ such that $\mathcal{W}$ covers a dense set.

By Theorem 2, $X$ is a Baire space. By Theorem 3, the set $D$ is dense in $X$ where $D=\left\{x \in X: \mathcal{V} \text{ is locally finite at } x \right\}$ . Let $\mathcal{W}$ be the collection of all $V \in \mathcal{V}$ such that $V \cap D \ne \varnothing$ . Note that $\bigcup \mathcal{W}$ is open and dense in $X$ . Furthermore, it is straightforward to show that $\mathcal{W}$ is locally finite at each point $x \in D$ . By Theorem 4, $\mathcal{W}$ is finite. $\blacksquare$

Corollary 6
Let $X$ be a pseudocompact and metacompact space. Then $X$ is compact.

Reference

Engelking, R., General Topology, Revised and Completed edition, Heldermann Verlag, Berlin, 1989.
Scott, B., M., Pseudocompact Metacompact Spaces are Compact, Topology Proc., 4, 577-587, 1979.
Watson, W. S., Pseudocompact Metacompact Spaces are Compact, Proc. Amer. Math. Soc., 81, 151-152, 1981.
Willard, S., General Topology, Addison-Wesley Publishing Company, 1970.

Baire Category Theorem and the Finite Intersection Property

Posted on August 12, 2012 by Dan Ma

A Baire space is a topological space in which the intersection of any countable family of open and dense sets is dense (equivalently every non-empty open subset is of second category). One version of the Baire category theorem states that every complete metric space is a Baire space. Another common version states that every compact Hausdorff space is a Baire space. Another version states that every locally compact Hausdorff space is a Baire space. The commonality among these versions is the finite intersection property (whenever a collection of a certain type of sets satisfies the property that any finite subcollection has non-empty intersection, the whole collection has non-empty intersection). For each of these classes of spaces, in addition to countably compact spaces and pseudocompact spaces, Baire category theorem is derived from having one specific form of the finite intersection property. In this post, we explore this relationship.

In each of the following theorem pairs, the B Theorem follows from the A theorem. The A theorem is a form of the finite intersection property and the B theorem is a version of Baire category theorem.

Another interesting observation is that the finite intersection properties discussed here can give a stronger property than being a Baire space. This stronger property is defined by the Banach-Mazur game.

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Theorem 1A
Let $(X, \rho)$ be a metric space. Then the following conditions are equivalent.

$(X, \rho)$ is a complete metric space.
For each decreasing sequence $C_1 \supset C_2 \supset C_3 \supset \cdots$ of non-empty closed subsets of $X$ such that the diameters of the sets $C_n$ converge to zero, we have $\bigcap \limits_{n=1}^\infty C_n \ne \varnothing$ .

Theorem 1B
Every complete metric space is a Baire space.

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Theorem 2A
Let $X$ be a Hausdorff space. Then the following conditions are equivalent.

$X$ is a compact space.
For every family $\mathcal{F}$ consisting of non-empty closed subsets of $X$ , if $\mathcal{F}$ has the finite intersection property, then $\mathcal{F}$ has non-empty intersection.

Theorem 2B

Every compact Hausdorff space is a Baire space.
Every locally compact Hausdorff space is a Baire space.

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Theorem 3A
Let $X$ be a Hausdorff space. Then the following conditions are equivalent.

$X$ is a countably compact space.
For every countable family $\mathcal{F}$ consisting of non-empty closed subsets of $X$ , if $\mathcal{F}$ has the finite intersection property, then $\mathcal{F}$ has non-empty intersection.
For each decreasing sequence $C_1 \supset C_2 \supset C_3 \supset \cdots$ of non-empty closed subsets of $X$ , we have $\bigcap \limits_{n=1}^\infty C_n \ne \varnothing$ .

Theorem 3B
Every countably compact Hausdorff space is a Baire space.

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Theorem 4A
Let $X$ be a regular space. The following conditions are equivalent:

The space $X$ is pseudocompact.
If $\mathcal{O}=\left\{O_1,O_2,O_3,\cdots \right\}$ is a family of non-empty open subsets of $X$ such that $O_n \supset O_{n+1}$ for each $n$ , then $\bigcap \limits_{n=1}^\infty \overline{O_n} \ne \varnothing$ .
If $\mathcal{V}=\left\{V_1,V_2,V_3,\cdots \right\}$ is a family of non-empty open subsets of $X$ such that $\mathcal{V}$ has the finite intersection property, then $\bigcap \limits_{n=1}^\infty \overline{V_n} \ne \varnothing$ .

Theorem 4B
Every regular pseudocompact space is a Baire space.

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Remark
Theorem 1A (the Cantor Theorem) can be found in Engelking (page 269 in [1]). Theorem 2A and Theorem 3A can also be found in Engelking (they are also proved in this post). Theorem 4B is also found in Engelking (Theorem 3.10.23 in page 207 of [1]) and is proved this post.

We would like to explicitly point out that between Thoerem 1A and Theorem 2A, none of the two theorems implies the other. For example, even though both complete metric spaces and compact Hausdorff spaces are Baire spaces, complete metric spaces are not necessarily compact and there are compact spaces that are not even metrizable. However, the finite intersection property of Theorem 2A implies that of Theorem 3A, which in turn implies the finite intersection property of Theorem 4A.
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Baire Category Theorem

The proofs of all four B theorems are amazingly similar. It is a matter of exploiting the fact that whenever a decreasing sequence of open sets satisfying the condition that each closure is a subset of the previous open set (and satisfying some other condition), the sequence of open sets has non-empty intersection. For example, for complete metric space, make sure that the closures of the open sets have diameters going to zero. For any reader who is new to this material, it will be very instructive to walk through the arguments of these Baire category theorems. The proof of Theorem 1A can be found this post. We prove Theorem 4B.

Recall that $X$ is a Baire space if $\left\{U_1,U_2,U_3,\cdots \right\}$ is a countable family of open and dense sets in $X$ , $\bigcap \limits_{i=1}^\infty U_i$ is dense in $X$ , or equivalently every non-empty open subset of $X$ is of second category in $X$ . For more background about the concepts of Baire space and category (see [1] or this post).

Proof of Theorem 4B
Let $X$ be a regular pseudocompact space. Let $\left\{U_1,U_2,U_3,\cdots \right\}$ be a countable family of open and dense sets in $X$ . Let $O$ be a non-empty open subset of $X$ . We show that $O$ has to contain points of $\bigcap \limits_{n=1}^\infty U_n$ . We let $O_1=O \cap U_1$ . We find open $O_2$ such that $O_2 \subset U_2$ and $\overline{O_2} \subset O_1$ (using regularity). Continue this inductive process, we have for each $n$ , an open $O_n$ such that $O_n \subset U_n$ and $\overline{O_n} \subset O_{n-1}$ . Then we have a decreasing sequence of open sets $O_n$ as in condition 2 of Theorem 4A. Then we have $\bigcap \limits_{n=1}^\infty \overline{O_n} \ne \varnothing$ . Since $\overline{O_{n+1}} \subset O_n$ for each $n$ , we also have $\bigcap \limits_{n=1}^\infty O_n \ne \varnothing$ . It is clear that $\bigcap \limits_{n=1}^\infty O_n \subset \bigcap \limits_{n=1}^\infty U_n$ . $\blacksquare$

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Banach-Mazur Game

In proving the above versions of Baire category theorem, we can exploit the appropriate version of the finite intersection property – the situation that any nested decreasing sequence of open sets (under some specified conditions) has non-empty intersection. In fact, the finite intersection property offers more than just Baire category theorem; it can endow the space in question a type of completeness property stronger than Baire space. This completeness property is defined using the Banach-Mazur game.

The Banach-Mazur game is a two-person game played on a topological space. Let $X$ be a space. There are two players, $\alpha$ and $\beta$ . They take turn choosing nested decreasing nonempty open subsets of $X$ as follows. The player $\beta$ goes first by choosing a nonempty open subset $U_0$ of $X$ . The player $\alpha$ then chooses a nonempty open subset $V_0 \subset U_0$ . At the nth play where $n \ge 1$ , $\beta$ chooses an open set $U_n \subset V_{n-1}$ and $\alpha$ chooses an open set $V_n \subset U_n$ . The player $\alpha$ wins if $\bigcap \limits_{n=0}^\infty V_n \ne \varnothing$ . Otherwise the player $\beta$ wins. For more detailed discussion of the game, see this post.

One interesting point that we like to make about the finite intersection property ranging from Theorem 1A to Theorem 4A is that the player $\alpha$ can always win the Banach-Mazur game as long as he/she plays the game according to each specific version of the finite intersection. For example, playing the game in a complete metric space, player $\alpha$ always wins as long as he/she makes the diameters of the closures of the open sets going to zero. In a regular pseudocompact space, player $\alpha$ can always win by making the closure of each of his/her open sets a subset of the previous move of other player.

A topological space in which the player $\alpha$ has a winning strategy is said to be a weakly $\alpha$ -favorable space. Thus complete metric spaces, compact Hausdorff spaces, locally compact Hausdorff spaces, countably compact Hausdorff spaces, regular pseudocompact spaces are all weakly $\alpha$ -favorable.

There is characterization of Baire spaces in terms of the Banach-Mazur game. A space $X$ is a Baire space if and only if the player $\beta$ has no winning strategy in the Banach-Mazur game played on the space $X$ (see theorem 1 in this post). If the player $\alpha$ can always win, then player $\beta$ can never win. In terms of game terminology, if player $\alpha$ has a winning strategy, then the other player (player $\beta$ ) has no winning strategy. Thus a space is weakly $\alpha$ -favorable implies that it is a Baire space. But the implication is not reversible (see example in this post).

So all the spaces discussed from Theorem 1A to Theorem 4A are all weakly $\alpha$ -favorable, a property stronger than Baire spaces. These observations are summarized in the following theorems.

Theorem 1C
Every complete metric space is a weakly $\alpha$ -favorable space.

Theorem 2C

Every compact Hausdorff space is a weakly $\alpha$ -favorable space.
Every locally compact Hausdorff space is a weakly $\alpha$ -favorable space.

Theorem 3C
Every countably compact Hausdorff space is a weakly $\alpha$ -favorable space.

Theorem 4C
Every regular pseudocompact space is a weakly $\alpha$ -favorable space.

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Reference

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

When is a Pseudocompact Space Metrizable?

Posted on August 8, 2012 by Dan Ma

Compactness, countably compactness and pseudocompactness are three successively weaker properties. It follows easily from definitions that

$(A) \ \ \ \ \ \text{compact} \Rightarrow \text{countably compact} \Rightarrow \text{pseudocompact}$

None of these arrows can be reversed. It is well known that either compactness or countably complactness plus having a $G_\delta$ -diagonal implies metrizability. We have:

$(B) \ \ \ \ \ \text{compact} + \text{having a } G_\delta \text{-diagonal} \Rightarrow \text{metrizable}$

$(C) \ \ \ \ \ \text{countably compact} + \text{having a } G_\delta \text{-diagonal} \Rightarrow \text{metrizable}$

A question can be asked whether these results can be extended to pseudocompactness.

Question $(D) \ \ \ \ \ \text{pseudocompact compact} + \text{having a } G_\delta \text{-diagonal} \Rightarrow \text{metrizable?}$

The answer to this question is no. The space defined using a maximal almost disjoint family of subsets of $\omega$ is an example of a non-metrizable pseudocompact space with a $G_\delta$ -diagonal (discussed in this post). In this post we show that if we strengthen “having a $G_\delta$ -diagonal” to being submetrizable, we have a theorem. Specifically, we show:

$(E) \ \ \ \ \ \text{pseudocompact} + \text{submetrizable} \Rightarrow \text{metrizable}$

For the result of $(B)$ , see this post. For the result of $(C)$ , see this post. In this post, we discuss the basic properties of pseudocompactness that build up to the result of $(E)$ . All spaces considered here are at least Tychonoff (i.e. completely regular). For any basic notions not defined here, see [1] or [2].

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Pseudocompact Spaces

A space $X$ is said to be pseudocompact if every real-valued continuous function defined on $X$ is a bounded function. Any real-valued continuous function defined on a compact space must be bounded (and is thus pseudocomppact). If there were an unbounded real-valued continuous function defined on a space $X$ , then $X$ would have a countably infinite discrete set (thus not countably compact). Thus countably compact implies pseudocompact, as indicated by $(A)$ .

A space $X$ is submetrizable if there is a coarser (i.e. weaker) topology that is a metrizable topology. Specifically the topological space $(X,\tau)$ is submetrizable if there is another topology $\tau^*$ that can be defined on $X$ such that $\tau^* \subset \tau$ and $(X,\tau^*)$ is metrizable. The Sorgenfrey line is non-metrizable and yet the Sorgenfrey topology has a weaker topology that is metrizable, namely the Euclidean topology of the real line.

The following two theorems characterizes pseudocompact spaces in terms of locally finite open family of open sets (Theorem 1) and the finite intersection property (Theorem 2). Both theorems are found in Engelking (Theorem 3.10.22 and Theorem 3.10.23 in page 207 of [1]). Theorem 3 states that in a pseudocompact space, closed domains are pseudocompact (the definition of closed domain is stated before the theorem). Theorem 4 is the main theorem (result $E$ stated above).

Theorem 1
Let $X$ be a space. The following conditions are equivalent:

The space $X$ is pseudocompact.
If $\mathcal{V}$ is a locally finite family of non-empty open subsets of $X$ , then $\mathcal{V}$ is finite.
If $\mathcal{V}$ is a locally finite open cover of $X$ , then $\mathcal{V}$ is finite.
If $\mathcal{V}$ is a locally finite open cover of $X$ , then $\mathcal{V}$ has a finite subcover.

Proof
$1 \Rightarrow 2$
Suppose that condition $2$ does not hold. Then there is an infinite locally finite family of non-empty open sets $\mathcal{V}$ such that $\mathcal{V}=\left\{V_1,V_2,V_3,\cdots \right\}$ . We wish to define an unbounded continuous function using $\mathcal{V}$ .

The directions $2 \Rightarrow 3$ and $3 \Rightarrow 4$ are clear.

$4 \Rightarrow 1$
Let $g:X \rightarrow \mathbb{R}$ be a continuous function. We want to show that $g$ is a bounded function. Consider the open family $\mathcal{O}=\left\{\cdots,O_{-3},O_{-2},O_{-1},O_0,O_1,O_2,O_3,\cdots \right\}$ where each $O_n=g^{-1}((n,n+2))$ . Note that $\mathcal{O}$ is a locally finite family in $X$ since its members $O_n=g^{-1}((n,n+2))$ are inverse images of members of a locally finite family in the range space $\mathbb{R}$ . By condition $4$ , $\mathcal{O}$ has a finite subcover, leading to the conclusion that $g$ is a bounded function. $\blacksquare$

Theorem 2
Let $X$ be a space. The following conditions are equivalent:

The space $X$ is pseudocompact.
If $\mathcal{O}=\left\{O_1,O_2,O_3,\cdots \right\}$ is a family of non-empty open subsets of $X$ such that $O_n \supset O_{n+1}$ for each $n$ , then $\bigcap \limits_{n=1}^\infty \overline{O_n} \ne \varnothing$ .
If $\mathcal{V}=\left\{V_1,V_2,V_3,\cdots \right\}$ is a family of non-empty open subsets of $X$ such that $\mathcal{V}$ has the finite intersection property, then $\bigcap \limits_{n=1}^\infty \overline{V_n} \ne \varnothing$ .

Proof
$1 \Rightarrow 2$
Suppose that $X$ is pseudocompact. Suppose $\mathcal{O}=\left\{O_1,O_2,O_3,\cdots \right\}$ satisfies the hypothesis of condition $2$ . If there is some positive integer $m$ such that $O_n=O_m$ for all $n \ge m$ , then we are done. So assume that $O_n$ are distinct for infinitely many $n$ . According to condition $2$ in Theorem 1, $\mathcal{O}$ must not be a locally finite family. Then there exists a point $x \in X$ such that every open set containing $x$ must meet infinitely many $O_n$ . This implies that $x \in \overline{O_n}$ for infinitely many $n$ . Thus $x \in \bigcap \limits_{n=1}^\infty \overline{O_n}$ .

$2 \Rightarrow 3$
Suppose $\mathcal{V}=\left\{V_1,V_2,V_3,\cdots \right\}$ is a family of non-empty open sets with the finite intersection property as in the hypothesis of $3$ . Then let $O_1=V_1$ , $O_2=V_1 \cap V_2$ , $O_3=V_1 \cap V_2 \cap V_3$ , and so on. By condition $2$ , we have $\bigcap \limits_{n=1}^\infty \overline{O_n} \ne \varnothing$ , which implies $\bigcap \limits_{n=1}^\infty \overline{V_n} \ne \varnothing$ .

$3 \Rightarrow 1$
Let $g:X \rightarrow \mathbb{R}$ be a continuous function such that $g$ is unbounded. For each positive integer $n$ , let $V_n=\left\{x \in X: \lvert g(x) \lvert > n \right\}$ . Clearly the open sets $V_n$ have the finite intersection property. Because $g$ is unbounded, it follows that $\bigcap \limits_{n=1}^\infty \overline{V_n} = \varnothing$ . $\blacksquare$

Let $X$ be a space. Let $A \subset X$ . The interior of $A$ , denoted by $\text{int}(A)$ , is the set of all points $x \in X$ such that there exists an open set $O$ with $x \in O \subset A$ . Points of $\text{int}(A)$ are called the interior points of $A$ . A subset $C \subset X$ is said to be a closed domain if $C=\overline{\text{int}(C)}$ . It is clear that $C$ is a closed domain if and only if $C$ is the closure of an open set.

Theorem 3
The property of being a pseudocompact space is hereditary with respect to subsets that are closed domains.

Proof
Let $X$ be a pseudocompact space. We show that $\overline{U}$ is pseudocompact for any nonempty open set $U \subset X$ . Let $Y=\overline{U}$ where $U$ is a non-empty open subset of $X$ . Let $S_1 \supset S_2 \supset S_3 \supset \cdots$ be a decreasing sequence of open subsets of $Y$ . Note that each $S_i$ contains points of the open set $U$ . Let $O_i=S_i \cap U$ for each $i$ . Note that the open sets $O_i$ form a decreasing sequence of open sets in the pseudocompact space $X$ . By Theorem 2, we have $\bigcap \limits_{n=1}^\infty \overline{O_n} \ne \varnothing$ (closure here is with respect to $X$ ). Note that points in $\bigcap \limits_{n=1}^\infty \overline{O_n}$ are also points in $\bigcap \limits_{n=1}^\infty \overline{S_n}$ (closure with respect to $Y$ ). By Theorem 2, $Y=\overline{U}$ is pseudocompact. $\blacksquare$

Theorem 4 (Statement $E$ above)
Let $X$ be a pseudocompact submetrizable space. Then $X$ is metrizable.

Proof
Let $(X,\tau)$ be a pseudocompact submetrizable space. Then there exists topology $\tau^*$ on $X$ such $(X,\tau^*)$ is metrizable and $\tau^* \subset \tau$ . We show that $\tau \subset \tau^*$ , leading to the conclusion that $(X,\tau)$ is also metrizable. If $A \subset X$ , we denote the closure of $A$ in $(X,\tau)$ by $cl_{\tau}(A)$ and the closure of $A$ in $(X,\tau^*)$ by $cl_{\tau^*}(A)$ .

To show that $\tau \subset \tau^*$ , we show any closed set with respect to the topology $\tau$ is also a closed set with respect to the topology $\tau^*$ . Let $C$ be a closed set in $(X,\tau)$ . Consider the family $\mathcal{W}=\left\{cl_{\tau}(U): U \in \tau \text{ and } C \subset U \right\}$ . We make the following claims.

Claim 1. $C=\bigcap \left\{W: W \in \mathcal{W} \right\}$ .

Claim 2. Each $W \in \mathcal{W}$ is pseudocompact in $(X,\tau)$ .

Claim 3. Each $W \in \mathcal{W}$ is pseudocompact in $(X,\tau^*)$ .

Claim 4. Each $W \in \mathcal{W}$ is compact in $(X,\tau^*)$ .

We now discuss each of these four claims. For Claim 1, it is clear that $C \subset \bigcap \left\{W: W \in \mathcal{W} \right\}$ . The reverse set inclusion follows from the fact that $X$ is a regular space. Claim 2 follows from Theorem 3. Note that sets in $\mathcal{W}$ are closed domains in the pseudocompact space $(X,\tau)$ .

If sets in $\mathcal{W}$ are pseudocompact in the larger topology $\tau$ , they would be pseudocompact in the weaker topology $\tau^*$ too. Thus Claim 3 is established. In a metrizable space, compactness and weaker notions such as countably compactness and pseudocompactness coincide. Because they are pseudocompact subsets, sets in $\mathcal{W}$ are compact in the metrizable space $(X,\tau^*)$ . Thus Claim 4 is established.

It follows that $C$ is closed in $(X,\tau^*)$ since it is the intersection of compact sets in $(X,\tau^*)$ . Thus $(X,\tau)$ is identical to $(X,\tau^*)$ , implying that $(X,\tau)$ is metrizable. $\blacksquare$

Reference

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

Sequentially compact spaces, II

Posted on September 4, 2010 by Dan Ma

All spaces under consideration are Hausdorff. Countably compactness and sequentially compactness are notions related to compactness. A countably compact space is one in which every countable open cover has a finite subcover, or equivalently, every countably infinite subset has a limit point. For a space $X$ , the point $p \in X$ is a limit point of $A \subset X$ if every open subset of $X$ containing $p$ contains a point of $A$ distinct from $p$ . On the other hand, a space $X$ is sequentially compact if every sequence $\left\{x_n:n=1,2,3,\cdots\right\}$ of points of $X$ has a subsequence that converges. Any sequentially compact space is countably compact. The converse is not true. The product space $2^I$ where $I=[0,1]$ is not sequentially compact (see Sequentially compact spaces, I) . However, for sequential spaces (first countable spaces in particular), the notion of sequentially compactness and countably compactness are equivalent. For previous discussion in this blog about sequential spaces, see the links below.

Lemma
Any countably compact space that is countable in size is metrizable and thus first countable.

Proof. Let $X$ be countably compact such that $\lvert X \lvert=\aleph_0$ . Then $X$ is compact (any Lindelof countably compact space is compact). In any countable space, the set of all singleton sets is a countable network. Any compact Hausdorff space with a countable network is metrizable and thus first countable. See Spaces With Countable Network. $\blacksquare$

Theorem
Let $X$ be a sequential space. Then $X$ is countably compact if and only if $X$ is sequentially compact.

Proof. The direction $\Leftarrow$ always holds without the space being sequential.

$\Rightarrow$ Suppose $X$ is countably compact. Suppose that $X$ is not sequentially compact. Then there is a sequence $\left\{x_n\right\}$ of points of $X$ with no convergent subsequence. Let $A$ be the set of all terms in this sequence, i.e. $A=\left\{x_n:n=1,2,3,\cdots\right\}$ . Note that $A$ is sequentially closed. Since $X$ is sequential, $A$ is closed in $X$ . As a closed subset of a countably compact space, $A$ is countably compact. By the lemma, $A$ is first countable. Since $A$ is an infinite compact space, $A$ has a non-isolated point $x$ . This means some sequence of points of $A$ converges to $x$ , contradicting the assumption that $\left\{x_n\right\}$ has no convergent subsequence. Therefore $X$ must be sequentially compact. $\blacksquare$

Previous posts on sequential spaces and k-spaces:
Sequential spaces, I
Sequential spaces, II
Sequential spaces, III
Sequential spaces, IV
Sequential spaces, V
k-spaces, I
k-spaces, II
A note about the Arens’ space
An observation about sequential spaces

Reference

Engelking, R. General Topology, Revised and Completed edition, 1989, Heldermann Verlag, Berlin.
Henkel, D. Solution to Monthly Problem 5698, American Mathematical Monthly 77, p. 896, 1970
Willard, S., General Topology, 1970, Addison-Wesley Publishing Company.

Dan Ma's Topology Blog

Expository Articles In General Topology

Tag Archives: Countably compact space

The product of a normal countably compact space and a metric space is normal

Countably paracompact spaces

Normality in the powers of countably compact spaces

One way to find collectionwise normal spaces

C*-Embedding Property and Stone-Cech Compactification

Completely Regular Spaces and Pseudocompact Spaces

Pseudocompact + Metacompact implies Compact

Baire Category Theorem and the Finite Intersection Property

When is a Pseudocompact Space Metrizable?

Sequentially compact spaces, II