# Countably paracompact spaces

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

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

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

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Normal Countably Paracompact Spaces

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

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

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

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

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

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

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

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

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

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

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Some Examples

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

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

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

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

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

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

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

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

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

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

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

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

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

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More on Countably Paracompactness

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

The part about decreasing follows from:

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

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

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

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Reference

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

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

# The product of uncountably many factors is never hereditarily normal

The space $Y=\prod_{\alpha<\omega_1} \left\{0,1 \right\}=\left\{0,1 \right\}^{\omega_1}$ is the product of $\omega_1$ many copies of the two-element set $\left\{0,1 \right\}$ where $\omega_1$ is the first uncountable ordinal. It is a compact space by Tychonoff’s theorem. It is a normal space since every compact Hausdorff space is normal. A space is hereditarily normal if every subspace is normal. Is the space $Y$ hereditarily normal? In this post, we give two proofs that it is not hereditarily normal. It then follows that any product space $\prod X_\alpha$ cannot be hereditarily normal as long as there are uncountably many factors and every factor has at least two point.

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The connection with a theorem of Katetov

It turns out that there is a connection with a theorem of Katetov. For any compact space, knowing hereditary normality of the first several self product spaces can reveal a great deal of information about the compact space. More specifically, for any compact space $X$, knowing whether $X$, $X^2$ and $X^3$ are hereditarily normal can tell us whether $X$ is metrizable. If all three are hereditarily normal, then $X$ is metrizable. If one of the three self products is not hereditarily normal, then $X$ is not metrizable. This fact is based on a theorem of Katetov (see this previous post). The space $Y=\left\{0,1 \right\}^{\omega_1}$ is not metrizable since it is not first countable (see Problem 1 below). Thus one of its first three self products must fail to be hereditarily normal.

These two proofs are not direct proof in the sense that a non-normal subspace is not explicitly produced. Instead the proofs use other theorem or basic but important background results. One of the two proofs (#2) uses a theorem of Katetov on hereditarily normal spaces. The other proof (#1) uses the fact that the product of uncountably many copies of a countable discrete space is not normal. We believe that these two proofs and the required basic facts are an important training ground for topology. We list out these basic facts as exercises. Anyone who wishes to fill in the gaps can do so either by studying the links provided or by consulting other sources.

The theorem of Katetov mentioned earlier provides a great exercise – for any non-metrizable compact space $X$, determine where the hereditary normality fails. Does it fail in $X$, $X^2$ or $X^3$? This previous post examines a small list of compact non-metrizable spaces. In all the examples in this list, the hereditary normality fails in $X$ or $X^2$. The space $Y=\left\{0,1 \right\}^{\omega_1}$ can be added to this list. All the examples in this list are defined using no additional set theory axioms beyond ZFC. A natural question: does there exist an example of compact non-metrizable space $X$ such that the hereditary normality holds in $X^2$ and fails in $X^3$? It turns out that this was a hard problem and the answer is independent of ZFC. This previous post provides a brief discussion and has references for the problem.

All spaces under consideration are Hausdorff spaces.

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Exercises

Problem 1
Let $X$ be a compact space. Show that $X$ is normal.

Problem 2
For each $\alpha<\omega_1$, let $A_\alpha$ be a set with cardinality $\le \omega_1$. Show that $\lvert \bigcup_{\alpha<\omega_1} A_\alpha \lvert \le \omega_1$.

Problem 2 holds for any infinite cardinal, not just $\omega_1$. One reference for Problem 2 is Lemma 10.21 on page 30 of Set Theorey, An Introduction to Independence Proofs by Kenneth Kunen.

Problem 3
For each $\alpha<\omega_1$, let $X_\alpha$ be a space with at least two points. Show that for every point $p \in \prod_{\alpha<\omega_1} X_\alpha$, there does not exist a countable base at the point $p$. In other words, the product space $\prod_{\alpha<\omega_1} X_\alpha$ is not first countable at every point. It follows that product space $\prod_{\alpha<\omega_1} X_\alpha$ is not metrizable.

Problem 4
In any space, a $G_\delta$-set is a set that is the intersection of countably many open sets. When a singleton set $\left\{ x \right\}$ is a $G_\delta$-set, we say the point $x$ is a $G_\delta$-point. For each $\alpha<\omega_1$, let $X_\alpha$ be a space with at least two points. Show that every point $p$ in the product space $\prod_{\alpha<\omega_1} X_\alpha$ is not a $G_\delta$-point.

Note that Problem 4 implies Problem 3.

For Problem 3 and Problem 4, use the fact that there are uncountably many factors and that a basic open set in the product space is of the form $\prod_{\alpha<\omega_1} O_\alpha$ and that it has only finitely many coordinates at which $O_\alpha \ne X_\alpha$.

Problem 5
For each $\alpha<\omega_1$, let $X_\alpha=\left\{0,1,2,\cdots \right\}$ be the set of non-negative integers with the discrete topology. Show that the product space $\prod_{\alpha<\omega_1} X_\alpha$ is not normal.

See here for a discussion of Problem 5.

Problem 6
Let $\displaystyle Y=\left\{0,1 \right\}^{\omega_1}$. Show that $Y$ has a countably infinite subspace

$W=\left\{y_0,y_1,y_2,y_3\cdots \right\}$

such that $W$ is relatively discrete. In other words, $W$ is discrete in the subspace topology of $W$. However $W$ is not discrete in the product space $Y$ since $Y$ is compact.

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Proof #1

Let $\displaystyle Y=\left\{0,1 \right\}^{\omega_1}$. We show that $Y$ is not hereditarily normal.

Note that the product space $\displaystyle Y=\left\{0,1 \right\}^{\omega_1}$ can be written as the product of $\omega_1$ many copies of itself:

$\displaystyle \left\{0,1 \right\}^{\omega_1} \cong \left\{0,1 \right\}^{\omega_1} \times \left\{0,1 \right\}^{\omega_1} \times \left\{0,1 \right\}^{\omega_1} \times \cdots \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (1)$

The fact (1) follows from the fact that the union of $\omega_1$ many pairwise disjoint sets, each of which has cardinality $\omega_1$, has cardinality $\omega_1$ (see Problem 2). The space $\left\{0,1 \right\}^{\omega_1}$ has a countably infinite subspace that is relatively discrete (see Problem 6). In other words, it has a subspace that is homemorphic to $\omega=\left\{0,1,2,\cdots \right\}$ where $\omega$ has the discrete topology. Thus the following is homeomorphic to a subspace of $\displaystyle Y=\left\{0,1 \right\}^{\omega_1}$.

$\displaystyle \omega^{\omega_1} = \omega \times \omega \times \omega \times \cdots \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (2)$

By Problem 5, the space $\omega^{\omega_1}$ is not normal. Hence the compact space $\displaystyle Y=\left\{0,1 \right\}^{\omega_1}$ contains the non-normal space $\omega^{\omega_1}$ and is thus not hereditarily normal. $\blacksquare$

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Proof #2

Let $\displaystyle Y=\left\{0,1 \right\}^{\omega_1}$. We show that $Y$ is not hereditarily normal. This proof uses a theorem of Katetov, discussed in this previous post and stated below.

Theorem 1
If $X_1 \times X_2$ is hereditarily normal (i.e. every one of its subspaces is normal), then one of the following condition holds:

• The factor $X_1$ is perfectly normal.
• Every countable and infinite subset of the factor $X_2$ is closed.

First, $Y$ can be written as the product of two copies of itself:

$\displaystyle \left\{0,1 \right\}^{\omega_1} \cong \left\{0,1 \right\}^{\omega_1} \times \left\{0,1 \right\}^{\omega_1} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (3)$

This is because the union of two disjoints sets, each of which has cardinality $\omega_1$, has carinality $\omega_1$. Note that the countably infinite subset $W$ from Problem 6 is not a closed subset of $Y$. If it were, the compact space $Y$ would contain an infinite set with no limit point. Thus the second condition of Theorem 1 is not satisfied. If $Y \cong Y \times Y$ were to be hereditarily normal, then the first condition must be satisfied, i.e. $Y$ is perfectly normal (meaning that $Y$ is normal and that every closed subset of it is a $G_\delta$-set). However, Problem 4 indicates that no point in $Y$ can be a $G_\delta$ point. Therefore $Y$ cannot be hereditarily normal. $\blacksquare$

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Corollary

The product of uncountably many spaces, each one of which has at least two points, contains a homeomorphic copy of the space $\displaystyle Y=\left\{0,1 \right\}^{\omega_1}$. Thus such a product space can never be hereditarily normal. We state this more formally below.

Theorem 2
Let $\kappa$ be any uncountable cardinal. For each $\alpha<\kappa$, let $X_\alpha$ be a space with at least two points. Then $\prod_{\alpha<\kappa} X_\alpha$ is not hereditarily normal.

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

# A useful lemma for proving normality

In this post we discuss a lemma (Lemma 1 below) that is useful for proving normality. In some cases, it is more natural using this lemma to prove that a space is normal than using the definition of normality. The proof of Lemma 1 is not difficult. Yet it simplifies some proofs of normality. One reason is that the derivation of two disjoint open sets that are to separate two disjoint closed sets is done in the lemma, thus simplifying the main proof at hand. The lemma is well known and is widely used in the literature. See Lemma 1.5.15 in [1]. Two advanced examples of applications are [2] and [3]. After proving the lemma, we give three elementary applications of the lemma. One of the applications is a characterization of perfectly normal spaces. This characterization is, in some cases, easier to use, e.g. making it easy to show that perfectly normal implies hereditarily normal.

In this post, we only consider spaces that are regular and $T_1$. A space $X$ is regular if for each open set $U \subset X$ and for each $x \in U$, there exists an open $V \subset X$ with $x \in V \subset \overline{V} \subset U$. A space is $T_1$ if every set with only one point is a closed set.

Lemma 1
A space $Y$ is a normal space if the following condition (Condition 1) is satisfied:

1. For each closed subset $L$ of $Y$, and for each open subset $M$ of $Y$ with $L \subset M$, there exists a sequence $M_1,M_2,M_3,\cdots$ of open subsets of $Y$ such that $L \subset \bigcup_{i=1}^\infty M_i$ and $\overline{M_i} \subset M$ for each $i$.

Proof of Lemma 1
Suppose the space $Y$ satisfies condition 1. Let $H$ and $K$ be disjoint closed subsets of the space $Y$. Consider $H \subset U=Y \backslash K$. Using condition 1, there exists a sequence $U_1,U_2,U_3,\cdots$ of open subsets of the space $Y$ such that $H \subset \bigcup_{i=1}^\infty U_i$ and $\overline{U_i} \cap K=\varnothing$ for each $i$. Consider $K \subset V=Y \backslash H$. Similarly, there exists a sequence $V_1,V_2,V_3,\cdots$ of open subsets of the space $Y$ such that $K \subset \bigcup_{i=1}^\infty V_i$ and $\overline{V_i} \cap H=\varnothing$ for each $i$.

For each positive integer $n$, define the open sets $U_n^*$ and $V_n^*$ as follows:

$U_n^*=U_n \backslash \bigcup_{k=1}^n \overline{V_k}$

$V_n^*=V_n \backslash \bigcup_{k=1}^n \overline{U_k}$

Let $P=\bigcup_{n=1}^\infty U_n^*$ and $Q=\bigcup_{n=1}^\infty V_n^*$. It is clear $P$ and $Q$ are open and that $H \subset P$ and $K \subset Q$. We claim that $P$ and $Q$ are disjoint. Suppose $y \in P \cap Q$. Then $y \in U_n^*$ for some $n$ and $y \in V_m^*$ for some $m$. Assume that $n \le m$. The fact that $y \in U_n^*$ implies $y \in U_n$. The fact that $y \in V_m^*$ implies that $y \notin \overline{U_j}$ for all $j \le m$. In particular, $y \notin U_n$, a contradiction. Thus $P \cap Q=\varnothing$. This completes the proof that the space $Y$ is normal. $\blacksquare$

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Spaces with nice bases

One application is that spaces with a certain type of bases satisfy condition 1 and thus are normal. For example, spaces with bases that are countable and spaces with bases that are $\sigma$-locally finite. Spaces with these bases are metrizable. The proof that these spaces are metrizable will be made easier if they can be shown to be normal first. The Urysohn functions (the functions described in Urysohn’s lemma) can then be used to embed the space in question into some universal space that is known to be metrizable. Using regularity and Lemma 1, it is straightforward to verify the following three propositions.

Proposition 2
Let $X$ be a regular space with a countable base. Then $X$ is normal.

Proposition 3
Let $X$ be a regular space with a $\sigma$-locally finite base. Then $X$ is normal.

Proposition 4
Let $X$ be a regular space with a $\sigma$-discrete base. Then $X$ is normal.

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A characterization of perfectly normal spaces

Another application of Lemma 1 is that it leads naturally to a characterization of perfect normality. Recall that a space $X$ is perfectly normal if $X$ is normal and perfect. A space $X$ is perfect if every closed subset of $X$ is a $G_\delta$ set (i.e. the intersection of countably many open subsets of $X$). Equivalently a space $X$ is perfect if and only if every open subset of $X$ is an $F_\sigma$ set, i.e., the union of countably many closed subsets of $X$. We have the following theorem.

Theorem 5
A space $Y$ is perfectly normal if and only if the following condition holds.

1. For each open subset $M$ of $Y$, there exists a sequence $M_1,M_2,M_3,\cdots$ of open subsets of $Y$ such that $M \subset \bigcup_{i=1}^\infty M_i$ and $\overline{M_i} \subset M$ for each $i$.

Clearly, condition 2 is strongly than condition 1.

Proof of Theorem 5
$\Longrightarrow$
Suppose that the space $Y$ is perfectly normal. Let $M$ be a non-empty open subset of $Y$. Then $M=\bigcup_{n=1}^\infty P_n$ where each $P_n$ is a closed subset of $Y$. Using normality of $Y$, for each $n$, there exists open subset $M_n$ of $Y$ such that $P_n \subset M_n \subset \overline{M_n} \subset M$. Then consition 2 is satisfied.

$\Longleftarrow$
Suppose condition 2 holds, which implies condition 1 of Lemma 1. Then $Y$ is normal. It is clear that condition 2 implies that every open subset of $Y$ is an $F_\sigma$ set. $\blacksquare$

The characterization of perfectly normal spaces in Theorem 5 is hereditary. This means that any subspace of a perfectly normal space is also perfectly normal. In particular, perfectly normal implies hereditarily normal. Thus we have the following theorem.

Theorem 6
Condition 2 in Theorem 5 is hereditary, i.e., if a space satisfies Condition 2, every subspace satisfies Condition 2. Therefore if the space $Y$ is a perfectly normal space, then every subspace of $Y$ is also perfectly normal. In particular, if $Y$ is perfectly normal, then $Y$ is hereditarily normal (i.e. every subspace of $Y$ is normal).

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Normality is hereditary with respect to $F_\sigma$ subsets

Normality is not a hereditary notion. Lemma 1 can used to show that normality is hereditary with respect to $F_\sigma$ subspaces.

Theorem 7
Let $Y$ be a normal space. Then every $F_\sigma$ subspace of $Y$ is normal.

Proof of Theorem 7
Let $H$ be a subspace of $Y$ such that $H=\bigcup_{n=1}^\infty P_n$ where each $P_n$ is closed subset of $Y$. Let $L$ be a closed subset of $H$ and let $M$ be an open subset of $H$ such that $L \subset M$. We need to find $M_1,M_2,M_3,\cdots$, open in $H$, such that $L \subset \bigcup_{i=1}^\infty M_i$ and $\overline{M_i} \subset M$ for all $i$ (closure of $M_i$ is within $H$).

Let $U$ be an open subset of $Y$ such that $M=U \cap H$. For each positive integer $n$, let $H_n=P_n \cap L$. Obviously $H_n$ is closed in $H$. It is also the case that $H_n$ is closed in $Y$. To see this, let $p \in Y$ be a limit point of $H_n$. Then $p$ is a limit point of $P_n$. Hence $p \in P_n$ since $P_n$ is closed in $Y$. We now have $p \in H$. The point $p$ is also a limit point of $L$. Thus $p \in L$ since $L$ is closed in $H$. Now we have $p \in H_n=P_n \cap L$, proving that $H_n$ is closed in $Y$.

Now we have $H_n \subset U$ for all $n$. By Lemma 1, for each $n$, there exists a sequence $U_{n,1},U_{n,2},U_{n,3},\cdots$ of open subsets of $Y$ such that $H_n \subset \bigcup_{j=1}^\infty U_{n,j}$ and $\overline{U_{n,j}} \subset U$ for all $j$. Note that $L=\bigcup_{n=1}^\infty H_n$. Rename $M_{n,j}=U_{n,j} \cap H$ over all $n,j$ by the sequence $M_1,M_2,M_3,\cdots$. Then $L \subset \bigcup_{i=1}^\infty M_i$. It also follows that $\overline{M_i} \subset M$ for all $i$ (closure of $M_i$ is within $H$). This completes the proof that the $F_\sigma$ set $H$ is normal. $\blacksquare$

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Reference

1. Engelking, R., General Topology, Revised and Completed edition, Heldermann Verlag, Berlin, 1989.
2. Gruenhage, G., Normality in $X^2$ for complete $X$, Trans. Amer. Math. Soc., 340 (2), 563-586, 1993.
3. Nyikos, P., A compact nonmetrizable space $P$ such that $P^2$ is completely normal, Topology Proc., 2, 359-363, 1977.

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$\copyright \ 2014-2015 \text{ by Dan Ma}$ Revised April 14, 2015

# Bing’s Example H

In a previous post we introduced Bing’s Example G, a classic example of a normal but not collectionwise normal space. Other properties of Bing’s Example G include: completely normal, not perfectly normal and not metacompact. This is an influential example introduced in an influential paper of R. H. Bing in 1951 (see [1]). In the same paper, another example called Example H was introduced. This space has some of the same properties of Example G, except that it is perfectly normal. In this post, we define and discuss Example H.

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Defining Bing’s Example H

Throughout the discussion in this post, we use $\omega$ to denote the first infinite ordinal, i.e., $\omega =\left\{0,1,2,3,\cdots \right\}$. Let $P$ be any uncountable set. Let $Q$ be the set of all subsets of the set $P$, i.e., it is the power set of $P$. Let $H$ be the set of all functions $f:Q \rightarrow \omega$. In other words, the set $H$ is the Cartesian product $\prod \limits_{q \in Q} \omega$. But the topology on $H$ is not the product topology.

For each $p \in P$, consider the function $f_p:Q \rightarrow 2=\left\{0,1 \right\}$ such that for each $q \in Q$:

$f_p(q) = \begin{cases} 1, & \mbox{if } p \in q \\ 0, & \mbox{if } p \notin q \end{cases}$

Let $H_P=\left\{f_p: p \in P \right\}$. Now define a topology on the set $H$ by the following:

• Each point of $H-H_P$ is an isolated point.
• Each point $f_p \in H_P$ has basic open sets of the form $U(p,W,n)$ defined as follows:

$U(p,W,n)=\left\{f_p \right\} \cup D(p,W,n)$

$D(p,W,n)=\left\{f \in H: \forall q \in Q, f(q) \ge n \text{ and } \forall q \in W, f(q) \equiv f_p(q) \ (\text{mod} \ 2) \right\}$

where $p \in P$, $W \subset Q$ is finite, and $n \in \omega$.

If $a$ and $b$ are integers, the $a \equiv b \ (\text{mod} \ 2)$ means that $a-b$ is divisible by $2$. The congruence equation $f(q) \equiv f_p(q) \ (\text{mod} \ 2)$ means that $f(q)$ is an even integer if $f_p(q)=0$. On the other hand, $f(q) \equiv f_p(q) \ (\text{mod} \ 2)$ means that $f(q)$ is an odd integer if $f_p(q)=1$.

The set $D(p,W,n)$ seems to mimic a basic open set of the point $f_p$ in the product topology: for each point in $D(p,W,n)$, the value of each coordinate is an integer $\ge n$ and the values for finitely many coordinates are fixed to agree with the function $f_p$ modulo $2$. Adding the point $f_p$ to $D(p,W,n)$, we have a basic open set $U(p,W,n)$.

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Basic Discussion

The points in $H-H_P$ are the isolated points in the space $H$. The points in $H_P$ are the non-isolated points (limit points). The space $H$ is a Hausdorff space. Another interesting point is that the set $H_P$ is a closed and discrete set in the space $H$.

To see that $H$ is Hausdorff, let $h_1, h_2 \in H$ with $h_1 \ne h_2$. Consider the case that $h_1$ is an isolated point and $h_2=f_p$ for some $p \in P$. Let $n$ be the minimum of all $h_1(q)$ over all $q \in Q$. Let $O_1=\left\{h_1 \right\}$ and $O_2=U(p,W,n+1)$ where $W \subset Q$ is any finite set. Then $O_1$ and $O_2$ are disjoint open set containing $h_1$ and $h_2$, respectively.

Now consider the case that $h_1=f_p$ and $h_2=f_{p'}$ where $p \ne p'$. Let $O_1=U(p,W,0)$ and $O_2=U(p',W,0)$ where $W=\left\{ \left\{ p \right\},\left\{ p' \right\} \right\}$. Then $O_1$ and $O_2$ are disjoint open set containing $h_1$ and $h_2$, respectively.

The set $H_P$ is a closed and discrete set in the space $H$. It is closed since $H-H_P$ consists of isolated points. To see that $H_P$ is discrete, note that $U(p,W,0)$, where $W=\left\{ \left\{ p \right\} \right\}$, is an open set with $f_p \in U(p,W,0)$ and $f_{p'} \notin U(p,W,0)$ for all $p' \ne p$.

In the sections below, we show that the space $H$ is normal, completely normal (thus hereditarily normal), and is perfectly normal. Furthermore, we show that it is not collectionwise Hausdorff (hence not collectionwise normal) and not meta-lindelof (hence not metacompact).

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Bing’s Example H is Normal

In the next section, we show that Bing’s Example H is completely normal (i.e. any two separated sets can be separated by disjoint open sets). Note that any two disjoint closed sets are separated sets.

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Bing’s Example H is Completely Normal

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

Let $S$ and $T$ be separated sets in the space $H$, i.e.,

$S \cap \overline{T}=\varnothing=\overline{S} \cap T$

We consider two cases. Case 1 is that one of the sets consists entirely of isolated points. Assume that $S \subset H-H_P$. Let $O_1=S$. For each $x \in T$, choose an open set $V_x$ with $x \in V_x$ and $V_x \cap \overline{S}=\varnothing$. Let $O_2=\bigcup \limits_{x \in T} V_x$. Then $O_1$ and $O_2$ are disjoint open sets containing $S$ and $T$ respectively.

Now consider Case 2 where $S_1=S \cap H_P \ne \varnothing$ and $T_1=T \cap H_P \ne \varnothing$. Consider the sets $q_1$ and $q_2$ defined as follows:

$q_1=\left\{p \in P: f_p \in S_1 \right\}$

$q_2=\left\{p \in P: f_p \in T_1 \right\}$

Let $W=\left\{q_1,q_2 \right\}$. Let $Y_1$ and $Y_2$ be the following open sets:

$Y_1=\bigcup \limits_{p \in q_1} U(p,W,0)$

$Y_2=\bigcup \limits_{p \in q_2} U(p,W,0)$

Immediately, we know that $S_1 \subset Y_1$, $T_1 \subset Y_2$ and $Y_1 \cap Y_2=\varnothing$. Let $S_2=S \cap (H-H_P)$ and $T_2=T \cap (H-H_P)$ (both of which are open). Let $O_1$ and $O_2$ be the following open sets:

$O_1=(Y_1 \cup S_2)-\overline{T}$

$O_2=(Y_2 \cup T_2)-\overline{S}$

Then $O_1$ and $O_2$ are disjoint open sets containing $S$ and $T$ respectively.

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Bing’s Example H is Perfectly Normal

A space is perfectly normal if it is normal and that every closed subset is a $G_\delta$-set (i.e. the intersection of countably many open subsets). All we need to show here is that every closed subset is a $G_\delta$-set.

Let $C \subset H$ be a closed set. Of course, if $C$ consists entirely of isolated points, then we are done. So assume that $C \cap H_P \ne \varnothing$. Let $q*=\left\{p \in P: f_p \in C \right\}$. Let $O=C \cap (H-H_P)$, which is open. For each positive integer $n$, define the open set $Y_n$ as follows:

$Y_n=O \cup \biggl( \bigcup \limits_{p \in q*} U(p,\left\{q* \right\},n) \biggr)$

Immediately we have $C \subset Y_n$ for each $n$. Let $g \in \bigcap \limits_{n=1}^\infty Y_n$. We claim that $g \in C$. Suppose $g \notin C$. Then $g \notin O$. It follows that for each $n$ $g \in U(p_n,\left\{q* \right\},n)$ for some $p_n \in q*$. Recall that $U(p_n,\left\{q* \right\},n)=\left\{f_{p_n} \right\} \cup D(p_n,\left\{q* \right\},n)$.

The assumption that $g \notin C$ implies that $g \ne f_{p_n}$ for all $n$. Then $g \in D(p_n,\left\{q* \right\},n)$ for all $n$. By the definition of $D(p_n,\left\{q* \right\},n)$, it follows that for all $q \in Q$, $g(q) \ge n$ for all positive integer $n$. This is a contradiction. So it must be the case that $g \in C$. This completes the proof that Bing’s Example H is perfectly normal.

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Collectionwise Normal Spaces

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

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

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

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Bing’s Example H is not Collectionwise Hausdorff

To prove that Bing’s Example H is not collectionwise Hausdorff, we need an intermediate result (Lemma 1) that is based on an infinitary combinatorial result called the Delta-system lemma.

A family $\mathcal{A}$ of sets is called a Delta-system (or $\Delta$-system) if there exists a set $r$, called the root of the $\Delta$-system, such that for any $A,B \in \mathcal{A}$ with $A \ne B$, we have $A \cap B=r$. The following is a version of the Delta-system lemma (see Theorem 1.5 in p. 49 of [2]).

Delta-System Lemma

Let $\mathcal{A}$ be an uncountable family of finite sets. Then there exists an uncountable $\mathcal{B} \subset \mathcal{A}$ such that $\mathcal{B}$ is a $\Delta$-system.
Lemma 1

Let $P_0 \subset P$ be any uncountable subset. For each $p \in P_0$, let $U(p,W_p,n_p)$ be a basic open subset containing $f_p$. Then there exists an uncountable $P_1 \subset P_0$ such that $\bigcap \limits_{p \in P_1} U(p,W_p,n_p) \ne \varnothing$.

Proof of Lemma 1
Let $\mathcal{A}=\left\{W_p: p \in P_0 \right\}$. We need to break this up into two cases – $\mathcal{A}$ is a countable family of finite sets or an uncountable family of finite sets. The first case is relatively easy to see. The second case requires using the Delta-system lemma.

Suppose that $\mathcal{A}$ is countable. Then there exists an uncountable $R \subset P_0$ such that for all $p,t \in R$ with $p \ne t$, we have $W_p=W_t=W$ and $n_p=n_t=n$. Suppose that $W=\left\{q_1,q_2,\cdots,q_m \right\}$. By inductively working on the sets $q_j$, we can obtain an uncountable set $P_1 \subset R$ such that for all $p,t \in P_1$ with $p \ne t$, we have $f_p(q_j)=f_t(q_j)$ for each $j=1,2,\cdots,m$. Clearly, we have:

$\bigcap \limits_{p \in P_1} U(p,W,n) \ne \varnothing$

To show the above, just define a function $h:Q \rightarrow \left\{n,n+1,n+2,\cdots \right\}$ such that $h(q_j)=f_p(q_j)$ for all $j=1,2,\cdots,m$ for one particular $p \in P_1$. Then $h$ belongs to the intersection.

Suppose that $\mathcal{A}$ is uncountable. By the Delta-system lemma, there is an uncountable $R \subset P_0$ and there exists a finite set $r \subset Q$ such that for all $p,t \in R$ with $p \ne t$, we have $W_p \cap W_t=r$. Suppose that $r=\left\{q_1,q_2,\cdots,q_m \right\}$. As in the previous case, work inductively on the sets $q_j$, we can obtain an uncountable $S \subset R$ such that for all $p,t \in S$ with $p \ne t$, we have $f_p(q_j)=f_t(q_j)$ for each $j=1,2,\cdots,m$. Now narrow down to an uncountable $P_1 \subset S$ such that $n_p=n_t=n$ for all $p,t \in P_1$ with $p \ne t$. We now show that

$\bigcap \limits_{p \in P_1} U(p,W_p,n) \ne \varnothing$

To define a function $h:Q \rightarrow \left\{n,n+1,n+2,\cdots \right\}$ that belongs to the above intersection, we define $h$ so that $h$ matches $f_t$ (mod 2) with one particular $t \in P_1$ on the set $r=\left\{q_1,q_2,\cdots,q_m \right\}$. Note that $W_p-r$ are disjoint over all $p \in P_1$. So $h$ can be defined on $W_p-r$ to match $f_p$ (mod 2). For any remaining values in the domain, define $h$ freely to be at least the integer $n$. Then the function $h$ belongs to the intersection.

With the two cases established, the proof of Lemma 1 is completed. $\blacksquare$

The fact that Example H is not collectionwise Hausdorff is a corollary of Lemma 1. The set $H_P$ is a discrete collection of points in the space $H$. It follows that $H_P$ cannot be separated by disjoint open sets. For each $p \in P$, let $U(p,W_p,n_p)$ be a basic open set containing the point $f_p$. By Lemma 1, there is an uncountable $P_1 \subset P$ such that $\bigcap \limits_{p \in P_1} U(p,W_p,n_p) \ne \varnothing$. Thus there can be no disjoint collection of open sets in $H$ that separate the points in $H_P$.

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Bing’s Example H is not Metacompact

Let $X$ be a space. A collection $\mathcal{A}$ of subsets of $X$ is said to be a point-finite (point-countable) collection if every point of $X$ belongs to only finitely (countably) many sets in $\mathcal{A}$. A space $X$ is said to be a metacompact space if every open cover $\mathcal{U}$ of $X$ has a point-finite open refinement $\mathcal{V}$. A space $X$ is said to be a meta-Lindelof space if every open cover $\mathcal{U}$ of $X$ has a point-countable open refinement $\mathcal{V}$. Clearly, every metacompact space is meta-Lindelof.

It follows from Lemma 1 that Example H is not meta-Lindelof. Thus Example H is not metacompact. To see that it is not meta-Lindelof, for each $f_p \in H_P$, let $U_{f_p}=U(p,\left\{\left\{p \right\} \right\},0)$, and for each $x \in H-H_P$, let $U_x=\left\{x \right\}$. Let $\mathcal{U}$ be the following open cover of $H$:

$\mathcal{U}=\left\{U_x: x \in H \right\}$

Each $f_p \in H_P$ belongs to only one set in $\mathcal{U}$, namely $U_{f_p}$. So for any open refinement $\mathcal{V}$ of $\mathcal{U}$ (consisting of basic open sets), we have uncountably many open sets of the form $U(p,W_p,n_p)$. By Lemma 1, we can find uncountably many such open sets with non-empty intersection. So no open refinement of $\mathcal{U}$ can be point-countable.

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Reference

1. Bing, R. H., Metrization of Topological Spaces, Canad. J. Math., 3, 175-186, 1951.
2. Kunen, K., Set Theory, An Introduction to Independence Proofs, North-Holland, Amsterdam, 1980.

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

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# A subspace of Bing’s example G

Bing’s Example G is the first example of a topological space that is normal but not collectionwise normal (see [1]). Example G was an influential example from an influential paper. The Example G and its subspaces had been extensively studied. In addition to being normal and not collectionwise normal, Example G is not perfectly normal and not metacompact. See the previous post “Bing’s Example G” for a basic discussion of Example G. In this post we focus on one subspace of Example G examined by Michael in [3]. This subspace is normal, not collectionwise normal and not perfectly normal just like Example G. However it is metacompact. In [3], Michael proved that any metacompact collectionwise normal space is paracompact (metacompact was called pointwise paracompact in that paper). This subspace of Example G demonstrates that collectionwise normality in Michael’s theorem cannot be replaced by normality.

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Bing’s Example G

For a more detailed discussion of Bing’s Example G in this blog, see the blog post “Bing’s Example G”. For the sake of completeness, we repeat the definition of Example G. Let $P$ be any uncountable set. Let $Q$ be the set of all subsets of $P$. Let $F=2^Q$ be the set of all functions $f: Q \rightarrow 2=\left\{0,1 \right\}$. Obviously $2^Q$ is simply the Cartesian product of $\lvert Q \lvert$ many copies of the two-point discrete space $\left\{0,1 \right\}$, i.e., $\prod \limits_{q \in Q} \left\{0,1 \right\}$. For each $p \in P$, define the function $f_p: Q \rightarrow 2$ by the following:

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

Let $F_P=\left\{f_p: p \in P \right\}$. Let $\tau$ be the set of all open subsets of $2^Q$ in the product topology. The following is another topology on $2^Q$:

$\tau^*=\left\{U \cup V: U \in \tau \text{ and } V \subset 2^Q \text{ with } V \cap F_P=\varnothing \right\}$

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

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Michael’s Subspace of Example G

For each $f \in F$, let $supp(f)$ be the support of $f$, i.e., $supp(f)=\left\{q \in Q:f(q) \ne 0 \right\}$. Michael in [3] considered the following subspace of $F$.

$M=F_P \cup \left\{f \in F: supp(f) \text{ is finite} \right\}$

Michael in [3] used the letter $G$ to denote the space $M$. We choose another letter to distinguish it from Example G. The subspace $M$ consists of all points $f_p \in F_P$ and all other $f \in F$ such that $f(q)=1$ for only finitely many $q \in Q$. The space $M$ is normal and not collectionwise Hausdorff (hence not collectionwise normal and not paracompact). By eliminating points $f \in F$ that have values of $1$ for infinitely many $q \in Q$, we obtain a subspace that is metacompact. We discuss the following points:

• The space $M$ is normal.
• The space $M$ is not collectionwise Hausdorff and hence not collectionwise normal.
• The space $M$ is metacompact.
• The space $M$ is not perfectly normal.

The space $M$ is normal since the space $F$ that is Example G is hereditarily normal (see the section called Bing’s Example G is Completely Normal in the post “Bing’s Example G”).

To show that the space $M$ is not collectionwise Hausdorff, it is helpful to first look at $M$ as a subspace of the product space $2^Q$. The product space $2^Q$ has the countable chain condition (CCC) since it is a product of separable spaces. Note that $M$ is dense in the product space $2^Q$. Thus $M$ as a subspace of the product space has the CCC.

In the space $M$, the set $F_P$ is still a closed and discrete set. In the space $M$, open sets containing points of $F_P$ are the same as product open sets in $2^Q$ relative to the set $M$. Since $M$ has CCC (as a subspace of the product space $2^Q$), $M$ cannot have uncountably many pairwise disjoint open sets containing points of $F_P$ (in either the product topology or the Example G subspace topology). It follows that $M$ is not collectionwise Hausdorff. If it were, there would be uncountably many pairwise disjoint product open sets separating points in $F_P$, which is not possible.

To see that $M$ is metacompact, let $\mathcal{U}$ be an open cover of $M$. For each $p \in P$, choose $U_p \in \mathcal{U}$ such that $f_p \in U_p$. For each $p \in P$, let $W_p=\left\{f \in M: f(\left\{p \right\})=1 \right\}$. Let $\mathcal{V}$ be the following:

$\mathcal{V}=\left\{U_p \cap W_p: p \in P \right\} \cup \left\{\left\{x \right\}: x \in M-F_P \right\}$

Note that $\mathcal{V}$ is a point-finite open refinement of $\mathcal{U}$. Each $U_p \cap W_p$ contains only one point of $F_P$, namely $f_p$. On the other hand, each $f \in M$ with finite support can belong to at most finitely many $U_p \cap W_p$.

The space $M$ is not perfectly normal. This point is alluded to in [3] by Michael and elsewhere in the literature, e.g. in Bing’s paper (see [1]) and in Engelking’s general topology text (see 5.53 on page 338 of [2]). In fact Michael indicated that one can obtain a perfectly normal example with the aforementioned properties using Example H defined in [1] instead of using the subspace $M$ defined here in this post.

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Reference

1. Bing, R. H., Metrization of Topological Spaces, Canad. J. Math., 3, 175-186, 1951.
2. Engelking, R., General Topology, Revised and Completed edition, Heldermann Verlag, Berlin, 1989.
3. Michael, E., Point-finite and locally finite coverings, Canad. J. Math., 7, 275-279, 1955.
4. Willard, S., General Topology, Addison-Wesley Publishing Company, 1970.

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$\copyright \ \ 2012$

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# Bing’s Example G

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

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

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Bing’s Example G – Definition

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

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

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

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

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

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Bing’s Example G – Initial Discussion

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

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

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

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

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

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Bing’s Example G is Normal

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

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

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

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

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Collectionwise Normal Spaces

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

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

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

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Bing’s Example G is not Collectionwise Hausdorff

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

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Bing’s Example G is Completely Normal

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

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

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

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

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

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

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Bing’s Example G is not Perfectly Normal

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

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

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

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

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Bing’s Example G is not Metacompact

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

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

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

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Reference

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

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$\copyright \ \ 2012$

# Cartesian Products of Two Paracompact Spaces – Continued

Consider the real line $\mathbb{R}$ with a topology finer than the usual topology obtained by isolating each point in $\mathbb{P}$ where $\mathbb{P}$ is the set of all irrational numbers. The real line with this finer topology is called the Michael line and we use $\mathbb{M}$ to denote this topological space. It is a classic result that $\mathbb{M} \times \mathbb{P}$ is not normal (see “Michael Line Basics”). Even though the Michael line $\mathbb{M}$ is paracompact (it is in fact hereditarily paracompact), $\mathbb{M}$ is not perfectly normal. Result 3 below will imply that the Michael line cannot be perfectly normal. Otherwise $\mathbb{M} \times \mathbb{P}$ would be paracompact (hence normal). Result 3 is the statement that if $X$ is paracompact and perfectly normal and Y is a metric space then $X \times Y$ is paracompact and perfectly normal. We also use this result to show that if $X$ is hereditarily Lindelof and $Y$ is a separable metric space, then $X \times Y$ is hereditarily Lindelof (see Result 4 below).

This post is a continuation of the post “Cartesian Products of Two Paracompact Spaces”. In that post, four results are listed. They are:

Result 1

If $X$ is paracompact and $Y$ is compact, then $X \times Y$ is paracompact.

Result 2

If $X$ is paracompact and $Y$ is $\sigma$-compact, then $X \times Y$ is paracompact.

Result 3

If $X$ is paracompact and perfectly normal and $Y$ is metrizable, then $X \times Y$ is paracompact and perfectly normal.

Result 4

If $X$ is hereditarily Lindelof and $Y$ is a separable metric space, then $X \times Y$ is hereditarily Lindelof.

Result 1 and Result 2 are proved in the previous post “Cartesian Products of Two Paracompact Spaces”. Result 3 and Result 4 are proved in this post. All spaces are assumed to be regular.

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Paracompact Spaces, Lindelof Spaces and Other Information

A paracompact space is one in which every open cover has a locally finite open refinement. The previous post “Cartesian Products of Two Paracompact Spaces” has a basic discussion on paracompact spaces. For the sake of completeness, we repeat here some of the results discussed in that post. A proof of Proposition 1 can be found in [1] (Theorem 5.1.11 in page 302) or in [2] (Theorem 20.7 in page 146).. For a proof of Proposition 2, see Theorem 3 in the previous post “Cartesian Products of Two Paracompact Spaces”. We provide a proof for Proposition 3.

Proposition 1
Let $X$ be a regular space. Then $X$ is paracompact if and only if every open cover $\mathcal{U}$ of $X$ has a $\sigma$-locally finite open refinement.

Proposition 2
Every $F_\sigma$-subset of a paracompact space is paracompact.

Proposition 3
Any paracompact space with a dense Lindelof subspace is Lindelof.

Proof of Proposition 3
Let $L$ be a paracompact space. Let $M \subset L$ be a dense Lindelof subspace. Let $\mathcal{U}$ be an open cover of $L$. Since we are working with a regular space, let $\mathcal{V}$ be an open cover of $L$ such that $\left\{\overline{V}: V \in \mathcal{V} \right\}$ refines $\mathcal{U}$. Let $\mathcal{W}$ be a locally finite open refinement of $\mathcal{V}$. Choose $\left\{W_1,W_2,W_3,\cdots \right\} \subset \mathcal{W}$ such that it is a cover of $M$. Since $M \subset \bigcup \limits_{i=1}^\infty W_i$, $\overline{\bigcup \limits_{i=1}^\infty W_i}=L$.

Since the sets $W_i$ come from a locally finite collection, they are closure preserving. Hence we have:

$\overline{\bigcup \limits_{i=1}^\infty W_i}=\bigcup \limits_{i=1}^\infty \overline{W_i}=L$

For each $i$, choose some $U_i \in \mathcal{U}$ such that $\overline{W_i} \subset U_i$. Then $\left\{U_1,U_2,U_3,\cdots \right\}$ is a countable subcollection of $\mathcal{U}$ covering the space $L$. $\blacksquare$

A space is said to be a perfectly normal if it is a normal space with the additional property that every closed subset is a $G_\delta$-set in the space (equivalently every open subset is an $F_\sigma$-set). We need two basic results about hereditarily Lindelof spaces. A space is Lindelof if every open cover of that space has a countable subcover. A space is hereditarily Lindelof if every subspace of that space is Lindelof. Proposition 4 below, stated without proof, shows that to prove a space is hereditarily Lindelof, we only need to show that every open subspace is Lindelof.

Proposition 4
Let $L$ be a space. Then $L$ is hereditarily Lindelof if and only if every open subspace of $L$ is Lindelof.

Proposition 5
Let $L$ be a Lindelof space. Then $L$ is hereditarily Lindelof if and only if $L$ is perfectly normal.

Proof of Proposition 5
$\Rightarrow$ Suppose $L$ is hereditarily Lindelof. It is well known that regular Lindelof space is normal. Thus $L$ is normal. It remains to show that every open subset of $L$ is $F_\sigma$. Let $U \subset L$ be an non-empty open set. For each $x \in U$, let $V_x$ be open such that $x \in V_x$ and $\overline{V_x} \subset U$ (the space is assumed to be regular). By assumption, the open set $U$ is Lindelof. The open sets $V_x$ form an open cover of $U$. Thus $U$ is the union of countably many $\overline{V}_x$.

$\Leftarrow$ Suppose $L$ is perfectly normal. To show that $L$ is hereditarily Lindelof, it suffices to show that every open subset of $L$ is Lindelof (by Proposition 4). Let $U \subset L$ be non-empty open. By assumption, $U=\bigcup \limits_{i=1}^\infty F_i$ where each $F_i$ is a closed set in $L$. Since the Lindelof property is hereditary with respect to closed subsets, $U$ is Lindelof. $\blacksquare$

Another important piece of information that we need is the following metrization theorem. It shows that being a metrizable space is equivalent to have a base that is $\sigma$-locally finite. In proving Result 3, we will assume that the metric factor has such a base. This is a classic metrization theorem (see [1] or [2] or any other standard topology text).

Theorem 6
Let $X$ be a space. Then $X$ is metrizable if and only if $X$ has a $\sigma$-locally finite base.

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Result 3

Result 3 is the statement that:

If $X$ is paracompact and perfectly normal and Y is a metric space then $X \times Y$ is paracompact and perfectly normal.

Result 3 follows from the following two lemmas.

Lemma 7
If the following two conditions hold:

• every open subset of $X$ is an $F_\sigma$-set in $X$,
• $Y$ is a metric space,

then every open subset of $X \times Y$ is an $F_\sigma$-set in $X \times Y$.

Proof of Lemma 7
Let $U$ be a open subset of $X \times Y$. If $U=\varnothing$, then $U$ is certainly the union of countably many closed sets. So assume $U \ne \varnothing$. Let $\mathcal{B}=\bigcup \limits_{i=1}^\infty \mathcal{B}_i$ be a base for $Y$ such that each $\mathcal{B}_i$ is locally finite in $Y$ (by Theorem 6, such a base exists since $Y$ is metrizable).

Consider all non-empty $B \in \mathcal{B}$ such that we can choose nonempty open set $W_B \subset X$ with $W_B \times \overline{B} \subset U$. Since $U$ is non-empty open, such pairs $(B, W_B)$ exist. Let $\mathcal{B}^*$ be the collection of all non-empty $B \in \mathcal{B}$ for which there is a matching non-empty $W_B$. For each $i$, let $\mathcal{B}_i^*=\mathcal{B}^* \cap \mathcal{B}_i$. Of course, each $\mathcal{B}_i^*$ is still locally finite.

Since every open subset of $X$ is an $F_\sigma$-set in $X$, for each $W_B$, we can write $W_B$ as

$W_B=\bigcup \limits_{j=1}^\infty W_{B,j}$

where each $W_{B,i}$ is closed in $X$.

For each $i=1,2,3,\cdots$ and each $j=1,2,3,\cdots$, consider the following collection:

$\mathcal{V}_{i,j}=\left\{W_{B,j} \times \overline{B}: B \in \mathcal{B}_i^* \right\}$

Each element of $\mathcal{V}_{i,j}$ is a closed set in $X \times Y$. Since $\mathcal{B}_i^*$ is a locally finite collection in $Y$, $\mathcal{V}_{i,j}$ is a locally finite collection in $X \times Y$. Define $V_{i,j}=\bigcup \mathcal{V}_{i,j}$. The set $V_{i,j}$ is a union of closed sets. In general, the union of closed sets needs not be closed. However, $V_{i,j}$ is still a closed set in $X \times Y$ since $\mathcal{V}_{i,j}$ is a locally finite collection of closed sets. This is because a locally finite collection of sets is closure preserving. Note the following:

$\overline{V_{i,j}}=\overline{\bigcup \mathcal{V}_{i,j}}=\overline{\bigcup \left\{W_{B,j} \times \overline{B}: B \in \mathcal{B}_i^* \right\}}=\bigcup \left\{\overline{W_{B,j} \times \overline{B}}: B \in \mathcal{B}_i^* \right\}$

$=\bigcup \left\{W_{B,j} \times \overline{B}: B \in \mathcal{B}_i^* \right\}=V_{i,j}$

Finally, we have $U=\bigcup \limits_{i=1}^\infty \bigcup \limits_{j=1}^\infty V_{i,j}$, which is the union of countably many closed sets. $\blacksquare$

Lemma 8
If $X$ is a paracompact space satisfying the following two conditions:

• every open subset of $X$ is an $F_\sigma$-set in $X$,
• $Y$ is a metric space,

then $X \times Y$ is paracompact.

Proof of Lemma 8
As in the proof of the above lemma, let $\mathcal{B}=\bigcup \limits_{i=1}^\infty \mathcal{B}_i$ be a base for $Y$ such that each $\mathcal{B}_i$ is locally finite in $Y$. Let $\mathcal{U}$ be an open cover of $X \times Y$. Assume that elements of $\mathcal{U}$ are of the form $A \times B$ where $A$ is open in $X$ and $B \in \mathcal{B}$.

For each $B \in \mathcal{B}$, consider the following two items:

$\mathcal{W}_B=\left\{A: A \times B \in \mathcal{U} \right\}$

$W_B=\bigcup \mathcal{W}_B$

To simplify matter, we only consider $B \in \mathcal{B}$ such that $\mathcal{W}_B \ne \varnothing$. Each $W_B$ is open in $X$ and hence by assumption an $F_\sigma$-set in $X$. Thus by Proposition 2, each $W_B$ is paracompact. Note that $\mathcal{W}_B$ is an open cover of $W_B$. Let $\mathcal{H}_B$ be a locally finite open refinement of $\mathcal{W}_B$. Consider the following two items:

For each $j=1,2,3,\cdots$, let $\mathcal{V}_j=\left\{A \times B: A \in \mathcal{H}_B \text{ and } B \in \mathcal{B}_j \right\}$

$\mathcal{V}=\bigcup \limits_{j=1}^\infty \mathcal{V}_j$

We observe that $\mathcal{V}$ is an open cover of $X \times Y$ and that $\mathcal{V}$ refines $\mathcal{U}$. Furthermore each $\mathcal{V}_j$ is a locally finite collection. The open cover $\mathcal{U}$ we start with has a $\sigma$-locally finite open refinement. Thus $X \times Y$ is paracompact. $\blacksquare$

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Result 4

Result 4 is the statement that:

If $X$ is hereditarily Lindelof and $Y$ is a separable metric space, then $X \times Y$ is hereditarily Lindelof.

Proof of Result 4
Suppose $X$ is hereditarily Lindelof and that $Y$ is a separable metric space. It is well known that regular Lindelof spaces are paracompact. Thus $X$ is paracompact. By Proposition 5, $X$ is perfectly normal. By Result 3, $X \times Y$ is paracompact and perfectly normal.

Let $D$ be a countable dense subset of $Y$. We can think of $D$ as a $\sigma$-compact space. The product of any Lindelof space with a $\sigma$-compact space is Lindelof (see Corollary 3 in the post “The Tube Lemma”). Thus $X \times D$ is Lindelof. Furthermore $X \times D$ is a dense Lindelof subspace of $X \times Y$. By Proposition 3, $X \times Y$ is Lindelof. By Proposition 5, $X \times Y$ is hereditarily Lindelof. $\blacksquare$

Remark
In the previous post “Bernstein Sets and the Michael Line”, a non-normal product space where one factor is Lindelof and the other factor is a separable metric space is presented. That Lindelof space is not hereditarily Lindelof (it has uncountably many isolated points). Note that by Result 4, for any such non-normal product space, the Lindelof factor cannot be hereditarily Lindelof.

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Reference

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

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$\copyright \ \ 2012$