# 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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$\copyright \ \ 2012$