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.

___________________________________________________________________________________

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.

___________________________________________________________________________________

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

___________________________________________________________________________________

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.

___________________________________________________________________________________

Reference

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

___________________________________________________________________________________

\copyright \ \ 2012

Advertisements

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s