A Note About Countably Compact Spaces

This is a discussion on several additional conditions that would turn a countably compact space into a compact space. For example, a countably compact space having a G_\delta- diagonal is compact (proved in this post). Each of the following properties, if possessed by a countably compact space, would lead to compactness: (1) having a G_\delta- diagonal, (2) being metrizable, (3) being a Moore space, (4) being paracompact, and (5) being metacompact. All spaces are at least Hausdorff. We have the following theorem. Some relevant definitions and links to posts in this blog are given below. For any terms that are not defined here, see Engelking ([1]).

Theorem. Let X be a countably compact space. If X possesses any one of the following conditions, then X is compact.

  1. Having a G_\delta- diagonal.
  2. Being a metrizable space.
  3. Being a Moore space.
  4. Being a paracompact space.
  5. Being a metacompact space.

The proof of 1 has already been presented in another post in this blog. Since metrizable spaces are Moore spaces, between 2 and 3 we only need to prove 3. Between 4 and 5, we only need to prove 5 (since paracompact compact spaces are metacompact).

Proof of 3. A Moore space is a regular space that has a development (see this post for the definition). In this post, I showed that a space X has a G_\delta-diagonal if and only it has a G_\delta-diagonal sequence. It is easy to verify that the development for a Moore space is a G_\delta-diagonal sequence. Thus any Moore space has a G_\delta-diagonal and any countably compact Moore space is compact (and metrizable). Saying in another way, in the class of Moore spaces, countably compactness is equivalent to compactness.

Proof of 5. A space X is metacompact if every open cover of X has a point-finite open refinement. Let X be metacompact. Let \mathcal{U} be an open cover of X. By the metacompactness, \mathcal{U} has a point-finite open refinement \mathcal{O}. We are done if we can show \mathcal{O} has a finite subcover. This finite subcover is obtained through the following claims.

Claim 1. There is a set M \subset X such that \lvert M \cap O \lvert \thinspace \leq 1 for each O \in \mathcal{O} and such that M is maximal. That is, by adding an additional point x \notin M, \lvert (M \cup \lbrace{x}\rbrace) \cap O \lvert \thinspace \ge 2 for some O \in \mathcal{O}.

Such a set can be obtained by using the Zorn’s Lemma.

Claim 2. Let \mathcal{W}=\lbrace{O \in \mathcal{O}:O \cap M \neq \phi}\rbrace. We claim that \mathcal{W} is an open cover of X.

To see this, let x \in X. If x \in M, then x \in O for some O \in \mathcal{W}. If x \notin M, then by the maximality of M, M \cup \lbrace{x}\rbrace intersects with some O \in \mathcal{O} with at least 2 points. This means that x and at least one point of M are in O. Then O \in \mathcal{W}.

Since each open set in \mathcal{W} contains at most one point of M, M is a closed and discrete set in X. By the countably compactness of X, M must be finite. Since each point of M is in at most finitely many open sets in \mathcal{O}, \mathcal{W} is finite. Thus \mathcal{W} is a finite subcover of \mathcal{O}.

Reference

  1. Engelking, R., General Topology, Revised and Completed Edition, 1989, Heldermann Verlag, Berlin.
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