Pseudocompact + Metacompact implies Compact

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Reference

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