# 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.