# 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 ). 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  and . 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  and  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.