# Metacompact and countably compact spaces

A space $X$ is said to be countably compact if every countable open cover of $X$ has a finite subcover. There are several equivalent characterizations of countably compactness, one of which is that every infinite subset has a limit point. In this blog, we have started to explore various additional conditions that turn a countably compact space into a compact space (that every open cover of the space has a finite subcover). See the previous post A Note About Countably Compact Spaces. One of the conditions discussed in that post is the notion of metacompactness. In this post we present a proof that in the presence of metacompactness, countably compactness implies compactness. This is an opportunity of showcasing an application of Zorn’s lemma. All spaces in consideration are at least Hausdirff (T2). For any basic notions not defined here, see [1] (Steen and Seebach) and [2] (Willard).

Let $X$ be a space. We say a collection $\mathcal{U}$ of subsets of $X$ is a cover of $X$ if every point of $X$ belongs to some member of $\mathcal{U}$. We say $\mathcal{U}$ is an open cover of $X$ if $\mathcal{U}$ is a cover of $X$ and $\mathcal{U}$ consists entirely of open subsets of $X$. We say $\mathcal{V}$ is a subcover of $\mathcal{U}$ if $\mathcal{V} \subset \mathcal{U}$ and $\mathcal{V}$ is also a cover of $X$. The space $X$ is said to be compact if every open cover of $X$ has a subcover that has only finitely many members. The space $X$ is said to be countably compact if every countable open cover of $X$ has a finite subcover. It is well known that countably compactness is equivalent to the condition that every infinite subset has a limit point. Let $A \subset X$ and let $p \in X$. The point $p$ is said to be a limit point of the set $A$ if every open set containing $p$ contains a point of $A$ different from $p$, i.e. for every open set $O$ with $p \in O$, $O \cap (A-\left\{p\right\}) \ne \phi$.

A collection $\mathcal{U}$ of subsets of $X$ is said to be point-finite if every point of $X$ belongs to only finitely many sets in $\mathcal{U}$. In contrast, a collection $\mathcal{U}$ of subsets of $X$ is said to be locally finite if for each $x \in X$, there is an open set $O$ with $x \in O$ such that $O$ has nonempty intersection with only finitely many sets in $\mathcal{U}$. Clearly any locally finite collection is a point-finite collection.

Let $\mathcal{U}$ and $\mathcal{V}$ be collections of subsets of the space $X$. The collection $\mathcal{V}$ is said to be a refinement of $\mathcal{U}$ if for every $V \in \mathcal{V}$, $V \subset U$ for some $U \in \mathcal{U}$. If, in addition, both $\mathcal{U}$ and $\mathcal{V}$ are also open covers, then $\mathcal{V}$ is said to be an open refinement of $\mathcal{U}$.

A space $X$ is said to be metacompact if every open cover of $X$ has a point-finite open refinement. A space $X$ is said to be paracompact if every open cover of $X$ has a locally finite open refinement. Clearly, any paracompact space is metacompact. The converse is not true (an example is indicated at the end of the post). We prove the following theorem.

Theorem
Let $X$ be a metacompact space. Then if $X$ is countably compact, then $X$ is compact.

The theorem will be established by the following lemmas. An open cover $\mathcal{U}$ of $X$ is said to be irreducible if by removing any open set from $\mathcal{U}$, the resulting subcollection fails to be a cover of the space $X$.

Lemma 1
In any space $X$, any point-finite open cover has a subcover that is irreducible.

Proof. This is an application of Zorn’s lemma (chapter 1 of [2]).

Let $\mathcal{U}$ be a point-finite open cover of $X$. Consider the following partially ordered set where the reverse set inclusion is the partial order:

$\displaystyle \mathcal{P}=\left\{\mathcal{W}:\mathcal{W} \subset \mathcal{U} \text{ and }\mathcal{W} \text{ is a cover of } X\right\}$

We define this partial order on $\mathcal{P}$: $\mathcal{W}_1 \le \mathcal{W}_2$ if $\mathcal{W}_1 \supset \mathcal{W}_2$

Clearly $\mathcal{P} \ne \phi$ as $\mathcal{U} \in \mathcal{P}$. The partial order $\le$ is indeed a partial order (a relation on $\mathcal{P}$ that is reflexive, antisymmetric and transitive; see [1] for definitions).

Let $\mathcal{C} \subset \mathcal{P}$ be a chain (i.e. for any $\mathcal{W}_1, \mathcal{W}_2 \in \mathcal{C}$, we have $\mathcal{W}_1 \le \mathcal{W}_2$ or $\mathcal{W}_2 \le \mathcal{W}_1$). We want to show that the chain $\mathcal{C}$ has a maximal element (i.e. there exists $\mathcal{M} \in \mathcal{P}$ such that $\mathcal{W} \le \mathcal{M}$ for all $\mathcal{W} \in \mathcal{C}$). We claim that $\mathcal{M}=\bigcap \limits_{\mathcal{W} \in \mathcal{C}} \mathcal{W}$.

It is clear that $\mathcal{W} \le \mathcal{M}$ for all $\mathcal{W} \in \mathcal{C}$. But we need to show that $\mathcal{M} \in \mathcal{P}$. To this end, we show that $\mathcal{M}$ is a cover of $X$. This is where we require $\mathcal{U}$ to be point-finite.

Let $x \in X$. Then $x$ belongs to only finitely many open sets in $\mathcal{U}$, say, $U_1, \cdots, U_n$. We claim that some $U_i \in \mathcal{M}$. Suppose not. Then for each $i$, there exists $\mathcal{W}_i \in \mathcal{C}$ such that $U_i \notin \mathcal{W}_i$. Since $\mathcal{C}$ is a chain, we can order the $\mathcal{W}_i$ from smallest to largest. We assume that $\mathcal{W}_n$ is the maximum:

$\displaystyle \mathcal{W}_i \le \mathcal{W}_n$ or $\displaystyle \mathcal{W}_i \supset \mathcal{W}_n$ for $i=1,2,3,\cdots,n$

Since $\mathcal{W}_n$ is an open cover and $U_n \notin \mathcal{W}_n$, some $U_j \in \mathcal{W}_n$. This means $U_j \in \mathcal{W}_j$, contradicting that $U_j \notin \mathcal{W}_j$. Thus some $U_i \in \mathcal{M}$, showing that $\mathcal{M}$ is an open cover.

We just show that every chain in $\mathcal{P}$ has a maximal element in $\mathcal{P}$. By Zorn’s lemma, the partial ordered set $\mathcal{P}$ has a maximal element $\mathcal{V}$. Note $\mathcal{V}$ is the irreducible subcover that we are looking for. For any $\mathcal{W} \subset \mathcal{U}$ that is also an open cover, we have $\mathcal{W} \supset \mathcal{V}$. Thus no proper subset of $\mathcal{V}$ can be an open cover. $\blacksquare$

Lemma 2
In any countably compact space $X$, every irreducible point-finite open cover is finite.

Proof. Let $\mathcal{U}$ be an irreducible point-finite open cover of $X$, which is countably compact. For each $U \in \mathcal{U}$, there is some point $x(U) \in U$ such that $x(U)$ is in no other open set in $\mathcal{U}$. If not, then $\mathcal{U}$ is not irreducible. Consider the following set:

$\displaystyle A=\left\{x(U): U \in \mathcal{U}\right\}$

The set $A$ is a discrete set in the space $X$. Note that every point of $X$ belongs to one open set $U \in \mathcal{U}$ that contains at most one point of $A$. Since $X$ is countably compact, all discrete sets must be finite. Thus $\mathcal{U}$ is finite (otherwise, $A$ is infinite). $\blacksquare$

Proof of Theorem. Let $\mathcal{U}$ be an open cover of $X$, which is countably compact and metacompact. By the metacompactness, let $\mathcal{V}$ be a point-finite open refinement of $\mathcal{U}$. By lemma 1, it is OK to assume that $\mathcal{V}$ is an irreducible open cover. By lemma 2, $\mathcal{V}$ is finite. It follows that $\mathcal{U}$ has a finite subcover. $\blacksquare$

Remark
As a corollary, any paracompact countably compact space is compact. Another corollary is that any countably compact non-compact space is not metacompact. One handy example of a countably compact non-compact space is the space of all countable ordinals $\omega_1$. See the post The First Uncountable Ordinal.

Reference

1. Steen, L. A. and Seebach, J. A., Counterexamples in Topology, 1995, Dover Publications, Inc, New York.
2. Willard, S., General Topology, 1970, Addison-Wesley Publishing Company.