A space is said to be countably compact if every countable open cover of 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  (Steen and Seebach) and  (Willard).
Let be a space. We say a collection of subsets of is a cover of if every point of belongs to some member of . We say is an open cover of if is a cover of and consists entirely of open subsets of . We say is a subcover of if and is also a cover of . The space is said to be compact if every open cover of has a subcover that has only finitely many members. The space is said to be countably compact if every countable open cover of 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 and let . The point is said to be a limit point of the set if every open set containing contains a point of different from , i.e. for every open set with , .
A collection of subsets of is said to be point-finite if every point of belongs to only finitely many sets in . In contrast, a collection of subsets of is said to be locally finite if for each , there is an open set with such that has nonempty intersection with only finitely many sets in . Clearly any locally finite collection is a point-finite collection.
Let and be collections of subsets of the space . The collection is said to be a refinement of if for every , for some . If, in addition, both and are also open covers, then is said to be an open refinement of .
A space is said to be metacompact if every open cover of has a point-finite open refinement. A space is said to be paracompact if every open cover of 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.
Let be a metacompact space. Then if is countably compact, then is compact.
The theorem will be established by the following lemmas. An open cover of is said to be irreducible if by removing any open set from , the resulting subcollection fails to be a cover of the space .
In any space , any point-finite open cover has a subcover that is irreducible.
Proof. This is an application of Zorn’s lemma (chapter 1 of ).
Let be a point-finite open cover of . Consider the following partially ordered set where the reverse set inclusion is the partial order:
We define this partial order on : if
Clearly as . The partial order is indeed a partial order (a relation on that is reflexive, antisymmetric and transitive; see  for definitions).
Let be a chain (i.e. for any , we have or ). We want to show that the chain has a maximal element (i.e. there exists such that for all ). We claim that .
It is clear that for all . But we need to show that . To this end, we show that is a cover of . This is where we require to be point-finite.
Let . Then belongs to only finitely many open sets in , say, . We claim that some . Suppose not. Then for each , there exists such that . Since is a chain, we can order the from smallest to largest. We assume that is the maximum:
Since is an open cover and , some . This means , contradicting that . Thus some , showing that is an open cover.
We just show that every chain in has a maximal element in . By Zorn’s lemma, the partial ordered set has a maximal element . Note is the irreducible subcover that we are looking for. For any that is also an open cover, we have . Thus no proper subset of can be an open cover.
In any countably compact space , every irreducible point-finite open cover is finite.
Proof. Let be an irreducible point-finite open cover of , which is countably compact. For each , there is some point such that is in no other open set in . If not, then is not irreducible. Consider the following set:
The set is a discrete set in the space . Note that every point of belongs to one open set that contains at most one point of . Since is countably compact, all discrete sets must be finite. Thus is finite (otherwise, is infinite).
Proof of Theorem. Let be an open cover of , which is countably compact and metacompact. By the metacompactness, let be a point-finite open refinement of . By lemma 1, it is OK to assume that is an irreducible open cover. By lemma 2, is finite. It follows that has a finite subcover.
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 . See the post The First Uncountable Ordinal.
- Steen, L. A. and Seebach, J. A., Counterexamples in Topology, 1995, Dover Publications, Inc, New York.
- Willard, S., General Topology, 1970, Addison-Wesley Publishing Company.