It is a classic result in general topology that any compact space with a diagonal is metrizable ([3]). This theorem also holds for countably compact spaces (due to Chaber in [2]). The goal of this post is to present a proof of this theorem. We prove that if is countably compact and has a diagonal, then is compact and thus metrizable. All spaces are at least Hausdorff. This post has a discussion on the theorem on compact spaces with diagonal. This post has a discussion on some metrizaton theorems for compact spaces.

If is a collection of subsets of a space , then for each , define . A sequence of open covers of the space is a diagonal sequence for if for each , we have . We use the following lemma (due to Ceder, [1]). This lemma was proved in this previous post.

* Lemma*. The space has a diagonal if and only if it has a diagonal sequence.

* Theorem*. Let be a countably compact space that has a diagonal. Then is compact.

* Proof*. Let be a countably compact space. Let be a diagonal sequence for . If is Lindelof, then we are done. Suppose we have an open cover of that has no countable subcover. From this open cover , we derive a contradiction.

We inductively, for each , choose a point and an integer with the following properties:

For each ,

- , and
- the open cover does not have a countable subcollection that covers .

To start off, choose . There is an integer such that no countable subcollection of covers . Suppose this integer does not exist. Then for each , we have a countable such that covers . Then would be a countable subcollection of that covers . This would mean that has a countable subcover of .

Suppose that and have been chosen such that conditions (1) and (2) are satisfied for each . We have the following claim. Proving this claim allows us to choose and .

* Claim*. No countable subcollection of covers .

Suppose we do have a countable such that covers . Then is a countable open cover of and thus has a finite subcover . Let be the largest ordinal such that is in this finite subcover . Then is a counntable subcollection of that covers . This violates condition (2) above for the ordinal . This proves the claim.

Now, pick . There must be some integer such that conditon (2) above is satisfied for . If not, for each , there is some countable such that covers . Then would be a countable subcollection of that covers . This would mean that has a countable subcover of . This violates the above claim. Now the induction process is completed.

To conclude the proof of the theorem, note that there is some and there is some uncountable such that for each , . Then is an uncountable closed and discrete set in . Note that each open set in contains at most one point of . Thus must be Lindelof. With being countably compact, is compact.

**Reference**

- Ceder, J. G. Some generalizations of metric spaces,
*Pacific J. Math*., 11 (1961), 105-125. - Chaber,
*Conditions which imply compactness in countably compact spaces*, Bull. Acad. Pol. Sci. Ser. Math., 24 (1976), 993-998. - Sneider, V.,
*Continuous images of Souslin and Borel sets: metrization theorems*,*Dokl. Acad. Nauk USSR*, 50 (1945), 77-79.

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