Countably Compact Spaces with G-delta Diagonals

It is a classic result in general topology that any compact space with a G_\delta-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 X is countably compact and has a G_\delta-diagonal, then X is compact and thus metrizable. All spaces are at least Hausdorff. This post has a discussion on the theorem on compact spaces with G_\delta-diagonal. This post has a discussion on some metrizaton theorems for compact spaces.

If \mathcal{T} is a collection of subsets of a space X, then for each x \in X, define st(x,\mathcal{T})=\bigcup\lbrace{T \in \mathcal{T}:x \in T}\rbrace. A sequence of open covers \lbrace{\mathcal{T}_n:n \in \omega}\rbrace of the space X is a G_\delta-diagonal sequence for X if for each x \in X, we have \lbrace{x}\rbrace=\bigcap_{n<\omega} st(x,\mathcal{T}_n). We use the following lemma (due to Ceder, [1]). This lemma was proved in this previous post.

Lemma. The space X has a G_\delta-diagonal if and only if it has a G_\delta-diagonal sequence.

Theorem. Let X be a countably compact space that has a G_\delta-diagonal. Then X is compact.

Proof. Let X be a countably compact space. Let \lbrace{\mathcal{T}_n:n \in \omega}\rbrace be a G_\delta-diagonal sequence for X. If X is Lindelof, then we are done. Suppose we have an open cover \mathcal{V} of X that has no countable subcover. From this open cover \mathcal{V}, we derive a contradiction.

We inductively, for each \alpha < \omega_1, choose a point x_\alpha \in X and an integer m(\alpha) \in \omega with the following properties:

For each \alpha < \omega_1,

  1. x_\alpha \in X-\bigcup\lbrace{st(x_\beta,\mathcal{T}_{m(\beta)}): \beta < \alpha}\rbrace, and
  2. the open cover \mathcal{V} does not have a countable subcollection that covers X-\bigcup_{\beta \leq \alpha} st(x_\beta,\mathcal{T}_{m(\beta)}).

To start off, choose x_0 \in X. There is an integer m(0) \in \omega such that no countable subcollection of \mathcal{V} covers X-st(x_0,\mathcal{T}_{m(0)}). Suppose this integer m(0) does not exist. Then for each n \in \omega, we have a countable \mathcal{V}_n \subset \mathcal{V} such that \mathcal{V}_n covers X-st(x_0,\mathcal{T}_n). Then \bigcup_{n<\omega} \mathcal{V}_n would be a countable subcollection of \mathcal{V} that covers X-\lbrace{x_0}\rbrace. This would mean that \mathcal{V} has a countable subcover of X.

Suppose that \lbrace{x_\beta:\beta<\alpha}\rbrace and \lbrace{m(\beta):\beta<\alpha}\rbrace have been chosen such that conditions (1) and (2) are satisfied for each \beta<\alpha. We have the following claim. Proving this claim allows us to choose x_\alpha and m(\alpha).

Claim. No countable subcollection of \mathcal{V} covers X-\bigcup_{\beta<\alpha} st(x_\beta,\mathcal{T}_{m(\beta)}).

Suppose we do have a countable \mathcal{W} \subset \mathcal{V} such that \mathcal{W} covers X-\bigcup_{\beta<\alpha} st(x_\beta,\mathcal{T}_{m(\beta)}). Then \mathcal{S}=\lbrace{st(x_\beta,\mathcal{T}_m(\beta)):\beta < \alpha}\rbrace \cup \mathcal{W} is a countable open cover of X and thus has a finite subcover \mathcal{F}. Let \delta be the largest ordinal <\alpha such that st(x_\delta,\mathcal{T}_m(\delta)) is in this finite subcover \mathcal{F}. Then \mathcal{W} is a counntable subcollection of \mathcal{V} that covers X-\bigcup_{\beta \leq \delta} st(x_\beta,\mathcal{T}_{m(\beta)}). This violates condition (2) above for the ordinal \delta. This proves the claim.

Now, pick x_\alpha \in X-\bigcup\lbrace{st(x_\beta,\mathcal{T}_{m(\beta)}): \beta < \alpha}\rbrace. There must be some integer m(\alpha) \in \omega such that conditon (2) above is satisfied for \alpha. If not, for each n \in \omega, there is some countable \mathcal{V}_n \subset \mathcal{V} such that \mathcal{V}_n covers X-\bigcup_{\beta \leq \alpha} st(x_\beta,\mathcal{T}_n). Then \bigcup_{n<\omega} \mathcal{V}_n would be a countable subcollection of \mathcal{V} that covers X-\biggl(\bigcup_{\beta<\alpha} st(x_\beta,\mathcal{T}_{m(\beta)}) \biggr) \bigcup \lbrace{x_\alpha}\rbrace. This would mean that \mathcal{V} has a countable subcover of X-\bigcup_{\beta<\alpha} st(x_\beta,\mathcal{T}_{m(\beta)}). This violates the above claim. Now the induction process is completed.

To conclude the proof of the theorem, note that there is some n \in \omega and there is some uncountable D \subset \omega_1 such that for each \alpha \in D, n=m(\alpha). Then Y=\lbrace{x_\alpha:\alpha \in D}\rbrace is an uncountable closed and discrete set in X. Note that each open set in \mathcal{T}_n contains at most one point of Y. Thus X must be Lindelof. With X being countably compact, X is compact.


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

2 thoughts on “Countably Compact Spaces with G-delta Diagonals

  1. Pingback: Looking for spaces in which every compact subspace is metrizable | Dan Ma's Topology Blog

  2. Pingback: Pseudocompact spaces with regular G-delta diagonals | Dan Ma's Topology Blog

Leave a Reply

Fill in your details below or click an icon to log in: Logo

You are commenting using your account. Log Out /  Change )

Google photo

You are commenting using your Google account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )

Connecting to %s