# 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.

Reference

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.