# Metrization Theorems for Compact Spaces

In this blog I have already presented two metrization theorems for compact spaces: (1) any compact space with a countable network is metrizable (see the post), (2) any compact space with a $G_\delta-$diagonal is metrizable (see the post). I now present another classic theorem: any countably compact space with a point-countable base is metrizable. This theorem is a classic result of Miscenko ([1]). All spaces are at least Hausdorff and regular. We have the following three metrization theorems for compact spaces. In subsequent posts, I will discuss generalizations of these theorems and discuss related concepts.

Thoerem 1. Any compact space with a countable network is metrizable.
The proof is in this post.

Thoerem 2. Any compact space with a $G_\delta-\text{diagonal}$ is metrizable.
The proof is in this post.

Thoerem 3. Any countably compact space with a point-countable base is metrizable.

A base $\mathcal{B}$ for a space $X$ is a point-countabe base if every point in $X$ belongs to at most countably elements of $\mathcal{B}$.

Proof of Theorem 3. Let $\mathcal{B}$ be a point-countable base for the countably compact space $X$. We show that $X$ is separable. Once we have a countable dense subset, the base $\mathcal{B}$ has to be a countable base. So we inductively define a sequence of countable sets $\lbrace{D_0,D_1,...}\rbrace$ such that $D=\bigcup_{n<\omega}D_n$ is dense in $X$.

Let $D_0=\lbrace{x_0}\rbrace$ be a one-point set to start with. For $n>0$, let $E_n=\bigcup_{i. Let $\mathcal{B}_n=\lbrace{B \in \mathcal{B}:B \cap E_n \neq \phi}\rbrace$. For each finite $T \subset \mathcal{B}_n$ such that $X - \bigcap T \neq \phi$, choose a point $x(T) \in X - \bigcup T$. Let $D_n$ be the union of $E_n$ and the set of all points $x(T)$. Let $D=\bigcup_{n<\omega}D_n$.

We claim that $\overline{D}=X$. Suppose we have $x \in X-\overline{D}$. Let $\mathcal{A}=\lbrace{B \in \mathcal{B}:B \cap D \neq \phi \phantom{X} \text{and} \thinspace x \notin B}\rbrace$. We know that $\mathcal{A}$ is countable since every element of $\mathcal{A}$ contains points of the countable set $D$. We also know that $\mathcal{A}$ is an open cover of $\overline{D}$. By the countably compactness of $\overline{D}$, we can find a finite $T \subset \mathcal{A}$ such that $\overline{D} \subset \bigcup T$. The finite set $T$ must have appeared during the induction process of selecting points for $D_n$ for some $n$ (i.e. $T \subset \mathcal{B}_n$). So a point $x(T)$ has been chosen such that $x(T) \notin \bigcup T$ (thus we have $x(T) \in D_n \subset \overline{D}$). On the other hand, since $\overline{D} \subset \bigcup T$, we observe that $x(T) \notin \overline{D}$, producing a contradiction. Thus the countable set $D$ is dense in $X$, making the point-countable base $\mathcal{B}$ a countable base.

Reference

1. Miscenko, A., Spaces with a point-countable base, Dokl. Acad. Nauk SSSR, 144 (1962), 985-988. (English translation: Soviet Math. Dokl. 3 (1962), 1199-1202).