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<n}D_n. 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).
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