# Application of Urysohn’s metrization theorem

By | June 14, 2015

Theorem. Let be a compact metric space, a Hausdorff space, and a continuous function from onto . Then is metrizable.

Proof. Note that since and is continuous and is compact, is compact. Since is Hausdorff space, is . So is . It suffices to show is second
countable. Since is compact, it is second countable. Let be a countable basis for . Define Clearly, is countable. For , define . Since is open, is closed. Since is closed map, is open. So if we collect these , say , we claim that is basis for .

Let be an open set in and let . Then is open and is closed. So is closed since is continuous. Since is compact space, is compact. So there is a finite collection of such that since is a basis for and is open in . Let . Then by definition and . Moreover, . For any point , . So . This shows . Since , then . This implies for each open set in and , there is a member of such that The second condition of basis easily comes from . This shows is a countable basis for So by Urysohn’s metrization theorem, is metrizable.