Application of Urysohn’s metrization theorem

By | June 14, 2015

Theorem. Let X be a compact metric space, Y a Hausdorff space, and f:X\rightarrow Y a continuous function from X onto Y. Then Y is metrizable.

Proof. Note that since f\left(X\right)=Y and f is continuous and X is compact, Y is compact. Since Y is Hausdorff space, Y is T_{4}. So Y is T_{3}. It suffices to show Y is second
countable. Since X is compact, it is second countable. Let \mathcal{B} be a countable basis for X. Define

    \[ \mathcal{A}=\left\{ \bigcup_{n\in F}B_{n}:F\subset\mathbb{N},\text{ finite subset}\right\} . \]

Clearly, \mathcal{A} is countable. For A\in\mathcal{A}, define A^{*}=Y\setminus f\left(X\setminus A\right). Since A is open, X\setminus A is closed. Since f is closed map, Y\setminus f\left(X\setminus A\right)=A^{*} is open. So if we collect these A^{*}, say \mathcal{A}^{*}, we claim that \mathcal{A}^{*} is basis for Y.

Let U be an open set in Y and let y\in U. Then f^{-1}\left(U\right) is open and \left\{ y\right\} is closed. So f^{-1}\left(\left\{ y\right\} \right) is closed since f is continuous. Since X is compact space, f^{-1}\left(\left\{ y\right\} \right) is compact. So there is a finite collection \left\{ B_{i}\right\} _{i=1}^{N} of \mathcal{B} such that

    \[ f^{-1}\left(\left\{ y\right\} \right)\subset\bigcup_{i=1}^{N}B_{i}\subset f^{-1}\left(U\right) \]

since \mathcal{B} is a basis for X and f^{-1}\left(U\right) is open in X. Let A=\bigcup_{i=1}^{N}B_{i}. Then A\in\mathcal{A} by definition and A^{*}\in\mathcal{A}^{*}. Moreover, y\in\left(Y\setminus f\left(X\setminus A\right)\right)=A^{*}. For any point f\left(x\right)\in A^{*}, f\left(x\right)\notin f\left(X\setminus A\right). So x\notin X\setminus A. This shows x\in A. Since A\subset f^{-1}\left(U\right), then f\left(x\right)\in U. This implies for each open set U in Y and y\in U, there is a member A^{*} of \mathcal{A}^{*} such that

    \[ y\in A^{*}\subset U. \]

The second condition of basis easily comes from \mathcal{B}. This shows \mathcal{A}^{*} is a countable basis for Y. So by Urysohn’s metrization theorem, Y is metrizable.

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