Answer to Question #213052 in Functional Analysis for smi

Question

Let X and Y be metric spaces, X compact, and T: X →Y bijective

and continuous. Show that T is a homeomorphism.

Answer 1

Since $T: X\to Y$ bijective, the mapping $S=T^{-1}: Y \to X$ exists. We are to prove that S is continuous. For this purpose it is sufficient to show that $S^{-1}(F)$ is a closed subset of Y for any closed subset F of X.

Since X is a compact metric space, any closet subset F of X is also compact.

Since $S=T^{-1}$ is bijective, $S^{-1}(F)=T(F)$ (in this expression $S^{-1}$ is the inverse image of a set).

Since F is compact and T is continuous, T(F) is a compact subset of Y, hence, $S^{-1}(F)$ is a compact subset of Y.

Any compact subset of metric space is a closet subspace. Therefore, $S^{-1}(F)$ is closed.

We have proved that $S^{-1}(F)$ is closed subset of Y for any closed subset F of X.

It is equivalent to the assertion that $S^{-1}(U)$ is an open subset of Y for any open subset U of X (by taking $U=X\setminus F$). Therefore, S is continuous and, hence, T is a homomorphism.