Answer to Question #213052 in Functional Analysis for smi

Question #213052

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

and continuous. Show that T is a homeomorphism.


1
Expert's answer
2021-07-05T17:40:04-0400

Since T:XYT: X\to Y bijective, the mapping S=T1:YXS=T^{-1}: Y \to X exists. We are to prove that S is continuous. For this purpose it is sufficient to show that S1(F)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=T1S=T^{-1} is bijective, S1(F)=T(F)S^{-1}(F)=T(F) (in this expression S1S^{-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, S1(F)S^{-1}(F) is a compact subset of Y.

Any compact subset of metric space is a closet subspace. Therefore, S1(F)S^{-1}(F) is closed.

We have proved that S1(F)S^{-1}(F) is closed subset of Y for any closed subset F of X.

It is equivalent to the assertion that S1(U)S^{-1}(U) is an open subset of Y for any open subset U of X (by taking U=XFU=X\setminus F). Therefore, S is continuous and, hence, T is a homomorphism.


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