Let X and Y be metric spaces, X compact, and T: X →Y bijective
and continuous. Show that T is a homeomorphism.
Since bijective, the mapping exists. We are to prove that S is continuous. For this purpose it is sufficient to show that 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 is bijective, (in this expression is the inverse image of a set).
Since F is compact and T is continuous, T(F) is a compact subset of Y, hence, is a compact subset of Y.
Any compact subset of metric space is a closet subspace. Therefore, is closed.
We have proved that is closed subset of Y for any closed subset F of X.
It is equivalent to the assertion that is an open subset of Y for any open subset U of X (by taking ). Therefore, S is continuous and, hence, T is a homomorphism.
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