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**hunkydory19** If f: X ---> Y is a continuous map of topological spaces and X is compact, then f(X) is compact.

Suppose that $\displaystyle \mho $ is an open cover of f(X) by sets open in Y.

By continuity of f, $\displaystyle f^-1(U)$ is open in X for every U in $\displaystyle \mho.$

The collection $\displaystyle {f^{-1}(U) : U \in \mho} $ is an open cover of X, since for any x in X, f(X) must belong to some U in $\displaystyle \mho.$

By compactness of X, there is a finite subcover, $\displaystyle {f^{-1}(U_1), f^{-1}(U_2)....f^{-1}(U_r)} $ hence $\displaystyle {U_1, U_r....}$ is a finite subcover of $\displaystyle \mho $for f(X).

I'm preparing proofs for my exam, and I was just wondering if anyone could please explain how this proof would alter if the question was:

Show that if K is a compact subset of X and f is continuous, then f(K) is compact in Y.

I'm thinking the proof would just stay the same since it is just a subset?

Thanks in advance!