If f: X ---> Y is a continuous map of topological spaces and X is compact, then f(X) is compact.
Suppose that is an open cover of f(X) by sets open in Y.
By continuity of f, is open in X for every U in
The collection is an open cover of X, since for any x in X, f(X) must belong to some U in
By compactness of X, there is a finite subcover, hence is a finite subcover of 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!