# compact space

• May 2nd 2009, 06:39 AM
hunkydory19
compact space
If f: X ---> Y is a continuous map of topological spaces and X is compact, then f(X) is compact.

Suppose that $\mho$ is an open cover of f(X) by sets open in Y.
By continuity of f, $f^-1(U)$ is open in X for every U in $\mho.$
The collection ${f^{-1}(U) : U \in \mho}$ is an open cover of X, since for any x in X, f(X) must belong to some U in $\mho.$
By compactness of X, there is a finite subcover, ${f^{-1}(U_1), f^{-1}(U_2)....f^{-1}(U_r)}$ hence ${U_1, U_r....}$ is a finite subcover of $\mho$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?

• May 2nd 2009, 09:23 AM
Gamma
Yeah the same basic structure works, but you gotta specify about being open in which topology, with subspaces running around.

For instance take any open covering of F(K) with sets open in Y (this is important because continuity of f is w.r.t. the topology on Y). Pull back, its still an open cover by continuity and set inclusion invariance under pullbacks of continuous maps. K is compact so yeah just pick the finite subcover, then send that stuff back into Y and it will still cover K, but it is finite now, and they are still a subset of the original collection of open sets. (perhaps an argument should be made here about the open sets still being the same you know like why is $f(f^{-1}(U))=U$).

But yeah it is pretty much exactly the same, just want to stress that you just gotta at least state where the sets are open when dealing with subsets like this generally.
• May 3rd 2009, 05:43 AM
aliceinwonderland
Quote:

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

Suppose that $\mho$ is an open cover of f(X) by sets open in Y.
By continuity of f, $f^-1(U)$ is open in X for every U in $\mho.$
The collection ${f^{-1}(U) : U \in \mho}$ is an open cover of X, since for any x in X, f(X) must belong to some U in $\mho.$
By compactness of X, there is a finite subcover, ${f^{-1}(U_1), f^{-1}(U_2)....f^{-1}(U_r)}$ hence ${U_1, U_r....}$ is a finite subcover of $\mho$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?

$K = \bigcup_{i=1}^{n}f^{-1}(O_i)$, where $\{O_i\}_{i=1}^{n}$ is a finite collection of open sets in Y.
Then, $f(K) = f( \bigcup_{i=1}^{n}f^{-1}(O_i)) = \bigcup_{i=1}^{n}f(f^{-1}(O_i)) \subset \bigcup_{i=1}^{n}O_i$.