Compactness is Preserved Under Continuous Functions

One key topology homeomorphism invariant is compactness, a fact fundamental to the extreme value theorem.

Theorem

Let \(f : X \to Y\) be a continuous function between topological spaces \(X\) and \(Y\), and let \(X'\) be a compact subset of \(X\). Then, \(f(X')\) is compact as a subset of \(Y\)

Proof

The main strategy here is to take an arbitrary open cover of \(f(X')\), use continuity with images to induce an open cover of \(X\), which has a finite subcover, which can then be used to find a corresponding finite subcover of \(f(X')\) in the codomain.

Consider an arbitrary open cover of \(f(X')\), that is

\[ \bigcup_{i \in I} V_i \supseteq f(X').\]

Then, since pre-images preserve unions,

\[ f^{-1}\left(\bigcup_{i \in I} V_i\right) = \bigcup_{i \in I} f^{-1}\left(V_i\right).\]

First we show that this set is an open cover of \(X'\).

Each set is open from continuity of \(f\) and openness of \(V_i\) for each \(i\).

Then because pre-images preserve inclusions and any set is a subset of the pre-image of its image under some function, we have that

\[ \bigcup_{i \in I} V_i \supseteq f(X') \implies f^{-1}\left(\bigcup_{i \in I} V_i\right) \supseteq f^{-1}(f(X')) \supseteq X'\]

and hence the desired set covers \(X'\), this cover has a finite subcover, namely a finite subset \(J \subseteq I\) such that

\[ \bigcup_{j \in J} f^{-1}(V_j)\]

is an open cover of \(X'\).

We will now prove that the corresponding indices give a subcover of \(f(X')\), that is that

\[ \bigcup_{j \in J} V_j\]

is a finite cover of \(f(X')\) which is a subcover of our original cover.

Clearly this cover is open, and finite, we simply must prove that it does indeed cover \(f(X')\).

Now, for each \(y \in f(X')\), there exists and \(x \in X'\) such that \(f(x) = y\). Then, since \(\bigcup_{j \in J} f^{-1}(V_j)\) covers \(X'\), we know that \(x \in V_j\) for some \(j \in J\) and hence \(f(x) = y \in f(V_j)\) for the same \(j\). This means for any \(y \in f(X')\), our finite subcover contains \(y\), and thus it covers \(f(X')\).