Show that if \(f: X \to Y\) is continuous, where \(X\) is compact and \(Y\) is Hausdorff, then \(f\) is a closed map (that is, \(f\) carries closed sets to closed sets).
Solution
1. Let \(C \subseteq X\) be an arbitrary closed subset.
setup.
2. \(C\) is a compact subspace of \(X\).
2.1. \(X\) is compact by hypothesis.
hypothesis.
2.2. Every closed subset of a compact topological space is compact.
if \(\mathcal{U}\) is an open cover of \(C\), then \(\mathcal{U} \cup \{X \setminus C\}\) is an open cover of \(X\); the finite subcover of \(X\) yields a finite subcover of \(C\).
2.3. Hence \(C\) is compact.
1 and 2.
3. The image \(f(C)\) is a compact subspace of \(Y\).
3.1. \(f: X \to Y\) is continuous.
hypothesis.
3.2. The continuous image of any compact space is compact.
if \(\{V_\alpha\}\) is an open cover of \(f(C)\), then \(\{f^{-1}(V_\alpha)\}\) is an open cover of \(C\); a finite subcover of \(C\) maps under \(f\) to a finite subcover of \(f(C)\).
3.3. Hence \(f(C)\) is compact in \(Y\).
2, 1, and 2.
4. \(f(C)\) is a closed subset of \(Y\).
4.1. \(Y\) is Hausdorff by hypothesis.
hypothesis.
4.2. Every compact subset of a Hausdorff space is closed.
for any \(y_0 \notin f(C)\), Hausdorff separation gives disjoint open neighborhoods \(U_x\) of \(x \in f(C)\) and \(V_x\) of \(y_0\); compactness of \(f(C)\) yields a finite subcover \(\bigcup_{i=1}^n U_{x_i} \supset f(C)\), and the intersection \(\bigcap_{i=1}^n V_{x_i}\) is an open neighborhood of \(y_0\) disjoint from \(f(C)\).
4.3. Hence \(f(C)\) is closed in \(Y\).
3, 1, and 2.
5. Conclusion: \(f\) maps every closed set \(C \subseteq X\) to a closed set \(f(C) \subseteq Y\), so \(f\) is a closed map.
1 and 4.
Q.E.D.