Let \(X\) be a compact topological space and let \(A \subseteq X\) be a closed subset of \(X\). Prove that \(A\) is compact in the subspace topology.
Solution
1. Let \(\{V_i\}_{i\in I}\) be an open cover of \(A\) in the subspace topology. For each \(i\), choose an open set \(U_i\subseteq X\) with\[ V_i=U_i\cap A. \]
2. Since \(A\) is closed, \(X\setminus A\) is open. Hence\[ \{U_i:i\in I\}\cup\{X\setminus A\} \]is an open cover of \(X\).
3. Compactness of \(X\) gives finitely many indices \(i_1,\dots,i_m\) such that\[ X\subseteq U_{i_1}\cup\cdots\cup U_{i_m}\cup(X\setminus A). \]Intersecting with \(A\) yields\[ A\subseteq V_{i_1}\cup\cdots\cup V_{i_m}. \]Thus every open cover of \(A\) has a finite subcover, so \(A\) is compact.