Let \(X\) be a compact space and let \(A\) be a closed subspace. Show that \(A\) is compact.
Solution
Let \(X\) be compact, \(A\subset X\) closed, and \(\theset{U_\alpha} \covers A\) be an open cover. By definition of the subspace topology, each \(U_\alpha = V_\alpha \intersect A\) for some open \(V_\alpha \subset X\), and \(A\subset \union_\alpha V_\alpha\). Since \(A\) is closed in \(X\), \(X\setminus A\) is open. Then \(\theset{V_\alpha}\union \theset{X\setminus A}\covers X\) is an open cover, since every point is either in \(A\) or \(X\setminus A\). By compactness of \(X\), there is a finite subcover \(\theset{U_j \suchthat j\leq N}\union \theset{X\setminus A}\) Then \(\qty{\theset{U_j} \union \theset{X\setminus A}} \intersect A \definedas \theset{V_j}\) is a finite cover of \(A\).