- Let \(X\) and \(Y\) be topological spaces, and let \(f: X \to Y\) be a continuous map. Prove that if \(X\) is compact, then its image \(f(X) \subseteq Y\) is compact.
- Let \((X, d)\) be a metric space. Prove that if \(X\) is compact, then \((X, d)\) is complete (every Cauchy sequence in \(X\) converges to a limit in \(X\)).
Solution
1. If \(X\) is compact and \(f:X\to Y\) is continuous, then \(f(X)\) is compact.
Let \(\{V_\alpha\}\) be an open cover of \(f(X)\). Then \(\{f^{-1}(V_\alpha)\}\) is an open cover of \(X\). Compactness gives finitely many indices \(\alpha_1,\dots,\alpha_r\) with \[ X=\bigcup_{j=1}^r f^{-1}(V_{\alpha_j}). \] Applying \(f\) shows \[ f(X)\subseteq\bigcup_{j=1}^r V_{\alpha_j}, \] so the original cover has a finite subcover.
2. Every compact metric space is complete.
Let \((x_n)\) be Cauchy in compact metric \(X\). Compact metric spaces are sequentially compact, so some subsequence \(x_{n_k}\) converges to a point \(x\in X\).
Given \(\varepsilon>0\), choose \(N\) so that \[ d(x_m,x_n)<\varepsilon/2\qquad(m,n\ge N). \] Choose \(k\) with \(n_k\ge N\) and \(d(x_{n_k},x)<\varepsilon/2\). Then for every \(n\ge N\), \[ d(x_n,x)\le d(x_n,x_{n_k})+d(x_{n_k},x)<\varepsilon. \] Thus \(x_n\to x\), so every Cauchy sequence converges in \(X\).