Show that every compact metric space \((X, d)\) is complete (every Cauchy sequence in \(X\) converges to a point in \(X\)).
Solution
1. Let \((x_n)\) be a Cauchy sequence in the compact metric space \(X\). Compactness of a metric space implies sequential compactness, so some subsequence satisfies\[ x_{n_k}\longrightarrow x\in X. \]
2. Fix \(\varepsilon>0\). Since \((x_n)\) is Cauchy, choose \(N\) such that\[ m,n\ge N\implies d(x_m,x_n)<\varepsilon/2. \]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 \(x_n\to x\).
3. Every Cauchy sequence in \(X\) therefore converges to a point of \(X\), so \(X\) is complete.