Assume \((X,d)\) is a compact metric space.
Prove that \(X\) is both complete and separable.
Suppose \(\{x_k\}_{k=1}^\infty \subseteq X\) is a sequence such that the series \(\sum_{k=1}^\infty d(x_k, x_{k+1})\) converges. Prove that the sequence \(\{x_k\}_{k=1}^\infty\) converges in \(X\).
Solution
Part 1a: completeness.
1. Every sequence in \(X\) has a convergent subsequence.
compact metric spaces are sequentially compact.
2. Every Cauchy sequence \((x_n)\) in \(X\) converges.
2.1. \((x_n)\) has a convergent subsequence \(x_{n_k} \to x\).
1.
1.2. \(x_n \to x\).
for \(\eps > 0\), choose \(N\) with \(d(x_n, x_m) < \eps/2\) for \(n, m \ge N\) (Cauchy), and \(k\) with \(n_k \ge N\) and \(d(x_{n_k}, x) < \eps/2\); then \(d(x_n, x) \le d(x_n, x_{n_k}) + d(x_{n_k}, x) < \eps\) for \(n \ge N\).
3. Q.E.D.: \(X\) is complete.
2 is the definition.
Part 1b: separability.
4. For each \(m \in \NN\), \(X\) has a finite \(1/m\)-net: finitely many points \(x^{(m)}_1, \ldots, x^{(m)}_{N_m}\) with \(X = \bigcup_j B(x^{(m)}_j, 1/m)\).
\(\{B(x, 1/m)\}_{x \in X}\) is an open cover of the compact space \(X\), so it has a finite subcover; take the centers.
5. \(D = \bigcup_{m \in \NN}\{x^{(m)}_1, \ldots, x^{(m)}_{N_m}\}\) is countable and dense.
\(D\) is a countable union of finite sets, hence countable; given \(x \in X\) and \(\eps > 0\), choose \(m > 1/\eps\) and \(j\) with \(x \in B(x^{(m)}_j, 1/m)\); then \(d(x, x^{(m)}_j) < 1/m < \eps\).
Part 2.
6. \((x_k)\) is Cauchy: for \(n < m\), \(d(x_n, x_m) \le \sum_{k=n}^{m-1} d(x_k, x_{k+1})\).
iterated triangle inequality.
7. \(d(x_n, x_m) \to 0\) as \(n \to \infty\) uniformly in \(m > n\).
6 bounds \(d(x_n, x_m)\) by the tail \(\sum_{k \ge n} d(x_k, x_{k+1})\) of a convergent series, which tends to \(0\).
8. Q.E.D.: \((x_k)\) converges in \(X\).
6 and 7 show \((x_k)\) is Cauchy; completeness (<1>3) gives convergence.