Show that a metric space \((X, d)\) is complete if and only if for every nested sequence \(A_1 \supset A_2 \supset A_3 \supset \cdots\) of non-empty closed subsets of \(X\) such that \(\lim_{n \to \infty} \operatorname{diam}(A_n) = 0\), the intersection \(\bigcap_{n=1}^\infty A_n\) is non-empty (and in fact consists of a unique point).
Solution
1. If \((X, d)\) is complete, then every nested sequence of nonempty closed sets with diameters tending to \(0\) has intersection a single point.Proof:
1.1. Let \(A_1 \supset A_2 \supset A_3 \supset \cdots\) be a nested sequence of non-empty closed subsets of \(X\) with \(\lim_{n \to \infty} \operatorname{diam}(A_n) = 0\).
1.2. For each \(n \ge 1\), since \(A_n \ne \emptyset\), choose a point \(x_n \in A_n\).
1.3. We claim that \(\{x_n\}_{n \ge 1}\) is a Cauchy sequence in \((X, d)\):- Let \(\varepsilon > 0\).- Since \(\operatorname{diam}(A_n) \to 0\), choose \(N\) such that \(\operatorname{diam}(A_N) < \varepsilon\).- For any \(n, m \ge N\), since the sequence of sets is nested (\(A_n \subseteq A_N\) and \(A_m \subseteq A_N\)), we have \(x_n \in A_N\) and \(x_m \in A_N\).- By definition of diameter:\[d(x_n, x_m) \le \operatorname{diam}(A_N) < \varepsilon.\]- Thus \(\{x_n\}\) is a Cauchy sequence.
1.4. Since \((X, d)\) is complete, there exists a limit point \(x = \lim_{n \to \infty} x_n \in X\).
1.5. For any fixed \(k \ge 1\), the tail sequence \(\{x_n\}_{n \ge k}\) is entirely contained in \(A_k\) (since \(x_n \in A_n \subseteq A_k\) for all \(n \ge k\)).
1.6. Since \(A_k\) is a closed set, the limit of the sequence lies in \(A_k\):\[x = \lim_{n \to \infty} x_n \in A_k.\]
1.7. Since this holds for all \(k \ge 1\), we have:\[x \in \bigcap_{n=1}^\infty A_n \ne \emptyset.\]
1.8. Uniqueness: If \(x, y \in \bigcap_{n=1}^\infty A_n\), then \(d(x, y) \le \operatorname{diam}(A_n)\) for all \(n \ge 1\). Taking \(n \to \infty\) gives \(d(x, y) \le 0 \implies x = y\).
2. If every nested sequence of nonempty closed sets with diameters tending to \(0\) has nonempty intersection, then \((X, d)\) is complete.Proof:
2.1. Let \(\{x_n\}_{n \ge 1}\) be any Cauchy sequence in \((X, d)\).
2.2. For each \(n \ge 1\), define the tail set \(T_n \coloneqq \{ x_k \mid k \ge n \}\).
2.3. Define \(A_n \coloneqq \overline{T_n}\) to be the closure of the \(n\)-th tail.
2.4. Each \(A_n\) is a closed nonempty subset of \(X\), and \(T_{n+1} \subseteq T_n\) gives \(A_{n+1} \subseteq A_n\), so the sequence is nested.
2.5. We show that \(\operatorname{diam}(A_n) \to 0\):- For any set \(S \subseteq X\), its closure has the same diameter: \(\operatorname{diam}(\overline{S}) = \operatorname{diam}(S)\).- Thus \(\operatorname{diam}(A_n) = \operatorname{diam}(T_n) = \sup_{j, k \ge n} d(x_j, x_k)\).- Since \(\{x_n\}\) is Cauchy, for any \(\varepsilon > 0\) there exists \(N\) such that \(d(x_j, x_k) < \varepsilon\) for all \(j, k \ge N\), so \(\operatorname{diam}(A_N) \le \varepsilon\).- Therefore, \(\lim_{n \to \infty} \operatorname{diam}(A_n) = 0\).
2.6. By the hypothesis, the intersection is non-empty:\[\bigcap_{n=1}^\infty A_n \ne \emptyset.\]
2.7. Let \(x \in \bigcap_{n=1}^\infty A_n\).
2.8. We show that \(x_n \to x\):- For any \(\varepsilon > 0\), choose \(N\) such that \(\operatorname{diam}(A_N) < \varepsilon\).- Since \(x \in A_N\) and for all \(n \ge N\) we have \(x_n \in T_N \subseteq A_N\):\[d(x_n, x) \le \operatorname{diam}(A_N) < \varepsilon \quad \text{for all } n \ge N.\]- Hence \(\lim_{n \to \infty} x_n = x \in X\).
2.9. Thus every Cauchy sequence converges in \(X\), which proves that \((X, d)\) is complete.
3. Q.E.D.Steps 1 and 2 give the two implications.