Exercise · ZW-SEQUENTIAL-COMPACTNESS-METRIC-SPACES-009 · ★★★ Advanced
Let \(K\) be a sequentially compact metric space and \(F\subseteq K\) closed . Is \(F\) sequentially compact ?
Choose one option.
A No: \(F\) may lose limit points B Yes, but only if \(F\) is also open C Yes: a sequence in \(F\) has a subsequence converging in \(K\), and the limit is in \(F\) because \(F\) is closed D No: \(F\) may be unbounded Show answer C. Yes: a sequence in \(F\) has a subsequence converging in \(K\), and the limit is in \(F\) because \(F\) is closed← Back to the question
Explanation A sequence in \(F\) is a sequence in \(K\); some subsequence converges to a point of \(K\), and since \(F\) is closed it contains that limit . A closed subset of \(K\) cannot lose limits nor be unbounded, since \(K\) is bounded.
A metric space \(X\) is sequentially compact if every sequence \((x_n)\) in \(X\) has a subsequence \((x_{n_k})\) converging to a point \(x\in X\).