Show that if \(X\) is second-countable and Hausdorff (or a metric space), then the following are equivalent: (1) \(X\) is compact (every open cover has a finite subcover). (2) \(X\) is limit-point compact (every infinite subset \(A \subseteq X\) has a limit point in \(X\)). (3) \(X\) is sequentially compact (every sequence in \(X\) has a convergent subsequence in \(X\)).
Solution
1. \((1)\Rightarrow(2)\). Suppose \(X\) is compact and \(A\subseteq X\) is infinite. If \(A\) had no limit point, then for every \(x\in X\) there would be an open neighborhood \(U_x\) with\[ U_x\cap A\subseteq\{x\}. \]A finite subcover \(U_{x_1},\dots,U_{x_m}\) would then imply\[ A\subseteq\{x_1,\dots,x_m\}, \]a contradiction.
2. \((2)\Rightarrow(3)\). Assume \(X\) is second-countable Hausdorff; the metric case is identical because metric spaces are first-countable and Hausdorff.
2.1. Let \((x_n)\) be a sequence. If its range is finite, some value occurs infinitely often and gives a constant convergent subsequence.
2.2. Otherwise its range \(S\) is infinite, so it has a limit point \(x\).
2.3. Second countability implies first countability, so choose a decreasing countable neighborhood basis\[ B_1\supseteq B_2\supseteq\cdots \]at \(x\).
2.4. Since Hausdorff spaces are \(T_1\), every neighborhood of a limit point of the infinite set \(S\) contains infinitely many points of \(S\): if one contained only finitely many points of \(S\setminus\{x\}\), deleting those finitely many closed points would produce a neighborhood of \(x\) meeting \(S\) in at most \(\{x\}\).
2.5. Hence one can choose \(n_1<n_2<\cdots\) with \(x_{n_k}\in B_k\). Then \(x_{n_k}\to x\).
3. \((3)\Rightarrow(1)\). Assume first that \(X\) is second-countable.
3.1. Every open cover has a countable subcover: if \(\mathcal B\) is a countable basis, for each basis element \(B\in\mathcal B\) lying in some member of the cover choose one such member; these chosen sets cover \(X\).
3.2. Let \(U_1,U_2,\dots\) be a countable open cover with no finite subcover. Choose\[ x_N\in X\setminus\bigcup_{n=1}^N U_n. \]Sequential compactness gives a subsequence \(x_{N_k}\to x\). Choose \(m\) with \(x\in U_m\). Eventually \(x_{N_k}\in U_m\), while for large \(k\) one has \(N_k\ge m\) and hence \(x_{N_k}\notin U_m\), a contradiction.
4. For a metric space, \((3)\Rightarrow(1)\) follows by the standard metric argument: sequential compactness implies total boundedness and completeness, and every complete totally bounded metric space is compact. Equivalently, one may use the Lebesgue-number proof for an arbitrary open cover.
5. Therefore compactness, limit-point compactness, and sequential compactness are equivalent in every second-countable Hausdorff space and in every metric space.