Show that if \(X\) is second countable and Hausdorff, or a metric space, then TFAE:
\(X\) is compact;
Every infinite subset \(A\subseteq X\) has a limit point in \(X\);
Every sequence in \(X\) has a convergent subsequence in \(X\).
Solution
1. If \(X\) is compact and Hausdorff (hence \(T_1\)), every infinite subset \(A\subseteq X\) has a limit point.
Assume \(A\) has no limit point. For each \(x\in X\) choose an open neighborhood \(U_x\) meeting \(A\) in at most the point \(x\) itself. A finite subcover \(U_{x_1},\dots,U_{x_r}\) would then meet \(A\) in at most finitely many points, contradicting infinitude of \(A\).
2. If \(X\) is first countable and every infinite subset has a limit point, then every sequence has a convergent subsequence.
Given \((x_n)\), if some value occurs infinitely often, take a constant subsequence. Otherwise its range contains an infinite subset \(A\) of distinct values. Let \(x\) be a limit point of \(A\), and choose a nested countable neighborhood basis \(U_1\supseteq U_2\supseteq\cdots\) at \(x\). Inductively choose distinct sequence terms \(x_{n_k}\in U_k\) with \(n_1<n_2<\cdots\). Then \(x_{n_k}\to x\).
3. If \(X\) is second countable and every sequence has a convergent subsequence, then \(X\) is compact.
Second countability implies Lindelöf, so every open cover has a countable subcover \(U_1,U_2,\dots\). If no finite subfamily covers \(X\), choose \[x_n\in X\setminus(U_1\cup\cdots\cup U_n).\] A convergent subsequence \(x_{n_k}\to x\) must eventually lie in some \(U_m\) containing \(x\), but for \(n_k\ge m\) the construction forces \(x_{n_k}\notin U_m\), a contradiction.
4. Therefore, in every second-countable Hausdorff space, compactness, limit-point compactness, and sequential compactness are equivalent.
Second countability implies first countability, so <1>1–<1>3 form a cycle of implications.
5. The same equivalence holds for every metric space.
The implications compact \(\Rightarrow\) limit-point compact and limit-point compact \(\Rightarrow\) sequentially compact follow from <1>1–<1>2 because metric spaces are Hausdorff and first countable. Conversely, a sequentially compact metric space is totally bounded: otherwise one can construct an \(\varepsilon\)-separated sequence with no convergent subsequence. For each \(m\) choose a finite \(1/m\)-net; the union of these nets is countable and dense, so the metric space is second countable. Now apply 3.