Show that for \(X\) metrizable, the following are equivalent:
\(X\) is compact;
\(X\) is limit point compact;
\(X\) is sequentially compact.
The original exercise read “Show that if \(X\) is metrizable, then \(X\) is compact”, which is false (\(\RR\) is metrizable and not compact).
Solution
Fix a metric \(d\) inducing the topology of \(X\).
1. Compactness implies limit-point compactness.
Let \(A\subseteq X\) be infinite. If \(A\) had no limit point, then for each \(x\in X\) there would be an open neighborhood \(U_x\) with \[ U_x\cap A\subseteq\{x\}. \] A finite subcover \(U_{x_1},\dots,U_{x_r}\) would then give \[ A\subseteq\{x_1,\dots,x_r\}, \] a contradiction.
2. Limit-point compactness implies sequential compactness.
Let \((x_n)\) be a sequence. If its range is finite, one value occurs infinitely often and gives a constant subsequence. If its range is infinite, let \(x\) be a limit point of the range.
Every ball about \(x\) contains infinitely many terms of the sequence: otherwise a smaller ball would meet the range in only finitely many points, contradicting that \(x\) is a limit point. Inductively choose \[ n_1<n_2<\cdots, \qquad d(x_{n_k},x)<1/k. \] Then \(x_{n_k}\to x\).
3. Sequential compactness implies total boundedness.
If some \(\varepsilon>0\) admitted no finite cover by \(\varepsilon\)-balls, choose inductively \(x_{n+1}\) outside \[ \bigcup_{j=1}^n B(x_j,\varepsilon). \] Then \(d(x_i,x_j)\ge\varepsilon\) for \(i\ne j\), so the sequence has no convergent subsequence, a contradiction.
4. Sequential compactness implies the Lebesgue-number property: every open cover \(\mathcal U\) has some \(\delta>0\) such that every ball \(B(x,\delta)\) lies in a member of \(\mathcal U\).
If not, for each \(n\) choose \(x_n\) such that \(B(x_n,1/n)\) is contained in no member of \(\mathcal U\). Pass to a subsequence \(x_{n_k}\to x\). Choose \(U\in\mathcal U\) and \(r>0\) with \(B(x,r)\subseteq U\). For large \(k\), \[ d(x_{n_k},x)<r/2, \qquad 1/n_k<r/2, \] so \(B(x_{n_k},1/n_k)\subseteq B(x,r)\subseteq U\), a contradiction.
5. Sequential compactness implies compactness.
Let \(\mathcal U\) be an open cover and choose a Lebesgue number \(\delta>0\). By total boundedness, finitely many balls \[ B(x_1,\delta),\dots,B(x_m,\delta) \] cover \(X\). Each such ball lies in some member of \(\mathcal U\), so those finitely many members form a finite subcover.
6. Hence compactness, limit-point compactness, and sequential compactness are equivalent for metrizable spaces.
Source Collections
Backlinks
E-6MBWQA metrizable space is compact if and only if it is sequentially compactrelated-to