Give an example of a space that is compact but not sequentially compact, and vice versa.
Solution
1. A compact space need not be sequentially compact.
Let \[X=\{0,1\}^{\mathcal P(\mathbb N)}\] with the product topology. It is compact by Tychonoff’s theorem. For each \(n\in\mathbb N\), define \(x_n\in X\) by \[x_n(A)=\mathbf 1_{\{n\in A\}}.\] Given any subsequence \(x_{n_k}\), take \(A=\{n_{2k}:k\ge1\}\). Then the \(A\)-coordinate of \(x_{n_k}\) alternates \(0,1,0,1,\dots\), so the subsequence cannot converge. Thus \(X\) is not sequentially compact.
2. A sequentially compact space need not be compact.
Let \(\omega_1\) be the space of all countable ordinals with the order topology. Every sequence in \(\omega_1\) is bounded above by a countable ordinal \(\alpha<\omega_1\); its range lies in the compact metrizable ordinal interval \([0,\alpha]\), hence has a convergent subsequence. Thus \(\omega_1\) is sequentially compact. It is not compact because the open cover \[\{[0,\alpha):\alpha<\omega_1\}\] (or equivalently a suitable increasing family of initial open sets) has no finite subcover.