(June 2010 #2b) Prove that \(\{1/n:n\in\mathbb Z\setminus\{0\}\}\cup\{0\}\) is compact using the above definition.
Solution
1. Let \(\{U_\alpha\}\) be any open cover of \(K = \{1/n : n \in \mathbb{Z}\setminus\{0\}\} \cup \{0\}\) in \(\mathbb{R}\).
2. One open set covers a neighborhood of \(0\) and hence all but finitely many points of \(K\).Proof: \(0 \in K\), so some \(U_{\alpha_0}\) contains \(0\); since \(U_{\alpha_0}\) is open, \((-\epsilon, \epsilon) \subseteq U_{\alpha_0}\) for some \(\epsilon > 0\).Choose \(N\) with \(1/N < \epsilon\); then every point \(1/n\) with \(|n| \ge N\) lies in \((-\epsilon, \epsilon) \subseteq U_{\alpha_0}\).
3. The finitely many remaining points are covered by finitely many sets of the cover.Proof: the remaining points \(\{\pm 1, \pm 1/2, \ldots, \pm 1/(N-1)\}\) form a finite set; for each such point \(p\) choose some \(U_{\alpha_p}\) containing \(p\).Together with \(U_{\alpha_0}\) this gives a finite subcover.
4. Q.E.D. Proof: every open cover of \(K\) has a finite subcover, so \(K\) is compact by the definition.