Show that every compact set is closed and bounded.
Show that if a subset of a metric space is complete and totally bounded, then it is compact.
Show that if \(K\) is compact and \(F\) is closed with \(K, F\) disjoint then \(\dist(K, F) > 0\).
Solution
Work in a metric space \((X, d)\).
1. A compact \(K \subseteq X\) is closed and bounded.
For \(x \notin K\) and \(y \in K\), the balls \(U_y = B(y, d(x,y)/3)\) and \(V_y = B(x, d(x,y)/3)\) are disjoint. Finitely many \(U_{y_1}, \ldots, U_{y_n}\) cover \(K\), and \(\bigcap_i V_{y_i}\) is an open ball about \(x\) disjoint from \(K\), so \(X \setminus K\) is open. For boundedness, fix \(x_0\); finitely many of the balls \(B(x_0, n)\) cover \(K\), so \(K \subseteq B(x_0, N)\) for some \(N\). See E-FFARP.
2. A complete and totally bounded \(E \subseteq X\) is compact.
Given a sequence in \(E\), cover \(E\) by finitely many balls of radius \(1\), pass to a subsequence in one of them, then cover by finitely many balls of radius \(1/2\) and pass to a further subsequence, and so on. The diagonal subsequence has its \(m\)th and later terms in one ball of radius \(1/m\), so it is Cauchy, and it converges in \(E\) because \(E\) is complete. So \(E\) is sequentially compact, which for metric spaces is equivalent to compactness. See E-4CL6A.
3. If \(K\) is compact, \(F\) is closed, and \(K \cap F = \emptyset\), then \(\dist(K, F) > 0\).
\(x \mapsto \dist(x, F)\) satisfies \(|\dist(x, F) - \dist(y, F)| \le d(x, y)\), so it is continuous and attains its minimum on \(K\) at some \(k_0\), assuming \(K\) and \(F\) nonempty. If \(\dist(k_0, F) = 0\), then \(k_0 \in \overline F = F\), contradicting \(K \cap F = \emptyset\). See E-SA3HI.