Let \((X,d)\) be a metric space, \(K\subset X\) be compact, and \(F\subset X\) be closed. If \(K\cap F=\emptyset\), prove that there exists an \(\epsilon>0\) so that \(d(k,f)\geq \epsilon\) for all \(k\in K\) and \(f\in F\).
Proof. We prove this by contrapositive. Suppose for all \(\epsilon >0\), there exists \(k \in K\), \(f \in F\) such that \(d(k,f)< \epsilon\). Then for all \(n \in \mathbb{N}\), we can choose \(k_n \in K\), \(f_n \in F\) such that \(d(k_n, f_n) < \frac{1}{n}\).
Since \(k_n\) is a sequence in \(K\), which is compact (and therefore sequentially compact), there exists a subsequence \(k_{n_j} \subseteq k_n\) with the property that \(k_{n_j}\) converges to some \(k_0 \in K\). Find \(N \in \mathbb{N}\) such that for \(n \geq N\), \(d(k_{n_j}, k_0) < \frac{\epsilon}{2}\) and \(\frac{1}{n} < \frac{\epsilon}{2}\). Then \[d(f_{n_j}, k_0) \leq d(f_{n_j}, k_{n_j}) + d(k_{n_j}, k_0) < \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon\]
Thus, \(f_{n_j}\) also converges to \(k_0\), and since \(F\) is closed, \(k_0 \in F\). So \(K \cap F \neq \emptyset\). ◻
Solution
1. The function \(\delta(x) \da d(x,F) = \inf_{f\in F} d(x,f)\) is \(1\)-Lipschitz, hence continuous.
for \(x, x' \in X\) and any \(f \in F\), \(d(x,f) \le d(x,x') + d(x',f)\), so \(\delta(x) \le d(x,x') + \delta(x')\); symmetrically \(\delta(x') \le d(x,x') + \delta(x)\), hence \(|\delta(x) - \delta(x')| \le d(x,x')\).
2. \(\delta\) attains its minimum on the compact set \(K\).
\(\delta\) is continuous (<1>1), and a continuous function on a compact set attains its minimum: choose \(k_n \in K\) with \(\delta(k_n) \to \inf_K \delta\); a subsequence \(k_{n_j} \to k \in K\) (sequential compactness in the metric space), and continuity gives \(\delta(k) = \inf_K\delta\).
3. \(\inf_{k\in K} d(k,F) > 0\).
by 2 the infimum is \(\delta(k)\) for some \(k \in K\).
4. There is \(\eps > 0\) with \(d(k,f) \ge \eps\) for all \(k \in K\), \(f \in F\).
take \(\eps \da \inf_{k\in K} d(k,F) > 0\) from 3; then \(d(k,f) \ge d(k,F) \ge \eps\) for all \(k \in K\), \(f \in F\).
5. Q.E.D.
(The card already contains a proof by contrapositive using sequential compactness; the argument above is the direct version via continuity of the distance function.)