- Show that if \(K\) is compact and \(F\) is closed with \(K, F\) disjoint then \(\dist(K, F) > 0\).
Compactness of \(K\) cannot be weakened to closedness. In \(\RR\), \(K = \NN\) and \(F = \theset{n + 2^{-n} : n \in \NN}\) are closed and disjoint, and \(\dist(K, F) = \inf_n 2^{-n} = 0\).
Solution
Work in a metric space \((X,d)\) with \(K\) and \(F\) nonempty; if either is empty, \(\dist(K,F) = \infty\).
1. The function \(x \mapsto \dist(x, F)\) is \(1\)-Lipschitz, hence continuous.
For \(z \in F\), \(\dist(x,F) \leq d(x,z) \leq d(x,y) + d(y,z)\). Taking the infimum over \(z\) gives \(\dist(x,F) \leq d(x,y) + \dist(y,F)\), and the same holds with \(x\) and \(y\) exchanged.
2. There is \(k_0 \in K\) with \(\dist(k_0, F) = \dist(K, F)\).
\(\dist(K,F) = \inf_{k \in K}\dist(k,F)\), and a continuous function on a nonempty compact set attains its infimum; step 1 gives continuity.
3. \(\dist(k_0, F) > 0\).
If \(\dist(k_0, F) = 0\), then \(k_0\) is a limit of points of \(F\); since \(F\) is closed, \(k_0 \in F\), contradicting \(K \cap F = \emptyset\).
4. Q.E.D.
Steps 2 and 3.