Show that if \(f: A\to B\) is a continuous map between metric spaces and \(K\subset A\) is compact, then \(\restrictionof{f}{K}\) is uniformly continuous.
Solution
1. Fix \(\varepsilon>0\). For each \(x\in K\), continuity of \(f\) at \(x\) gives \(r_x>0\) such that\[ d_A(z,x)<r_x\implies d_B(f(z),f(x))<\varepsilon/2. \]The balls \(B_A(x,r_x/2)\) cover \(K\).
2. By compactness, choose \(x_1,\dots,x_m\in K\) such that\[ K\subseteq\bigcup_{i=1}^m B_A(x_i,r_{x_i}/2). \]Set\[ \delta=\min_{1\le i\le m}r_{x_i}/2>0. \]
3. Let \(x,y\in K\) with \(d_A(x,y)<\delta\). Choose \(i\) with \(x\in B_A(x_i,r_{x_i}/2)\). Then\[ d_A(x,x_i)<r_{x_i}/2<r_{x_i} \]and\[ d_A(y,x_i)\le d_A(y,x)+d_A(x,x_i) <\delta+r_{x_i}/2\le r_{x_i}. \]Therefore\[ d_B(f(x),f(x_i))<\varepsilon/2, \qquad d_B(f(y),f(x_i))<\varepsilon/2. \]The triangle inequality gives\[ d_B(f(x),f(y))<\varepsilon. \]Thus \(f|_K\) is uniformly continuous.