Uniform continuity on compact sets

1. Suppose, toward a contradiction, that \(f\) is not uniformly continuous on \(K\).Proof: then there is \(\epsilon > 0\) such that for every \(\delta > 0\) there are \(x, y \in K\) with \(|x - y| < \delta\) but \(|f(x) - f(y)| \ge \epsilon\).Taking \(\delta = 1/n\), we get sequences \(x_n, y_n \in K\) with \(|x_n - y_n| < 1/n\) and \(|f(x_n) - f(y_n)| \ge \epsilon\) for all \(n\).

2. Extract a convergent subsequence.Proof: \(K\) is compact, hence sequentially compact (in \(\mathbb{R}^n\) compact ⟹ every sequence has a convergent subsequence), so a subsequence \(x_{n_j} \to x \in K\).Since \(|x_{n_j} - y_{n_j}| < 1/n_j \to 0\), the corresponding subsequence \(y_{n_j}\) also converges to \(x\).

3. Contradict continuity.Proof: \(f\) is continuous at \(x\), so \(f(x_{n_j}) \to f(x)\) and \(f(y_{n_j}) \to f(x)\); hence \(|f(x_{n_j}) - f(y_{n_j})| \to 0\).But by construction \(|f(x_{n_j}) - f(y_{n_j})| \ge \epsilon\) for all \(j\), contradiction.

4. Q.E.D. Proof: hence \(f\) is uniformly continuous on \(K\).