Prove the Heine–Cantor Theorem: let \((X, d_X)\) and \((Y, d_Y)\) be metric spaces. If \(X\) is compact and \(f: X \to Y\) is continuous, then \(f\) is uniformly continuous.
Solution
Fix \(\varepsilon>0\). For each \(p\in X\), continuity gives \(r_p>0\) such that \[ d_X(x,p)<r_p \quad\Longrightarrow\quad d_Y(f(x),f(p))<\frac\varepsilon2. \] The balls \(B(p,r_p/2)\) cover \(X\). By compactness, choose a finite subcover \[ B(p_1,r_1/2),\dots,B(p_N,r_N/2). \] Set \[ \delta=\frac12\min_{1\le j\le N}r_j>0. \]
If \(d_X(x,y)<\delta\), choose \(j\) with \(x\in B(p_j,r_j/2)\). Then \[ d_X(y,p_j)\le d_X(y,x)+d_X(x,p_j) <\delta+\frac{r_j}{2}\le r_j. \] Thus \[ d_Y(f(x),f(p_j))<\frac\varepsilon2, \qquad d_Y(f(y),f(p_j))<\frac\varepsilon2. \] By the triangle inequality, \[ d_Y(f(x),f(y))<\varepsilon. \] Hence \(f\) is uniformly continuous.