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Exercise 1782Topology › Metric and normed spacesL1Added 3 Nov 2026

Heine–Cantor theorem

#Heine-Cantor#uniform continuity#compact

Statement

State and prove that continuous maps from a compact metric space are uniformly continuous.

Hint

Contradiction via sequences.

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Solution

Theorem. Continuous f:K→Yf:K\to Y with KK compact is uniformly continuous.

Proof. Suppose not: ∃ε>0\exists\varepsilon>0 with xn,yn∈Kx_n,y_n\in K, d(xn,yn)→0d(x_n,y_n)\to0 but d(f(xn),f(yn))≥εd(f(x_n),f(y_n))\ge\varepsilon. Compactness of KK gives a subsequence xnk→xx_{n_k}\to x; then ynk→xy_{n_k}\to x. Continuity: f(xnk),f(ynk)→f(x)f(x_{n_k}),f(y_{n_k})\to f(x), contradiction.

Conclusion. Heine–Cantor holds in general metric spaces.