{
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  "body": "\n## Statement\n\nLet $X$ and $Y$ be [[metric spaces|063W5JMY85]] and let $f \\colon X \\to Y$ be a [[continuous map|0SY9ECQ19Q]].\nIf $X$ is [[compact|0QPHW703PV]], then $f$ is [[uniformly continuous|0FHSDCQACY]].\n\n## Remarks\n\nProof by contradiction: if $f$ is not uniformly continuous, there are $\\varepsilon \u003e 0$ and points $x_n, y_n$ with $d(x_n, y_n) \\to 0$ and $d(f(x_n), f(y_n)) \\ge \\varepsilon$. By [[sequential compactness|0K8SGQSV7G]], a subsequence of $(x_n)$ converges to some $x$, and the matching subsequence of $(y_n)$ also converges to $x$. Continuity at $x$ gives a contradiction.\nFor example, each continuous $f \\colon [a, b] \\to \\mathbb{R}$ is uniformly continuous. This fact is used to show that $f$ is Riemann integrable.\nThe theorem fails on $\\mathbb{R}$: $x \\mapsto x^2$ is continuous but not uniformly continuous.\n\n## Explanations\n\n### L1 human\n\nOn a compact space, continuity at each point is enough to get one $\\delta$ for the whole space.\nCompactness rules out regions where the function becomes steeper and steeper without end.\n",
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