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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 [[Lipschitz map|0TMZ21WKCN]].\nThen $f$ is [[uniformly continuous|0FHSDCQACY]].\n\n## Remarks\n\nProof: let $K \u003e 0$ be a Lipschitz constant of $f$. For $\\varepsilon \u003e 0$, take $\\delta = \\varepsilon / K$. If $d(x, y) \u003c \\delta$, then $d(f(x), f(y)) \\le K d(x, y) \u003c \\varepsilon$. If $f$ is Lipschitz with $K = 0$, then $f$ is constant.\nThe converse fails: $x \\mapsto \\sqrt{x}$ on $[0, 1]$ is uniformly continuous, but it is not Lipschitz.\nIn particular, each Lipschitz map is continuous.\n\n## Explanations\n\n### L1 human\n\nFor a Lipschitz map, you can give an explicit $\\delta$ for each $\\varepsilon$: divide $\\varepsilon$ by the constant $K$.\nThis $\\delta$ does not depend on the point, so the map is uniformly continuous.\n",
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