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  "body": "\n## Proof\n\nWe give an explicit $\\delta$ for each $\\varepsilon$.\n\n1. Let $(X, d_X)$ and $(Y, d_Y)$ be [[metric spaces|063W5JMY85]]. Let $f \\colon X \\to Y$ be [[Lipschitz|0TMZ21WKCN]] with a constant $K \\ge 0$.\n2. So $d_Y(f(x), f(y)) \\le K \\, d_X(x, y)$ for all $x, y \\in X$.\n3. Let $\\varepsilon \u003e 0$ be real, and put $\\delta = \\varepsilon / (K + 1)$. Then $\\delta \u003e 0$, because $K + 1 \\ge 1$.\n4. Let $x, y \\in X$ with $d_X(x, y) \u003c \\delta$.\n5. By step 2 and $K \\ge 0$, we get $d_Y(f(x), f(y)) \\le K \\, d_X(x, y) \\le K \\delta = \\varepsilon K / (K + 1)$.\n6. Since $K / (K + 1) \u003c 1$ and $\\varepsilon \u003e 0$, we get $\\varepsilon K / (K + 1) \u003c \\varepsilon$. So $d_Y(f(x), f(y)) \u003c \\varepsilon$.\n7. The number $\\delta$ depends only on $\\varepsilon$ and $K$, not on $x$ or $y$.\n8. So condition 1 of [[uniform continuity|0FHSDCQACY]] holds, and $f$ is uniformly continuous.\n\n## Idea\n\nThe Lipschitz inequality turns a bound on the input distance into a bound on the output distance, with the same factor everywhere.\nDividing $\\varepsilon$ by $K + 1$ instead of $K$ avoids a special case when $K = 0$.\n",
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