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  "body": "\n## Statement\n\nLet $X$ be a [[normal space|0AJ646N339]] and let $A, B \\subseteq X$ be disjoint [[closed sets|0G5AV9Q4RR]].\nThen there is a [[continuous map|0SY9ECQ19Q]] $f \\colon X \\to [0, 1]$ with $f(a) = 0$ for all $a \\in A$ and $f(b) = 1$ for all $b \\in B$.\n\n## Remarks\n\nThe converse also holds: if such a function exists for all disjoint closed $A$ and $B$, then $X$ is normal. Take $U = f^{-1}([0, 1/2))$ and $V = f^{-1}((1/2, 1])$.\nThe proof builds open sets $U_r$ for the dyadic rationals $r \\in [0, 1]$, nested by the order of $r$. Then $f(x)$ is the infimum of the $r$ with $x \\in U_r$.\nIn a metric space, the function $f(x) = d(x, A) / (d(x, A) + d(x, B))$ works.\nThe [[Tietze extension theorem|0X52TNMTCJ]] extends this result to all continuous functions on closed sets.\n\n## Explanations\n\n### L1 human\n\nIn a normal space, a continuous function can separate two disjoint closed sets: it is $0$ on one set and $1$ on the other.\nSo normal spaces have many continuous real functions.\n",
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    "name": "Urysohn's lemma",
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  "msc": [
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  "name": "Urysohn's lemma",
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