{
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  "body": "\n## Statement\n\nLet $X$ be a [[normal space|0AJ646N339]], let $A \\subseteq X$ be [[closed|0G5AV9Q4RR]], and let $f \\colon A \\to \\mathbb{R}$ be [[continuous|0SY9ECQ19Q]] for the subspace topology on $A$.\nThen there is a continuous $F \\colon X \\to \\mathbb{R}$ whose [[restriction|0D2MXBV21T]] to $A$ is $f$.\nIf also $|f(a)| \\le M$ for all $a \\in A$, then one can choose $F$ with $|F(x)| \\le M$ for all $x \\in X$.\n\n## Remarks\n\nThe proof applies [[Urysohn's lemma|01EY3VR7ZD]] many times and sums a uniformly convergent series of functions.\nThe converse holds: if each continuous function from each closed subset to $\\mathbb{R}$ extends, then $X$ is normal.\nThe closedness of $A$ is needed: the function $\\sin(1/x)$ on $(0, 1]$ has no continuous extension to $[0, 1]$.\n\n## Explanations\n\n### L1 human\n\nOn a normal space, each continuous real function on a closed set can be extended to the whole space, without losing continuity.\nYou can also keep the same bound on the function.\n",
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