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  "body": "\n## Proof\n\nWe first treat bounded $f$ by a series of Urysohn functions, then reduce the general case to it.\n\n1. Notation: $X$ is a [[normal space|0AJ646N339]], $A \\subseteq X$ is [[closed|0G5AV9Q4RR]], and $A$ has the [[subspace topology|0D5YHQGDJY]]. \"Continuous\" means [[continuous|0SY9ECQ19Q]] as a map of topological spaces.\n2. Criterion (C). Let $Z$ be a topological space and $h \\colon Z \\to \\mathbb{R}$. Then $h$ is continuous if and only if for each $z_0 \\in Z$ and $\\varepsilon \u003e 0$ there is an open $W \\ni z_0$ with $|h(z) - h(z_0)| \u003c \\varepsilon$ for all $z \\in W$.\n3. Proof of (C). If $h$ is continuous, take $W = h^{-1}((h(z_0) - \\varepsilon, h(z_0) + \\varepsilon))$, the preimage of an [[open ball|0B2305GNAZ]].\n4. Conversely, let $V \\subseteq \\mathbb{R}$ be open and $z_0 \\in h^{-1}(V)$. By [[the ball criterion|0A26X79RTB]], some $(h(z_0) - \\varepsilon, h(z_0) + \\varepsilon)$ lies in $V$. The set $W$ for this $\\varepsilon$ lies in $h^{-1}(V)$.\n5. So $h^{-1}(V)$ is a union of open sets, hence open by the [[axioms of a topology|0CNQXVBHCX]].\n6. By (C), if $g, h \\colon Z \\to \\mathbb{R}$ are continuous, then $g + h$ and $g - h$ are continuous. Given $\\varepsilon$, intersect the two sets $W$ for $\\varepsilon / 2$.\n7. The restriction to $A$ of a continuous $g \\colon X \\to \\mathbb{R}$ is continuous. It is $g \\circ \\iota$ with $\\iota$ the [[inclusion|065J822CKN]], and [[compositions are continuous|05NK26G0DR]].\n8. If $B \\subseteq A$ is closed in $A$, then $B$ is closed in $X$. Indeed $A \\setminus B = A \\cap U$ with $U$ open in $X$, so $X \\setminus B = (X \\setminus A) \\cup U$ is open.\n9. Lemma. Let $r \u003e 0$ and let $h \\colon A \\to \\mathbb{R}$ be continuous with $|h| \\le r$ on $A$. Then there is a continuous $g \\colon X \\to [-r/3, r/3]$ with $|h - g| \\le 2r/3$ on $A$.\n10. Proof of the lemma. Let $B = h^{-1}([-r, -r/3])$ and $D = h^{-1}([r/3, r])$. They are closed in $A$ by [[continuity via closed sets|0B8GPXFSP7]], so closed in $X$ by step 8. They are disjoint.\n11. By [[Urysohn's lemma|01EY3VR7ZD]] there is a continuous $u \\colon X \\to [0, 1]$ with $u = 0$ on $B$ and $u = 1$ on $D$. Let $g = (r/3)(2u - 1)$, which is continuous by (C) and takes values in $[-r/3, r/3]$.\n12. On $B$: $g = -r/3$ and $h \\in [-r, -r/3]$, so $|h - g| \\le 2r/3$. On $D$: $g = r/3$ and $h \\in [r/3, r]$, so $|h - g| \\le 2r/3$.\n13. On $A \\setminus (B \\cup D)$: $|h| \u003c r/3$ and $|g| \\le r/3$, so $|h - g| \u003c 2r/3$. This proves the lemma.\n14. Bounded case. Let $|f| \\le M$ on $A$. If $M = 0$, then $f = 0$ and $F = 0$ works. Let $M \u003e 0$.\n15. We build continuous $g_0, g_1, \\ldots \\colon X \\to \\mathbb{R}$ with $|g_n| \\le \\tfrac{M}{3}(\\tfrac{2}{3})^n$ on $X$ and $|f - s_n| \\le M(\\tfrac{2}{3})^{n+1}$ on $A$. Here $s_n = g_0 + \\cdots + g_n$.\n16. The lemma with $h = f$ and $r = M$ gives $g_0$. Given $g_0, \\ldots, g_n$, the function $f - s_n|_A$ is continuous on $A$ by steps 6 and 7. Apply the lemma to it with $r = M(\\tfrac{2}{3})^{n+1}$ to get $g_{n+1}$.\n17. Then $|g_{n+1}| \\le \\tfrac{M}{3}(\\tfrac{2}{3})^{n+1}$, and $|f - s_{n+1}| = |(f - s_n) - g_{n+1}| \\le M(\\tfrac{2}{3})^{n+2}$ on $A$.\n18. Fix $x \\in X$. For $m \u003e n$, $|s_m(x) - s_n(x)| \\le \\sum_{k = n+1}^{m} \\tfrac{M}{3}(\\tfrac{2}{3})^k \u003c M(\\tfrac{2}{3})^{n+1}$, by the [[finite geometric sum|05APWSX53S]]. So $(s_n(x))$ is Cauchy, and it has a limit $F(x)$ by the [[completeness of the real numbers|E3CPZNJZNG]].\n19. Letting $m \\to \\infty$ and using that [[limits preserve non-strict inequalities|0N32V0E3XJ]] gives $|F(x) - s_n(x)| \\le M(\\tfrac{2}{3})^{n+1}$ for all $x \\in X$ and all $n$.\n20. Also $|s_n(x)| \\le \\sum_{k=0}^{n} \\tfrac{M}{3}(\\tfrac{2}{3})^k \u003c M$, so $|F(x)| \\le M$ for all $x \\in X$.\n21. For $a \\in A$, $|f(a) - F(a)| \\le |f(a) - s_n(a)| + |s_n(a) - F(a)| \\le 2M(\\tfrac{2}{3})^{n+1}$ for each $n$. So $F(a) = f(a)$, and $F$ restricts to $f$.\n22. $F$ is continuous. Let $x_0 \\in X$ and $\\varepsilon \u003e 0$. Choose $n$ with $M(\\tfrac{2}{3})^{n+1} \u003c \\varepsilon / 3$. Each $s_n$ is continuous by step 6, so (C) gives an open $W \\ni x_0$ with $|s_n(x) - s_n(x_0)| \u003c \\varepsilon / 3$ on $W$.\n23. For $x \\in W$, step 19 gives $|F(x) - F(x_0)| \\le |F(x) - s_n(x)| + |s_n(x) - s_n(x_0)| + |s_n(x_0) - F(x_0)| \u003c \\varepsilon$. By (C), $F$ is continuous. This proves the bounded part of the theorem.\n24. General case. Let $\\varphi \\colon \\mathbb{R} \\to (-1, 1)$, $\\varphi(t) = t / (1 + |t|)$, and $\\psi \\colon (-1, 1) \\to \\mathbb{R}$, $\\psi(s) = s / (1 - |s|)$.\n25. The map $t \\mapsto |t|$ is continuous, since $||t| - |t_0|| \\le |t - t_0|$ by the [[reverse triangle inequality|0C9S13WKMF]]. So $\\varphi$ and $\\psi$ are continuous, as [[quotients|0GR2APWTAC]] with nonzero denominators.\n26. For $t \\in \\mathbb{R}$, $1 - |\\varphi(t)| = 1/(1 + |t|)$, so $\\psi(\\varphi(t)) = t$. For $s \\in (-1, 1)$, $1 + |\\psi(s)| = 1/(1 - |s|)$, so $\\varphi(\\psi(s)) = s$.\n27. The map $\\varphi \\circ f \\colon A \\to \\mathbb{R}$ is continuous, by (C) and the [[epsilon-delta continuity|0W8YHHGCJG]] of $\\varphi$. It satisfies $|\\varphi \\circ f| \u003c 1$ on $A$.\n28. The bounded case with $M = 1$ gives a continuous $G \\colon X \\to [-1, 1]$ with $G = \\varphi \\circ f$ on $A$.\n29. Let $E = G^{-1}(\\{-1, 1\\})$. It is closed in $X$ by continuity via closed sets. It is disjoint from $A$, because $|G| \u003c 1$ on $A$.\n30. By Urysohn's lemma there is a continuous $v \\colon X \\to [0, 1]$ with $v = 0$ on $E$ and $v = 1$ on $A$. Let $H = vG$.\n31. $H$ is continuous. For $x$ near $x_0$, $|H(x) - H(x_0)| \\le |v(x)||G(x) - G(x_0)| + |G(x_0)||v(x) - v(x_0)| \\le |G(x) - G(x_0)| + |v(x) - v(x_0)|$. Apply (C) to $G$ and $v$.\n32. $|H| \u003c 1$ on $X$. On $E$, $H = 0$. Off $E$, $|G| \u003c 1$ and $0 \\le v \\le 1$, so $|H| \\le |G| \u003c 1$.\n33. Let $F = \\psi \\circ H \\colon X \\to \\mathbb{R}$. It is continuous by (C) and the epsilon-delta continuity of $\\psi$ at $H(x_0)$.\n34. For $a \\in A$, $v(a) = 1$, so $H(a) = G(a) = \\varphi(f(a))$, and $F(a) = \\psi(\\varphi(f(a))) = f(a)$ by step 26.\n35. So $F$ is a continuous map $X \\to \\mathbb{R}$ whose [[restriction|0D2MXBV21T]] to $A$ is $f$. With the bounded case, this proves both parts.\n\n## Idea\n\nUrysohn's lemma lets us approximate a bounded function on $A$ by a function on all of $X$, with an error two thirds as large. Repeating this on the errors gives a geometric series of functions that converges uniformly to an extension. For unbounded $f$, squeeze $\\mathbb{R}$ into $(-1, 1)$, extend, and use Urysohn again to keep the extension away from $\\pm 1$.\n",
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