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  "body": "\n## Proof\n\nWe show that the closure is complete and totally bounded, and use [[compact iff complete and totally bounded|054Q5PW5GX]]. Write $d$ for the metric of $X$, $\\norm{f - g} = d_\\infty(f, g)$, and $G$ for the closure of $\\mathcal{F}$ in $C(X)$.\n\nSetting up.\n\n1. If $X$ is empty, $C(X)$ has one element. Then $G$ is finite, hence compact, and $\\mathcal{F}$ is vacuously equicontinuous and bounded with $M = 0$. From now on let $X$ be nonempty.\n2. Let $f \\in C(X)$. Since $X$ is [[compact|0QPHW703PV]], $f$ [[attains its bounds|0V4DET59NW]], so $f$ is bounded. Thus $d_\\infty$ is the [[uniform metric|06HE9J7DXF]] restricted to $C(X)$, a metric.\n3. The space $\\R$ is complete by the [[completeness of the real numbers|E3CPZNJZNG]]. By step 2, $C(X)$ is the space of bounded continuous functions $X \\to \\R$. So $C(X)$ is [[complete|09GRCZVR5E]].\n4. By [[the closure is the smallest closed superset|0SHC6Z304Z]], $\\mathcal{F} \\subseteq G$. By [[closure is closed|02TZDKG50Q]], $G$ is closed in $C(X)$.\n5. If $\\mathcal{F}$ is empty, then $G$ is empty and both sides hold. From now on let $\\mathcal{F}$ be nonempty.\n\nFirst, let $G$ be compact.\n\n6. By [[compact iff complete and totally bounded|054Q5PW5GX]], $G$ is [[totally bounded|0WE5BBHGM6]]. By the [[epsilon-net characterisation|0NMGJJ19HD]], for each $\\varepsilon \u003e 0$ there are finitely many $g_1, \\ldots, g_m \\in G$ such that each $h \\in G$ has $\\norm{h - g_k} \u003c \\varepsilon$ for some $k$.\n7. Uniform bound. Take $\\varepsilon = 1$ in step 6. By step 2, choose $M_k$ with $\\abs{g_k(x)} \\le M_k$ for all $x$, and let $M = \\max_k M_k + 1$.\n8. Let $f \\in \\mathcal{F}$ and $x \\in X$. Since $f \\in G$, some $k$ has $\\norm{f - g_k} \u003c 1$. Then $\\abs{f(x)} \\le \\abs{f(x) - g_k(x)} + \\abs{g_k(x)} \u003c 1 + M_k \\le M$.\n9. Equicontinuity. Let $a \\in X$ and $\\varepsilon \u003e 0$. Take $g_1, \\ldots, g_m$ as in step 6 for $\\varepsilon / 3$.\n10. Each $g_k$ is a continuous map $X \\to \\R$. By [[continuity via open sets|0YF82GBTND]] at the point $a$, there is $\\delta_k \u003e 0$ with $\\abs{g_k(x) - g_k(a)} \u003c \\varepsilon / 3$ when $d(x, a) \u003c \\delta_k$. Let $\\delta = \\min_k \\delta_k \u003e 0$.\n11. Let $f \\in \\mathcal{F}$ and $d(x, a) \u003c \\delta$. Choose $k$ with $\\norm{f - g_k} \u003c \\varepsilon / 3$. Then $\\abs{f(x) - f(a)} \\le \\abs{f(x) - g_k(x)} + \\abs{g_k(x) - g_k(a)} + \\abs{g_k(a) - f(a)} \u003c \\varepsilon$.\n12. The number $\\delta$ does not depend on $f$. So $\\mathcal{F}$ is [[equicontinuous|0MW2WXTHMC]] at $a$, for each $a \\in X$.\n\nConversely, let $\\mathcal{F}$ be equicontinuous, with $\\abs{f(x)} \\le M$ for all $f \\in \\mathcal{F}$ and $x \\in X$.\n\n13. We first show that $\\mathcal{F}$ is totally bounded. Fix $\\varepsilon \u003e 0$ and let $\\eta = \\varepsilon / 5$.\n14. For each $a \\in X$, equicontinuity gives $\\delta_a \u003e 0$ with $\\abs{f(x) - f(a)} \u003c \\eta$ for all $f \\in \\mathcal{F}$ and all $x$ with $d(x, a) \u003c \\delta_a$.\n15. The balls $B(a, \\delta_a)$ are [[open|0B2305GNAZ]] and cover $X$. By compactness, there are $a_1, \\ldots, a_p \\in X$ with $X = B(a_1, \\delta_{a_1}) \\cup \\cdots \\cup B(a_p, \\delta_{a_p})$.\n16. Since [[the reals are Archimedean|0A3GWZ86AY]], choose an integer $N \\ge 0$ with $N \\eta \\ge M$. Let $\\Lambda = \\set{j \\eta : j \\in \\Z, \\ \\abs{j} \\le N}$, a finite set.\n17. Let $t \\in [-M, M]$ and $j = \\floor{t / \\eta}$, the [[floor|0YVRNEDSVG]]. Since $-N \\le t / \\eta \\le N$, we get $-N \\le j \\le N$. Also $j \\eta \\le t \u003c j \\eta + \\eta$, so $\\abs{t - j \\eta} \u003c \\eta$ with $j \\eta \\in \\Lambda$.\n18. For each map $\\varphi \\colon \\set{1, \\ldots, p} \\to \\Lambda$, let $\\mathcal{F}_\\varphi$ be the set of $f \\in \\mathcal{F}$ with $\\abs{f(a_k) - \\varphi(k)} \u003c \\eta$ for each $k$. There are finitely many maps $\\varphi$.\n19. Each $f \\in \\mathcal{F}$ lies in some $\\mathcal{F}_\\varphi$. Indeed, $f(a_k) \\in [-M, M]$, so step 17 gives $\\varphi(k) \\in \\Lambda$ with $\\abs{f(a_k) - \\varphi(k)} \u003c \\eta$.\n20. Let $f, g \\in \\mathcal{F}_\\varphi$ and $x \\in X$. By step 15, $d(x, a_k) \u003c \\delta_{a_k}$ for some $k$.\n21. Then $\\abs{f(x) - g(x)}$ is at most $\\abs{f(x) - f(a_k)} + \\abs{f(a_k) - \\varphi(k)} + \\abs{\\varphi(k) - g(a_k)} + \\abs{g(a_k) - g(x)}$. By steps 14 and 18, this is less than $4 \\eta$.\n22. Taking the supremum over $x$, $\\norm{f - g} \\le 4 \\eta \u003c \\varepsilon$. So each nonempty $\\mathcal{F}_\\varphi$ has [[diameter|0692CGQF00]] less than $\\varepsilon$.\n23. By steps 19 and 22, the finitely many nonempty sets $\\mathcal{F}_\\varphi$ cover $\\mathcal{F}$. So $\\mathcal{F}$ is totally bounded.\n24. Next, $G$ is totally bounded. Let $\\varepsilon \u003e 0$. By step 23 and the [[epsilon-net characterisation|0NMGJJ19HD]], there is a finite $H \\subseteq \\mathcal{F}$ with each $f \\in \\mathcal{F}$ within $\\varepsilon / 2$ of some $h \\in H$.\n25. Let $g \\in G$. By [[closure in a metric space|068P0P32DW]], some $f \\in \\mathcal{F}$ has $\\norm{g - f} \u003c \\varepsilon / 2$. Choose $h \\in H$ with $\\norm{f - h} \u003c \\varepsilon / 2$. Then $\\norm{g - h} \u003c \\varepsilon$.\n26. Since $H \\subseteq \\mathcal{F} \\subseteq G$, step 25 gives a finite $\\varepsilon$-net of $G$ inside $G$. By the epsilon-net characterisation, $G$ is totally bounded.\n27. Next, $G$ is complete. Let $(g_n)$ be a Cauchy sequence in $G$. By step 3, $g_n \\to g$ for some $g \\in C(X)$.\n28. Since $G$ is closed, [[closed sets contain their limits|09GQ3ZTQJ9]], so $g \\in G$. Thus $G$ is complete.\n29. By steps 26 and 28 and [[compact iff complete and totally bounded|054Q5PW5GX]], $G$ is compact.\n\n## Idea\n\nIn the complete space $C(X)$, a closed set is compact exactly when it is totally bounded. Equicontinuity lets finitely many sample points control a function everywhere, up to a small error. Uniform boundedness lets finitely many grid values control the samples, so finitely many small groups cover the family. Conversely, a finite net of continuous functions passes its boundedness and its continuity to the whole family.\n",
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