In $C(K,\RR^m)$, compactness is exactly closedness, boundedness, and equicontinuity.Suggest a slogan
Let \(n,m\geq1\), let \(K\subseteq\RR^n\) be compact, and equip \(C(K,\RR^m)\) with the metric \(d(f,g)\coloneqq\sup_{x\in K}\norm{f(x)-g(x)}\). A subset \(\mathcal F\subseteq C(K,\RR^m)\) is compact if and only if \(\mathcal F\) is closed, bounded, and equicontinuous at every point of \(K\), that is, \[ \forall x\in K\quad\forall\varepsilon>0\quad\exists\delta>0\quad\forall y\in K\quad\forall f\in\mathcal F:\quad\norm{x-y}<\delta\implies\norm{f(x)-f(y)}<\varepsilon. \] [@Mun00].