> [!definition] > > Let $X$ be a [[Compactness|compact]] [[Hausdorff Space|Hausdorff]] space and $\cf \subset C(X)$ be a set of [[Space of Continuous Functions|continuous functions]] on $X$. For any $x \in X$, $\cf$ is **equicontinuous at** $x$ if for every $\eps > 0$, there exists a [[Neighbourhood|neighbourhood]] $U \in \cn^o(x)$ such that $\abs{f(y) - f(x)} < \eps$ for all $y \in U$. If $\cf$ is equicontinuous at all $x \in X$, then $\cf$ is **equicontinuous**. If for every $x \in X$, $\bracs{f(x): f \in \cf}$ is bounded for each $x$, then $\cf$ is **pointwise bounded**. > [!theoremb] Arzelà-Ascoli > > Let $X$ be a compact Hausdorff space and $\cf \subset C(X)$, then the following are equivalent: > 1. $\cf$ is equicontinuous and pointwise bounded. > 2. $\cf$ is [[Totally Bounded|totally bounded]] with respect to the [[Uniform Norm|uniform norm]]. > > If any of the above holds, $\cf$ is [[Precompact|precompact]], and every sequence in $\cf$ admits a convergent subsequence. > > *Proof*. Suppose that $\cf$ is equicontinuous and uniformly bounded. Let $\eps > 0$. For each $x \in X$, there exists $U_x \in \cn^o(x)$ such that $\abs{f(y) - f(x)} < \eps/4$ for all $y \in U_x$ and $f \in \cf$. Since $X$ is [[Compactness|compact]], there exists $\seqf{x_j}$ such that $\bigcup_{j = 1}^n U_{x_j} = X$. > > By pointwise boundedness, $E = \bracs{f(x_j): f \in \cf, j \in [n]} \subset \complex$ is bounded, and there exists a finite covering $E \supset \bigcup_{j = 1}^n B(z_j, \eps/4)$ using balls of radius $\eps/4$. > > Let $S = \bracs{\phi: \seqf{x_j} \to \seqf{z_j}}$ be the space of functions from the original sample points to the covering centres, and define > $ > \cf_\phi = \bracs{f \in \cf: \abs{f(x_j) - \phi(x_j)} < 1/4\eps \forall j \in [n]} > $ > Firstly, for any $f, g \in \cf_\phi$ and $x \in X$, there exists $j \in [n]$ such that $x \in U_{x_j}$. In which case, > $ > \begin{align*} > \abs{f(x) - g(x)} &\le \abs{f(x) - f(x_j)} + \abs{g(x) - g(x_j)} \\ > &+ \abs{f(x_j) - \phi(x_j)} + \abs{g(x_j) - \phi(x_j)} \le \eps > \end{align*} > $ > so $\cf_\phi$ is contained in a ball of radius $\eps$. On the other hand, for any $f \in \cf$, there exists $\phi \in S$ such that $f \in \cf_\phi$ as $\seqf{B(z_j, \eps/4)}$ is a covering of $E$. > > Since $S$ itself is finite, $\cf$ is totally bounded with respect to the uniform norm. > > Now suppose that $\cf$ is totally bounded. Since the pointwise evaluation map is continuous with respect to the uniform norm, $\{f(x): f \in \ol{\cf}\}$ is compact and thus bounded for all $x \in X$. On the other hand, let $f \in C(X)$, $\eps > 0$, $x \in X$, and $g \in B(f, \eps/3)$, then there exists $U \in \cn^o(x)$ such that $\abs{f(x) - f(y)} < \eps/3$ for all $y \in U$. In this case, > $ > \begin{align*} > \abs{g(x) - g(y)} &\le \abs{f(x) - g(x)} + \abs{f(x) - f(y)} + \abs{f(y) - g(y)} \\ > &\le 2\norm{f - g}_u + \abs{f(x) - f(y)} < \eps > \end{align*} > $ > Therefore $B(f, \eps/3)$ is equicontinuous. Since $\cf$ is totally bounded, picking the intersection of all relevant neighbourhoods with an $\eps/3$-covering yields the desired result.