Past exam of the mathematics course of the University of Cambridge / 2025 / ii / Paper 3 / 21I / a / Solution

The Arzela-Ascoli theorem in sequential form states: if $K$ is a compact metric space and $(f_n)$ is a uniformly bounded, equicontinuous sequence in $C(K)$, then it has a uniformly convergent subsequence. Equivalently, a subset of $C(K)$ is relatively compact in the uniform norm exactly when it is uniformly bounded and equicontinuous.

For sufficiency, choose a finite $1/m$-net $E_m\subset K$ for every $m$. The union $D={\bigcup }_m{E}_m$ is countable and dense. Uniform boundedness makes each scalar sequence $(f_n(x){)}_n$, $x\in D$, bounded. Successive applications of Bolzano-Weierstrass followed by a diagonal choice give a subsequence $(g_n)$ for which $g_n(x)$ converges at every $x\in D$.

Given $\epsilon >0$, equicontinuity supplies $\delta >0$ such thatfor every $n$. Choose a finite $\delta$-net from $D$. Pointwise convergence on those finitely many points makes $(g_n)$ uniformly Cauchy on the net, and the two equicontinuity estimates make it uniformly Cauchy on all of $K$. Since $C(K)$ is complete, $g_n$ converges uniformly.

Conversely, a relatively compact family is uniformly bounded because its closure is compact in the normed space $C(K)$. Given $\epsilon >0$, cover that compact closure by finitely many uniform balls of radius $\epsilon /3$, centred at continuous functions $h_1,\dots ,h_r$. Uniform continuity of the finitely many $h_i$ gives one $\delta$ that works for all of them. Approximating any family member by one $h_i$ proves equicontinuity.