Arzelà–Ascoli Theorem

Let $(K,d)$ be a compact metric space and consider $C(K,\mathbb{R})$ with the sup metric $d_\infty(f,g)=\sup_{x\in K}|f(x)-g(x)|.$

A subset $\mathcal{F}\subseteq C(K,\mathbb{R})$ is relatively compact if its closure in $(C(K,\mathbb{R}),d_\infty)$ is compact.

Arzelà–Ascoli Theorem (real-valued, compact metric domain): For $\mathcal{F}\subseteq C(K,\mathbb{R})$, the following are equivalent:

• $\mathcal{F}$ is relatively compact in $(C(K,\mathbb{R}),d_\infty)$.
• $\mathcal{F}$ is equicontinuous on $K$ and pointwise bounded on $K$.

Equivalently (sequential form): $\mathcal{F}$ is relatively compact if and only if every sequence in $\mathcal{F}$ has a uniformly convergent subsequence (with respect to $d_\infty$).

This theorem is the main compactness criterion for families of continuous functions and is central in existence proofs and approximation theory.

Proof sketch: ($\Rightarrow$) If $\overline{\mathcal{F}}$ is compact, then any sequence in $\mathcal{F}$ has a uniformly convergent subsequence. Pointwise boundedness follows because uniform convergence implies pointwise boundedness of a convergent subsequence and compactness gives uniform boundedness over the compact set. Equicontinuity follows by contradiction: if not equicontinuous, there exist $\varepsilon_0>0$, functions $f_n\in\mathcal{F}$, and points $x_n,y_n\in K$ with $d(x_n,y_n)\to 0$ but $|f_n(x_n)-f_n(y_n)|\ge \varepsilon_0$. Extract a uniformly convergent subsequence $f_{n_k}\to f$ and convergent subsequences $x_{n_k}\to x$, $y_{n_k}\to x$ (compactness of $K$). Then uniform convergence and continuity of $f$ force $|f_{n_k}(x_{n_k})-f_{n_k}(y_{n_k})|\to 0$, contradicting $\varepsilon_0$.

($\Leftarrow$) Assume $\mathcal{F}$ is equicontinuous and pointwise bounded. Choose a countable dense subset $D=\{x_1,x_2,\dots\}\subseteq K$. By pointwise boundedness and Bolzano–Weierstrass in $\mathbb{R}$, from any sequence $(f_n)$ in $\mathcal{F}$ one can extract a subsequence that converges at $x_1$; then extract a further subsequence that converges at $x_2$; continue and take a diagonal subsequence $(g_n)$ that converges at every $x_j\in D$. Equicontinuity then upgrades convergence on the dense set $D$ to uniform convergence on $K$ (using compactness of $K$ to pass from local equicontinuity to global control). The uniform limit is continuous, so the subsequence converges in $C(K,\mathbb{R})$. Hence every sequence has a uniformly convergent subsequence, so $\mathcal{F}$ is relatively compact.