---
tag: 0381S8Z5P0
kind: statement
role: theorem
name: Arzelà–Ascoli theorem
aliases: [ascoli-theorem]
msc: [54C35]
uses: [0MW2WXTHMC, 06HE9J7DXF, 0QPHW703PV, 054Q5PW5GX]
sources:
  - url: https://en.wikipedia.org/wiki/Arzel%C3%A0%E2%80%93Ascoli_theorem
    locator: lead section
  - url: https://en.wikipedia.org/wiki/Arzel%C3%A0%E2%80%93Ascoli_theorem
    locator: Statement and first consequences
same_as:
  wikidata: Q1477053
  wikipedia: Arzelà–Ascoli theorem
---

## Statement

Let $X$ be a [[compact|0QPHW703PV]] [[metric space|063W5JMY85]] and let $C(X)$ be the set of continuous functions $X \to \mathbb{R}$, with the [[uniform metric|06HE9J7DXF]] $d_\infty(f, g) = \sup_{x \in X} |f(x) - g(x)|$.
Let $\mathcal{F} \subseteq C(X)$.
Then the closure of $\mathcal{F}$ in $C(X)$ is compact if and only if $\mathcal{F}$ is [[equicontinuous|0MW2WXTHMC]] and uniformly bounded.
Here uniformly bounded means that there is a real $M$ with $|f(x)| \le M$ for all $f \in \mathcal{F}$ and $x \in X$.

## Remarks

A sequential form: each uniformly bounded and equicontinuous sequence of functions $[a, b] \to \mathbb{R}$ has a subsequence that converges uniformly on $[a, b]$.
The proof shows that $\mathcal{F}$ is [[totally bounded|0WE5BBHGM6]] in the complete space $C(X)$, and then uses [[compact iff complete and totally bounded|054Q5PW5GX]].
For example, the functions $f \colon [0, 1] \to [0, 1]$ that are [[Lipschitz|0TMZ21WKCN]] with constant $1$ form a compact set in $C([0, 1])$.
The theorem is a main tool for existence proofs, for example in the Peano existence theorem for ordinary differential equations.

## Explanations

### L1 human

In $\mathbb{R}^n$, bounded sequences have convergent subsequences. The Arzelà–Ascoli theorem gives the same result for sequences of functions, but you also need equicontinuity.
Equicontinuity stops the functions from becoming more and more steep, as $x^n$ does near $1$.