LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-02

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Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
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An equicontinuous family on a compact metric space is uniformly equicontinuous

Statement

Let $(K, d)$ be a nonempty compact metric space and let $F \subseteq C (K, R)$ be equicontinuous. For every $ε > 0$ there is $δ > 0$ such that for every $f \in F$ and $x, y \in K$ , $d (x, y) < δ$ implies $∣ f (x) - f (y) ∣ < ε$ .

Facts & Assumptions

Given: A positive real $ε$ and an equicontinuous family $F$ on $K$ .

[L1]

Equicontinuity at each $a \in K$ gives a radius $r_{a} > 0$ such that $d (x, a) < r_{a}$ implies $∣ f (x) - f (a) ∣ < ε / 2$ for every $f \in F$ (Equicontinuity, pointwise boundedness, and uniform boundedness for families in $C (K, R)$ ).

[L2]

Compactness means that every open cover has a finite subcover (Open cover, subcover, compact metric space, and compact subset of a metric space).

Proof

technique · direct

1.1

The balls $B (a, r_{a} / 2)$ for $a \in K$ cover $K$ , so choose finitely many centres $a_{0}, \dots, a_{N}$ whose balls cover $K$ .

L1L2choose

2.1

Let $δ$ be the least of the finitely many positive radii $r_{a_{i}} / 2$ . If $d (x, y) < δ$ , choose $i$ with $x \in B (a_{i}, r_{a_{i}} / 2)$ ; then both $x$ and $y$ lie in $B (a_{i}, r_{a_{i}})$ .

step 1.1algebra

3.1

The two estimates from [L1] and the triangle inequality give $∣ f (x) - f (y) ∣ < ε$ for every $f \in F$ .

step 2.1L1algebra∎

Depends on

Used by

Dependency tree · two levels

9 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

The Ascoli--Arzelà Theorem (MIT) (standard reference, not scraped)