LemmaStatement: Literature-sourcedProof: AI-adaptedverified 2026-09-26 (gpt-6-sol)

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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]

For each $a \in K$ there exists some radius $r > 0$ such that $d (x, a) < r$ 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

1.1L1L2construct

Form the set of all pairs $(a, r)$ with $a \in K$ , $r > 0$ , and the property in [L1]. The balls $B (a, r / 2)$ for these pairs form an open cover of $K$ by [L1]. Compactness gives finitely many such pairs $(a_{0}, r_{0}), \dots, (a_{N}, r_{N})$ whose balls cover $K$ . No radius was selected simultaneously for every point of $K$ .

2.1step 1.1algebra

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

3.1step 2.1L1algebra∎

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

Depends on

Dependency tree · 2 levels

Sources

The Ascoli--Arzelà Theorem (MIT)