LemmaStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-02

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

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.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

An equicontinuous family on a compact metric space is uniformly equicontinuous

Statement

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

Facts & Assumptions

Given: A positive real $\varepsilon$ and an equicontinuous family $\mathcal F$ on $K$ .

[L1]

Equicontinuity at each $a\in K$ gives a radius $r_a>0$ such that $d(x,a)<r_a$ implies $|f(x)-f(a)|<\varepsilon/2$ for every $f\in\mathcal F$ (Equicontinuity, pointwise boundedness, and uniform boundedness for families in $C(K,\mathbb R)$ ).

[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 $a_0,\ldots,a_N$ whose balls cover $K$ .

L1L2choose

2.1

Let $\delta$ be the least of the finitely many positive radii $r_{a_i}/2$ . If $d(x,y)<\delta$ , 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 $|f(x)-f(y)|<\varepsilon$ for every $f\in\mathcal F$ .

step 2.1L1algebra∎

Depends on

Used by

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 33 results over 12 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources

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