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LemmaStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-02
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An equicontinuous pointwise-bounded family in C(K,R)C(K,\mathbb R) has a finite net in the supremum metric

Statement

An equicontinuous pointwise-bounded family FC(K,R)\mathcal F\subseteq C(K,\mathbb R) is totally bounded for the supremum metric.

Facts & Assumptions

Given: A positive real ε\varepsilon and an equicontinuous pointwise-bounded family F\mathcal F.

[L1]

Uniform equicontinuity gives a finite set EKE\subseteq K such that agreement within ε/3\varepsilon/3 at every point of EE forces agreement within ε\varepsilon everywhere (An equicontinuous family on a compact metric space is uniformly equicontinuous).

[L3]

Totally bounded means that every positive radius admits a finite covering by metric balls (Finite ε\varepsilon-net and totally bounded metric space).

Proof

technique · constructive
1.1

Choose a finite δ\delta-net E={a0,,aN}E=\{a_0,\ldots,a_N\} in KK from the uniform equicontinuity radius for ε/3\varepsilon/3.

L1construct
1.2

By [L2], every vector (f(a0),,f(aN))(f(a_0),\ldots,f(a_N)) lies in one bounded box in RN+1\mathbb R^{N+1}; cover that box by finitely many coordinate cubes of side less than ε/3\varepsilon/3.

L2construct
2.1

Choose one member of F\mathcal F from each nonempty inverse image of such a cube. Every fFf\in\mathcal F and its chosen representative differ by less than ε/3\varepsilon/3 on EE.

step 1.2construct
3.1

For any xKx\in K, choose aiEa_i\in E with d(x,ai)<δd(x,a_i)<\delta and use equicontinuity for both functions and step 2.1 to obtain f(x)g(x)<ε|f(x)-g(x)|<\varepsilon.

step 1.1step 2.1L1algebra
4.1

The finitely many representatives form an ε\varepsilon-net, so F\mathcal F is totally bounded.

step 3.1L3discharge-construct

Depends on

Used by

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