Alphabeta Math
LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-02
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  • 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 pointwise-bounded family in C(K,R) has a finite net in the supremum metric

Statement

An equicontinuous pointwise-bounded family F⊆C(K,R) is totally bounded for the supremum metric.

Facts & Assumptions

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

[L1]

Uniform equicontinuity gives a finite set E⊆K such that agreement within ε/3 at every point of E forces agreement within ε 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-net and totally bounded metric space).

Proof

technique · constructive
1.1

Choose a finite δ-net E={a0,…,aN} in K from the uniform equicontinuity radius for ε/3.

L1construct
1.2

By [L2], every vector (f(a0),…,f(aN)) lies in one bounded box in RN+1; cover that box by finitely many coordinate cubes of side less than ε/3.

L2construct
2.1

Choose one member of F from each nonempty inverse image of such a cube. Every f∈F and its chosen representative differ by less than ε/3 on E.

step 1.2construct
3.1

For any x∈K, choose ai∈E with d(x,ai)<δ and use equicontinuity for both functions and step 2.1 to obtain ∣f(x)−g(x)∣<ε.

step 1.1step 2.1L1algebra
4.1

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

step 3.1L3discharge-construct∎

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