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

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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 $F \subseteq C (K, R)$ 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 $E \subseteq K$ 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).

[L2]

The family is uniformly bounded (Equicontinuity and pointwise boundedness on a compact metric space imply uniform boundedness).

[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

1.1L1construct

Choose a finite $δ$ -net $E = {a_{0}, \dots, a_{N}}$ in $K$ from the uniform equicontinuity radius for $ε / 3$ .

1.2L2construct

By [L2], every vector $(f (a_{0}), \dots, f (a_{N}))$ lies in one bounded box in $R^{N + 1}$ ; cover that box by finitely many coordinate cubes of side less than $ε / 3$ .

2.1step 1.2construct

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

3.1step 1.1step 2.1L1algebra

For any $x \in K$ , choose $a_{i} \in E$ with $d (x, a_{i}) < δ$ and use equicontinuity for both functions and step 2.1 to obtain $∣ f (x) - g (x) ∣ < ε$ .

4.1step 3.1L3discharge-construct∎

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

Depends on

Dependency tree · 2 levels

Sources

The Ascoli--Arzelà Theorem (MIT)

An equicontinuous pointwise-bounded family in C(K,R)C(K,R) has a finite net in the supremum metric