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

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Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
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Equicontinuity and pointwise boundedness on a compact metric space imply uniform boundedness

Statement

An equicontinuous, pointwise-bounded family $F \subseteq C (K, R)$ on a nonempty compact metric space is uniformly bounded.

Facts & Assumptions

Given: An equicontinuous and pointwise-bounded family $F$ .

[L1]

There is a common radius $δ > 0$ such that $d (x, y) < δ$ implies $∣ f (x) - f (y) ∣ < 1$ for all $f \in F$ (An equicontinuous family on a compact metric space is uniformly equicontinuous).

[L2]

Pointwise boundedness and uniform boundedness have the quantified meanings in Equicontinuity, pointwise boundedness, and uniform boundedness for families in $C (K, R)$ .

Proof

technique · direct

1.1

The $δ$ -balls cover $K$ ; compactness supplies finitely many centres $a_{0}, \dots, a_{N}$ whose $δ$ -balls cover it.

L1choose

2.1

By pointwise boundedness, choose $M_{i} \geq 0$ with $∣ f (a_{i}) ∣ \leq M_{i}$ for every $f \in F$ , and let $M : = 1 + \max_{i} M_{i}$ .

L2step 1.1choose

3.1

For $x \in K$ , choose $i$ with $d (x, a_{i}) < δ$ ; then $∣ f (x) ∣ \leq ∣ f (x) - f (a_{i}) ∣ + ∣ f (a_{i}) ∣ < 1 + M_{i} \leq M$ .

step 1.1step 2.1L1algebra

4.1

Thus $M$ uniformly bounds $F$ .

step 3.1L2∎

Depends on

Used by

Dependency tree · two levels

4 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

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