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

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Equicontinuity and pointwise boundedness on a compact metric space imply uniform boundedness

Statement

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

Facts & Assumptions

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

[L1]

There is a common radius $\delta>0$ such that $d(x,y)<\delta$ implies $|f(x)-f(y)|<1$ for all $f\in\mathcal F$ (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 $C(K,\mathbb R)$ .

Proof

technique · direct

1.1

The $\delta$ -balls cover $K$ ; compactness supplies finitely many centres $a_0,\ldots,a_N$ whose $\delta$ -balls cover it.

L1choose

2.1

By pointwise boundedness, choose $M_i\ge0$ with $|f(a_i)|\le M_i$ for every $f\in\mathcal F$ , and let $M:=1+\max_i M_i$ .

L2step 1.1choose

3.1

For $x\in K$ , choose $i$ with $d(x,a_i)<\delta$ ; then $|f(x)|\le|f(x)-f(a_i)|+|f(a_i)|<1+M_i\le M$ .

step 1.1step 2.1L1algebra

4.1

Thus $M$ uniformly bounds $\mathcal F$ .

step 3.1L2∎

Depends on

Used by

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

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 13 results over 8 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)