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

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Every pointwise-bounded equicontinuous sequence in $C(K,\mathbb R)$ has a uniformly convergent subsequence

Statement

Assume the Axiom of Countable Choice (The Axiom of Countable Choice ( $\mathrm{AC}_\omega$ )) and the Axiom of Dependent Choice (The axiom of dependent choice: a relation in which every element is related to something admits an $\mathbb{N}$ -indexed chain). Let $K$ be a nonempty compact metric space. Every equicontinuous pointwise-bounded sequence in $C(K,\mathbb R)$ has a subsequence converging uniformly to a member of $C(K,\mathbb R)$ .

Facts & Assumptions

Given: A nonempty compact metric space $K$ , the stated choice principles, and an equicontinuous pointwise-bounded sequence $(f_n)$ in $C(K,\mathbb R)$ .

[L1]

For a nonempty compact metric space $K$ , an equicontinuous pointwise-bounded family in $C(K,\mathbb R)$ has compact closure in the supremum metric (Arzelà--Ascoli for real $C(K)$ under Countable Choice and Dependent Choice: compact closure iff equicontinuous and pointwise bounded).

[L2]

Assuming Countable Choice and Dependent Choice, a compact metric space is sequentially compact (For a metric space, compact, countably compact, limit point compact, sequentially compact, and complete together with totally bounded are all equivalent, given countable choice and dependent choice).

Proof

technique · direct

1.1

The sequence lies in its compact closure by [L1].

2.1

By [L2], it has a subsequence converging in the supremum metric to a point of that closure.

step 1.1L2choose

3.1

Supremum-metric convergence is uniform convergence, so the claimed subsequence converges uniformly.

step 2.1algebra∎

Depends on

Used by

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Sources

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

Every pointwise-bounded equicontinuous sequence in C(K,R)C(K,\mathbb R)C(K,R) has a uniformly convergent subsequence