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CorollaryStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-02
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Every pointwise-bounded equicontinuous sequence in C(K,R) has a uniformly convergent subsequence

Statement

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

Facts & Assumptions

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

[L1]

For a nonempty compact metric space K, an equicontinuous pointwise-bounded family in C(K,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
1.1L1

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

2.1step 1.1L2choose

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

3.1step 2.1algebra∎

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

Depends on

Dependency tree · 2 levels

Sources