TheoremStatement: AI-adaptedProof: AI-adaptedverified 2026-09-26 (gpt-6-sol)

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

Arzelà--Ascoli for real $C (K)$ under Countable Choice and Dependent Choice: compact closure iff equicontinuous and pointwise bounded

Statement

Assume the Axiom of Countable Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )) 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 and $F \subseteq C (K, R)$ . Its closure in the supremum metric is compact if and only if $F$ is equicontinuous and pointwise bounded.

Facts & Assumptions

Given: The Axiom of Countable Choice, the Axiom of Dependent Choice, and a family $F \subseteq C (K, R)$ .

[L1]

Every open cover of the compact metric space $K$ has a finite subcover (Open cover, subcover, compact metric space, and compact subset of a metric space).

[L2]

$C (K, R)$ is complete in the supremum metric ( $C (K, R)$ is complete in the supremum metric for every nonempty compact metric space $K$ ).

[L3]

A subspace of a complete metric space is complete exactly when it is closed; assuming Countable Choice and Dependent Choice, in a metric space compactness is equivalent to completeness together with total boundedness (Closed subspaces of complete metric spaces are complete; the converse under countable choice, 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).

[L4]

Equicontinuity and pointwise boundedness are as defined in Equicontinuity, pointwise boundedness, and uniform boundedness for families in $C (K, R)$ .

[L5]

A continuous function on a compact metric space is uniformly continuous (Heine-Cantor: a continuous map from a compact metric space to any metric space is uniformly continuous).

Proof

technique · direct

1.1L1L4algebra

If $F = \emptyset$ , its closure is empty and compact, while both family properties hold vacuously. Hence assume $F \neq \emptyset$ . Suppose first that $F$ is equicontinuous and pointwise bounded, and fix $ε > 0$ . Let $U$ contain every ball $B (a, r)$ with $a \in K$ , $r > 0$ , and $∣ f (x) - f (a) ∣ < ε / 3$ for all $f \in F$ whenever $d (x, a) < 2 r$ . Equicontinuity makes $U$ an open cover without selecting a radius for every $a$ . By [L1] take a finite subcover $B (a_{i}, r_{i})$ , $i = 1, \dots, N$ . Pointwise boundedness supplies finite bounds $M_{i}$ for the values $∣ f (a_{i}) ∣$ . Partition the bounded box $\prod_{i = 1}^{N} [- M_{i}, M_{i}]$ into finitely many boxes of coordinate diameter less than $ε / 3$ , and choose one member of $F$ from each box met by an evaluation vector $(f (a_{i}))_{i}$ . For any $f \in F$ , a chosen $g$ in the same box satisfies $∣ f (a_{i}) - g (a_{i}) ∣ < ε / 3$ for all $i$ ; if $x \in B (a_{i}, r_{i})$ , the two equicontinuity bounds give $∣ f (x) - g (x) ∣ < ε$ . Thus these finitely many representatives form an $ε$ -net. The closure is totally bounded as well, by using an $ε / 2$ -net for $F$ and approximating each point of its closure by a member of $F$ .

1.2L3L5

Conversely suppose the closure is compact. For $ε > 0$ , [L3] supplies a finite $ε / 3$ -net $g_{1}, \dots, g_{N}$ in the closure. By [L5], a common positive radius makes every $g_{i}$ vary by less than $ε / 3$ .

2.1L2L3step 1.1

The closure of $F$ is closed in the complete space of [L2], hence complete by [L3]. It is totally bounded by step 1.1, so [L3] makes it compact.

2.2step 1.2algebra

For $f \in F$ , choose $g_{i}$ within $ε / 3$ in supremum distance. The two uniform-distance bounds and step 1.2 give $∣ f (x) - f (y) ∣ < ε$ whenever $d (x, y)$ is below the common radius.

2.3step 1.2L4algebra

The same finite net bounds $∣ f (a) ∣$ at each fixed $a \in K$ , so $F$ is pointwise bounded.

3.1step 2.1step 2.2step 2.3L4∎

Step 2.1 proves the forward implication; steps 2.2 and 2.3 give equicontinuity and pointwise boundedness in the converse.

Depends on

Dependency tree · 2 levels

Sources

The Ascoli--Arzelà Theorem (MIT)

Arzelà--Ascoli for real C(K)C(K) under Countable Choice and Dependent Choice: compact closure iff equicontinuous and pointwise bounded