LemmaStatement: AI-adaptedProof: AI-generatedPipeline-generated

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.

The Axiom of Choice implies countable choice

Statement

Assume the full Axiom of Choice. For every sequence $(A_{n})_{n \in N}$ of nonempty sets there is a sequence $(a_{n})_{n \in N}$ with $a_{n} \in A_{n}$ for every $n \in N$ . Thus the Axiom of Choice implies the pair-local countable choice principle ${A C}_{ω}$ (The countable-choice principle used in the foliation pair).

Facts & Assumptions

Given: The Axiom of Choice and a sequence $(A_{n})_{n \in N}$ of nonempty sets.

[F1]

The Axiom of Choice states that every family of nonempty sets has a choice function: there is a function $g$ with domain $F$ such that $g (S) \in S$ for all $S \in F$ (The Axiom of Choice).

[F2]

The pair-local countable choice principle ${A C}_{ω}$ states that for every sequence $(A_{n})_{n \in N}$ of nonempty sets there is a function $c$ with domain $N$ such that $c (n) \in A_{n}$ for every $n$ (The countable-choice principle used in the foliation pair).

[F3]

A sequence indexed by $N$ is a function on $N$ , and the composition of functions is a function with the appropriate domains (A function is a relation $f$ with $(a, b) \in f$ and $(a, c) \in f$ implying $b = c$ ; $f : A \to B$ , the value $f (a)$ , domain and codomain).

Proof

technique · direct

1.1F1F3

Let $F : = {A_{n} : n \in N}$ be the family of sets occurring in the sequence; every member of $F$ is nonempty, so by the Axiom of Choice there is a choice function $g$ with domain $F$ and $g (S) \in S$ for all $S \in F$ [F1]. Define $c (n) : = g (A_{n})$ for $n \in N$ . This is a composite of the function $n \mapsto A_{n}$ with $g$ , hence a function with domain $N$ [F3].

2.1F2step 1.1∎

For every $n$ one has $c (n) = g (A_{n}) \in A_{n}$ , since $A_{n} \in F$ and $g$ is a choice function on $F$ . Thus $(c (n))_{n \in N}$ is a sequence with $c (n) \in A_{n}$ for every $n$ , which is exactly the witness required by ${A C}_{ω}$ ; the sequence of nonempty sets was arbitrary, so the Axiom of Choice implies the countable choice principle.

Depends on

Used by

Dependency tree · two levels

10 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

D. H. Fremlin, Measure Theory, Chapter 56 (standard reference, not scraped)
Axiom of countable choice (Wikipedia) (standard reference, not scraped)