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.

Countable choice is equivalent to nonempty countable products

Statement

For every sequence $(A_{n})_{n \in N}$ of nonempty sets, there exists a indexed choice function $c$ with domain $N$ and $c (n) \in A_{n}$ for every $n \in N$ if and only if the product $\prod_{n \in N} A_{n}$ is nonempty. Thus the sequence formulation of The countable-choice principle used in the foliation pair and the nonempty-product formulation of the same principle are equivalent.

Facts & Assumptions

Given: A sequence $(A_{n})_{n \in N}$ of nonempty sets.

[F1]

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

[F2]

The product of an indexed family $(A_{i})_{i \in I}$ is the set of functions $f$ with domain $I$ such that $f (i) \in A_{i}$ for every $i$ ; in particular an element of $\prod_{n \in N} A_{n}$ is a function with domain $N$ taking its value at $n$ inside $A_{n}$ (The product $\prod_{i \in I} A_{i} : = {f : I \to ⋃_{i \in I} A_{i} ∣ f (i) \in A_{i} for every i \in I}$ ).

[F3]

A sequence indexed by $N$ is a function on $N$ , and an indexed choice function for $(A_{n})$ has domain $N$ and selects an element of $A_{n}$ at $n$ ; this differs from a family choice function, whose domain is the set of factors (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, Choice function).

Proof

technique · direct

1.1F1F2F3

(Forward direction.) Assume there is an indexed choice function $c$ with $c (n) \in A_{n}$ for every $n$ [F1]. Then $c$ is a function with domain $N$ whose value at each $n$ lies in $A_{n}$ [F3], so by the defining description of the product $c \in \prod_{n \in N} A_{n}$ [F2]. In particular the product is nonempty.

1.2F1F2

(Reverse direction.) Assume the product $\prod_{n \in N} A_{n}$ is nonempty and choose an element $x$ of it; this is one existential instantiation. By [F2], $x$ is a function with domain $N$ and $x (n) \in A_{n}$ for every $n$ , that is, a choice function for the sequence $(A_{n})_{n \in N}$ in the sense of [F1]. Hence a choice function with $c (n) \in A_{n}$ for all $n$ exists.

2.1step 1.1step 1.2∎

The two directions identify the same objects: a function with domain $N$ whose value at $n$ belongs to $A_{n}$ is at once the indexed choice function of the sequence formulation and the element of the product of the product formulation. Hence the sequence formulation and the nonempty-product formulation are equivalent for every sequence of nonempty sets.

Depends on

Used by

Nothing in the library uses this result yet.

Cited to discharge well-definedness by The countable-choice principle used in the foliation pair.

Dependency tree · two levels

15 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)