LemmaStatement: AI-adaptedProof: AI-adaptedprecheck passverified 2026-08-06 (claude-opus-5)

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.

For an indexed family $(A_{i})_{i \in I}$ the collection of functions $f$ with domain $I$ and $f (i) \in A_{i}$ for every $i \in I$ is a set

Statement

Let $(A_{i})_{i \in I}$ be an indexed family. Then there is a set whose elements are exactly the functions $f$ with $dom f = I$ and $f (i) \in A_{i}$ for every $i \in I$ , and it is a subset of $C^{I}$ where $C : = ⋃_{i \in I} A_{i}$ .

Facts & Assumptions

Given: an indexed family $(A_{i})_{i \in I}$ , and $C : = ⋃_{i \in I} A_{i}$ .

[L1]

An indexed family with index set $I$ is a function $A$ with $dom A = I$ (An indexed family $(A_{i})_{i \in I}$ is a function with domain $I$ ; ${A_{i} : i \in I}$ is its range).

[L2]

$z \in ⋃_{i \in I} A_{i}$ holds if and only if $z \in A_{i}$ for some $i \in I$ ( $⋃_{i \in I} A_{i} : = ⋃ {A_{i} : i \in I}$ , and $⋂_{i \in I} A_{i} : = ⋂ {A_{i} : i \in I}$ for $I \neq \emptyset$ ).

[L3]

$f \in B^{A}$ holds if and only if $f : A \to B$ (The set $B^{A}$ of all functions $A \to B$ ).

[L4]

We write $f : A \to B$ , and say $f$ is a function from $A$ to $B$ , when $f$ is a function with $dom f = A$ and $ran f \subseteq B$ (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).

[L5]

For any parameters $\overset{ˉ}{p}$ and any set $x$ , there is a set $y$ whose elements are exactly the elements $z$ of $x$ for which $φ (z, \overset{ˉ}{p})$ holds (The Axiom Schema of Separation: for each formula $φ$ , $\forall \overset{ˉ}{p} \forall x \exists y \forall z (z \in y \leftrightarrow (z \in x \land φ (z, \overset{ˉ}{p})))$ ).

[L6]

$x \subseteq y$ means that every element of $x$ is an element of $y$ (Subset $x \subseteq y$ , proper subset $x ⊊ y$ , and the separation notation ${z \in x : φ (z)}$ ).

[L7]

$ran R : = {b : \exists a (a, b) \in R}$ (Relation, $dom R$ , $ran R$ , $fld R$ , and the specialisations "relation from $A$ to $B$ " and "relation on $A$ ").

Proof

technique · direct

1.1L1L2L3L4L6L7

Let $f$ be a function with $dom f = I$ and $f (i) \in A_{i}$ for every $i \in I$ . Every element of $ran f$ is $f (i)$ for some $i \in I$ , hence lies in $A_{i}$ and therefore in $C$ ; so $ran f \subseteq C$ and $f : I \to C$ , that is, $f \in C^{I}$ .

2.1L3L5L6step 1.1∎

Separating inside $C^{I}$ with the formula saying that $z (i) \in A_{i}$ for every $i \in I$ , with parameters $I$ and the family, gives a set whose elements are exactly the members of $C^{I}$ with that property; by step 1.1 every function of the kind described already lies in $C^{I}$ , so this set has exactly the intended elements and is included in $C^{I}$ .

Depends on

Used by

The product ∏_i ∈ I Aᵢ := { f : I → ⋃_i ∈ I Aᵢ ∣ f(i) ∈ Aᵢ for every i ∈ I } Definition

Dependency tree · two levels

18 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

B. Kaya, MATH 320 Set Theory (METU), Def. 21 (standard reference, not scraped)
Cartesian product (Wikipedia) (standard reference, not scraped)
Indexed family (Wikipedia) (standard reference, not scraped)

For an indexed family (Ai)i∈I(Ai​)i∈I​ the collection of functions ff with domain II and f(i)∈Aif(i)∈Ai​ for every i∈Ii∈I is a set