PropositionStatement: 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.

$\prod_{i \in \emptyset} A_{i} = {\emptyset}$ ; if $A_{j} = \emptyset$ for some $j \in I$ then $\prod_{i \in I} A_{i} = \emptyset$ ; and for $I = {j}$ the evaluation $f \mapsto f (j)$ is a bijection $\prod_{i \in I} A_{i} \to A_{j}$

Statement

Let $(A_{i})_{i \in I}$ be an indexed family. Then

(i) if $I = \emptyset$ then $\prod_{i \in I} A_{i} = {\emptyset}$ ;
(ii) if $A_{j} = \emptyset$ for some $j \in I$ then $\prod_{i \in I} A_{i} = \emptyset$ ;
(iii) if $I = {j}$ then the evaluation $e : = {(f, a) \in (\prod_{i \in I} A_{i}) \times A_{j} : a = f (j)}$ is a bijection $\prod_{i \in I} A_{i} \to A_{j}$ .

Clauses (i) and (ii) pull in opposite directions and are the two cases most often mis-stated: an empty index set gives a product with one element, while a single empty member collapses the product entirely.

Facts & Assumptions

Given: an indexed family $(A_{i})_{i \in I}$ .

[L1]

an element of $\prod_{i \in I} A_{i}$ is a function with domain $I$ that takes its value at each index inside the member carried by that index (The product $\prod_{i \in I} A_{i} : = {f : I \to ⋃_{i \in I} A_{i} ∣ f (i) \in A_{i} for every i \in I}$ ).

[L2]

A function is a relation $f$ such that $(a, b) \in f$ and $(a, c) \in f$ imply $b = c$ (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).

[L3]

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

[L4]

$f$ is injective (one-to-one) if $f (x) = f (y)$ implies $x = y$ , for all $x, y \in A$ (Injection, surjection, bijection).

[L5]

$f$ is surjective (onto) if for every $b \in B$ there is some $x \in A$ with $f (x) = b$ (Injection, surjection, bijection).

[L6]

$f = g$ if and only if $dom f = dom g$ and $f (x) = g (x)$ for every $x \in dom f$ (Functions $f$ and $g$ are equal if and only if $dom f = dom g$ and $f (x) = g (x)$ for every $x$ in that common domain).

[L7]

There is exactly one set with no elements, written $\emptyset$ (There is exactly one set with no elements, written $\emptyset$ ).

[L8]

${x, y}$ is the set whose elements are exactly $x$ and $y$ , and ${x} : = {x, x}$ (The unordered pair ${x, y}$ and the singleton ${x} = {x, x}$ ).

[L9]

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 $φ$ , $\forall \overset{ˉ}{p} \forall x \exists y \forall z (z \in y \leftrightarrow (z \in x \land φ (z, \overset{ˉ}{p})))$ ).

[L10]

$z \in A \times B$ holds if and only if $z = (a, b)$ for some $a \in A$ and some $b \in B$ (The Cartesian product $A \times B : = {z \in P (P (A \cup B)) : \exists a \in A \exists b \in B z = (a, b)}$ ).

[L11]

$z \in ⋃_{i \in I} A_{i}$ 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 $I \neq \emptyset$ ).

[L12]

If every $z$ satisfies $z \in x$ if and only if $z \in y$ , then $x = y$ (The Axiom of Extensionality: $\forall x \forall y (\forall z (z \in x \leftrightarrow z \in y) \to x = y)$ ).

Proof

technique · direct

1.1

Claim (i): let $I = \emptyset$ . A function with domain $\emptyset$ has no elements, since each of its elements would be a pair whose first coordinate lies in its domain, so the only such function is $\emptyset$ ; and it satisfies the condition " $f (i) \in A_{i}$ for every $i \in I$ " vacuously. Hence $\emptyset$ is the only element of the product, and the product is ${\emptyset}$ .

L1L2L3L7L8L12

1.2

Claim (ii): let $j \in I$ with $A_{j} = \emptyset$ . An element $f$ of the product would satisfy $f (j) \in A_{j}$ , and $\emptyset$ has no elements; so the product has no elements.

L1L7L12

1.3

Claim (iii), the map: let $I = {j}$ and write $P : = \prod_{i \in I} A_{i}$ . Each $f \in P$ has $f (j) \in A_{j}$ , so $e$ as displayed is a set by separation inside $P \times A_{j}$ , is single valued because $f (j)$ is, has domain $P$ , and has range inside $A_{j}$ ; thus $e : P \to A_{j}$ with $e (f) = f (j)$ .

L1L2L3L9L10

2.1

Claim (iii), injectivity: if $e (f) = e (g)$ for $f, g \in P$ , then $f$ and $g$ have the same domain ${j}$ and agree at $j$ , hence are equal.

L4L6L8step 1.3

2.2

Claim (iii), surjectivity: let $a \in A_{j}$ and put $f : = {(j, a)}$ . This is a function with domain ${j}$ and $f (j) = a \in A_{j}$ , and $a$ lies in $⋃_{i \in I} A_{i}$ , so $f \in P$ and $e (f) = a$ .

L1L2L3L5L8L11step 1.3

3.1

Claims (i), (ii) and (iii) are established, which is the statement.

step 1.1step 1.2step 1.3step 2.1step 2.2∎

Remarks

Where the general question is settled, and where it is not. These three computations use no choice principle: in (i) and (ii) nothing is selected, and in (iii) the single value $f (j)$ is determined. None of them is an instance of the general question, because each names its elements outright. Whether a product of nonempty members over an arbitrary index set is nonempty is that general question, and it is exactly the product formulation of the Axiom of Choice, stated earlier on this page at The Axiom of Choice; clause (ii) is the reason that formulation carries the hypothesis that every member is nonempty.

Depends on

Used by

Dependency tree · two levels

25 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 and Exercise 13 (standard reference, not scraped)
Cartesian product (Wikipedia) (standard reference, not scraped)
Empty product (Wikipedia) (standard reference, not scraped)

∏i∈∅Ai={∅}∏i∈∅​Ai​={∅}; if Aj=∅Aj​=∅ for some j∈Ij∈I then ∏i∈IAi=∅∏i∈I​Ai​=∅; and for I={j}I={j} the evaluation f↦f(j)f↦f(j) is a bijection ∏i∈IAi→Aj∏i∈I​Ai​→Aj​