Equivalence of Formulations of Axiom of Choice

Theorem
The statements of the Axiom of Choice:


 * $(1): \quad$ For every set, we can provide a mechanism for choosing one element of any non-empty subset of the set:


 * $\forall x \in a: \exists y: P \left({x, y}\right) \implies \exists y: \forall x \in a: P \left({x, y \left({x}\right)}\right)$


 * $(2): \quad$ The Cartesian product of a non-empty family of non-empty sets is non-empty

are equivalent.

Proof
Suppose that $(2)$ holds.

That is, the Cartesian product of a non-empty family of non-empty sets is non-empty

Let $\mathcal C$ be a non-empty set of non-empty sets.

$\mathcal C$ may be converted into an indexed set by using $\mathcal C$ itself as the index and using the identity mapping on $\mathcal C$ to do the indexing.

Then the Cartesian product of all the sets in $\mathcal C$ has at least one element.

An element of such a Cartesian product is a mapping, that is a family whose domain is the indexing set which in this context is $\mathcal C$.

The value of this mapping at each index is an element of the set which bears that index.

So we have a mapping $f$ whose domain is $\mathcal C$ such that:
 * $A \in \mathcal C \implies f \left({A}\right) \in A$

Now let $\mathcal C$ be the set of all non-empty subsets of some set $X$.

Then the assertion means that there exists a mapping $f$ whose domain is $\mathcal P \left({X}\right) \setminus \left\{{\varnothing}\right\}$ such that:
 * $A \in \mathcal P \left({X}\right) \setminus \left\{{\varnothing}\right\} \implies f \left({A}\right) \in A$

That is, $(1)$ holds.

The argument can be seen to be reversible.