Equivalence of Formulations of Axiom of Choice

Theorem
The following formulations of the Axiom of Choice are equivalent:

Formulation 2 implies Formulation 1
Suppose that Formulation 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, Formulation 1 holds.

Formulation 1 implies Formulation 2
The argument can be seen to be reversible.