Equivalence of Versions of Axiom of Choice/Formulation 3 implies Formulation 1

Theorem
The following formulation of the Axiom of Choice:

Formulation 3
implies the following formulation of the Axiom of Choice:

Proof
Let $\BB$ be a non-empty indexed family of non-empty sets indexed by $\II$.

Consider sets of the following form:


 * $\CC = \set {\tuple {B_i, x}: i \in \II, B_i \in \BB, x \in B_i}$

That is, it is the set of ordered pairs of which the first coordinate is a set $B_i \in \BB$ and the second coordinate is an element of $B_i$.

This is, by construction, a subset of the cartesian product:


 * $\CC \subseteq \ds \BB \times \bigcup_{i \mathop \in \II} B_i$

By hypothesis all sets $B$ are non-empty.

Thus there exists at least one $\tuple {B_i, x}$ element in $\CC$ for each $B_i \in \CC$.

By Equality of Ordered Pairs, if $B_j \ne B_k$ in $\BB$, then $\tuple {B_i, x} \ne \tuple {B_j, x}$ for all such pairs in $\CC$.

So any two sets $\set {\tuple {B_j, x}: x \in B_j}$ and $\set {\tuple {B_k, x} :x \in B_k}$ are disjoint, by the inequality of their first coordinates.

Then $\CC$ is an indexed family of disjoint non-empty sets in $\BB \times \bigcup_{i \mathop \in \II} B_i$.

Hence $\CC$ satisfies the hypotheses of the third formulation of the Axiom of Choice.

Let $c$ be a set satisfying:
 * $\forall r \in \CC: c \cap r = \set t$

All elements of $c$ are ordered pairs:
 * the first coordinate of which is a set in $B_i \in \BB$

and:
 * the second coordinate of which is an element $x \in B_i$.

Viewed as a relation, each set in $\BB$ bears $c$ to an element in that set.

Every set in $\BB$ bears $c$ to exactly one element inside that set.

So $c$ is in fact a mapping and satisfies the criteria of a choice function as expounded in formulation $(1)$.