# Proving Equivalence of Two Version of Axiom of Choice

I am working on an assignment that requires proving the equivalence of two versions of the axiom of choice.

(1st form): For any relation $R$, there is a function $H \subseteq R$ with dom $H =$ dom $R$.

(2nd form): For any set $I$ and any function $H$ with domain $I$, if $H(i) \neq \emptyset$ for all $i \in I$, then $\prod_{i \in I} H(i) \neq \emptyset$.

To clarify, $\prod_{i \in I} H(i) = \{f | f\text{ is a function with domain }I \ \& \ (\forall i \in I) f(i) \in H(i)\}$.

$I$ is supposed to be like an Index Set.

Here is what I have so far.

\

(1st form) $\implies$ (2nd form):

(1) We assume the 1st form and that $h(i) \neq \emptyset$ for all $i \in I$.

(2) Considering $\prod_{i \in I} H(i)$, this itself is a relation since it is a set of ordered pairs (i.e. - function are sets of ordered pairs $<x,y>$ where whenever $<x,y>$ and $<x,y'>$ then $y = y'$).

(3) By the 1st form, there is a function $H \subseteq X_{i \in I} H(i)$ with $\operatorname{dom}H =\operatorname{dom}\prod_{i \in I} H(i)$.

(4) Then, since dom $h =\operatorname{dom}\prod_{i \in I} H(i) =\operatorname{dom}I$, and $H(i) \neq \emptyset$ for all $i \in I$, we have $H$ itself as a member of $\prod_{i \in I} H(i)$.

(5) So $\prod_{i \in I} H(i) \neq \emptyset$.

I am still working on the other direction but any help or hints anyone can provide would be greatly appreciated.

If $R\subseteq I\times B$ is a relation whose domain is $I$, for each $i\in I$ let $H(i)=\{b\in B\mid \langle i,b\rangle\in R\}$. Then $H(i)$ is non-empty. What sort of elements live in the product $\prod_{i\in I}H(i)$?