As a little project for myself this winter break, I'm trying to go through as much of Enderton's Elements of Set Theory as I can. I hit a snag trying to show two forms of the Axiom of Choice are equivalent. This is exercise 31 on page 55.
The first form is:
For any relation $R$ there is a function $G\subseteq R$ with $\text{dom}\ G=\text{dom}\ R$.
and the second form is:
For any set $I$ and any function $H$ with domain $I$, if $H(i)\neq\emptyset$ for all $i\in I$, then $\times_{i\in I}H(i)\neq\emptyset$.
Here is what I have so far:
Assume the first form. Take any set $I$ and let $H$ be a function with domain $I$ such that $H(i)\neq\emptyset$ for all $i\in I$. This function $H$ is a relation, so by the Axiom of Choice, there exists a function $G\subseteq H$ such that $\text{dom}\ G=\text{dom}\ H=I$. Since $\text{dom}\ G=I$, for each $i\in I$, there exists some $G(i)$ such that $(i,G(i))\in G$. But since $G\subseteq H$, $(i,G(i))=(j,H(j))$ for some $j\in I$. Since these are ordered pairs, $i=j$ and $G(i)=H(j)$? I suppose I want to be able to show that for all $i\in I$, I can have $G$ "choose" some element $G(i)\in H(i)$, and thus $G\in\times_{i\in I}H(i)$, showing that $\times_{i\in I}H(i)\neq\emptyset$, but I don't see how the first form allows one to do that. Instead, all I see is that $G(i)=H(i)$.
Conversely, I assume the second form. I take any relation $R$, and denote $\text{dom}\ R=I$. Let $H$ be any function with domain $I$. Now if $H(i)\neq\emptyset$ for all $i$, then $\times_{i\in I}H(i)\neq\emptyset$, so then I could take some $f\in\times_{i\in I}H(i)$, so by definition, $\text{dom} f=I$, and for all $i$, $f(i)\in H(i)$. If it is the case that $(i,H(i))\in R$, then $f\subseteq R$, and the first form would be proven. Again, I suppose I want $H$ to be a function that, for each $i\in I$, $H$ takes the value of exactly one $y_i$ such that $iRy_i$, but again, I don't see how the assumed axiom allows one to do this.
Can anyone explain how to get around these two issues? Thank you.