# Proof of one instance of the Axiom of Choice from another

I'm trying to show (i) $\implies$ (ii):

(i) For any relation $R$, there exists a function $H\subseteq R$, with $\newcommand{\dom}{\mathrm{dom} \ } \dom H = \dom R$.

(ii) For any set $I$ and any function $H$ with domain $I$, if $H(i)\neq\emptyset$ for each $i\in I$, then there exists a function $F$ with domain $I$, and for every $i\in I$, we have $F(i)\in H(i)$.

Pf: Assume (i) and let $I$ be a set, $H$ a function with $\dom H = I$ and $H(i)\neq\emptyset$ for each $i \in I$. Define

$$R = \left\{x \in {\cal P}{\cal P}\left(I\cup \bigcup_{i\in I} H(i)\right) | x = (i,h), i \in I, h \in H(i) \right\}.$$

Then, $R$ is a relation, and by (i), there exists some function $F\subseteq R$ with $\dom F = \dom R = I$, and for each $i\in I$, we have $F(i) \in H(i)$, which is what we want.

Is this fine? Is $R$ a set? That is, if I understand correctly, that ${\cal P}{\cal P}\left(I\cup \bigcup_{i\in I} H(i)\right)$ is a set comes from power set and union axioms, and this set contains exactly the $(i,h)$'s we want; but is the use of one of these subset axioms here done correct?

Your definition of $R$ is extremely cumbersome. You don't have to mention the set over which you apply separation if the definition is obviously bounded. Writing $R=\{(i,h)\mid h\in H(i)\}$ is far clearer.