Consider the following statement: for any set $E$ and $G\subseteq E\times E$, you can to get a function $f:A\rightarrow B$ where $A=dom G$, $B=ran G$ and $f\subseteq G$.
I want to show that this implies axiom of choice.
My attempt:I will prove that the statement above implies the existence of a choice function. Indeed, let A an arbitrary set and consider $E=\mathcal{P}(A)\setminus\{\emptyset\}$ and take $G:=\{(B,\{x\})\subseteq E\times E:x\in B\}$. So, by hypotesis exists a function $f:E\rightarrow A$ that it is, in fact, a choice function.
My doubts in the proof is with the set $G$, I don't know if in your definition I use axiom of choice, because it seems that I do infinitely many choices.
Best Answer
Yes, you have a problem there. Specifically, your function returns singletons of elements of $A$, not elements of $A$.
The obvious way to correct this is to take $E=A\cup\mathcal P(A)$ and $G=\{(B,x)\mid x\in B\subseteq A\}$.