For the theorem,
Let $E$ be the set and suppose $G\subseteq E\times E$. Let $A=domG$ and $B=ranG$, then there exists a function $f:A\rightarrow B$ such that $f\subseteq G$.
The theorem imples Axiom of choice.
I'm proving the statement, but I think my proof is going wrong. So Could you tell me how can I correct the proof?
Proof) Need to show that arbitrary set E has a choice function $\gamma$.
Assume the theorem, and Let $E$ be a set and let $Q_D=\text{{$(D,d)|d\in D$}}$ for all $D \in \mathscr P'(E)$. Let $Q = \cup_{D\in \mathscr P'(E)} Q_D$
Then, $Q \subseteq E\times E$? I'm very confused.
If the above thing is right, let $A=domQ$ and $B=ranQ$. Then there exist a function $f:A\rightarrow B$ such that $f\subseteq Q$.
Since $Q = \cup_{D\in \mathscr P'(E)} Q_D$, $A = domQ = \mathscr P'(E)$
Since $f:A\rightarrow B$ is a fuction, there exist a $d\in D$ for all $D \in \mathscr P'(E)$ by F2.
Then, Let $\gamma: \mathscr P'(E) \rightarrow E$ such that $\gamma(D)=f(D)=d\in D$ for all $D \in \mathscr P'(E)$.
Hence, Axiom of choice holds
Thanks
First, some unclarified assumptions, $\mathcal P'(E)$ is the set of non-empty subsets of $E$, and the axiom of choice states that $\mathcal P'(E)$ admits a choice function (when you say that $E$ has a choice function, you mean there is a function choosing from the elements of $E$, but it seems that you mean a function choosing from subsets of $E$).
The essence of your proof is correct, but some parts are a bit murky. Let's work backwards, okay?
Instead of using $E$, let's use $X$ for our set. This is important because mentally you want to aim for $E$ to be the set on which the relation $G$ is given, and you don't necessarily know that this would be $X$ itself.
We want a choice function from $\mathcal P'(X)$, so this means that it would be a function with domain $\mathcal P'(X)$ and its range would be a subset of $X$.
So if we want a set $E$ such that $f$ is a subset of $E\times E$, it means that $E$ has to include both $X$ and its power set. Therefore, taking $E=X\cup\mathcal P(X)$ would easily do the job.
Now, to simplify your definition, we can just define $G$ to be $\{(D,d)\mid d\in D\subseteq X\}$. Now you don't need to define $\gamma$ anymore, because $f$ is already a choice function.