I'm trying to prove the following statement is equivalent to the Axiom of Choice:
"For any set $A$, there exists a function $F$ with dom $F = ⋃A$ and for each $x ∈ ⋃A$, $x ∈ F(x) ∈ A$." (1)
The resemblance is uncanny to the choice function version of the Axiom of Choice, which I interpret loosely to say that for any collection of nonempty sets, you can pick a member from each set in the collection. This statement (1) seems to be saying that for any set $S$, for each $T ∈ S$, you can pick exactly one "representative" $x ∈ T$.
I managed to show that the Axiom of Choice proves (1), but can't get that (1) proves the Axiom of Choice. I was given advice to try to use this specific form of the Axiom of Choice: "For any relation $R$, there exists a function $f$ with $f ⊆ R$ and dom $f =$ dom $R$." However, I'm not sure how exactly to use (1) to refine any relation $R$ to a function.
Any hints or input would be much appreciated.
If we have an instance $G$ of the axiom of choice, replace each set $B\in G$ with the set of all pairs $B' = \{ \{(0,B), (1,x)\}: x \in B\}$. Note $B'$ is a set of unordered pairs of ordered pairs.
Let $$G' = \bigcup \{ B' : B \in G\} = \{ \{(0,B), (1,x)\} : x \in B \in G\}.$$ Apply your principle to $G'$ to obtain a function $f$ as described in the question.
Then, given $B$, we can define an $x \in B$ as the unique $x$ such that $f(B) = \{(0,B),(1,x)\}$. This gives us a choice function for $G$, which was the goal.