# Using the Axiom of Choice.

Use the axiom of choice to prove that for any set $B$ there is a function $g:\mathcal{P}(B)-\{{\emptyset}\} \to B$ such that $g(A) \in A$ for every nonempty subset $A$ of $B$.

I see that the axiom of choice definition closely relates to the question at hand but I am not seeing how this can be used in the proof.

• Do you mean "any non empty set $B$"? – ajotatxe Mar 28 '17 at 22:42
• There are 3000 provably equivalent versions of AC. What you wish to prove is one of them. Which version are you taking as your axiom? – DanielWainfleet Mar 28 '17 at 23:20
• @ajotatxe: If $B$ is empty, then $g$ needs to be $\varnothing\to\varnothing$, and the empty function clearly satisfies the desired property. – hmakholm left over Monica Apr 27 '18 at 17:29

Note that $\emptyset \not \in \mathcal{P}(B) \setminus \{\emptyset\}$. Therefore there is a choice function $$f: \mathcal{P}(B) \setminus \{\emptyset\} \to \bigcup \left[\mathcal{P}(B) \setminus \{\emptyset\}\right]$$ such that for all $A \in \mathcal{P}(B) \setminus \{\emptyset\}$ we have $f(A) \in A$. (That's precisely the axiom of choice.)
But since $\bigcup \left[\mathcal{P}(B) \setminus \{\emptyset\}\right] = B$, that's equivalent to saying there is a function $f: \mathcal{P}(B) \setminus \{\emptyset\} \to B$ such that for all $A \subseteq B$ nonempty, we have $f(A) \in A$.
Recall that a choice function on a family of nonempty sets is a function satisfying $f(A)\in A$ for all $A$ in its domain.
In this case the domain is $\mathcal P(B)\setminus\{\varnothing\}$, this is indeed a family of nonempty sets. So if $A$ is in the family, by definition $A$ is a subset of $B$. So if $f(A)\in A$, then $f(A)\in B$.