1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Do you mean "any non empty set $B$"? $\endgroup$ – ajotatxe Mar 28 '17 at 22:42
  • $\begingroup$ 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? $\endgroup$ – DanielWainfleet Mar 28 '17 at 23:20
  • $\begingroup$ @ajotatxe: If $B$ is empty, then $g$ needs to be $\varnothing\to\varnothing$, and the empty function clearly satisfies the desired property. $\endgroup$ – hmakholm left over Monica Apr 27 '18 at 17:29
1
$\begingroup$

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 $f(A) \in A$.

$\endgroup$
1
$\begingroup$

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 $f(A)\in B$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.