I'm trying to prove that the Axiom of Choice is equivalent to the following statement:

Given a non-empty set $X, \ \exists \ f:\mathcal{P}(X) \setminus\{\emptyset\} \rightarrow X$ so that $f(B) \in B \ \ \forall \ B \in \mathcal{P}(X) \setminus \{\emptyset\}$

Proving $\implies$ is easy enough (I think):

Suppose we have a set $X \neq \emptyset$. Let $\{X\}_{\lambda \in \Lambda}$ denote the collection of sets given by $\mathcal{P}(X) \setminus \{\emptyset\}$. Then by the Axiom of Choice, there exists a choice function $$ \{f: \mathcal{P}(X) \setminus \{\emptyset\} \rightarrow \cup_{\lambda \in \Lambda} X_{\lambda} : f(\lambda) \in X_{\lambda} \ \ \forall \ \ \lambda \in \Lambda\} $$ and then we notice that the union of the $X_{\lambda}$'s is simply $X$.

But I'm having a hard time proving $\impliedby$. I've tried taking the Axiom of Choice assumptions and allowing $X = \cup_{\lambda \in \Lambda} X_{\lambda}$ so that I can apply the statement above with $X$, but this power set is obviously larger than the union of my $X_{\lambda}$'s, so I'm not sure where to go from there with that approach.

My question is, how can I prove that the statement above implies the Axiom of Choice? A clarification on whether my idea would work or a hint for another approach would be appreciated.


Let $X=\bigcup_{\lambda\in\Lambda}X_\lambda$, and apply the statement to $\wp(X)$ to get $f:\wp(X)\to X$ such that $f(A)\in A$ for each non-empty $A\subseteq X$. In particular, $f(X_\lambda)\in X_\lambda$ for each $\lambda\in\Lambda$. Let $g$ be the restriction of $f$ to $\{X_\lambda:\lambda\in\Lambda\}$. Let $\ell$ be the indexing map $\lambda\mapsto X_\lambda$; then $g\circ\ell$ is the desired choice function.

  • $\begingroup$ :( I was nearly there, thanks! $\endgroup$ – Eric Hansen Sep 18 '16 at 18:30
  • $\begingroup$ @Eric: Yes, you were definitely looking in the right direction. You’re welcome! $\endgroup$ – Brian M. Scott Sep 18 '16 at 18:31

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