# How to prove that you can choose an element from a set in ZF? [duplicate]

Let's say we're in ZF. How would we prove the following theorem?

Let $$S$$ be a set such that $$S \neq \emptyset$$. There exists some function $$f$$ such that $$f(S) \in S$$.

Let $$a \in S$$, define $$f = \{S\} \times \{a\}$$.
• So each nonempty $S$ is allowed a different $f$, which needn't work for other sets, right? – J.G. Apr 19 at 19:44
• It looks fine to me. ZF certainly proves that whenever $x$ is a set then $\{x\}$ is also a set, and it also proves that Cartesian products exist. – Nate Eldredge Apr 19 at 20:20
• Just a minor comment about interpretation, though I am sure that interpretation you've already given it is intended: Since $S$ is a set, $f(S)$ could also mean the set $\{f(s)\mid s \in S\}$. In that case, such an $f$ would only be possible if there is a set $A \in S$ whose cardinality is $\le$ the cardinality of $S$. – Paul Sinclair Apr 20 at 3:28