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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 $f(S) \in S$.

It seems easy if you can start with

Let $a \in S$, define $f = \{S\} \times \{a\}$.

But I'm not sure how ZF allows you to do that.

Andrés E. Caicedo
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  • So each nonempty $S$ is allowed a different $f$, which needn't work for other sets, right? – J.G. Apr 19 '20 at 19:44
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    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 '20 at 20:20
  • Also https://math.stackexchange.com/questions/1839913/axiom-of-choice-where-does-my-argument-for-proving-the-axiom-of-choice-fail-he – Eric Wofsey Apr 19 '20 at 21:14
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    This really boils down to the underlying logic (classical first-order logic) rather than ZF itself, specifically existential instantiation. Our ability to "for free" pick an object satisfying a given property is given to us even before we write down the ZF axioms. – Noah Schweber Apr 19 '20 at 21:54
  • 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 '20 at 03:28

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