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I read the Jech, Set theory, and saw following proposition.

(☆) If S is a finite family of nonempty sets, existence of choice function of S can be proved without axiom of choice.

I tried to prove the proposition using axioms in ZF... But how can I pick the element of X in S explicitly despite I don't know information about X in S?

Edit: There seems to be confusions on what exactly I am asking.

I think ZFC Axioms instruct when we call a object set.(If object is not set, the object is called urelement or proper class) So I think we must show that nonemptyset has a element that is set. For example, if S is singleton, US is element of S and is set by axiom of union. That is, we can find a element of S that is set. Then using other axioms in ZF, we can construct a set we call choice function of S. But, If S has two elements, how I can know the element of S is set??

PPPiRi
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  • Choice for finite families of sets is provable in ZF by induction on the size of the family. – Noah Schweber Apr 02 '16 at 05:04
  • Everything in ZF is a set. You don't have to ask whether every element of $S$ is a set because there is nothing else such an element could be other than a set. – Eric Wofsey Apr 03 '16 at 05:39

2 Answers2

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We cannot necessarily construct a choice function, but be can show that a choice function exists, i.e., that the set $C$ of choice functions is non-empty (in other words: that the direct product of finitely many non-empty sets is non-empty). Once we know this, we can pick a choice function from this non-empty set and argue with it, e.g., if we have $C\ne\emptyset$ (aka. $\exists f\colon f\in C$) and $\forall f\colon (f\in C\to \Phi)$ then we may infer $\Phi$.

Hagen von Eitzen
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Sometimes it's possible to explicitly pick an element, sometimes it is not. It depends on what you know about your sets.

For all $x$ in $S$, $\neq\emptyset$ by assumption. Therefore $\exists y\in x$. Then consider the pair $(x,y)$. The set of all such pairs is a choice function on $S$.

If $y$ can be explicitly chosen, then $C=\{(x,y)\}$ can be. If not, then not.

Stella Biderman
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  • Statement (☆) insist that we can construct choice function of S without additional information about elements of S. – PPPiRi Apr 02 '16 at 04:53