TITLE: Finite axiom of choice: how do you prove it from just ZF?
QUESTION [24 upvotes]: The axiom of choice asserts the existence of a choice function for any family of sets F. Suppose, however, that F is finite, or even that F just has one set. Then how do we prove the existence of a choice function? 
The usual answer is that we just go from set to set, picking an element from each set. Since F is finite, this process will terminate. What I'm really wondering is how we can always choose from a single set. The informal answer seems to be just that it's possible... but this isn't an axiom, so it must be justified some other way. 
So: how do you prove from the axioms of just ZF without choice, that for any nonempty x there exists a function f:{x}->x?

REPLY [8 votes]: The Axiom of Choice does not allow to "choose" a choice function, it only says that a choice function exists. To show that a choice function for a single nonempty set exists, you do not need to "choose" an element in the set, it is enough to show that at least one element exists (i.e. the set is nonempty). Every element in the set will give a different choice function.
What did you mean by "choosing" an element from a set? ;)