Axiom of choice question (small index set)

[Jerbearrrrrr]

Hi there,
just wondering what this could mean:

"Remark: AC is trivial for |I|=1, since A non-empty means $$\exists x$$ st $$x \in A$$. Similarly for |I| finite (induction on |I|)."

I is the index set. As in, our sets are {A_i with i in I}.

I'm really not sure what this explanation is getting at.

Thanks

 

[g_edgar]

For axiom of choice you need to define a function (choice function) with domain I . If I is finite you can define a function by listing all its values. This already can be done based on the other axioms of set theory, so AC is not needed in this case.

 

[Jerbearrrrrr]

Oh right. I thought there was some problem in choosing what the output of the function was.
Okay then, thanks.
If I is countably infinite, you can also specify the function with a list though, surely?
You need a way of deciding what each f(i)=. But even if I is finite, you still need a way of deciding what each f(i)=.
Need help D:

(I understand that the AC isn't needed in choosing from well-ordered sets, so I thought the issue was just deciding how to pick the f(i))

 

[Hurkyl]

Try working through the details of the one-element index set. How, exactly, do you prove that a choice function exists?

Or, to phrase it differently, how do you prove the set of choice functions is non-empty?