# Axiom of choice question (small index set)

1. May 7, 2010

### 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

2. May 7, 2010

### 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.

3. May 7, 2010

### 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))

4. May 7, 2010

### Hurkyl

Staff Emeritus
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?