# Axiom of Choice equivalents

• Oct 26th 2010, 07:20 AM
kzxrr
Axiom of Choice equivalents
So apparently for all pairwise disjoint sets S not containing the empty set

$\displaystyle \exists f [\exists Y (f:S \rightarrow Y) and \forall x (x \in S \implies f(x) \in x)] \iff \exists t [\forall z z \in S \implies \exists! w (w \in z \cap t)]$

I had no problems proving the first version from the second, and assuming the first one and taking f(S) (image of the whole set S under f) for t, then obviously for all z in S f(z) is in z and f(S). The reason I'm posting is because I can't for the life of me prove that it's the only element of z intersection f(S)!
• Oct 26th 2010, 08:16 AM
Plato
Here is a useful website.
• Oct 26th 2010, 10:42 AM
MoeBlee
Suppose w in z/\range(f), and v in z/\range(f).

Show w=v:

w=f(x) for some x in S, and f(x) in x. But S is pairwise disjoint, so z=x, so w = f(z).

v=f(p) for some p in S, and f(p) in p. But S is pairwise disjoint, so z=p, so v = f(z).

So w=v.
• Oct 26th 2010, 01:19 PM
kzxrr
I wasn't seeing that z and x were in S which was disjoint so their intersection must be empty or they must be equal. Ty so much it's so clear now.
• Oct 26th 2010, 01:40 PM
MoeBlee
Yeah, in trying to prove stuff, if you don't see an intuitive path, it often helps just to start pulling out all the "information" or "implications" you can from the given premises. Extract everything you can from the premises, then lay it out in front of you to then see whether there might be a way to combine all that stuff back into a proof.