# Math Help - Axiom of choice on finite sets

1. ## Axiom of choice on finite sets

Hello all,

First: Have a wonderful new year ahead!!

I have a fundamental question about axiom of choice. Let me quote wiki over a variant of axiom of choice

Given any set X of pairwise disjoint non-empty sets, there exists at least one set C that contains exactly one element in common with each of the sets in X
or
Given any set X of pairwise disjoint non-empty sets, there exists at least one set C that contains exactly one element in common with each of the sets in X
Is this not a variation or a special case of axiom of schema which states that

Given any set A, there is a set B such that, given any set x, x is a member of B if and only if x is a member of A and φ holds for x.
where φ is a predicate.

My question is that it appears axiom of choice is a special case of the axiom of schema where predicate is a choice function. But axiom of choice is considered as a basic axiom and it is quoted that it cannot be proved in ZF. Can some one clarify this contradiction??

Best Regards,

Sudeep

2. ## Re: Axiom of choice on finite sets

Originally Posted by skanur
First: Have a wonderful new year ahead!!
Thanks, and the same to you!

Originally Posted by skanur
My question is that it appears axiom of choice is a special case of the axiom of schema where predicate is a choice function.
But the existence of a choice function is what is guaranteed by the axiom of choice (AC). If you can construct a choice function without AC, then AC is indeed unnecessary and you can use the axiom schema of specification, or restricted comprehension, to construct the set that contains exactly one element in common with each of the sets in X. A choice function can be constructed, in particular, when X is finite.