Math Help - Equivalent forms of axiom of choice

1. Equivalent forms of axiom of choice

Hello! I am new to this forum and I hope I will soon familiarize with posting new threads. Now, during the winter break I started to study for myself the set theory and especially the axiom of choice. I found the next problem in a book:
Prove that the 3 statements of the axiom of choice are equivalent :

1) For any non-empty collection X of pairwise disjoint non-empty sets, there exists a choice set.

2) For any non-empty collection of non-empty sets X there is a choice function.

3) For any non-empty set X, there exists a function f:P(X)\{∅}→X so that for any non-empty set A⊆X, f(A) ∈ A.
Now, I have already tried and also succeeded to prove 2 of the 6 possible implications between the statements. But I simply can't realize how to prove the next implications: 1=>2, 2=>3 and 3=>1.
Best regards,
Alice

2. Re: Equivalent forms of axiom of choice

Originally Posted by Alice10
Prove that the 3 statements of the axiom of choice are equivalent :
1) For any non-empty collection X of pairwise disjoint non-empty sets, there exists a choice set.
2) For any non-empty collection of non-empty sets X there is a choice function.
3) For any non-empty set X, there exists a function f:P(X)\{∅}→X so that for any non-empty set A⊆X, f(A) ∈ A.
You may find this webpage useful.

3. Re: Equivalent forms of axiom of choice

Indeed, the page is useful, but what if I want to prove the implications using the Zermelo-Fraenkel axioms instead of explaining with words the facts..this is actually what I am yearning for.

4. Re: Equivalent forms of axiom of choice

Originally Posted by Alice10
prove the implications using the Zermelo-Fraenkel axioms instead of explaining with words the facts
The reason I posted that link is that is proves my concern with your question. There is no standard way of discussing the axiom of choce because there is no standard statement of the axiom of choce.
For example, Charles Pinter uses the statement #3 in the OP as his statement of a choice function.

What is a choice set? That is not a standard term, is it?

Enderton's axiom of choce is $(\forall\text{ relation }R)(\exists\text{ function }F)[F\subseteq R~\&~\text{Dom}F=\text{Dom}R]$

Halmos: Axiom of choice. The Cartesian product of any collection of nonempty sets is nonempty.

So how do we work going 1 to 2 if the definitions vary from author to author?

5. Re: Equivalent forms of axiom of choice

Now I truly understand. Well, the book provides the next statement for the choice set the next statement:
X - non-empty collection of pairwise disjoint non-empty sets. Then Y is a choice set for X if Y is included in the reunion of all the elements of X and for any A, A included in X, there is a set H so that A ∩ Y = {H}. I suppose this is resembles more like what Halmos claimed. What I am trying to do is to prove all these implication using a formal language, with logical notations .