Re: Axiom of Choice for classes

David McAnally wrote: 
Nath Rao <RnNaDthOrMao@xxxxxxxxx> writes:


David McAnally wrote:

Could anybody please tell me whether or not the following statement is provable is VNB set theory:

Let A be a nonempty set, and suppose that there is a rule associating with each element a of A, a nonempty class F(a).
Then there exists a function f with domain A such that f(a) is an element of F(a) for all a in A.


Assuming that you are keeping the axiom of foundation, you can use the Scott trick: Define G(a) to be the class of all elements of F(a) of minimal rank. Then each G(a) is a set.


I had thought of this, but I started to feel a bit wary of it, as far as justification was concerned. I wanted to be sure that I could justify that step.

I assumed that by a 'rule' you mean a two-place predicate phi(x, y), with F(a) = {y | phi(a, y)}. Then
f(a) = {z | (phi(a, z)) and (phi(a,y) -> (rank(z) <= rank(y))}
You are not adding any class variables at all. So f(a) is defined.
If phi(a, y), then f(a) is the set of all sets of rank <= rank(y),
so f(a) is a set.

If you meant something else, then I have no idea.

Nath Rao
.