[FOM] questions re Axiom of Choice

[Mitchell Spector]

Paul Blain Levy wrote:
> .
 > .
 > .
> Question 2:
>
> Is the following statement provable in ZF?
>
> For any set A of Scott cardinals, there's a set B of sets, such that A = {card(X) | X in B}.
>
> Paul


Yes, you can set B equal to the union of A.

The Scott cardinal of any set x is the set of all sets of rank alpha that can be placed in 
one-to-one correspondence with x, where alpha is the least ordinal that makes this set non-empty.

So, for every Scott cardinal s, we have that s is non-empty and every member of s has Scott cardinal s.

So { card(X) | X is in B } = { card(X) | for some a in A, X is in a }
= { a | a is in A }
= A.


Mitchell