[FOM] questions re Axiom of Choice

Mitchell Spector spector at alum.mit.edu
Sat Feb 23 00:22:47 EST 2019

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.


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