[FOM] questions re Axiom of Choice

[Paul Blain Levy]

Yowch, you are quite right; I misstated my question.  It should be:

Is the following statement provable in ZF?

For any set A of Scott cardinals, there's a set B of sets and a 
bijection c : B --> A such that c(B) is the cardinal of B.

Paul


On 23/02/2019 05:22, Mitchell Spector wrote:
> 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