[FOM] questions re Axiom of Choice

[Paul Blain Levy]

Third attempt to formulate this correctly:

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, for all X in B, c(X) is the cardinal of X.

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