[FOM] questions re Axiom of Choice

[Mitchell Spector]

This statement is equivalent to: "Every set of Scott cardinals has a choice function."

According to Asaf Karagila's answer at https://math.stackexchange.com/q/423369 , Jech and Sochor 
showed that this is not provable in ZF, and then Pincus gave another proof of this fact.  I don't 
have specific references though.

Mitchell


Paul Blain Levy wrote:
> 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
>