[FOM] questions re Axiom of Choice
Mitchell Spector
spector at alum.mit.edu
Sun Feb 24 02:45:50 EST 2019
I need to correct a mistake I made in the previous post:
The statement you asked about looks like it may, in fact, be weaker than "Every set of Scott
cardinals has a choice function." (If your statement is strengthened by requiring that X belong to
c(X), then the two statements are equivalent.)
I would suggest looking at the Jech-Sochor proof or the Pincus proof to see if either method can
actually handle your apparently weaker proposition.
Mitchell
Mitchell Spector wrote:
> 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
>>
>
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