[FOM] CH in standard models
Harvey Friedman
friedman at math.ohio-state.edu
Sun May 25 04:59:44 EDT 2003
Reply to Jones 5/25/03 6:20AM.
>
> > I don't quite know what definition you want of "standard
>> model". Here are some possibilities.
>>
>> A. The model of ZFC is an initial segment of the cumulative
>> hierarchy of sets under epislon.
>>
>> B. The model of ZFC is a model of second order ZFC.
>>
>> C. The model of ZFC is a transitive set under membership.
>
>I did say "standard models of second order ZFC".
>I cannot see any difference between your A and B
>(unless the omission of "standard" from B is to be read
>as allowing non-standard models).
>Aren't both A and B the V(alpha) for alpha strongly inaccessible?
No. The least cumulative hierarchy that satisfies ZFC is much shorter
than the first strongly inaccessible cardinal (assuming it exists). I
think this goes back to Vaught.
>
>C on the other hand seems to exclude no models, and
>so is not what I intended.
It includes exactly the well founded models.
>
>> 1. The CH either holds in all A models or fails in all A
>> models. If there is an A model then the CH holds in it if and
>> only if the CH is true.
>>
>> 2. The CH either holds in all B models or fails in all B
>> models. If there is a B model then the CH holds in it if and
>> only if the CH is true.
>>
>> 3. If there is a C model then the CH model holds in some C
>> models and not in other C models.
>>
>> All three statements 1,2,3 are provable in a weak fragment of
>> ZFC without the power set axiom.
>
>So I take it that you are agreeing that with a suitable
>understanding of what "standard model of ZFC" means, there
>would be a consensus that the question of whether CH is true
>in standard models is meaningful?
No. The CH is outright equivalent to the question of whether the CH
holds in all or some A models or B models (assuming they exist). So
your use of standard models in connection with CH accomplishes
nothing.
>
>I wonder if anyone could say more about how much
>of the conflicting evidence for and against CH
>falls by the wayside if the more specific question
>of its truth in standard models is considered?
Since it is outright equivalent (in the sense stated above),
absolutely nothing whatsoever falls by the wayside.
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