[FOM] A proof that ZFC has no any omega-models
Monroe Eskew
meskew at math.uci.edu
Mon Feb 11 18:09:36 EST 2013
If alpha is the least ordinal such that L_alpha satisfies ZFC, then L_alpha must think there is no transitive model of ZFC, by the absoluteness of the Gödel operations. Shoenfield's absoluteness theorem does not apply to this situation because its hypothesis includes that the model in question contains all countable ordinals.
I'm not sure what V=M is supposed to say exactly, and I think there is some ambiguity if Bob Solovay has a different interpretation. In light of the above observation, I assumed it was ZFC + V=L + "There is no alpha such that L_alpha satisfies ZFC."
Monroe
On Feb 11, 2013, at 12:12 AM, Robert Solovay <solovay at gmail.com> wrote:
> The axiom V=M does not imply that ZFC has no omega-model.
>
> The assertion that "ZFC has an omega-model" is Sigma_1^1, and so holds
> in all well-founded models of ZFC (in particular M). And of course,
> the axiom V=M, formulated in the obvious way, holds in M as well.
>
> -- Bob Solovay
>
> On Fri, Feb 8, 2013 at 5:13 PM, Joe Shipman <JoeShipman at aol.com> wrote:
>> I would be very surprised if your result holds up, but not because I think
>> there must be an omega-model of ZFC; rather, I doubt ZFC is powerful enough
>> to show there isn't such a model. I have actually proposed an axiom denying
>> that such a model exists.
>>
>> Technically, I proposed that "V=L" be replaced by "V=M" where M is the
>> strongly constructible sets, which form Cohen's "minimal model" if there is
>> a standard model, but which are all of L if no standard model exists. V=M
>> is, in my opinion, a more principled axiom than V=L because it strengthens
>> the V=L notion "the only sets which exist are the ones which the axioms say
>> must exist given the ordinals" by not assuming more ordinals than necessary.
>> If there IS a standard model, then Cohen's minimal model M exists as a set
>> and satisfies "V=M" so I don't see the axiom as unnatural; to me it just
>> means that all the sets that must exist are all the sets there are.
>>
>> -- JS
>>
>> Sent from my iPhone
>>
>> On Feb 8, 2013, at 2:02 PM, Jaykov Foukzon <jaykovfoukzon at list.ru> wrote:
>>
>>
>> I am writing up a short sketch a proof that ZFC has no any omega-models
>> i.e.: ~con(ZFC+E(omega-model of ZFC)). A short sketch of this proof is
>> posted at
>>
>>
>> http://fs23.formsite.com//viXra/files/f-1-2-7439160_RgrjILnA_Inconsistentcountableset4..pdf
>>
>>
>>
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