[FOM] A proof that ZFC has no any omega-models

Joe Shipman JoeShipman at aol.com
Mon Feb 11 17:45:53 EST 2013


Thanks for correcting my misstatement, which arose from confusion between the terms "omega-model" and "standard model". If V=M, then there is no "standard model" of ZFC, meaning a set whose epsilon-relation satisfies ZFC, but there can still be an "omega-model", which is nonstandard as a model of ZFC but standard as far as the integers are concerned.

So my proposed axiom V=M does not imply that no omega-models exist. Nor is the converse true; if there are no omega models then there are no standard models, but there could still be nonconstructible sets.

-- JS

Sent from my iPhone

On Feb 11, 2013, at 3:12 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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