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

Robert Solovay solovay at gmail.com
Mon Feb 11 03:12:22 EST 2013

```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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```