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

[Joe Shipman]

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
   

_______________________________________________
FOM mailing list
FOM at cs.nyu.edu
http://www.cs.nyu.edu/mailman/listinfo/fom
-------------- next part --------------
An HTML attachment was scrubbed...
URL: </pipermail/fom/attachments/20130208/8423265b/attachment.html>