[FOM] A proof that ZFC has no any omega-models
Joe Shipman
JoeShipman at aol.com
Fri Feb 8 20:13:57 EST 2013
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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