[FOM] A proof that ZFC has no any omega-models (Joe Shipman)
Jaykov Foukzon
jaykovfoukzon at list.ru
Fri Feb 15 19:29:07 EST 2013
Yes of course ZFC isn't powerful enough to show there isn't such a model. But I deal under assumption ZFC+(SM of ZFC). The assumption there exists an standard model (SM) of ZFC is stronger than the assumption that there exists an model of ZFC.
On Feb 8, 2013, at 20:13:57 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.
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