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

Monroe Eskew meskew at math.uci.edu
Wed Feb 27 03:14:46 EST 2013

On Feb 25, 2013, at 4:10 PM, W.Taylor at math.canterbury.ac.nz wrote:

> But this is not at all the same sort of situation, is it?
> In the case of ZFC + MC (ZFC + exists a msrbl cardinal), we know that
> if ZFC (+ MC) is consistent then so is ZFC + ~MC, by consideration of
> the obvious sub-model.  But in the case of (ZFC + RH) vs (ZFC + ~RH)
> no such asymmetry applies.  So the lack of a discovered inconsistency
> in either of these, (i.e. a proof of the other one), is merely evidence
> that it's a pig of a problem.  Whereas the failure to find an inconsistency
> in ZFC + MC, even more so for ZFC + IC (any inaccessible cardinal at all),
> *is* some "evidence" that it is consistent.  Especially considering that
> most people who give it any thought can't see any way ZFC + IC
> could even *be* inconsistent, if ZFC were consistent.
> (Those last remarks may not fully apply to ZFC + MC, I suppose.)
> Bill Taylor

Asymmetry is not a good enough explanation.  There are many examples from set theory of statements whose consistency implies the consistency of their negation, yet we do not consider them axiom candidates.  Here is a famous one: Can aleph_omega be Jonsson?  Jonsson cardinals imply zero sharp exists, so if ZFC + "aleph_omega is Jonsson" is consistent, then so is ZFC + "There are no Jonsson cardinals."  The consistency of aleph_omega Jonsson is considered an open problem.  Several smart people have tried it.


More information about the FOM mailing list