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

W.Taylor at math.canterbury.ac.nz W.Taylor at math.canterbury.ac.nz
Mon Feb 25 19:10:27 EST 2013

Quoting Monroe Eskew <meskew at math.uci.edu>:

> ...why should we think it is reasonable to assume the consistency of
> ZFC + there exists a measurable cardinal?  We can't just say because  
> no one has found a contradiction in many  years, because that  
> applies to too many things, like the Riemann  Hypothesis and its  
> negation.

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

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