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

Robert Solovay solovay at gmail.com
Thu Feb 14 02:00:53 EST 2013

I have a number of comments concerning this thread.

First, I believe, (but perhaps my memory is misleading me) that Joe
Shipman has previously advocated the axiom that the continuum is
real-valued measurable. It seems worth remarking that this latter
axiom implies that "Zero sharp" exists, and a fortiori that M exists
and is countable.

Second, what I would mean by "V=M" is precisely what Mr. Eskew
formulates; The conjunction of V=L and the assertion that for no
ordinal alpha is L_alpha a model of ZFC. I am puzzled as to what I
have said that could make Eskew think I have some different
formulation in mind.

Finally, it seems to me that Mr. Koskensilta conflates the following
two propositions:

(A) ZFC is arithmetically sound.

(B) ZFC has an omega model.

While, of course, in ZFC, (A) follows from (B), the converse does not hold.

One way to see this is to note that in ZFC + (B) one can easily prove
that "ZFC + (A)" is consistent. Of course, by Godel's second
incompleteness theorem, ZFC + (A) is unable to prove this. So ZFC can
not prove "(A) implies (B)".

Of course, the results of this discussion can not be wholly carried
out in ZFC (which cannot even prove that ZFC is consistent.) But the
slightly stronger theory "ZFC = (B)" is perfectly adequate to carry
out the arguments I have sketched.

-- Bob Solovay

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