[FOM] A proof that ZFC has no any omega-models
Robert Solovay
solovay at gmail.com
Thu Feb 14 06:50:02 EST 2013
Correction to my last letter:
When I said "ZFC = (B)" I meant "ZFC + (B)"
-- Bob Solovay
On Feb 13, 2013 11:00 PM, "Robert Solovay" <solovay at gmail.com> wrote:
> 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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