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

[Joe Shipman]

Since I am the one who proposed "V=M", I can authoritatively say that Bob Solovay is correctly stating an equivalent version of it, although I prefer to follow Cohen's account, defining M in terms of operations to generate sets that are more restrictive than the operations used to build L. If that inductive process terminates, we get the minimal model, otherwise we get a proper class of "strongly constructible sets"; in that case the axiom says that all the sets that must exist are all the sets that there are.

I am of course aware that this proposed new axiom is incompatible with the alternative proposal RVM that I have supported  (which leads immediately to weak inaccessibles, and thus to M by taking L out to a weak inaccessible and then cutting down). I don't propose either axiom as being absolutely "true", but I think both are very useful in producing mathematics and either can be justified philosophically in a way that some would find compelling. 

I regard them as metaphysical poles: those who prefer the "definite", "intensional", ontologically minimal view of sets should be consistent and go all the way to V=M (not stop at V=L which makes the universe as narrow as possible but not as short as possible), while those who prefer the "combinatorial", "extensional", maximal picture should go at least as far as RVM (not stop at large cardinal axioms which make the universe taller but not as wide as possible).

If I had to make a metaphysical commitment here, I would say that care about arithmetical statements much more than I care about statements about infinite sets, and I would consider replacing RVM with "all the arithmetical consequences of RVM are true", which is compatible with V=M because V=M has no new arithmetical consequences. I am currently inclining to the view that we have no way of gaining reliable knowledge about non-arithmetical axioms because of Skolem's paradox, but that we can feel confidence in their arithmetical consequences. I cannot think of any pair of axioms extending ZF which were both seriously proposed but which have incompatible arithmetical consequences. (That is, ZFC & X proves A, ZFC & Y proves ~A where A is arithmetical).

Can anyone think of an example?

We axiomatize the arithmetical consequences of RVM or any other set-theoretical statement S, in the language of arithmetic, with the axiom scheme

If ZFC proves "S-->X_ZFC", then X

as X ranges over statements of arithmetic and X_ZFC is a standard coding of X in the language of set theory. 

Can anyone think of an example where a set theory's arithmetical consequences were axiomatized in the language of arithmetic in a better way than I just described, (i.e. in a way that did not require arithmetization of provability)?

-- JS


Sent from my iPhone

On Feb 14, 2013, at 2:00 AM, 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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