[FOM] A proof that ZFC has no any omega-models
Monroe Eskew
meskew at math.uci.edu
Sat Feb 16 20:11:42 EST 2013
Mr. Shipman,
Can you explain your formulation of "V=M" a bit more? What is the inductive process, and what does it mean for it to terminate? For instance it does not take very long to close under rudimentary functions, but surely we don't stop there. It seems like the only way to make sense of the process of building L is by referring to ordinals. And then we might as well just use Solovay's formulation of what it means to be in M. Keep building L_\alpha along the ordinals until you reach a point where ZF is satisfied.
Regarding your view of large cardinals, consider the fact that there are many examples where ZFC does not prove some natural statement, but it is only through large cardinals that we can see this. For example, does the continuum carry a total measure extending Lebesgue measure? From Gödel (plus Kunen's inner model theory, I guess) we know ZFC does not prove that it does, so now we ask, does ZFC prove that the continuum does *not* carry a total measure? Solovay showed that forcing over a model of ZFC plus a measurable cardinal can produce a model where the continuum is an RVM cardinal.
Note that he answered a purely arithmetical question: Does ZFC prove the continuum does not have a total measure? If we agree that this was a valuable result even though it used a large cardinal assumption, I think we can agree that this is an illustration of the following general principle: The use of large cardinals in set theory is needed to answer many natural questions which are essentially arithmetical.
Now we turn to the question of whether to adopt your scheme of reducing large cardinal axioms to axioms expressing their arithmetical consequences. I ask, does this not look like insignificant philosophical window dressing? Once we agree that large cardinals should be used in set theory and their arithmetical consequences believed, what difference does it make whether we adopt, after the fact, a more modest ontological interpretation? Set theory would have to go on the same way as before to derive consistency results. The scheme you propose would have no mathematical import, and would just be akin to a footnote about a possible meta-level interpretation like we find in Kunen's wonderful book on set theory.
Best,
Monroe
On Feb 14, 2013, at 10:08 PM, Joe Shipman <JoeShipman at aol.com> wrote:
> 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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