On Oct 26, 2014, at 1:43 PM, Sy David Friedman wrote:
On Fri, 24 Oct 2014, W Hugh Woodin wrote:
Here is a precise statement. Suppose every set X belongs to an inner model of a measurable Woodin (above X).
Suppose there is a club class C of cardinals such that for all ,
Then there is an inner model of V, N, such that and such that V is a class-generic extension of .
Very interesting! To make sure that I’ve got this right
please answer the following: Is amenable to and is the forcing definable over ?
and is amenable to but not necessarily to . The forcing is definable over and is generic over .
On Fri, 24 Oct 2014, W Hugh Woodin wrote:
Here is a precise statement. Suppose every set X belongs to an inner model of a measurable Woodin (above X). Suppose there is a club class C of cardinals such that for all ,
Then there is an inner model of V, N, such that and such that V is a class-generic extension of N.
Very interesting! To make sure that I’ve got this right please answer the following: Is C amenable to N and is the forcing definable over ?
Here is a precise statement. Suppose every set belongs to an inner model of a measurable Woodin (above X).
Suppose there is a club class of cardinals such that for all ,
Then there is an inner model of such that and such that is a class-generic extension of .
(Nothing special about GCH here). Though here one can require that is a fine-structure model if one wants.
The class can always be forced without adding sets.
If one has inner model theory for measurable Woodin cardinals then the class is unnecessary.
I’m repeating my query for the whole mailing list:
I confess I’m not sure precisely what you are claiming to prove. Let
us consider the case when we do not assume “inner model theory for a measurable Woodin”. Your theorem would say precisely: V is a class generic extension of a model of GCH if …?
Sitting in an airport and doing email is perhaps not such a good idea. To show that V is a generic extension of an inner model of GCH, the argument I have in mind needs that every set X belongs to an inner model of a measurable Woodin above X. And I have to either force a amenable club class first or have inner model theory at the level of a measurable Woodin.
You wrote to Pen:
But to turn to your second comment above: We already know why CH doesn’t have a determinate truth value, it is because there are and always will be axioms which generate good set theory which imply CH and others which imply not-CH. Isn’t this clear when one looks at what’s been going on in set theory? (Confession: I have to credit this e-mail discussion for helping me reach that conclusion; recall that I started by telling Sol that the HP might give a definitive refutation of CH! You told me that it’s OK to change my mind as long as I admit it, and I admit it now!)
ZF + AD will always generate “good set theory”… Probably also V = L…
This seems like a rather dubious basis for the indeterminateness of a problem.
I guess we have something else to put on our list of items we simply have to agree we disagree about.
So the best one can do with a problem like CH is to say: “Based on a certain Type of evidence, the truth value of CH is such and such.” As said above, Type 1 evidence (the development of set theory as an area of mathematics) will never yield a fixed truth value, we don’t know yet about Type 2 evidence (ST as a foundation) and I still conjecture that Type 3 evidence (based on the Maximality of the universe of sets in height and width) will imply that CH is false.
There will never be such a resolution of CH (for the reasons I gave above). The best one can do is to give a widely persuasive argument that CH (or not-CH) is needed for the foundations of mathematics or that CH (or not-CH) follows from the Maximality of the set-concept. But I would not expect either achievement to draw great acclaim, as nearly all set-theorists care only about the mathematical development of set theory and CH is not a mathematical problem.
This whole discussion about CH is of interest only to philosophers and a handful of philosophically-minded mathematicians. To find the leading open questions in set theory, one has to instead stay closer to what set-theorists are doing. For example: Provably in ZFC, is V generic over an inner model which satisfies GCH?
Why is this last question a leading question? If there is an inner model with a measurable Woodin cardinal it is true, V is a (class) generic extension of an inner model of GCH.
You must mean something else. Focusing on eliminating the assumption of there is an inner model of a measurable Woodin cardinal seems like a rather technical problem.