# Re: Paper and slides on indefiniteness of CH

Dear Hugh,

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 (N,C)?

$(N,V,C) \vDash \text{ZFC}$ and $C$ is amenable to $V$ but not necessarily to $N$.  The forcing is definable over $N$ and $(V,C)$ is generic over $N$.

If the forcing is definable over $N$ then it seems that the only role of $C$ is to define $N$ (over $(V,C)$). Right?

It is a very interesting result. Please send me the proof so that I can figure out what I should have posed as “a leading open question of set theory”.

Thanks,
Sy

# Re: Paper and slides on indefiniteness of CH

On Oct 26, 2014, at 1:43 PM, Sy David Friedman wrote:

Dear Hugh,

On Fri, 24 Oct 2014, W Hugh Woodin wrote:

Dear Bob,

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 $\gamma \in C$,

$(V_{\gamma},C\cap \gamma) \nvDash \text{ZFC}$

Then there is an inner model of V, N, such that $N \vDash \text{ZFC} + \text{GCH}$ 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 $(N,C)$?

$(N,V,C) \vDash \text{ZFC}$ and $C$ is amenable to $V$ but not necessarily to $N$.  The forcing is definable over $N$ and $(V,C)$ is generic over $N$.

Regards,
Hugh

# Re: Paper and slides on indefiniteness of CH

Dear Hugh,

On Fri, 24 Oct 2014, W Hugh Woodin wrote:

Dear Bob,

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 $\gamma \in C$,

$(V_{\gamma},C\cap \gamma) \nvDash \text{ZFC}$

Then there is an inner model of V, N, such that $N \vDash \text{ZFC} + \text{GCH}$ 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 $(N,C)$?

Thanks,
Sy

# Re: Paper and slides on indefiniteness of CH

Dear Bob,

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 $\gamma \in C$,

$(V_{\gamma},C\cap \gamma) \not\vDash \textsf{ZFC}$

Then there is an inner model $N$ of $V$ such that $N \vDash \textsf{ZFC + GCH}$ and such that $V$ is a class-generic extension of $N$.

(Nothing special about GCH here).  Though here one can require that $N$ is a fine-structure model if one wants.

The class $C$ can always be forced without adding sets.

If one has inner model theory for measurable Woodin cardinals then the class $C$ is unnecessary.

Regards,
Hugh

# Re: Paper and slides on indefiniteness of CH

I’m repeating my query for the whole mailing list:

Dear Hugh,

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 …?

Thanks,

Bob

# Re: Paper and slides on indefiniteness of CH

Dear Sy,

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.

Regards,
Hugh

# Re: Paper and slides on indefiniteness of CH

Dear Sy,

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.

Regards,
Hugh