[FOM] AI-completeness (and settling CH)

James Hirschorn James.Hirschorn at univie.ac.at
Sun May 27 21:38:05 EDT 2007

On Sunday 20 May 2007 17:35, Robbie Lindauer wrote:
> What about T_2 = ZFC + Not(CH)
Well, it is a nice exercise to show that Not(CH) is not generically complete 
for the \Sigma^2_1 theory, and thus T_1 = ZFC + Not(CH) does not work. I will 
attempt to solve it, but please note that I am (thus far) primarily a 
combinatorial set theorist. 

It is a theorem of Solovay (strengthening a similar theorem of Woodin) that 
one can force, with a small poset, p = aleph_2 together with a \Sigma^2_1 
well ordering of the reals (p is a standard cardinal characteristic. It is 
equal to the first cardinal \kappa where MA_\kappa for \sigma-centered posets 
fails. Thus p \le 2^\omega. See "Coding with ladders a well ordering of the 
reals", Abraham--Shelah (2002).) From the \Sigma^2_1 well ordering one can 
define a \Sigma^2_1 set of reals A that is not Lebesgue measurable. Thus 
there exists a generic extension M of V, preserving all known large 
cardinals, satisfying Not(CH) and the \Sigma^2_1 statement "A is not Lebesgue 

On the other hand, if I correctly understand Solovay's "A model of set-theory 
in which every set of reals is Lebesgue measurable", Lemma 3.8, there is a 
forcing extension N of V by a poset whose cardinality is the least 
inaccessible, such that in N: every OD(R) set of reals is Lebesgue 
measurable, and CH fails. My understanding (I'm not completely sure here) is 
that every \Sigma^2_n set of reals is in OD(R) (in fact, in HOD(R)). 
Therefore N satisfies Not(CH) and the negation of the \Sigma^2_1 statement "A 
is not Lebesgue measurable". N also preserves any known `substantial' large 
cardinal axiom. QED.

James Hirschorn

> On May 19, 2007, at 6:24 PM, James Hirschorn wrote:
> > In any case, we should probably keep in mind the current state of
> > progress. So
> > far (to my knowledge) there has only been one example of a theory T_1
> > satisfying the criterion, and no T_2 has been proven satisfactory.
> > The known example is T_1 = CH (or should it be ZFC + CH?).

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