[FOM] CH and forcing

[Andreas Blass]

Let me try to answer several of the questions and comments that have been 
raised here about Boolean-valued models and non-well-founded forcing 
extensions.

Roger Bishop Jones wrote:
> ------------------------------
> Consider specifically the interpretation of ZFC consisting of
> all pure well-founded collections of accessible rank.
>
> The only way to extend this interpretation is by adding sets
> which have inaccessible rank or which are not well-founded.
> In neither case could the extension change the truth value
> of CH, which is fixed as soon as we have all the well-founded
> sets of rank less than w+(a small natural number).
--------------------------------

The last sentence quoted here is incorrect.  The truth value of CH in the 
model under consideration depends also on the non-well-founded sets. 
Indeed, CH is about the collection of all subsets of the natural numbers, 
and, when interpreted in a model, this will refer to the natural numbers 
of that model, including non-standard (non-well-founded) ones.

He also wrote:
---------------------------------
> I am puzzled by the suggestion that a boolean valued model
> could make any difference.
> Surely if we want to use a boolean valued model to force the
> negation of CH we must have an appropriate forcing condition
> to determine the algebra of truth values, and this could
> only be the case if there were some set which could be added
> (to the model described above) which would change the truth
> value of CH?
----------------------------------

The sets being added in a Boolean-valued model are (in general) 
Boolean-valued sets.  The forcing extensions that produce violations of CH 
add new Boolean-valued sets, i.e., sets x such that, for all sets y in the 
ground model, the Boolean truth value of x=y is zero.

Drake O'Brien wrote:
----------------------------------
>> In this context, I would interpret "forcing extension" to mean a
> Boolean-valued
>> model.  With this interpretation, the assertion is correct.
>>
>
> Could you please show this, to the uninitiated?
----------------------------------

In the usual Boolean-valued models for proving consistency of an assertion 
A (like, for example, the negation of CH) with ZFC, all the following 
statements have Boolean truth-value 1:

The axioms of ZFC;
A;
V-check is a transitive class containing all the ordinals and satisfying 
ZFC;
[infinitary sentences expressing, for each set x in the original 
universe:] The members of x-check 
are exactly the y-check as y ranges over members of x.

In other words, with truth-value 1, the Boolean-valued model contains an 
exact copy V-check of the original universe (the ground model) and it 
satisfies ZFC plus A.

All this should be available in standard expositions of Boolean-valued 
models, for example Bell's book, or the older book by Rosser, or Jech's 
set-theory bible.

It may be helpful to add that the "exact copy" phenomenon remains true in 
the two-valued models (if any) obtained by reducing the Boolean-valued 
model by a *generic* ultrafilter (if such ultrafilters exist), but it need 
not remain true if one uses a non-generic ultrafilter.  Nevertheless, even 
with a non-generic ultrafilter, all the finitary, first-order consequences 
of "exact copy" remain correct; the reduction of V-check modulo any 
ultrafilter is an elementary extension of the ground model.

Roger Bishop Jones also presented an argument against Woodin's 
"weaker claim", namely:
>
> "This weak version simply asserts that if
> 	(M,E)	 |= ZFC,
> then there exists a structure
> 	(M**, E**) |= ZFC + "CH is false",
> such that M is contained in M** and such that
> 	E = E** intersection (M x M)"
>
The anti-Woodin argument contained several lemmas, including:
>
> Lemma2: ZFC |- (V(w+4) subset_of M) =>
> 		 ((M |= CH) <=> (V(w+4) |= CH))

This lemma is false.  In fact neither half of the bi-implication follows 
unless one has some additional information about M, like well-foundedness. 
In general, M could have new, non-standard natural numbers, in which case 
its "subsets of w" would be quite different from those of V.  (Like Jones, 
I abbreviate omeega as w.)  And its functions from the set of countable 
ordinals to the set of subsets of w would also be quite different from 
those of V.