I´m a bit puzzled by the non-ctm (countable transitive model) approach to forcing as described in Kunen (see here). Kunen calls this "forcing over the universe". I can see how the proofs of (a), (b) and (c) wold go (although I haven't yet sat down to work through them), and how, once we have them, we can get relative consistency results. However, I'm lost as to how one actually gets $\mathbb{1}\Vdash^* \neg $CH.
When looking at the proof involving ctms, we start with a ctm $M$, consider the poset (inside $M$) Add$(\aleph_2,\omega)$, and see that in the generic extension we have $\aleph_2$-many new reals, using the fact that this poset preserves cardinals since its ccc. This uses the definition of $\Vdash$ at several points, i.e. $p\Vdash \phi$ iff for every $M$-generic $G$ with $p\in G$, $M[G]\models\phi$ (with the required interpretations).
Now, if we never define $\Vdash$ nor talk about models, how can we get $\mathbb{1}\Vdash^* \neg $CH? The definition of $p\Vdash^*\phi$ is by recursion on the length of $\phi$, and unwinding something like $\neg$CH is not doable. Moreover, results like
Forcing with such or such poset preserves cardinals.
wouldn´t make sense if I don't have the models $M$ and $M[G]$ to talk about.
Plus, Kunen states it without further comment, so I assume there's something elementary which I' overlooking.
I know that the forcing over the universe approach is doable through Boolean valued models, but can we still make sense of the above without introducing them?
Best Answer
Here it something, very close to Boolean-valued models, that I learned from my advisor.
Let $\Bbb P$ be a forcing notion, and let $V^\Bbb P$ be the class of names. Note that the collection of dense open sets is a filter base, since intersection of finitely many dense open sets is dense open. So use choice to extend it to an ultrafilter on $\Bbb P$, say $U$. (Here we mean ultrafilter in the sense of a maximal filter on the partial order.)
Now use your filter to interpret the names. This defines a class $(M,E)$ where the objects in $M$ are the equivalence classes of $V^\Bbb P/U$ (using Scott's trick as necessary to turn those proper classes into sets again), with $\dot x =_U\dot y$ if and only if there is some $p\in U$ such that $p\Vdash\dot x=\dot y$. And Similarly $\dot x E\dot y$ if there is a condition forcing that.
We can now prove the usual truth lemma for the forcing relation. Finally, let's go back to your question.
We know that the forcing relation can be defined internally, so $1\Vdash\lnot\sf CH$ is something that you can define and prove internally to $V$ about $\Bbb P$. So now take $\Bbb P$ to be $\operatorname{Add}(\omega,\omega_2)$, and you're done.
Oh. Yeah. You want to ask now, if you have a filter that meets all the dense open sets, how is it not a generic filter? Well. Extending it to an ultrafilter makes it a bit more icky. And since $(M,E)$ is not well-founded, there's no reason for $U$ to be truly generic.