Is my intuition here regarding generic sets used in forcing correct?

Question

In Jech's Chapter 14, p. 218 (2nd edition), the proof for the generic model theorem (i.e. Theorem 14.5) begins with: "Let $(P, <) $ be a notion of forcing in the ground model $M$, and let $G \subset P$ be generic over $M$".

Afterwards, the corresponding boolean algebra $B=B(P)$ is constructed (for which $P$ is dense), followed by the boolean-valued model $M^P$ and finally the generic extension $M[G]$ for which $G \in M[G]$.

But from what I understand, the generic set $G \subset P$ over $M$ may not necessarily exist, whereas the proof begins with the assumption that $G$ does exist.

But if $M$ was assumed to be countable, then $G$ could be shown to exist (there's a proof in Jech to show this). However, as the book says in its Introduction to Chapter 14, this route is not ideal since a countable model of ZFC may not be proven within ZFC due to Godel's incompleteness theorem. Therefore, as per the book, the modern approach is to "pretend" that $G$ holds in the universe and the whole chapter 14 of Jech seems to rely on this premise...

But am I right to think that the we can also use this route (following my understanding of Hamkins and Seabold's paper): Given that any forcing condition set $P$ has a Boolean completion $B$ for which it is dense, we can first construct the Boolean-valued model $M^B$. Then we can pick any ultrafilter $H$ in $B$ and construct the generic extension $M[H]$. The ultrafilter $H$ is guaranteed to be related to some generic set $G$ such that their Boolean-valued models are equal, i.e. $M[G]=M[H]$, although we don't need to specify what $G$ is ...

Is my intuition correct or am I missing something here ? Any help is much appreciated.

Answer 1

You have the following formulation of the Baire Category Theorem:

Let $(P,<)$ be a partially ordered set. Let $\{D_n\}_{n=1}^\infty$ be subsets of $P$ each with the following property

• For all $p\in P$ there exists a $d\in D_n$ and $r\in P$ such that $r\le p$ and $r\le d$. $(*)$

and let $x\in P$. Then there exists a filter $G$ on $P$ such that $x\in G$ and $G\cap D_n\ne\emptyset$ for all $n$.

The proof uses the axiom of dependent choice (and I believe is known to be equivalent to it.) It goes as follows

• There exists a $d_1\in D_1$ and $r_1\in P$ such that $r_1\le x\land r_1\le d_1$.
• There exists a $d_2\in D_2$ and $r_2\in P$ such that $r_2\le r_1\land r_2\le d_2$.
• There exists a $d_3\in D_3$ and $r_3\in P$ such that $r_3\le r_2\land r_3\le d_3$.

and so on by induction $d_n, r_n$ for all $n$. Since $\{r_n\}_{n=1}^\infty$ is a descending sequence, it's a filter base. Let $G=\{p\in P\colon \exists n \,r_n\le p\}$. Then $G$ is a filter, $x\in G$ and $d_n\in G$ for all $n$, hence $G\cap D_n\ne\emptyset$ for all $n$. This completes the proof.

Now, as I understand, the reason for using a countable ground model $M$ is because then $M$ has only countably many sets, so it has only countably many sets $D$ satisfying the condition $(*)$ (which Jech calls 'predense'), so working in $\mathsf{ZFC}$, the theorem guarantees the existence of a filter $G$.