Models between a ground model and its forcing extension Jech’s book states and proves a standard result about forcing extensions:
Suppose $B$ is a Boolean algebra and let $G$ be $B$-generic over $V$. If $M$ is a model of ZFC such that $V \subset M \subset V [G]$, then there exists a complete subalgebra $D \subset B$ such that $M = V [D \cap G]$.
My aim is to prove an analogue result about partial orders. That is,
Suppose $P$ is a p.o. and let $G$ be $P$-generic over $V$. If $M$ is a model of ZFC such that $V \subset M \subset V [G]$, then there exists a poset $Q\subset P$ such that $M = V [H]$ for some $H$ which is $Q$-generic over $V$.
Kunen uses the c.t.m. approach to introduce forcing but I didn’t find a proof of this fact in his book.
Thank you.
 A: The most straightforward analogue of the result for the Boolean algebra approach can indeed fail in the poset approach. More specifically, we have the following: there is a forcing poset $\mathbb{P}$ and a generic filter $G\subseteq\mathbb{P}$ such that there is some $a\in V[G]$ such that $V(a)$, the smallest model of ZFC containing $V$ and $\{a\}$, is not a forcing extension of the form $V[G\cap\mathbb{Q}]$ where $\mathbb{Q}$ is a complete subposet of $\mathbb{P}$. By $\mathbb{Q}$ being a complete subposet of $\mathbb{P}$ ($\mathbb{Q}\subseteq_c\mathbb{P}$), I mean

*

*$\mathbb{Q}$ is a suborder of $\mathbb{P}$

*$\mathbb{P}$ and $\mathbb{Q}$ agree on the incompatibility of finitely many elements

*maximal antichains in $\mathbb{Q}$ remain maximal antichains in $\mathbb{P}$
Start with ground model $V$ and use $\mathbb{P}=\text{Col}(\omega,\kappa)$ where $\kappa$ is some cardinal bigger than $|\mathcal{P}(\mathbb{R})|$. This makes the ground model power set of the reals $(\mathcal{P}(\mathbb{R}))^V$ countable in $V[G]$, where $G$ is generic for this forcing. But now we can find a real $a$ in $V[G]$ that is random over $V$ by essentially performing random forcing $V$ (generics exist because from the point of view of $V[G]$, there are only countably many dense sets for the random forcing poset).
I claim that $V(a)$, though itself being a generic extension by the random forcing, is not of the form $V[G\cap\mathbb{Q}]$ where $\mathbb{Q}$ is a complete subposet of $\mathbb{P}$. To see this, suppose towards a contradiction that for some complete subposet $\mathbb{Q}\subseteq_c \mathbb{P}$ and $H=G\cap\mathbb{Q}$ we have $V(a)=V[H]$. Now let $f=\bigcup H$ be the canonical function added by forcing with $H$, so we have $\text{dom}(f)\subseteq\omega$ and $\text{ran}(f)\subseteq \kappa$. But since $V[H]=V(a)$ and the random forcing poset has the ccc, there is some $V$-countable set $S\in V$ that covers the range of $f$ (i.e., $\text{ran}(f)\subseteq S$).
But now $V[H]=V(a)$ can be obtained by the countable forcing $\text{Fn}(\omega,S)$, the set of finite functions from $\omega$ to $S$ ordered by reverse inclusion. Since every countable forcing poset is equivalent (i.e., has the same extension) to the Cohen poset $\text{Fn}(\omega,\omega)$, we can view $V[H]$ as a Cohen forcing extension. Now the contradiction: In a Cohen forcing extension, there is a real that is not dominated by any ground model real, whereas in a random forcing extension, every real is dominated by a ground model real (a proof of this can be found in Jech Lemma 15.30). So $V(a)$ cannot be equal to $V[H]$.
I believe this was left as an exercise in the newer Kunen.
