# 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.

• I have no idea about your question, but I guess using Boolean algebras is somewhat necessary: there is no analogue of these intermediate model theorem for class forcings, and class forcings do not have Boolean completion in general. Commented Oct 3, 2020 at 7:11

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.

• Two remarks: (1) It is enough to collapse the reals to be countable, no need to go that high with $\kappa$. (2) Why are you using $V(a)$ and not $V[a]$? (They are equal here, of course, since $a$ is a real.) Commented Dec 18, 2020 at 8:25
• @AsafKaragila no particular reason, either subconsciously I was trying to mimic Kunen's notation or just a random choice Commented Dec 18, 2020 at 18:12