# A poset oriented proof for the intermediate model theorem.

The intermediate model theorem:

If $M\subseteq N\subseteq M[G]$ are all models of $\sf ZFC$, with $G$ generic over $M$, then $N$ is a generic extension of $M$, and $M[G]$ is a generic extension of $N$.

The proofs I've seen (Jech lemma 15.43 and in Grigorieff's "Intermediate submodels and generic extensions in set theory") are somewhat roundabout, as they pass through complete boolean algebras. Is there a more "direct" proof using just posets?

Furthermore, given some poset $P$, a generic $G$ and $M\subseteq N\subseteq M[G]$, the theorem gives me some $Q\in M$ and $Q$-generic filter $H$ such that $N=M[H]$. I want to understand $Q$ in terms of $P$. What can I say about it? Can I construct it more or less explicitly?

• Also, both Jech and Grigorieff force using Boolean algebras. Try looking for the theorem in a source that doesn't. Like Kunen or Halbeisen. Jun 18, 2016 at 15:43
• Note that every forcing can be embedded into a collapse of some large enough piece of the universe to be countable. But while the collapse poset is very easy to describe, these embeddings are not. Jun 18, 2016 at 16:18
• I'll check Kunen. But are you implying that this might be hopeless? Jun 18, 2016 at 16:34
• P.S I thought maybe there's something like this math.stackexchange.com/a/1607749/63762 just on the other direction Jun 18, 2016 at 16:34
• Oh, there is probably a poset-based proof. But I doubt there is one that will make you go "Ohhhhhh!!!!!!" in any meaningful way. At best, it will make you go "Well, that's a poset-based proof alright..." Jun 18, 2016 at 16:40