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?

  • $\begingroup$ Also, both Jech and Grigorieff force using Boolean algebras. Try looking for the theorem in a source that doesn't. Like Kunen or Halbeisen. $\endgroup$
    – Asaf Karagila
    Jun 18, 2016 at 15:43
  • $\begingroup$ 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. $\endgroup$
    – Asaf Karagila
    Jun 18, 2016 at 16:18
  • $\begingroup$ I'll check Kunen. But are you implying that this might be hopeless? $\endgroup$
    – Ur Ya'ar
    Jun 18, 2016 at 16:34
  • $\begingroup$ P.S I thought maybe there's something like this math.stackexchange.com/a/1607749/63762 just on the other direction $\endgroup$
    – Ur Ya'ar
    Jun 18, 2016 at 16:34
  • 2
    $\begingroup$ 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..." $\endgroup$
    – Asaf Karagila
    Jun 18, 2016 at 16:40


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