Suppose that $M$ is a model of $\sf ZFC$, and we add some generic set $G$. Then it is not hard to see that for every $x\in M[G]$ it holds $M\subseteq M[x]\subseteq M[G]$.

Given $x\in M[G]$ such that $x\subseteq M$, can we find a forcing $P\in M[G]$ such that:

  1. If $H$ is $P$-generic over $M[G]$ then $H\in M[G]$ (triviality).
  2. If $H$ is $(P\cap M)$-generic over $M$, then $H\notin M$ (not so-triviality).
  3. $x$ is $(P\cap M)$-generic over $M$ (relevance).

Obviously we're not talking about minimal forcing like Sacks and friends, but on forcing which have a lot of non-trivial intermediate forcings (like Cohen and friends).


1 Answer 1


It is a standard result (e.g. Jech's Set Theory, lemma 15.43) that every intermediate model $W$ of ZFC with $M\subset W\subset M[G]$ has the form $W=M[G\cap\mathbb{C}]$ for some complete subalgebra $\mathbb{C}$ of the complete Boolean algebra $\mathbb{B}$ for the forcing giving rise to $G$.

In particular, since $x\subset M$ we know that $M[x]\models \text{ZFC}$, and so in particular there is a partial order $P\in M$ which (we can easily arrange) is nontrivial and adds $x$ via $G\cap P$. This achieves nontriviality and relevance (2 and 3).

Now, my observation is that your triviality condition 1 is also easy to obtain by modifying $P$, to add a single new atom below every element of $P$, but with a node that is not in $M$. For example, in $M[G]$, let $P^+$ be the poset $P$ together with a new atom, $G$, considered as a single point below every element of $P$. So $P^+$ is atomic, and hence trivial, but $P^+\cap M=P$, which satisfies your requirements in 2 and 3.

So the answer is yes.

  • $\begingroup$ I'm not fully sure that I understand the definition of $P^+$. $\endgroup$
    – Asaf Karagila
    Jun 10, 2013 at 13:14
  • 1
    $\begingroup$ @Asaf: Joel is just adjoining one new element to $P$, below all the existing ones. So the new poset is trivial as a forcing notion. But by choosing the new element to be outside $M$, he arranges that the new poset intersected with $M$ is the original one, which is nontrivial. Fancier versions of the same construction show that, in general, $P\cap M$ and $P$ may be essentially totally unrelated. Start with any desired poset of $M$ as $P\cap M$ and then obtain $P$ by putting whatever you want, using elements $\notin M$, below $P\cap M$. $\endgroup$ Jun 10, 2013 at 14:08
  • $\begingroup$ Yes, Andreas has exactly described what I had in mind. But perhaps this idea shows that you haven't asked what you intended? $\endgroup$ Jun 10, 2013 at 21:03
  • $\begingroup$ Joel, I gave it some thought, and I did ask what I intended to ask. I also think that your answer is very good for that. So it's all fine. Much like the previous question you answered regarding a subforcing, I was hoping to find some "nicely defined" poset, but would be impossible in the general case, and would require a bit of ad-hoc analysis of the set. I suppose I could do that. Thanks for the answer, I feel a bit silly for not seeing that on my own! :-) $\endgroup$
    – Asaf Karagila
    Jun 11, 2013 at 7:18

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