MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $P$ be a forcing notion. Let $B(P)$ be the boolean completion of $P$ and $i : P \rightarrow B(P)$ be the corresponding dense embedding (in $B(P)^{+}$). Let $G$ be $B(P)$-generic over $M$, the transitive ground model satisfying ZFC.

I know that if $N$ is a transitive model of ZFC such that $M \subset N \subset M[G]$, then $N = M[D \cap G]$ for some complete subalgebra $D$ of $B(P)$. But can we say anything about $(X :=) ran(i) \cap D$ and $(Y :=) i^{-1}[D]$? Is $X$ dense in $D^{+}$? Is $Y$ the range of some complete embedding into $P$? Are there any other interesting properties about them?

Thanks in advance.

share|cite|improve this question
Could you clarify what the superscript $+$ indicates here? – Joel David Hamkins Nov 19 '12 at 16:56
oh i meant it as the boolean algebra without the least element. – Zoorado Nov 19 '12 at 17:07
up vote 7 down vote accepted

It depends on the particular forcing, and in general, things may not work out so nicely.

On the one hand, it could be that $P=\mathbb{B}^+$, in which case for any intermediate model $N$ we have $X=Y=D^+$ and so $X$ is dense in $D^+$ and everything you want is true.

On the other hand, consider the case where $P$ is the forcing consisting of conditions $(s,t)$, where $s,t\in 2^{{\lt}\omega}$ are finite binary sequences of the same length $|s|=|t|$. The Boolean completion is $\mathbb{B}=\text{Add}(\omega,2)$ the forcing to add two Cohen reals $M[c,d]$. Consider the intermediate extension $M[c]$ to add just the first one. In this case, $D$ is the subalgebra consisting of conditions in $\mathbb{B}$ that do not determine any information about the second real $d$, although they may decide information about the first real $c$. This does not interact well with the image of $P$ inside $\mathbb{B}$, because conditions in the range of $P$ decide an equal number of bits for both $c$ and $d$. In particular, $D$ contains no members of $\text{ran}(i)$ except for the trivial condition $1$. So in this case, $X$ is not dense in $D^+$, and $Y$ has only the trivial condition.

share|cite|improve this answer
Wow the second example is very illuminating. Thanks a lot. Would it even be false to say in general that if M[G] is a P-generic extension, then an intermediate M[G'] is a Q-generic extension for some Q being a suborder of P? – Zoorado Nov 20 '12 at 14:25
Yes, that is false. You have to go the Boolean algebra, and get your subalgebra there. The same example works, since every subalgebra of my forcing $P$, where the conditions have the same length, which determines all the bits of the first Cohen real $c$, also determines all the bits of the second one, just because conditions in $P$ have the same length. So $P$ has no suborder adding just the first Cohen real. But meanwhile, $\mathbb{B}=RO(P)$ does have a subalgebra adding just the first real. – Joel David Hamkins Nov 20 '12 at 14:58
Ok I get why it is that way in the example. I think I messed up my second question. What I wanted to say is, if M[G] is a P-generic extension, then must an intermediate M[G'] be a Q-generic extension for some Q isomorphic to a suborder of P? My bad for the confusion. – Zoorado Nov 20 '12 at 16:15
If you want $Q$ to be a complete suborder, so that maximal antichains in Q are also maximal in P, then you can make a counterexample by using the forcing to collapse $\omega_1$ to $\omega$ via $P=\omega_1^{\lt\omega}$, which is a tree. In a tree order, the only complete suborders are the whole order. But forcing with P also adds a Cohen real (and lots of other stuff, which does not collapse $\omega_1$), and so the corresponding intermediate extensions do not arise from complete suborders of $P$. – Joel David Hamkins Nov 20 '12 at 17:27
It's not quite right. I should say that in this tree, the only (necessarily nontrivial) complete suborders are forcing equivalent to the whole order... – Joel David Hamkins Nov 20 '12 at 20:39

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.