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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.

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  • $\begingroup$ Could you clarify what the superscript $+$ indicates here? $\endgroup$
    – Joel David Hamkins
    Commented Nov 19, 2012 at 16:56
  • $\begingroup$ oh i meant it as the boolean algebra without the least element. $\endgroup$
    – Zoorado
    Commented Nov 19, 2012 at 17:07

1 Answer 1

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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.

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  • $\begingroup$ 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? $\endgroup$
    – Zoorado
    Commented Nov 20, 2012 at 14:25
  • $\begingroup$ 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. $\endgroup$
    – Joel David Hamkins
    Commented Nov 20, 2012 at 14:58
  • $\begingroup$ 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. $\endgroup$
    – Zoorado
    Commented Nov 20, 2012 at 16:15
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    $\begingroup$ 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 $P$. $\endgroup$
    – Joel David Hamkins
    Commented Nov 20, 2012 at 17:27
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    $\begingroup$ 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... $\endgroup$
    – Joel David Hamkins
    Commented Nov 20, 2012 at 20:39

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