My question is motivated by the following well-known fact regarding intermediate submodels of generic extensions. I would like to know if it can be proven using posets without the need for Boolean algebras.

Fact: Let $\mathbb{R}$, $\mathbb{S}$ be partial orders in $V$. Let $J$ be $\mathbb{R}$-generic over $V$ and let $K \in V[J]$ be $\mathbb{S}$-generic over V. Then there is some partial order $\mathbb{T} \in V[K]$ and $H$ that is $\mathbb{T}$-generic over $V[K]$ such that $$V[K][H]=V[J].$$

Remark: As I understand it, a standard way to prove this can be sketched as follows:

  1. By Jech03-Lem15.4, there is some Boolean algebra $\mathbb{B}$ which is a complete subalgebra of $RO(\mathbb{R})$ and such that $V[K]=V[\mathbb{B} \cap J']$, where $J'$ is the obvious counterpart of $J$ in $RO(\mathbb{R})$.
  2. Let $\mathbb{T} = RO(\mathbb{R})/(\mathbb{B} \cap J')$. (There are a number of ways of doing this but they all come to the same thing. I'm more comfortable treating the Boolean algebras as posets.)
  3. Let $H = J'/(\mathbb{B} \cap J')$. Then it can be shown that we may define $J'$ from $K$ and $H$ and vice versa; so, $$V[J]=V[J']=V[K][H].$$

Alternatively instead of steps 2. and 3., we might define $\mathbb{T}$ as the $\mathbb{S}$-name, $RO(\mathbb{R})/\mathbb{B}$, and show that $\mathbb{S} \times \mathbb{T}$ is forcing equivalent (or even isomorphic) to $\mathbb{R}$.

My question: Can the fact be demonstrated without recourse to Boolean algebras; i.e., with posets alone? Clearly steps, (2) and (3) above (or something like them) will work, but step (1) seems problematic.

If we want to follow the plan above, we want to find some $\mathbb{P}$ which is completely embedded in $\mathbb{R}$ and such that $V[K]=V[\mathbb{P} \cap J]$; thus, avoiding a detour through Boolean algebras.

However, Kunen provides a counter-example in his new Kunen11-ExV.4.50. Since the book is relatively new, I'll write it in:

"Exercise V.4.50 Assume in $M$ that $\mathbb{Q} = Fn(\omega,\kappa)$ with $\kappa > 2^\omega$, and let $\mathbb{H}$ be $\mathbb{Q}$-generic over $M$. Then there is a real number $A \in M[H]$ such that: $A$ is random over $M$; $A \subseteq I \in M$, where $I$ is the set of rationals; but there is no poset $\mathbb{P}$ such that $\mathbb{P} \subseteq_c \mathbb{Q}$ and $M[A]=M[H \cap \mathbb{P}]$."

This suggests that the path through 1. is blocked. Is there another way through?

In Kanamori, Proposition 10.10, we have the statement of similar fact. Kanamori seems to suggest that it can be established with partial orders.

A More Precise Problem: Can the fact above be established when V is a countable transitive model of $ZF-P$?

Remark: Without powerset we cannot guarantee the existence of $RO(\mathbb{R})$ so the obvious implementation of the strategy above is blocked.

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    $\begingroup$ I don't understand the goal here. A boolean algebra is a type of poset, so this theorem does use "posets alone." In this context, you'll have to consider the structure of the posets, and it may turn out that they have a boolean structure. Anyway, the partial order of regular open subsets of a separative partial order is a nice definable thing, just sayin. $\endgroup$ Jan 6, 2014 at 17:37
  • $\begingroup$ @MonroeEskew Sure. Perhaps it would be better to say, "with mere posets." So part of the goal is to avoid $RO(\mathbb{R})$. Of course, there's nothing wrong with $RO(\mathbb{R})$, but I've seen mention of the proof being able to be done without Boolean algebras and I'm interested to know how it's done. $\endgroup$ Jan 6, 2014 at 18:42
  • $\begingroup$ Kanamori's 10.10 is not strictly true. The claim that there is $P * \dot{Q} \cong R$ with the desired property has a counterexample given by Kunen. Rather, $ro(P * \dot{Q}) \cong ro(R)$. $\endgroup$ Jan 6, 2014 at 19:18
  • $\begingroup$ I still don't see what you mean by "mere posets." Suppose we are given a structure of the form $(X,\leq,c_0,c_1,...R_0,R_1,...,f_0,f_1,...)$, where the $c$'s are constants and the $R$'s are relations and the $f$'s are functions and $\leq$ defines a partial order on the domain $X$. We can always delete everything from the language besides $\leq$ and get something which is just a poset. So whatever you happen to construct in the course of proving this fact, just consider it "qua poset." You'll have to use some details of its structure in the proof though; not just any poset will work. $\endgroup$ Jan 6, 2014 at 19:33
  • $\begingroup$ @MonroeEskew Thanks. Do you mean the Kunen example above? If we relax isomorphism back to forcing equivalence, then 10.10 doesn't demand that $\dot{Q}_G \subseteq_c R$, in which case I can't see how Kunen's example would come in. With isomorphism (which is what Kanamori) used, then maybe I see the problem. In which case, I'm tempted to request considering the 10.10 revised to give mere forcing equivalence, but perhaps the credibility of the remark above 10.10 is diminished by these remarks. $\endgroup$ Jan 6, 2014 at 19:39

2 Answers 2


One possible way to prove the theorem without using the Boolean valued techniques is to use Bokovsky's theorem given in Characterization of generic extensions of models of set theory. I may mention there are several proofs of this fact without refering to Boolean algebras, see for example Schindler's paper The long extender algebra or the paper On the set-generic multiverse by Friedman-Fuchino-Sakai.

Note that the theorem is an immediate corollary of Bukovsky's theorem.


The "mere poset" approach has something in it. Of course a BA is a partial case of a poset. However the cba-hull of a poset may lack properties which the poset has.

Take eg the Cohen forcing $\mathbb C$. It is countable while its CBA-hull is not. Now, by the Solovay $\Sigma$-method, if $a$ is Cohen-generic over $V$ and $b$ a real in $V[a]$ then $V[a]$ is a $\Sigma$-generic extension of $V[b]$, where $\Sigma\subseteq\mathbb C$, hence, either still a Cohen extension of $V[b]$ or else $V[b]=V[a]$ -- just because any subset $\Sigma$ of $\mathbb C$ is countable and hence is either trivial of Cohen itself. Try to prove this old fact on the cba base and you will see the difference.

Further, $\Sigma$ itself is a $\mathbb Q$ generic set over $V$, where $\mathbb Q$ is a strengthening of $\mathbb C$ (the same domain but a $\subseteq$-larger order relation). Once again this implies that $V[b]=V[\Sigma]$ is a Cohen extension of $V$ (or else $V[b]=V[\Sigma]=V$). Again try to prove this on the cba base.


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