# Intermediate submodels without Boolean algebras

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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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. – Monroe Eskew Jan 6 '14 at 17:37
@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. – Toby Meadows Jan 6 '14 at 18:42
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, $ro(P * \dot{Q}) \cong ro(R)$. – Monroe Eskew Jan 6 '14 at 19:18
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. – Monroe Eskew Jan 6 '14 at 19:33
@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. – Toby Meadows Jan 6 '14 at 19:39