# proof that forcing notions are equivalent to their composition

This is a long and complicated proof, the more so because the meaning of $Q$ shifts depending on what generic subset of $P$ is being used. It is therefore broken into a number of steps. The core of the proof is to prove that, given any generic subset $G$ of $P$ and a generic subset $H$ of $\hat{Q}[G]$ there is a corresponding generic subset $G*H$ of $P*Q$ such that $\mathfrak{M}[G][H]=\mathfrak{M}[G*H]$, and conversely, given any generic subset $G$ of $P*Q$ we can find some generic $G_{P}$ of $P$ and a generic $G_{Q}$ of $\hat{Q}[G_{P}]$ such that $\mathfrak{M}[G_{P}][G_{Q}]=\mathfrak{M}[G]$.

We do this by constructing functions using operations which can be performed within the forced universes so that, for example, since $\mathfrak{M}[G][H]$ has both $G$ and $H$, $G*H$ can be calculated, proving that it contains $\mathfrak{M}[G*H]$. To ensure equality, we will also have to ensure that our operations are inverses; that is, given $G$, $G_{P}*G_{H}=G$ and given $G$ and $H$, $(G*H)_{P}=P$ and $(G*H)_{Q}=H$.

The remainder of the proof merely defines the precise operations, proves that they give generic sets, and proves that they are inverses.

Before beginning, we prove a lemma which comes up several times:

### Lemma: If $G$ is generic in $P$ and $D$ is dense above some $p\in G$ then $G\cap D\neq\emptyset$

Let $D^{\prime}=\{p^{\prime}\in P\mid p^{\prime}\in D\vee p^{\prime}\text{ is % incompatible with }p\}$. This is dense, since if $p_{0}\in P$ then either $p_{0}$ is incompatible with $p$, in which case $p_{0}\in D^{\prime}$, or there is some $p_{1}$ such that $p_{1}\leq p,p_{0}$, and therefore there is some $p_{2}\leq p_{1}$ such that $p_{2}\in D$, and therefore $p_{2}\leq p_{0}$. So $G$ intersects $D^{\prime}$. But since a generic set is directed, no two elements are incompatible, so $G$ must contain an element of $D^{\prime}$ which is not incompatible with $p$, so it must contain an element of $D$.

### $G*H$ is a generic filter

First, given generic subsets $G$ and $H$ of $P$ and $\hat{Q}[G]$, we can define:

 $G*H=\{\langle p,\hat{q}\rangle\mid p\in G\wedge\hat{q}[G]\in H\}$

#### $G*H$ is closed

Let $\langle p_{1},\hat{q}_{1}\rangle\in G*H$ and let $\langle p_{1},\hat{q}_{1}\rangle\leq\langle p_{2},\hat{q}_{2}\rangle$. Then we can conclude $p_{1}\in G$, $p_{1}\leq p_{2}$, $\hat{q}_{1}[G]\in H$, and $p_{1}\Vdash\hat{q}_{1}\leq\hat{q}_{2}$, so $p_{2}\in G$ (since $G$ is closed) and $\hat{q}_{2}[G]\in H$ since $p_{1}\in G$ and $p_{1}$ forces both $\hat{q}_{1}\leq\hat{q}_{2}$ and that $H$ is downward closed. So $\langle p_{2},\hat{q}_{2}\rangle\in G*H$.

#### $G*H$ is directed

Suppose $\langle p_{1},\hat{q}_{1}\rangle,\langle p_{1},\hat{q}_{1}\rangle\in G*H$. So $p_{1},p_{2}\in G$, and since $G$ is directed, there is some $p_{3}\leq p_{1},p_{2}$. Since $\hat{q}_{1}[G],\hat{q}_{2}[G]\in H$ and $H$ is directed, there is some $\hat{q}_{3}[G]\leq\hat{q}_{1}[G],\hat{q}_{2}[G]$. Therefore there is some $p_{4}\leq p_{3}$, $p_{4}\in G$, such that $p_{4}\Vdash\hat{q}_{3}\leq\hat{q}_{1},\hat{q}_{2}$, so $\langle p_{4},\hat{q}_{3}\rangle\leq\langle p_{1},\hat{q}_{1}\rangle,\langle p% _{1},\hat{q}_{1}\rangle$ and $\langle p_{4},\hat{q}_{3}\rangle\in G*H$.

#### $G*H$ is generic

Suppose $D$ is a dense subset of $P*\hat{Q}$. We can project it into a dense subset of $Q$ using $G$:

 $D_{Q}=\{\hat{q}[G]\mid\langle p,\hat{q}\rangle\in D\}\text{ for some }p\in G$

#### Lemma: $D_{Q}$ is dense in $\hat{Q}[G]$

Given any $\hat{q}_{0}\in\hat{Q}$, take any $p_{0}\in G$. Then we can define yet another dense subset, this one in $G$:

 $D_{\hat{q}_{0}}=\{p\mid p\leq p_{0}\wedge p\Vdash\hat{q}\leq\hat{q}_{0}\wedge% \langle p,\hat{q}\rangle\in D\}\text{ for some }\hat{q}\in\hat{Q}$

#### Lemma: $D_{\hat{q}_{0}}$ is dense above $p_{0}$ in $P$

Take any $p\in P$ such that $p\leq p_{0}$. Then, since $D$ is dense in $P*\hat{Q}$, we have some $\langle p_{1},\hat{q}_{1}\rangle\leq\langle p,\hat{q}_{0}\rangle$ such that $\langle p_{1},\hat{q}_{1}\rangle\in D$. Then by definition $p_{1}\leq p$ and $p_{1}\in D_{\hat{q}_{0}}$.

From this lemma, we can conclude that there is some $p_{1}\leq p_{0}$ such that $p_{1}\in G\cap D_{\hat{q}_{0}}$, and therefore some $\hat{q}_{1}$ such that $p_{1}\Vdash\hat{q}_{1}\leq\hat{q}_{0}$ where $\langle p_{1},\hat{q}_{1}\rangle\in D$. So $D_{Q}$ is indeed dense in $\hat{Q}[G]$.

Since $D_{Q}$ is dense in $\hat{Q}[G]$, there is some $\hat{q}$ such that $\hat{q}[G]\in D_{Q}\cap H$, and so some $p\in G$ such that $\langle p,\hat{q}\rangle\in D$. But since $p\in G$ and $\hat{q}\in H$, $\langle p,\hat{q}\rangle\in G*H$, so $G*H$ is indeed generic.

## $G_{P}$ is a generic filter

Given some generic subset $G$ of $P*\hat{Q}$, let:

 $G_{P}=\{p\in P\mid p^{\prime}\leq p\wedge\langle p^{\prime},\hat{q}\rangle\in G% \}\text{ for some }p^{\prime}\in P\text{ and some }\hat{q}\in Q$

#### $G_{P}$ is closed

Take any $p_{1}\in G_{P}$ and any $p_{2}$ such that $p_{1}\leq p_{2}$. Then there is some $p^{\prime}\leq p_{1}$ satisfying the definition of $G_{P}$, and also $p^{\prime}\leq p_{2}$, so $p_{2}\in G_{P}$.

#### $G_{P}$ is directed

Consider $p_{1},p_{2}\in G_{P}$. Then there is some $p^{\prime}_{1}$ and some $\hat{q}_{1}$ such that $\langle p^{\prime}_{1},\hat{q}_{1}\rangle\in G$ and some $p^{\prime}_{2}$ and some $\hat{q}_{2}$ such that $\langle p^{\prime}_{2},\hat{q}_{2}\rangle\in G$. Since $G$ is directed, there is some $\langle p_{3},\hat{q}_{3}\rangle\in G$ such that $\langle p_{3},\hat{q}_{3}\rangle\leq\langle p^{\prime}_{1},\hat{q}_{1}\rangle,% \langle p^{\prime}_{2},\hat{q}_{2}\rangle$, and therefore $p_{3}\in G_{P}$, $p_{3}\leq p_{1},p_{2}$.

#### $G_{P}$ is generic

Let $D$ be a dense subset of $P$. Then $D^{\prime}=\{\langle p,\hat{q}\rangle\mid p\in D\}$. Clearly this is dense, since if $\langle p,\hat{q}\rangle\in P*\hat{Q}$ then there is some $p^{\prime}\leq p$ such that $p^{\prime}\in D$, so $\langle p^{\prime},\hat{q}\rangle\in D^{\prime}$ and $\langle p^{\prime},\hat{q}\rangle\leq\langle p,\hat{q}\rangle$. So there is some $\langle p,\hat{q}\rangle\in D^{\prime}\cap G$, and therefore $p\in D\cap G_{P}$. So $G_{P}$ is generic.

## $G_{Q}$ is a generic filter

Given a generic subset $G\subseteq P*\hat{Q}$, define:

 $G_{Q}=\{\hat{q}[G_{P}]\mid\langle p,\hat{q}\rangle\in G\}\text{ for some }p\in P$

(Notice that $G_{Q}$ is dependant on $G_{P}$, and is a subset of $\hat{Q}[G_{P}]$, that is, the forcing notion inside $\mathfrak{M}[G_{P}]$, as opposed to the set of names $Q$ which we’ve been primarily working with.)

#### $G_{Q}$ is closed

Suppose $\hat{q}_{1}[G_{P}]\in G_{Q}$ and $\hat{q}_{1}[G_{P}]\leq\hat{q}_{2}[G_{P}]$. Then there is some $p_{1}\in G_{P}$ such that $p_{1}\Vdash\hat{q}_{1}\leq\hat{q}_{2}$. Since $p_{1}\in G_{P}$, there is some $p_{2}\leq p_{1}$ such that for some $\hat{q}_{3}$, $\langle p_{2},\hat{q}_{3}\rangle\in G$. By the definition of $G_{Q}$, there is some $p_{3}$ such that $\langle p_{3},\hat{q}_{1}\rangle\in G$, and since $G$ is directed, there is some $\langle p_{4},\hat{q}_{4}\rangle\in G$ and $\langle p_{4},\hat{q}_{4}\rangle\leq\langle p_{3},\hat{q}_{1}\rangle,\langle p% _{2},\hat{q}_{3}\rangle$. Since $G$ is closed and $\langle p_{4},\hat{q}_{4}\rangle\leq\langle p_{4},\hat{q}_{2}\rangle$, we have $\hat{q}_{2}[G_{P}]\in G_{Q}$.

#### $G_{Q}$ is directed

Suppose $\hat{q}_{1}[G_{P}],\hat{q}_{2}[G_{P}]\in G_{Q}$. Then for some $p_{1},p_{2}$, $\langle p_{1},\hat{q}_{1}\rangle,\langle p_{2},\hat{q}_{2}\rangle\in G$, and since $G$ is directed, there is some $\langle p_{3},\hat{q}_{3}\rangle\in G$ such that $\langle p_{3},\hat{q}_{3}\rangle\leq\langle p_{1},\hat{q}_{1}\rangle,\langle p% _{2},\hat{q}_{2}\rangle$. Then $\hat{q}_{3}[G_{P}]\in G_{Q}$ and since $p_{3}\in G$ and $p_{3}\Vdash\hat{q}_{3}\leq\hat{q}_{1},\hat{q}_{2}$, we have $\hat{q}_{3}[G_{P}]\leq\hat{q}_{1}[G_{P}],\hat{q}_{2}[G_{P}]$.

#### $G_{Q}$ is generic

Let $D$ be a dense subset of $Q[G_{P}]$ (in $\mathfrak{M}[G_{P}]$). Let $\hat{D}$ be a $P$-name for $D$, and let $p_{1}\in G_{P}$ be a such that $p_{1}\Vdash\hat{D}$ is dense. By the definition of $G_{P}$, there is some $p_{2}\leq p_{1}$ such that $\langle p_{2},\hat{q}_{2}\rangle\in G$ for some $q_{2}$. Then $D^{\prime}=\{\langle p,\hat{q}\rangle\mid p\Vdash\hat{q}\in D\wedge p\leq p_{2}\}$.

#### Lemma: $D^{\prime}$ is dense (in $G$) above $\langle p_{2},\hat{q}_{2}\rangle$

Take any $\langle p,\hat{q}\rangle\in P*Q$ such that $\langle p,\hat{q}\rangle\leq\langle p_{2},\hat{q}_{2}\rangle$. Then $p\Vdash\hat{D}$ is dense, and therefore there is some $\hat{q}_{3}$ such that $p\Vdash\hat{q}_{3}\in\hat{D}$ and $p\Vdash\hat{q}_{3}\leq\hat{q}$. So $\langle p,\hat{q}_{3}\rangle\leq\langle p,\hat{q}\rangle$ and $\langle p,\hat{q}_{3}\rangle\in D^{\prime}$.

Take any $\langle p_{3},\hat{q}_{3}\rangle\in D^{\prime}\cap G$. Then $p_{3}\in G_{P}$, so $\hat{q}_{3}\in D$, and by the definition of $G_{Q}$, $\hat{q}_{3}\in G_{Q}$.

## $G_{P}*G_{Q}=G$

If $G$ is a generic subset of $P*Q$, observe that:

 $G_{P}*G_{Q}=\{\langle p,\hat{q}\rangle\mid p^{\prime}\leq p\wedge\langle p^{% \prime},\hat{q}^{\prime}\rangle\in G\wedge\langle p_{0},\hat{q}\rangle\in G\}% \text{ for some }p^{\prime},\hat{q}^{\prime},p_{0}$

If $\langle p,\hat{q}\rangle\in G$ then obviously this holds, so $G\subseteq G_{P}*G_{Q}$. Conversely, if $\langle p,\hat{q}\rangle\in G_{P}*G_{Q}$ then there exist $p^{\prime},\hat{q}^{\prime}$ and $p_{0}$ such that $\langle p^{\prime},\hat{q}^{\prime}\rangle,\langle p_{0},\hat{q}\rangle\in G$, and since $G$ is directed, some $\langle p_{1},\hat{q_{1}}\rangle\in G$ such that $\langle p_{1},\hat{q}_{1}\rangle\leq\langle p^{\prime},\hat{q}^{\prime}\rangle% ,\langle p_{0},\hat{q}\rangle$. But then $p_{1}\leq p$ and $p_{1}\Vdash\hat{q}_{1}\leq\hat{q}$, and since $G$ is closed, $\langle p,\hat{q}\rangle\in G$.

## $(G*H)_{P}=G$

Assume that $G$ is generic in $P$ and $H$ is generic in $Q[G]$.

Suppose $p\in(G*H)_{P}$. Then there is some $p^{\prime}\in P$ and some $\hat{q}\in Q$ such that $p^{\prime}\leq p$ and $\langle p^{\prime},\hat{q}\rangle\in G*H$. By the definition of $G*H$, $p^{\prime}\in G$, and then since $G$ is closed $p\in G$.

Conversely, suppose $p\in G$. Then (since $H$ is non-trivial), $\langle p,\hat{q}\rangle\in G*H$ for some $\hat{q}$, and therefore $p\in(G*H)_{P}$.

## $(G*H)_{Q}=H$

Assume that $G$ is generic in $P$ and $H$ is generic in $Q[G]$.

Given any $q\in H$, there is some $\hat{q}\in Q$ such that $\hat{q}[G]=q$, and so there is some $p$ such that $\langle p,\hat{q}\rangle\in G*H$, and therefore $\hat{q}[G]\in H$.

On the other hand, if $q\in(G*H)_{Q}$ then there is some $\langle p,\hat{q}\rangle\in G*H$, and therefore some $\hat{q}[G]\in H$.

Title proof that forcing notions are equivalent to their composition ProofThatForcingNotionsAreEquivalentToTheirComposition 2013-03-22 12:54:26 2013-03-22 12:54:26 Henry (455) Henry (455) 5 Henry (455) Proof msc 03E40 msc 03E35 EquivalenceOfForcingNotions