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 $\widehat{Q}[G]$ there is a corresponding generic subset $G*H$ of $P*Q$ such that $𝔐[G][H]=𝔐[G*H]$, 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 $\widehat{Q}[G_P]$ such that $𝔐[G_P][G_Q]=𝔐[G]$.

We do this by constructing functions using operations which can be performed within the forced universes so that, for example, since $𝔐[G][H]$ has both $G$ and $H$, $G*H$ can be calculated, proving that it contains $𝔐[G*H]$. 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 $G\cap D\ne \mathrm{\varnothing }$

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 $p_1\le p,p_0$, and therefore there is some $p_2\le p_1$ such that $p_2\in D$, and therefore $p_2\le p_0$. 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 $\widehat{Q}[G]$, we can define:

$$G*H=\{⟨p,\widehat{q}⟩\mid p\in G\wedge \widehat{q}[G]\in H\}$$

$G*H$ is closed

Let $⟨p_1,{\widehat{q}}_1⟩\in G*H$ and let $⟨p_1,{\widehat{q}}_1⟩\le ⟨p_2,{\widehat{q}}_2⟩$. Then we can conclude $p_1\in G$, $p_1\le p_2$, ${\widehat{q}}_1[G]\in H$, and $p_1⊩{\widehat{q}}_1\le {\widehat{q}}_2$, so $p_2\in G$ (since $G$ is closed) and ${\widehat{q}}_2[G]\in H$ since $p_1\in G$ and $p_1$ forces both ${\widehat{q}}_1\le {\widehat{q}}_2$ and that $H$ is downward closed. So $⟨p_2,{\widehat{q}}_2⟩\in G*H$.

$G*H$ is directed

Suppose $⟨p_1,{\widehat{q}}_1⟩,⟨p_1,{\widehat{q}}_1⟩\in G*H$. So $p_1,p_2\in G$, and since $G$ is directed, there is some $p_3\le p_1,p_2$. Since ${\widehat{q}}_1[G],{\widehat{q}}_2[G]\in H$ and $H$ is directed, there is some ${\widehat{q}}_3[G]\le {\widehat{q}}_1[G],{\widehat{q}}_2[G]$. Therefore there is some $p_4\le p_3$, $p_4\in G$, such that $p_4⊩{\widehat{q}}_3\le {\widehat{q}}_1,{\widehat{q}}_2$, so $⟨p_4,{\widehat{q}}_3⟩\le ⟨p_1,{\widehat{q}}_1⟩,⟨p_1,{\widehat{q}}_1⟩$ and $⟨p_4,{\widehat{q}}_3⟩\in G*H$.

$G*H$ is generic

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

$$D_Q=\{\widehat{q}[G]\mid ⟨p,\widehat{q}⟩\in D\}\text{ for some }p\in G$$

Lemma: $D_Q$ is dense in $\widehat{Q}[G]$

Given any ${\widehat{q}}_0\in \widehat{Q}$, take any $p_0\in G$. Then we can define yet another dense subset, this one in $G$:

$$D_{{\widehat{q}}_0}=\{p\mid p\le p_0\wedge p⊩\widehat{q}\le {\widehat{q}}_0\wedge ⟨p,\widehat{q}⟩\in D\}\text{ for some }\widehat{q}\in \widehat{Q}$$

Lemma: $D_{{\widehat{q}}_0}$ is dense above $p_0$ in $P$

Take any $p\in P$ such that $p\le p_0$. Then, since $D$ is dense in $P*\widehat{Q}$, we have some $⟨p_1,{\widehat{q}}_1⟩\le ⟨p,{\widehat{q}}_0⟩$ such that $⟨p_1,{\widehat{q}}_1⟩\in D$. Then by definition $p_1\le p$ and $p_1\in D_{{\widehat{q}}_0}$.

From this lemma, we can conclude that there is some $p_1\le p_0$ such that $p_1\in G\cap D_{{\widehat{q}}_0}$, and therefore some ${\widehat{q}}_1$ such that $p_1⊩{\widehat{q}}_1\le {\widehat{q}}_0$ where $⟨p_1,{\widehat{q}}_1⟩\in D$. So $D_Q$ is indeed dense in $\widehat{Q}[G]$.

Since $D_Q$ is dense in $\widehat{Q}[G]$, there is some $\widehat{q}$ such that $\widehat{q}[G]\in D_Q\cap H$, and so some $p\in G$ such that $⟨p,\widehat{q}⟩\in D$. But since $p\in G$ and $\widehat{q}\in H$, $⟨p,\widehat{q}⟩\in G*H$, so $G*H$ is indeed generic.

$G_P$ is a generic filter

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

$$G_P=\{p\in P\mid p^{\prime }\le p\wedge ⟨p^{\prime },\widehat{q}⟩\in G\}\text{ for some }{p}^{\prime }\in P\text{ and some }\widehat{q}\in Q$$

$G_P$ is closed

Take any $p_1\in G_P$ and any $p_2$ such that $p_1\le p_2$. Then there is some $p^{\prime }\le p_1$ satisfying the definition of $G_P$, and also $p^{\prime }\le p_2$, so $p_2\in G_P$.

$G_P$ is directed

Consider $p_1,p_2\in G_P$. Then there is some $p_1^{\prime }$ and some ${\widehat{q}}_1$ such that $⟨p_1^{\prime },{\widehat{q}}_1⟩\in G$ and some $p_2^{\prime }$ and some ${\widehat{q}}_2$ such that $⟨p_2^{\prime },{\widehat{q}}_2⟩\in G$. Since $G$ is directed, there is some $⟨p_3,{\widehat{q}}_3⟩\in G$ such that $⟨p_3,{\widehat{q}}_3⟩\le ⟨p_1^{\prime },{\widehat{q}}_1⟩,⟨p_2^{\prime },{\widehat{q}}_2⟩$, and therefore $p_3\in G_P$, $p_3\le p_1,p_2$.

$G_P$ is generic

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

$G_Q$ is a generic filter

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

$$G_Q=\{\widehat{q}[G_P]\mid ⟨p,\widehat{q}⟩\in G\}\text{ for some }p\in P$$

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

$G_Q$ is closed

Suppose ${\widehat{q}}_1[G_P]\in G_Q$ and ${\widehat{q}}_1[G_P]\le {\widehat{q}}_2[G_P]$. Then there is some $p_1\in G_P$ such that $p_1⊩{\widehat{q}}_1\le {\widehat{q}}_2$. Since $p_1\in G_P$, there is some $p_2\le p_1$ such that for some ${\widehat{q}}_3$, $⟨p_2,{\widehat{q}}_3⟩\in G$. By the definition of $G_Q$, there is some $p_3$ such that $⟨p_3,{\widehat{q}}_1⟩\in G$, and since $G$ is directed, there is some $⟨p_4,{\widehat{q}}_4⟩\in G$ and $⟨p_4,{\widehat{q}}_4⟩\le ⟨p_3,{\widehat{q}}_1⟩,⟨p_2,{\widehat{q}}_3⟩$. Since $G$ is closed and $⟨p_4,{\widehat{q}}_4⟩\le ⟨p_4,{\widehat{q}}_2⟩$, we have ${\widehat{q}}_2[G_P]\in G_Q$.

$G_Q$ is directed

Suppose ${\widehat{q}}_1[G_P],{\widehat{q}}_2[G_P]\in G_Q$. Then for some $p_1,p_2$, $⟨p_1,{\widehat{q}}_1⟩,⟨p_2,{\widehat{q}}_2⟩\in G$, and since $G$ is directed, there is some $⟨p_3,{\widehat{q}}_3⟩\in G$ such that $⟨p_3,{\widehat{q}}_3⟩\le ⟨p_1,{\widehat{q}}_1⟩,⟨p_2,{\widehat{q}}_2⟩$. Then ${\widehat{q}}_3[G_P]\in G_Q$ and since $p_3\in G$ and $p_3⊩{\widehat{q}}_3\le {\widehat{q}}_1,{\widehat{q}}_2$, we have ${\widehat{q}}_3[G_P]\le {\widehat{q}}_1[G_P],{\widehat{q}}_2[G_P]$.

$G_Q$ is generic

Let $D$ be a dense subset of $Q[G_P]$ (in $𝔐[G_P]$). Let $\widehat{D}$ be a $P$-name for $D$, and let $p_1\in G_P$ be a such that $p_1⊩\widehat{D}$ is dense. By the definition of $G_P$, there is some $p_2\le p_1$ such that $⟨p_2,{\widehat{q}}_2⟩\in G$ for some $q_2$. Then $D^{\prime }=\{⟨p,\widehat{q}⟩\mid p⊩\widehat{q}\in D\wedge p\le p_2\}$.

Lemma: $D^{\prime }$ is dense (in $G$) above $⟨p_2,{\widehat{q}}_2⟩$

Take any $⟨p,\widehat{q}⟩\in P*Q$ such that $⟨p,\widehat{q}⟩\le ⟨p_2,{\widehat{q}}_2⟩$. Then $p⊩\widehat{D}$ is dense, and therefore there is some ${\widehat{q}}_3$ such that $p⊩{\widehat{q}}_3\in \widehat{D}$ and $p⊩{\widehat{q}}_3\le \widehat{q}$. So $⟨p,{\widehat{q}}_3⟩\le ⟨p,\widehat{q}⟩$ and $⟨p,{\widehat{q}}_3⟩\in D^{\prime }$.

Take any $⟨p_3,{\widehat{q}}_3⟩\in D^{\prime }\cap G$. Then $p_3\in G_P$, so ${\widehat{q}}_3\in D$, and by the definition of $G_Q$, ${\widehat{q}}_3\in G_Q$.

$G_P*G_Q=G$

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

$$G_P*G_Q=\{⟨p,\widehat{q}⟩\mid p^{\prime }\le p\wedge ⟨p^{\prime },{\widehat{q}}^{\prime }⟩\in G\wedge ⟨p_0,\widehat{q}⟩\in G\}\text{ for some }{p}^{\prime },{\widehat{q}}^{\prime },p_0$$

If $⟨p,\widehat{q}⟩\in G$ then obviously this holds, so $G\subseteq G_P*G_Q$. Conversely, if $⟨p,\widehat{q}⟩\in G_P*G_Q$ then there exist $p^{\prime },{\widehat{q}}^{\prime }$ and $p_0$ such that $⟨p^{\prime },{\widehat{q}}^{\prime }⟩,⟨p_0,\widehat{q}⟩\in G$, and since $G$ is directed, some $⟨p_1,\widehat{q_1}⟩\in G$ such that $⟨p_1,{\widehat{q}}_1⟩\le ⟨p^{\prime },{\widehat{q}}^{\prime }⟩,⟨p_0,\widehat{q}⟩$. But then $p_1\le p$ and $p_1⊩{\widehat{q}}_1\le \widehat{q}$, and since $G$ is closed, $⟨p,\widehat{q}⟩\in G$.

${(G*H)}_P=G$

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 $\widehat{q}\in Q$ such that $p^{\prime }\le p$ and $⟨p^{\prime },\widehat{q}⟩\in G*H$. 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), $⟨p,\widehat{q}⟩\in G*H$ for some $\widehat{q}$, 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 $\widehat{q}\in Q$ such that $\widehat{q}[G]=q$, and so there is some $p$ such that $⟨p,\widehat{q}⟩\in G*H$, and therefore $\widehat{q}[G]\in H$.

On the other hand, if $q\in {(G*H)}_Q$ then there is some $⟨p,\widehat{q}⟩\in G*H$, and therefore some $\widehat{q}[G]\in H$.

Title	proof that forcing notions are equivalent to their composition
Canonical name	ProofThatForcingNotionsAreEquivalentToTheirComposition
Date of creation	2013-03-22 12:54:26
Last modified on	2013-03-22 12:54:26
Owner	Henry (455)
Last modified by	Henry (455)
Numerical id	5
Author	Henry (455)
Entry type	Proof
Classification	msc 03E40
Classification	msc 03E35
Related topic	EquivalenceOfForcingNotions