# forcings are equivalent if one is dense in the other

Suppose $P$ and $Q$ are forcing  notions and that $f:P\rightarrow Q$ is a function such that:

• $p_{1}\leq_{P}p_{2}$ implies $f(p_{1})\leq_{Q}f(p_{2})$

• If $p_{1},p_{2}\in P$ are incomparable then $f(p_{1}),f(p_{2})$ are incomparable

• $f[P]$ is dense (http://planetmath.org/DenseInAPoset) in $Q$

## Proof

### $F(G)=H$ where $H$ is generic

Given a generic $G\subseteq P$, consider $H=\{q\mid f(p)\leq q\}$ for some $p\in G$.

If $q_{1}\in H$ and $q_{1}\leq q_{2}$ then $q_{2}\in H$ by the definition of $H$. If $q_{1},q_{2}\in H$ then let $p_{1},p_{2}\in P$ be such that $f(p_{1})\leq q_{1}$ and $f(p_{2})\leq q_{2}$. Then there is some $p_{3}\leq p_{1},p_{2}$ such that $p_{3}\in G$, and since $f$ is order preseving $f(p_{3})\leq f(p_{1})\leq q_{1}$ and $f(p_{3})\leq f(p_{2})\leq q_{2}$.

Suppose $D$ is a dense subset of $Q$. Since $f[P]$ is dense in $Q$, for any $d\in D$ there is some $p\in P$ such that $f(p)\leq d$. For each $d\in D$, assign (using the axiom of choice  ) some $d_{p}\in P$ such that $f(d_{p})\leq d$, and call the set of these $D_{P}$. This is dense in $P$, since for any $p\in P$ there is some $d\in D$ such that $d\leq f(p)$, and so some $d_{p}\in D_{P}$ such that $f(d_{p})\leq d$. If $d_{p}\leq p$ then $D_{P}$ is dense, so suppose $d_{p}\nleq p$. If $d_{p}\leq p$ then this provides a member of $D_{P}$ less than $p$; alternatively, since $f(d_{p})$ and $f(p)$ are compatible, $d_{p}$ and $p$ are compatible, so $p\leq d_{p}$, and therefore $f(p)=f(d_{p})=d$, so $p\in D_{P}$. Since $D_{P}$ is dense in $P$, there is some element $p\in D_{P}\cap G$. Since $p\in D_{P}$, there is some $d\in D$ such that $f(p)\leq d$. But since $p\in G$, $d\in H$, so $H$ intersects $D$.

### $G$ can be recovered from $F(G)$

Given $H$ constructed as above, we can recover $G$ as the set of $p\in P$ such that $f(p)\in H$. Obviously every element from $G$ is included in the new set, so consider some $p$ such that $f(p)\in H$. By definition, there is some $p_{1}\in G$ such that $f(p_{1})\leq f(p)$. Take some dense $D\in Q$ such that there is no $d\in D$ such that $f(p)\leq d$ (this can be done easily be taking any dense subset and removing all such elements; the resulting set is still dense since there is some $d_{1}$ such that $d_{1}\leq f(p)\leq d$). This set intersects $f[G]$ in some $q$, so there is some $p_{2}\in G$ such that $f(p_{2})\leq q$, and since $G$ is directed, some $p_{3}\in G$ such that $p_{3}\leq p_{2},p_{1}$. So $f(p_{3})\leq f(p_{1})\leq f(p)$. If $p_{3}\nleq p$ then we would have $p\leq p_{3}$ and then $f(p)\leq f(p_{3})\leq q$, contradicting the definition of $D$, so $p_{3}\leq p$ and $p\in G$ since $G$ is directed.

### $F^{-1}(H)=G$ where $G$ is generic

Given any generic $H$ in $Q$, we define a corresponding $G$ as above: $G=\{p\in P\mid f(p)\in H\}$. If $p_{1}\in G$ and $p_{1}\leq p_{2}$ then $f(p_{1})\in H$ and $f(p_{1})\leq f(p_{2})$, so $p_{2}\in G$ since $H$ is directed. If $p_{1},p_{2}\in G$ then $f(p_{1}),f(p_{2})\in H$ and there is some $q\in H$ such that $q\leq f(p_{1}),f(p_{2})$.

Consider $D$, the set of elements of $Q$ which are $f(p)$ for some $p\in P$ and either $f(p)\leq q$ or there is no element greater than both $f(p)$ and $q$. This is dense, since given any $q_{1}\in Q$, if $q_{1}\leq q$ then (since $f[P]$ is dense) there is some $p$ such that $f(p)\leq q_{1}\leq q$. If $q\leq q_{1}$ then there is some $p$ such that $f(p)\leq q\leq q_{1}$. If neither of these and $q$ there is some $r\leq q_{1},q$ then any $p$ such that $f(p)\leq r$ suffices, and if there is no such $r$ then any $p$ such that $f(p)\leq q$ suffices.

There is some $f(p)\in D\cap H$, and so $p\in G$. Since $H$ is directed, there is some $r\leq f(p),q$, so $f(p)\leq q\leq f(p_{1}),f(p_{2})$. If it is not the case that $f(p)\leq f(p_{1})$ then $f(p)=f(p_{1})=f(p_{2})$. In either case, we confirm that $H$ is directed.

Finally, let $D$ be a dense subset of $P$. $f[D]$ is dense in $Q$, since given any $q\in Q$, there is some $p\in P$ such that $p\leq q$, and some $d\in D$ such that $d\leq p\leq q$. So there is some $f(p)\in f[D]\cap H$, and so $p\in D\cap G$.

### $H$ can be recovered from $F^{-1}(H)$

Finally, given $G$ constructed by this method, $H=\{q\mid f(p)\leq q\}$ for some $p\in G$. To see this, if there is some $f(p)$ for $p\in G$ such that $f(p)\leq q$ then $f(p)\in H$ so $q\in H$. On the other hand, if $q\in H$ then the set of $f(p)$ such that either $f(p)\leq q$ or there is no $r\in Q$ such that $r\leq q,f(p)$ is dense (as shown above), and so intersects $H$. But since $H$ is directed, it must be that there is some $f(p)\in H$ such that $f(p)\leq q$, and therefore $p\in G$.

Title forcings are equivalent if one is dense in the other ForcingsAreEquivalentIfOneIsDenseInTheOther 2013-03-22 12:54:43 2013-03-22 12:54:43 Henry (455) Henry (455) 6 Henry (455) Result msc 03E35 msc 03E40