Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

(Throughout, fix a universe of set theory $V$ to serve as a base.) We can define a fairly natural preordering on forcing notions as follows. Let $\mathbb{P}_1$, $\mathbb{P}_2$ be two forcing notions. Then we have $\mathbb{P}_1\le\mathbb{P}_2$ iff for all $G$ $\mathbb{P}_2$-generic, there is some $\mathbb{P}_1$-generic $H$ such that $H\in V[G]$.

One question which occurred to me is the following: if $\mathbb{P}_1\le\mathbb{P}_2$, then is it the case that for every $\mathbb{P}_1$-generic $H$, there is some $\mathbb{P}_2$-generic $G$ such that $H\in V[G]$?

This is the wrong question to ask, since it has a trivial negative answer. Let $\mathbb{P}_2$ be, say, Sacks forcing, and let $\mathbb{P}_1$ consist of a copy of Sacks forcing and a copy of some countably closed forcing $\mathbb{Q}$ placed side by side, so that a generic for $\mathbb{P}_1$ is either Sacks generic or $\mathbb{Q}$ generic, but not both. Then clearly $\mathbb{P}_1\le\mathbb{P}_2$, but all the $\mathbb{P}_1$-generics which come from a $\mathbb{Q}$-generic need not live in any extension by Sacks forcing.

So my question has two parts. First, the following rephrasing of the wrong question:

Question 1: Suppose $\mathbb{P}_1\le\mathbb{P}_2$, and for all $p\in\mathbb{P}_1$, there exists a $\mathbb{P}_2$-generic $G$ and $\mathbb{P}_1$-generic $H$ with $p\in H\in V[G]$. Then is it the case that, for all $H$ $\mathbb{P}_1$-generic, there is some $\mathbb{P}_2$-generic $G$ such that $H\in V[G]$?

My second, more informal question is whether this is the "right" way to ask the question I'm trying to ask. I suppose an attempt to clarify this informal question would be:

Question 2: What sorts of assumptions on $\mathbb{P}_1$ and $\mathbb{P}_2$, besides $\mathbb{P}_1\le\mathbb{P}_2$, do we need to make to ensure that every $\mathbb{P}_1$-generic lives in the generic extension by some $\mathbb{P}_2$-generic?

share|cite|improve this question

You are looking for the concept of complete subalgebras (of complete Boolean algebras), and the associated concept of a complete embedding of one forcing notion into another, with their associated quotient forcing. The situation is somewhat nicer to explain in terms of Boolean algebras than in terms purely of posets, but every poset is forcing equivalent to its Boolean completion via the regular open algebra. (This material is in Jech's book.)

The basic situation for complete Boolean algebras is that $\mathbb{B}_1\leq \mathbb{B}_2$ just in case there is a condition $p\in\mathbb{B}_1$ such that the restriction of the forcing $\mathbb{B}_1\upharpoonright p$ below $p$ has a complete embedding into $\mathbb{B}_2$. This means that the forcing $\mathbb{B}_2$ is isomorphic to first forcing with $\mathbb{B}_1\upharpoonright p$, and then forcing with the quotient forcing of $\mathbb{B}_2$ over that subalgebra.

From this, it follows that two forcing notions $\mathbb{P}_1$ and $\mathbb{P}_2$ can have a forcing extension $V[G]=V[H]$ in common if and only if there are conditions $p\in G\subset\mathbb{P}_1$ and $q\in H\subset\mathbb{P}_2$, below which the forcing notions are isomorphic: $\text{RO}(\mathbb{P}_1\upharpoonright p)\cong\text{RO}(\mathbb{P}_2\upharpoonright q)$.

share|cite|improve this answer
Hi Joel! Off topic: I just saw the add for Cantor's attic. Very nice! Congratulations to Victoria and you. You should probably post an announcement on FOM. – Andrés E. Caicedo Jan 10 '12 at 2:54
@Andres: Thanks for the OT comment; I’d probably not have run across Cantor’s Attic without it. – Brian M. Scott Jan 10 '12 at 2:59
Thanks, Andres. We've just started with Cantor's attic---I hope you'll be able to make a contribution; we'd value it highly. And also Brian and other knowledgeable contributors. – JDH Jan 10 '12 at 11:47

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.