Separating atoms

Some of the first properties we learn about forcing notions are the notions of being atomless and being separative. Usually any kind of analysis of these terms are left out, as “all forcings we care about are atomless and separative”, so this post will be dedicated to taking a slightly closer look at these properties.

Let’s start off by remembering what we’re dealing with.

Definition. A forcing notion $\mathbb P$ is…

• atomless if every condition has two incompatible extensions.
• separative if for every $p,q\in\mathbb P$, $p\leq q$ implies that there’s an $r\leq p$ such that $r\perp q$.
• pruned if every condition has a proper extension.

Proposition. A forcing $\mathbb P$ is atomless iff $\mathbb P$ has no generic filters in $V$.

Proof. $(\Rightarrow)$: If $\mathbb P$ is atomless and $g\subseteq\mathbb P$ is a generic filter, then define $D:=\mathbb P-g$. Let $p\in\mathbb P$ be arbitrary. By atomlessness we get two incompatible extensions $q,r\leq p$, so that at least one of them has to be an element of $D$, making $D$ dense. But then $D\cap g=\emptyset$, a contradiction.

$(\Leftarrow)$: Assume $p\in\mathbb P$ is an atom, meaning that every two extensions of $p$ are compatible. Let $g$ be the set of all conditions comparable with $p$, which is then a filter since $p$ is an atom. It’s also generic since every dense set will contain a condition below $p$QED

This also shows that atomlessness is a forcing invariant: if two forcings are forcing equivalent then either they’re both atomless or neither is. Why we care about separativeness is a bit more elusive. Firstly we do have the following:

Theorem. A forcing $\mathbb P$ is separative iff $\mathbb P$ embeds densely into a complete boolean algebra $\mathbb B$.

Proof. $(\Leftarrow)$ is easy to check, so we’ll show $(\Rightarrow)$. Let a subset $U\subseteq\mathbb P$ be a cut if it’s downwards closed with respect to $\mathbb P$‘s ordering. To every $p\in\mathbb P$ let $U_p:=\{q\in\mathbb P\mid q\leq p\}$. Say that a cut $U$ is regular if $\forall p\in\mathbb P-U\exists q\leq p: U_q\cap U=\emptyset$. The topological analogue here is that cuts are open sets and regular cuts are clopen sets.

Now note that separativity of $\mathbb P$ implies that every $U_p$ is regular. Again, the analogue here is that when we’re dealing with trees then all the basic open sets are clopen. Now simply let $\mathbb B$ be the set of all regular cuts of $\mathbb P$ and note that to every cut $U$ there’s a least regular cut containing it; namely,

$\overline U:=\{p\in\mathbb P\mid\forall q\leq p: U\cap U_q\neq\emptyset\}$.

We can therefore let $U\cdot V:=U\cap V$, $U+V:=\overline{U\cup V}$ and $-U:=\{p\in\mathbb P\mid U_p\cap U=\emptyset\}$. Since any intersection of regular cuts is regular, $\mathbb B$ is complete. We can now define $i:\mathbb P\to\mathbb B$ as $i(p):=U_p$, which is clearly a dense embedding. QED

The only thing separativity is giving us in the above theorem is injectivity of the embedding however, as any forcing is strongly forcing equivalent to a separative one:

Theorem. To every forcing $\mathbb P$ there exists a separative forcing $\mathbb Q$, called the separative quotient, and a dense order-preserving map $\pi:\mathbb P\to\mathbb Q$ such that $p\perp q\Leftrightarrow\pi(p)\perp\pi(q)$ for all $p,q\in\mathbb P$.

Proof. Surprisingly, $\mathbb Q$ will be a quotient of $\mathbb P$. We define $p\sim q$ iff they are compatible with exactly the same things in $\mathbb P$. This can be shown to be an equivalence relation, so let $\mathbb Q:=\mathbb P/\sim$, which can be shown to be a separative forcing, and the quotient map $\pi:\mathbb P\to\mathbb Q$ satisfies the wanted. QED

This also shows that being separative is not a forcing invariant, in contrast with atomlessness.

As for the relationship between being atomless and being separative, note first of all that every pruned separative forcing is atomless. But aside from that we don’t really get any implications between the two:

Proposition. There exists a separative forcing which isn’t atomless, and an atomless forcing which isn’t separative.

Proof. For the first bit, let $\mathbb P$ to be the following forcing, which is easily seen to be separative and not atomless:

For the second one, let $\mathbb Q$ be the following forcing, which is the union of a full binary tree with an extra element $p\in\mathbb Q$ with the following relations:

It’s clearly atomless and note that $p\not\leq q$ but any condition below $p$ will be compatible with $q$, making it non-separative. QED

So, I suppose we can think of the atomless forcings as being the ‘non-trivial ones’, and the separative forcings as the ‘well-behaved ones’. In most cases we’d therefore want both, and we note that being separative and atomless is the same thing as being separative and pruned. If we take the boolean algebraic approach to forcing then restricting ourselves to separative pruned forcings would correspond to restricting ourselves to atomless complete boolean algebras, which seems quite reasonable.