Don't Force With Complete Boolean Algebras!

Use complete Boolean orders instead. January 25th, 2024

There are two different ways to approach learning (or teaching for that matter) forcing. Both come with their own strengths and weaknesses.

First, there is the "hands-on" approach:The slogan here is to "jump directly into action". Work with a general partial order, define dense sets, generic filters, names and construct the forcing extension as the evaluation of all names in the ground model by the generic filter. This is certainly the shortest path towards the forcing theorem and applications of forcing, say the consistency of $\neg\mathrm{CH}$ with $\mathrm{ZFC}$.

The other possibility is to take the road through the Boolean forest. Define complete Boolean algebras, Boolean valued models, construct the Boolean valued model associated to a forcing and mod out by a generic filter. This introduces some significant overhead and additional abstraction. However one benefits from a significantly smoother theory. Just as a basic example, in the definition of the internal forcing relation, the statement $$\{q\leq p\mid \exists \dot x\in V^{\mathbb P}\ q\Vdash \varphi(\dot x)\}\text{ is dense below }p$$ is less readable to me than $$p\leq \sup_{\dot x\in V^{\mathbb B}}\Vert \varphi(\dot x)\Vert.$$ But much more importantly, one can proof theorems about the theory of forcings which are simply not true for mere partial orders.

Though the biggest problem using this apporach is that nobody knows what a complete Boolean algebra is! Obviously, this is not exactly true, but chances are that if you ask a random mathematician what a Boolean algebra is, they will at least have to ponder for a short bit before coming up with the three operations and the 8 laws that they have to obey.

Even worse, some of these laws do not have an immediate motivation coming from forcing. Even worseer, Boolean algebras are technically trivial forcings as they have a minimal element $0_{\mathbb B}$, so one has to carry around a little $\times$ in the pocket and write $\mathbb B^\times$ instead.

Indeed, one must infer the order on a Boolean algebra from the other operations first before forcing. This is totally backwards! We only really care about the order! On the other hand, the order carries the full information about the Boolean algebra: We can define $$b\wedge c = \inf \{b, c\},\ b\vee c = \sup \{b, c\} \text{ and } \neg b =\sup\{c\in\mathbb B\mid c\perp b\}.$$ This suggests strongly that it is possible to characterize the partial orders which are induced by complete Boolean algebras. I would be surprised if this has not been done before, but I could not find anything in the literature, please let me know of a source if you know about one!

In any case, let us start with separativity. A partial order is seperative if whenever $p\not\leq q$ then there is some $r\leq p$ with $r\perp q$. This property is easily motivated from a forcing perspective, as any partial order is forcing equivalent to the corresponding separative quotient. Passing to this quotient gets rid of "unnecessary fluff" that does not carry any relevant information.

It is well-known that any Boolean algebra is seperative, so this should be part of our characterization. Clearly that does not suffice though!

If a supremum is tight then basically "no information of $X$ about $\dot G$ is lost when passing to $\sup X$". Here is an instructive example of a non-tight supremum: Consider the forcing $$\mathbb P=\{p\colon n\rightarrow \omega\mid n<\omega\}$$ ordered by extension. This is of course one representation of Cohen forcing. For $i=7, 42$, let $p_i$ be the condition with domain $\{0\}$ which sends $0$ to $i$ and let $X=\{p_7, p_{42}\}$. Now $X$ has some information about the generic Cohen real $\dot c$, namely "$\dot c(0)$ is either $7$ or $42$". Notice that $X$ has a supremum in $\mathbb P$, namely $\sup X=\emptyset$. But now all information of $X$ has been lost in translation, this is terrible!

From a forcing perspective, suprema are only interesting if they are tight. The supremum of some $X\subseteq\mathbb P$ should be $1_{\mathbb P}$ only if $X$ is predense (or in other words, if the downwards closure of $X$ is dense).

Complete Boolean algebras admit tight suprema: Suppose $X\subseteq\mathbb B$ is non-empty, $X$ has a supremum $b$ in $\mathbb B$ because $\mathbb B$ is complete. If $b$ was not a tight supremum of $X$ then there is some $c\leq b$ incompatible with all $x\in X$. But then $b\wedge\neg c$ would be an upper bound of $X$ strictly below $b$, contradiction! So let us add this property to our list of interesting properties of partial orders induced by cBa's.

It turns out that this is already it!

Only these two properties are necessary to complete characterize cBa's and both of them are naturally motivated via forcing!

The trick now is that one does not ever have to define what a complete Boolean algebra is in the first place, but one still gets all the benefits. Let's give a basic and another more involved argument. The following is a folklore fact, which is part of the reason why one may restrict to complete Boolean algebras when forcing.

I have never seen a complete proof of this fact. It is quite easy to define $\mathbb B$: The idea is related to taking Dedekind-cutes. We can take $\mathbb B$ to consists of the regular open subsets of $\mathbb P$. Equivalently, $U\subseteq\mathbb P$ is in $\mathbb B$ iff $U$ is nonempty, downwards closed and for any $p\in\mathbb P$, either $p\in U$ or there is $q\leq p$ so that the cone $C_q=\{r\in\mathbb P\mid r\leq q\}$ is disjoint from $U$. The idea is that $\mathbb B$ can be ordered by inclusion and embed $\mathbb P$ into the nonempty elements of $\mathbb B$ via $p\mapsto C_p$.

However, the problem is that verifying that $\mathbb B$ admits a compatible Boolean algebra structure is a bit annoying and hence usually left to the reader. But why bother with this at all? In terms of complete Boolean orders, the statement above is:

This is much simpler! We take $\mathbb C=\mathbb B^\times$, ordered as before, and show that $\mathbb C$ is a complete Boolean order. Seperativity is straightforward: If $V\not\leq_{\mathbb C} U$ then there is a point $p\in V$ which is not in $U$. As $U$ is regular open, we can find $q\leq p$ with $C_q\cap U=\emptyset$ and as $V$ is open, $C_q\leq_{\mathbb C} V$.

Next, let us show that $\mathbb C$ admits tight suprema. Suppose $X\subseteq \mathbb C$ is nonempty. The naive idea is to take $U=\bigcup X$, but this might not be regular open. Instead, we take the regular open set generated by $\bigcup X$, that is the interior of $\overline{\bigcup X}$. This can be expressed differently as $$U=\{p\in\mathbb P\mid \bigcup X\text{ is predense below }p\}.$$ Clearly, $\bigcup X\subseteq_{\mathbb C} U$, so $U$ is an upper bound of $X$. To check that $U$ is a tight supremum, it suffices to prove that $X$ is predense below $U$. That's easy as well! If $V\leq_{\mathbb C} U$, take any $p\in V$. By definition of $U$, $p$ is compatible with some $r\in \bigcup X$. Find $W\in X$ with $r\in W$. Then $C_r\leq_{\mathbb C} W, V$, so we are done.

In a future post, I plan to show some use cases of complete Boolean algebras/orders. I will explore what exactly it means for two forcings to be forcing equivalent. Moreover, I will show that $\sigma$-strategically forcings are exactly those partial orders which are forcing equivalent to some $\sigma$-closed forcing, based on an argument of Jensen.