I’d like to talk about some results pertaining to distributive lattices. In particular, there’s this one interesting theorem about Boolean algebras I’ve been thinking about lately. One direction is reasonably famous, pretty useful and not very hard to prove, so I’ll cover that. But what I really wanna talk about is the converse direction, which is a result that almost nobody I know has ever heard of, and is impossible to find anything about on the internet.

However, since lattice theory isn’t really all that popular among undergrads, there’s a lot of ground to cover, and I think I’ll have to spend a post or so winding up to it, so as to break it all up into digestible chunks. Some familiarity with the theory of lattices would be nice, but because I would otherwise be stuck in a limbo of “too much content for one post, not enough for two”, I’m going to mention everything I’ll use. So this should be more or less self-contained, though I might go a little fast.

A lattice is a set with a partial order $\le$ and well-defined greatest lower bound and least upper bound operations $\wedge$ and $\vee$, called meet and join, respectively. To wit, $x \le a \wedge b$ iff $x \le a$ and $x \le b$, and likewise $a \vee b \le y$ is iff $a \le y$ and $b \le y$. If there exist a global lower bound, we denote it by $\bot$ and say the lattice is bounded below. Similarly, the global upper bound is denoted by $\top$ if it exists and the lattice is bounded above. The lattice is bounded if it is bounded in both directions.

Here are a few handy elementary facts about lattices. If the operations $\wedge$ and $\vee$ are well-defined, they are unique. Also, meets and joins of any finite set are well-defined, by induction. Finally, note that $a \le b$ iff $a = a \wedge b$, and that $a \wedge (a \vee b) = a$.

A lattice is distributive if $\wedge$ distributes over $\vee$. Not every lattice is distributive, but I will only care about distributive lattices in what follows.

If $(L,{\le},{\wedge},{\vee})$ is a lattice, then $L^* = (L,{\ge},{\vee},{\wedge})$ is called the dual lattice of $L$. It is precisely the lattice $L$, flipped upside-down. It is a standard but somewhat unpleasant result of lattices that if $L$ is distributive, it is dual distributive, i.e. $\vee$ distributes over $\wedge$.

Proposition. If $L$ is distributive, so is $L^*$.

Proof. This is just a big disgusting computation. Let $a,b,c \in L$. First note that $a \le a \vee b$ and $b \wedge c \le b \le a \vee b$, so that $a \vee (b \wedge c) \le a \vee b$. By the symmetry in $b$ and $c$,

This is called weak dual distributivity. Now observe that if $x \le z$, then

by distributivity. This is called the modular law. We now compute

where $(*)$ is an application of the modular law to weak dual distributivity. ∎

Henceforth, let $L$ be a distributive bounded lattice.

$L$ is a Boolean algebra if every element is complemented, i.e. for every $x \in L$ there is a $y \in L$ such that $x \wedge y = \bot$ and $x \vee y = \top$. The familiar lattice $2^X$ of subsets of some set $X$, ordered by inclusion, is a Boolean algebra where complementation is given by the set complement $A \mapsto X \smallsetminus A$. I’ll call this the powerset algebra of $X$.

In a sense, powerset algebras are the prototypical Boolean algebras. It is fairly easy to see that every finite Boolean algebra is a powerset algebra.

Sure, not every Boolean algebra is a powerset algebra—infinite powerset algebras must have uncountably many elements, but the sublattice of $2^{\mathbb N}$ containing only finite and cofinite sets is a countable Boolean algebra—but thanks to Birkhoff we know every Boolean algebra must be a sublattice of a powerset algebra. So there’s no harm in thinking of them as collections of sets with the usual set-lattice operations.

We will also need to know about filters. A filter is a subset $F \subseteq L$ which is upward-closed and downward-directed. To be explicit: if $a \in F$ and $a \le a'$, then $a' \in F$; and if $a,b \in F$, then $a \wedge b \in F$. Filters can be defined more generally for any poset, but this way of stating it will be convenient for our purposes. A filter is nonempty iff it contains $\top$, and nonfull iff it does not contain $\bot$.

A nonfull filter $F$ is prime if whenever $a \vee b \in F$, either $a \in F$ or $b \in F$. A filter is an ultrafilter if it is maximal among nonfull filters, i.e. any filter properly containing it is full.

Proposition. Every ultrafilter is prime.

Proof. Let $F$ be an ultrafilter, and suppose for a contradiction that it is not prime. Then there are $a,b \notin F$ such that $a \vee b \in F$. Let $F' = F \cup \{ y : y \ge a \wedge x\ \text{for some}\ x \in F \}$. This is clearly1 a filter, and since $a = a \wedge (a \vee b) \in F'$, we have that $F \subsetneq F'$.

Since $F$ was maximal among nonfull filters, $F'$ is full, so in particular $b \in F'$. But $b \notin F$, so $b \ge a \wedge x$ for some $x \in F$. Then by distributivity,

Since $a \vee b$ and $x$ are both members of $F$, it follows that $b \in F$, which is a contradiction. ∎

Note that we did not use the boundedness of $L$.

Theorem 1. In a Boolean algebra, every nonempty prime filter is an ultrafilter.

Proof. Let $F$ be a nonempty prime filter and let $F' \supsetneq F$. Then there exists $x \in F' \smallsetminus F$. $x$ has a complement $\neg x$, and $x \vee \neg x = \top \in F$. Thus, $\neg x \in F$. But then both $x, \neg x \in F'$, so $\bot = x \wedge \neg x \in F'$. Every filter strictly containing $F$ is full, so $F$ is maximal among nonfull filters. ∎

This is the ostensibly famous theorem I mentioned. Granted, not everybody cares about filters, but definitely functional analysts do. And I’m sure there are still some ring and model theorists who study ultraproducts.

The converse to Theorem 1 that I’m thinking of is the following:

Theorem 2. Let $L$ be a lattice. If every nonempty prime filter in $L$ is an ultrafilter, then $L$ is a Boolean algebra.

This is substantially harder to show than Theorem 1, and I will need to introduce a bit of technology first. Well, it’s more than a bit, in all honesty, so I think I’ll save that for a second post.

I only know two people that have seen a proof of this result: myself and the logic professor who assigned it to me as homework. Hopefully this ‘blog post and its sequel will correct that.

This is the first post in a series. Here is the next.

1. My original proof here was not as clear as I thought, because I gave a bad definition of $F'$. Thanks to Niall Mitchell for catching that mistake.