# The Stone representation theorem

November 16, 2023

The Stone representation theorem asserts a perhaps surprising connection between two interesting but for me initially unfamiliar objects. On the one hand we have (for me countable) Boolean algebras which have flavors of logic and algebra, and on the other we have compact Hausdorff topological spaces which are totally disconnected – the primary example is the Cantor set. (Here countability is countability of a basis of clopen—closed and open—sets.) I want to describe an idiosyncratic point of view on these objects coming from the study of surfaces.

Anyway, here’s the statement of the theorem:

Theorem (Stone) Let $$B$$ be a Boolean algebra and $$S$$ its associated Stone space. Then $$B$$ is isomorphic to the algebra of clopen subsets of $$S$$.

There’s really some fun category-theoretic play at work here, but today I’m not interested in digging into that too deeply. Instead, let’s define what we’re talking about.

## 1. Boolean algebras

A Boolean algebra is a set $$B$$ equipped with two binary operations, $$\wedge$$ and $$\vee$$ called “meet” and “join” respectively. Both of them are associative and commutative (so $$a \wedge b = b \wedge a$$ and no matter how I parenthesize the expression $$a \vee b \vee c$$, I get the same result). Additionally there are two special elements $$0$$ and $$1$$ in $$B$$ such that $$0$$ is an identity for $$\vee$$—i.e. $$a \vee 0 = a$$ for all $$a$$—and $$1$$ is an identity for $$\wedge$$. Furthermore, we have that $$a \vee (a \wedge b) = a$$ and $$a \wedge (a \vee b) = a$$ for all $$a$$ and $$b$$, and that $$\vee$$ and $$\wedge$$ distribute over each other (e.g. $$a \vee (b \wedge c) = (a \vee b) \wedge (a \vee c)$$). Finally, each element $$a \in B$$ has a complement $$\lnot a$$ with the property that $$a \vee \lnot a = 1$$ and $$a \wedge \lnot a = 0$$.

Blegh, that’s a lot of algebra. Here’s a quick example: let $$X$$ be your favorite set and $$B$$ the collection of all subsets of $$X$$. Then $$B$$ is a Boolean algebra, where $$\vee = \cup$$ is the operation of taking unions, $$\wedge = \cap$$ is the operation of taking intersections, $$0 = \varnothing$$ is the empty subset, $$1 = X$$ is the whole set, and $$\lnot$$ is the operation of taking set-theoretic complements. It would be a great exercise to verify all of the axioms (there are five that I’ve listed) if they aren’t clear to you.

### 1.1. Homomorphisms

I’ve described for you a new kind of object, and a natural question for a mathematician to ask when faced with such a construction is “what are the maps?” They should preserve the structure in some way, so it would be reasonable to expect that if $$A$$ and $$B$$ are Boolean algebras, a homomorphism $$f\colon A \to B$$ would preserve meets and joins in the sense that, e.g. $$f(a \vee b) = f(a) \vee f(b)$$, and also preserve the special elements $$0$$ and $$1$$. It also would be reasonable to assume that $$f(\lnot a) = \lnot f(a)$$, but this turns out to be a consequence of the other asusmptions and the axioms.

The category (oh boy, here I go) of Boolean algebras and homomorphisms has an initial and a terminal object. The Boolean algebra with a single element $$\mathsf{1} = \{ 0 = 1\}$$ is terminal; there is exactly one function (of sets) from any Boolean algebra to $$\mathsf{1}$$, and it is clearly a homomorphism. The initial object is $$\mathsf{2} = \{ 0, 1\}$$, the Boolean algebra with only two special elements and nothing else.

### 1.2. Maps to $$\mathsf{2}$$

Now if $$B$$ is a Boolean algebra with at least two elements, then there are no maps from $$\mathsf{1}$$, but there may be interesting maps to $$\mathsf{2}$$. Indeed, suppose $$b \ne 0$$ is an element of $$B$$. We want to show that there exists a homomorphism $$f\colon B \to \mathsf{2}$$ which sends $$b$$ to $$1$$. Say that an element $$a$$ satisfies $$b \le a$$ if $$b \wedge a = b$$. If $$b \le a$$, then note that our desired homomorphism would have to satisfy $$f(b) \wedge f(a) = f(b) = 1$$. On the other hand, since $$1$$ is the identity for $$\wedge$$, we conclude that $$f(a) = 1$$. Neat!

Indeed, more generally if $$f(b) = f(c) = 1$$, then $$f(b \wedge c) = f(b) \wedge f(c) = 1$$

On the other hand, if there exists $$c$$ and $$d$$ such that $$c \vee d = b$$, note that $$f(c) \vee f(d) = 1$$, so since $$0$$ is the identity for $$\vee$$, we note that at least one of $$f(c)$$ and $$f(d)$$ must be $$1$$.

Now, of course, for every element $$b$$ of $$B$$, we must either have $$f(b) = 0$$ or $$f(b) = 1$$, and in fact either $$f(b) = 1$$ or $$f(\lnot b) = 1$$ in which case $$f(b) = 0$$. Therefore, to complete our hoped-for $$f$$ to an actual homomorphism, we must be able to make choices about the value of $$f(c)$$ for every $$c \le b$$ in some consistent way. It turns out that one can do this—assuming the Axiom of Choice.

## 2. The associated Stone space

So, we’ve determined that there are interesting homomorphisms to $$\mathsf{2}$$. It turns out the collection of all of them assemble into a topological space. Namely, let $$U_b = \{ f\colon B \to \mathsf{2} : f(b) = 1\}$$, and declare this subset open. Notice that its complement is $$U_{\lnot b}$$, which is also open, so these open sets are closed. More generally, we let a set $$U$$ be open when it contains some $$U_b$$.

We call $$S = \{ f\colon B \to \mathsf{2} : f \text{ is a homomorphism}\}$$ the Stone space associated to $$B$$.

### 2.1. Stone spaces are compact, Hausdorff and totally disconnected

It is clear that if $$f$$ and $$g$$ are different points of $$S = S(B)$$, then they disagree on some $$b \in B$$, and so one is contained in $$U_b$$, while the other is in $$U_{\lnot b}$$; these are disjoint open sets containing $$f$$ and $$g$$, proving that $$S(B)$$ is Hausdorff.

Having a basis of clopen sets implies that $$S(B)$$ is totally disconnected. Compactness is again the Axiom of Choice lurking in the background.

## 3. The point of the proof

is that if you start with $$B$$, then consider the space $$S(B)$$, it has an associated Boolean algebra, namely the algebra of clopen subsets. If one sends $$b$$ to $$U_b$$, one gets an isomorphism of Boolean algebras from $$B$$ to the Boolean algebra of clopen subsets of $$S(B)$$, proving the theorem.

## 4. Surfaces!

Essentially because of the Stone representation theorem, we have that if $$E$$ is a closed subset of the Cantor set, then $$E$$ is the Stone space of some countable Boolean algebra $$B(E)$$, and moreover the group of homeomorphisms of $$E$$ and the group of isomorphisms of $$B(E)$$ are naturally isomorphic. This is a neat consequence, since it says that $$\operatorname{Homeo}(E)$$ is a “non-Archimedean Polish group”—if that means nothing to you, that’s fine, just think “automorphism group of some countable structure, e.g. a countable graph.”

More generally, if $$F$$ is a closed subset of $$E$$, we can consider the group $$\operatorname{Homeo}(E,F)$$ of homeomorphisms of $$E$$ leaving the subset $$F$$ invariant. Then $$F$$ is a Stone space in its own right, and we obtain a map $$B(E) \to B(F)$$ by the rule that a clopen subset of $$E$$ yields by restriction a clopen subset of $$F$$. Homeomorphisms that preserve $$f$$ then yield automorphisms of $$B(E)$$ which, under this restriction, induce automorphisms of $$B(F)$$.

Anyway, the Classification of Infinite-Type Surfaces says that a surface $$\Sigma$$ of infinite type is determined up to homeomorphism by the following data:

• The genus of $$\Sigma$$, which may be finite or infinite.
• The end space of $$\Sigma$$, which is a closed subset of the Cantor set, and
• The subspace of ends accumulated by genus, which is a further closed subset, and which is nonempty if and only if the genus of $$\Sigma$$ is infinite.

Something I’m currently mumbling to myself about is the connection between separating simple closed curves on $$\Sigma$$ and elements of $$B(E)$$, where $$E$$ is the end space of $$\Sigma$$…