The Unapologetic Mathematician

Mathematics for the interested outsider

Stone’s Representation Theorem I

Today we start in on the representation theorem proved by Marshall Stone: every boolean ring \mathcal{B} is isomorphic (as a ring) to a ring of subsets of some set X. That is, no matter what B looks like, we can find some space X and a ring \mathcal{S} of subsets of X so that \mathcal{B}\cong\mathcal{S} as rings.

We start by defining the “Stone space” S(\mathcal{B}) of a Boolean ring \mathcal{B}. This is a representable functor, and the representing object is the two-point Boolean ring \mathcal{B}_0. That is, S(\mathcal{B})=\hom_\mathbf{Rng}(\mathcal{B},\mathcal{B}_0), the set of (boolean) ring homomorphisms from \mathcal{B} to \mathcal{B}_0. To be clear, \mathcal{B}_0 consists of the two points \{0,1\}, with the operations \Delta for addition and \cap for multiplication, and the obvious definitions of these operations. This is a contravariant functor — if we have a homomorphism of Boolean rings f:\mathcal{B}\to\hat{\mathcal{B}} we get a function between the Stone spaces S(f):S(\hat{\mathcal{B}})\to S(\mathcal{B}), which takes a function \lambda:\hat{\mathcal{B}}\to\mathcal{B}_0 to the function \lambda\circ f:\mathcal{B}\to\mathcal{B}_0.

The Stone space isn’t just a set, though; it’s a topological space. We define the topology by giving a base of open sets. That is, we’ll give a collection of sets — closed under intersections — which we declare to be open, and we define the collection of all open sets to be given by unions of these sets. For each element b\in\mathcal{B}, we define a basic set like so:

\displaystyle s(b)=\left\{\lambda\in S(\mathcal{B})\vert\lambda(b)=1\right\}

To see that this collection of sets is closed under intersection, consider two such sets s(b) and s(b'). I say that the intersection of these sets is the set s(b\cap b'). Indeed, if \lambda\in s(b) and \lambda\in s(b'), then

\displaystyle\lambda(b\cap b')=\lambda(b)\cap\lambda(b')=1\cap1=1

Conversely, if \lambda\in s(b\cap b'), then b\cap b'\subseteq b. Thus

\displaystyle1=\lambda(b\cap b')\subseteq\lambda(b)

and so \lambda(b)=1, and \lambda\in s(b). Similarly, \lambda\in s(b'). Thus we see that s(b)\cap s(b')=s(b\cap b').

In fact, this map from \mathcal{B} to the basic sets is exactly the mapping we’re looking for! We’ve already seen that our base is closed under intersection, and that the map s preserves intersections. I say that we also have s(b\Delta b')=s(b)\Delta s(b'). If \lambda\in s(b) but \lambda\notin s(b'), then \lambda(b)=1 and \lambda(b')=0. Then

\displaystyle\lambda(b\Delta b')=\lambda(b)\Delta\lambda(b')=1\Delta0=1

and similarly if \lambda\in s(b') but \lambda\notin s(b). Thus s(b)\Delta s(b')\subseteq s(b\Delta b'). Conversely, if \lambda\in s(b\Delta b'), then

\displaystyle\lambda(b)\Delta\lambda(b')=\lambda(b\Delta b')=1

and so either \lambda(b)=1 and \lambda(b')=0 or vice versa. Thus s(b\Delta b')=s(b)\Delta s(b').

So we know that s is a homomorphism of (boolean) rings. However, we don’t know yet that it’s an isomorphism. Indeed, it’s possible that s has a nontrivial kernel — s(b) could be \emptyset\subseteq S(\mathcal{B}) for some b. We must show that given any b there is some \lambda:\mathcal{B}\to\mathcal{B}_0 so that \lambda(b)=1.

For a finite boolean ring \mathcal{B} this is easy: we pick some minimal element b'\subseteq b and define \lambda(x)=1 if and only if b'\subseteq x. Such a b' exists because there’s at least one element below bb itself is one — and there can only be finitely many so we can just take their intersection. Clearly \lambda(b)=1 by definition, and it’s straightforward to verify that \lambda is a homomorphism of boolean rings using the fact that b' is an atom of \mathcal{B}.

For an infinite boolean ring, things are trickier. We define the set X^* of all functions \mathcal{B}\to\mathcal{B}_0, not just the ring homomorphisms. This is the product of one copy of \mathcal{B}_0 for every element of \mathcal{B}. Since each copy of \mathcal{B}_0 is a compact Hausdorff space, Tychonoff’s theorem tells us that X^* is a compact Hausdorff space. If \tilde{\mathcal{B}} is any finite subring of \mathcal{B} containing b, let X^*(\tilde{\mathcal{B}}) be the collection of those functions \lambda^*\in X^* which are homomorphisms when restricted to \tilde{\mathcal{B}} and for which \lambda^*(b)=1.

I say that the class of sets of the form X^*(\tilde{\mathcal{B}}) has the finite intersection property. That is, if we have some finite collection of finite subrings \tilde{\mathcal{B}}_1,\dots,\tilde{\mathcal{B}}_n and the finite subring \tilde{\mathcal{B}} they generate, then we have the relation

\displaystyle X^*(\tilde{\mathcal{B}})\subseteq\bigcap\limits_{i=1}^nX^*(\tilde{\mathcal{B}}_i)

Indeed, b is clearly contained in the generated ring \tilde{\mathcal{B}}. Further, if \lambda^* is a homomorphism on \tilde{\mathcal{B}} then it’s a homomorphism on each subring \tilde{\mathcal{B}}_i.

Okay, so since \tilde{\mathcal{B}} is a finite boolean ring, the proof given above for the finite case shows that X^*(\tilde{\mathcal{B}}) is nonempty. Thus the intersection of any finite collection of sets \{X^*(\tilde{\mathcal{B}}_i)\} is nonempty. And thus, since X^* is compact, the intersection of all of the \{X^*(\tilde{\mathcal{B}})\} is nonempty.

That is, there is some function \lambda^*:\mathcal{B}\to\mathcal{B}_0 which is a homomorphism of boolean rings on any finite boolean subring containing b, and with \lambda^*(b)=1. Given any other two points b_1 and b_2 there is some finite boolean subring containing b, b_1, and b_2, and so we must have \lambda^*(b_1\cap b_2)=\lambda^*(b_1)\cap\lambda^*(b_2) and \lambda^*(b_1\Delta b_2)=\lambda^*(b_1)\Delta\lambda^*(b_2) within this subring, and thus within the whole ring. Thus \lambda^* is a homomorphism of boolean rings sending b to 1, which shows that s(b)\neq\emptyset.

Therefore, the map s is a homomorphism sending the boolean ring \mathcal{B} isomorphically onto the identified base of the Stone space S(\mathcal{B}).

August 18, 2010 Posted by | Analysis, Measure Theory | 6 Comments