# The Unapologetic Mathematician

## 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 $b$$b$ 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