# 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

1. http://www-groups.dcs.st-and.ac.uk/~history/Biographies/Stone.html
Marshall Stone’s mother was Agnes Harvey. His father, Harlan Fiske Stone, was a distinguished lawyer who served as dean of Columbia Law school from 1910 to 1923 and was on the supreme court for 21 years, serving as chief justice for the last five of these from 1941 to 1946. The family tradition would have had Marshall follow his father’s into a law career and while he attended public schools in Englewood, New Jersey, it was assumed that he would become a lawyer. He entered Harvard in 1919 still intending to continue his studies at Harvard law school after first taking his bachelor’s degree, but he was so enthused by the mathematics courses that he took that, by the time he graduated in 1922, he began to think that it was mathematics, not law, that would be his life. An inspired, but extraordinary, arrangement by the Harvard Mathematics Department saw him appointed as an instructor for session 1922-23 to see whether he would enjoy teaching mathematics and whether he would decide to take his mathematical studies further….

Comment by Jonathan Vos Post | August 18, 2010 | Reply

2. […] Representation Theorem II We can extend yesterday’s result in the case that is a Boolean algebra. Now as a ring, has a unit. We adjust our definition of the […]

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3. […] Representation Theorem III We conclude our coverage of Stone’s representation theorem with a version for Boolean -algebra. Each such algebra is isomorphic to a -algebra of subsets of […]

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4. […] The Stone space functor we’ve been working with sends Boolean algebras to topological spaces. Specifically, it sends […]

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5. kindly update stone representation theorem for boolean algebra

Comment by nasir | January 9, 2011 | Reply

6. kindly update proof of stone representation theorem for boolean algebra

Comment by nasir | January 9, 2011 | Reply