# The Unapologetic Mathematician

## Stone’s Representation Theorem II

We can extend yesterday’s result in the case that $\mathcal{B}$ is a Boolean algebra. Now as a ring, $\mathcal{B}$ has a unit. We adjust our definition of the Stone space to define $\displaystyle S(\mathcal{B})=\hom_\mathbf{Ring}(\mathcal{B},\mathcal{B}_0)$

That is, we insist that the ring homomorphisms preserve the identity, sending the top element of $\mathcal{B}$ to $1\in\mathcal{B}_0$. This doesn’t really change anything we said yesterday, and it all goes through as before.

What is new is that the image of $s$ — the identified base for the topology on $S(\mathcal{B})$ — consists of all the subsets of $S(\mathcal{B})$ which are clopen — both open and closed. That is, elements of $\mathcal{B}$ correspond to unions of connected components of $S(\mathcal{B})$.

First, we must show that $s(b)$ is closed, since we already know that it’s open by definition. I say that the complement $s(b)^c$ is actually the open basic set $s(b^c)$. Indeed, $b^c=b\Delta1$, and we calculate \displaystyle\begin{aligned}s\left(b^c\right)&=s(b\Delta1)\\&=s(b)\Delta s(1)\\&=s(b)\Delta1\\&=s(b)^c\end{aligned}

Thus each set $s(b)$ is the complement of an open set, and is thus closed as well.

It also happens that our base is closed under finite unions. Indeed, we use DeMorgan’s laws and calculate \displaystyle\begin{aligned}s(b)\cup s(b')&=\left(s(b)^c\cap s(b')^c\right)^c\\&=\left(s\left(b^c\right)\cap s\left(b'^c\right)\right)^c\\&=s\left(b^c\cap b'^c\right)^c\\&=s\left(\left(b^c\cap b'^c\right)^c\right)\\&=s(b\cup b')\end{aligned}

And from there we can extend to any finite unions we want.

Now I say that if a base of clopen sets in a compact space is closed under finite unions, then it contains every clopen set in the space. Indeed, such a clopen set can be written as a union of sets in the base since it’s open. This union gives an open covering of the set. Since the set is closed, it is compact. And so the open covering we just found has a finite subcover. That is, we can write our clopen set as a finite union of basic sets, and so it is itself in the base by assumption.

Thus in our particular case, our base of $S(\mathcal{B})$ consists of all the clopen sets in the Stone space, as we asserted!

August 19, 2010 Posted by | Analysis, Measure Theory | 2 Comments