User:Dfeuer/Stone's Representation Theorem for Boolean Algebras

Theorem
Let $B$ be a Boolean algebra.

Let $S \left({B}\right)$ be the Stone space of $B$.

Then $B$ is isomorphic to the algebra of clopen subsets of $S \left({B}\right)$, where the isomorphism, $E$, sends an element $b \in B$ to the set of all ultrafilters that contain $b$.

Proof
For each $b \in B$, let $E\left({b}\right)$ be the set of ultrafilters in $B$ containing $b$.

Lemma
If $b \in B$, then $E \left({b}\right)$ is clopen in $S \left({B}\right)$.

Proof
$E \left({b}\right)$ and $E \left({\neg b}\right)$ are open by the definition of the Stone Space.

Then since an ultrafilter on a Boolean algebra contains either an element or its complement but not both:


 * $E \left({b}\right) \cup E \left({\neg b}\right) = S \left({B}\right)$

and
 * $E \left({b}\right) \cap E \left({\neg b}\right) = \varnothing$

Lemma
$E$ is injective.

Proof
Let $a$ and $b$ be elements of $B$.

If $c=(a \wedge \neg b)≠\bot$ or $(\neg a \wedge b) ≠ \bot$ then the principle ultrafilter based at one of these which is not $\bot$ contains either $a$ or $b$ but not both.

Otherwise, $a \wedge \neg b = \bot$ and $\neg a \wedge b = \bot$

Negating the latter shows that $a \vee \neg b = \top$.

Since we also have $a \wedge \neg b = \bot$ and the negation of an element is unique,

$a = \neg\neg b$, so $a =b$.

Thus if $a ≠ b$, there is an ultrafilter containing one but not the other, so $E(a) ≠ E(b)$.