Stone's Representation Theorem for Boolean Algebras

Theorem
Let $B$ be a Boolean algebra.

Let $S$ be the Stone space of $B$.

Then:
 * The set of clopen sets in $S$ is a Boolean algebra under union, intersection, and complementation in $S$.
 * That Boolean algebra is isomorphic to $B$.

Proof
First, the statement of the proof has to be shown to be equivalent to the form:

Let $B$ be a Boolean algebra.

Then there exists a set $X$ and a bijection $\phi: B \to \powerset X$, where $\powerset X$ denotes the power set of $X$, such that for all $a, b \in B$:
 * $\map \phi {\neg a} = X \setminus \map \phi a$, where $\neg a$ denotes the complement of $a$ in $B$
 * $\phi(a \wedge b) = \phi(a) \cap \phi(b)$ and $\phi(a \vee b) = \phi(a) \cup \phi(b)$

Furthermore, the ordering on $B$ induced by inclusion is isomorphic to the partial ordering on $\powerset X$ induced by inclusion.

...because the above is nothing like the statement of the problem as put together by the original author of this page.

Let $B$ be a Boolean algebra. Define $X$ to be the set of all ultrafilters of $B$, that is, the maximal filters containing no proper filter. We claim that $\phi : B \to \powerset X$ defined by $\phi(a) = { F \in X : a \in F }$ is the desired bijection.

First, we show that $\phi$ is a well-defined function. Let $a \in B$. Since $B$ is a Boolean algebra, it is a distributive lattice, so the set of all filters containing $a$ is closed under finite intersections and unions. Let $\mathcal{F}_a$ denote the set of all filters containing $a$. Since $B$ is a Boolean algebra, it has a least element $\bot$, which is contained in every filter, so $\mathcal{F}_a$ is nonempty. Furthermore, if $\mathcal{F}$ is a proper filter containing $a$, then $\mathcal{F}$ is not maximal, so there exists a filter $\mathcal{G}$ containing $\mathcal{F}$ and not containing $a$. But then $\mathcal{G}$ contains $\neg a$, so $\mathcal{G}$ is not in $\mathcal{F}_a$. Therefore, $\mathcal{F}_a$ is a proper filter containing $a$, and hence $\mathcal{F}_a$ is an ultrafilter.

Next, we show that $\phi$ is injective. Let $a, b \in B$ such that $\phi(a) = \phi(b)$. Then for all $F \in X$, we have $a \in F$ if and only if $b \in F$. In particular, if $F$ is an ultrafilter containing $a$, then $F$ contains $b$, so $a \leq b$. Similarly, if $F$ is an ultrafilter containing $b$, then $F$ contains $a$, so $b \leq a$. Therefore, $a = b$, and hence $\phi$ is injective.

To show that $\phi$ is surjective, let $A \subseteq X$. Define $a = \bigwedge_{F \in A} F$, where the intersection is taken in $B$. Since $B$ is a Boolean algebra, it is complete, so the infimum of any subset of $B$ exists. We claim that $\phi(a) = A$. To see this, let $F \in X$. Then $F$ is an ultrafilter, so either $a \in F$ or $\neg a \in F$. If $a \in F$, then $F$ contains every element of $A$, so $F \in \phi(a)$. If $\neg a \in F$, then for each $G \in A$, we have $G \nsubseteq F$, so $F \in \phi(\neg G)$. But then $F \in \phi(\bigvee_{G \in A} \neg G) = \phi(\neg a)$, so $F \notin \phi(a)$. Therefore, $\phi(a) = A$, and hence $\phi$ is surjective.

Finally, we show that $\phi$ preserves the Boolean algebra operations and the order relation. Let $a, b \in B$. Then for all $F \in X$:

\begin{align*} F \in \phi(\neg a) &\iff \neg a \in F \ &\iff a \notin F \ &\iff F \in \phi(a)^c \ &\iff F \in X \setminus \phi(a) \ &\iff F \in \phi(\neg \phi(a)). \end{align*}

Therefore, $\phi(\neg a) = X \setminus \phi(a)$. Similarly, for all $F \in X$:

\begin{align*} F \in \phi(a \wedge b) &\iff a \wedge b \in F \ &\iff a, b \in F \ &\iff F \in \phi(a) \cap \phi(b) \ &\iff F \in \phi(a \wedge b). \end{align*}

Therefore, $\phi(a \wedge b) = \phi(a) \cap \phi(b)$. Similarly, for all $F \in X$:

\begin{align*} F \in \phi(a \vee b) &\iff a \vee b \in F \ &\iff a \in F \text{ or } b \in F \ &\iff F \in \phi(a) \cup \phi(b) \ &\iff F \in \phi(a \vee b). \end{align*}

Therefore, $\phi(a \vee b) = \phi(a) \cup \phi(b)$. Finally, to show that $\phi$ preserves the order relation, let $a, b \in B$. Then $a \leq b$ if and only if $a \wedge \neg b = \bot$. Therefore, for all $F \in X$:

\begin{align*} F \in \phi(a) &\iff a \in F \ &\implies a \wedge \neg b \notin F \ &\iff b \in F \ &\iff F \in \phi(b) \ &\iff \phi(a) \subseteq \phi(b). \end{align*}

Therefore, $\phi$ preserves the order relation. This completes the proof.