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.
This proof will assume the Ultrafilter Lemma holds.
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 $B$
:$\map \phi {a \wedge b} = \map \phi a \cap \map \phi b$ and $\map \phi {a \vee b} = \map \phi a \cup \map \phi b$
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:
:$\map \phi a = {F \in X : a \in F}$
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 $\FF_a$ 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 $\FF_a$ is nonempty.
Furthermore, if $\FF$ is a proper filter containing $a$, then $\FF$ is not maximal.
So there exists a filter $\GG$ containing $\FF$ and not containing $a$.
But then $\GG$ contains $\neg a$.
So $\GG$ is not in $\FF_a$.
Therefore, $\FF_a$ is a proper filter containing $a$.
Hence $\FF_a$ is an ultrafilter.
Next, we show that $\phi$ is injective. 
Let $a, b \in B$ such that:
:$\map \phi a = \map \phi b$.
Then for all $F \in X$, we have:
:$a \in F \iff b \in F$
In particular, if $F$ is an ultrafilter containing $a$, then $F$ contains $b$, so $a \le b$.
Similarly, if $F$ is an ultrafilter containing $b$, then $F$ contains $a$, so $b \le a$. 
Therefore:
:$a = b$
and hence $\phi$ is injective.
To show that $\phi$ is surjective, let:
:$A \subseteq X$
Define:
:$\ds a = \bigwedge_{F \mathop \in A} F$
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:
:$\map \phi a = A$
To see this, let $F \in X$.
Then $F$ is an ultrafilter, so:
:$a \in F \lor \neg a \in F$
If $a \in F$, then $F$ contains every element of $A$.
So:
:$F \in \map \phi a$
If:
:$\neg a \in F$
then for each $G \in A$, we have:
:$G \nsubseteq F$
so:
:$F \in \map \phi {\neg G}$
But then:
:$F \in \map \phi {\bigvee_{G \in A} \neg G} = \map \phi {\neg a}$
so:
:$F \notin \map \phi a$
Therefore:
:$\map \phi a = A$
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 $F \in X$:
Therefore:
:$\map \phi {\neg a} = X \setminus \map \phi a$
Similarly, for all $F \in X$:
Therefore:
:$\map \phi {a \wedge b} = \map \phi a \cap \map \phi b$
Similarly, for all $F \in X$:
Therefore:
:$\map \phi {a \vee b} = \map \phi a \cup \map \phi b$
Finally, to show that $\phi$ preserves the order relation, let $a, b \in B$.
Then:
:$a \le b$
:
:$a \wedge \neg b = \bot$
Therefore, for all $F \in X$:
Therefore, $\phi$ preserves the order relation. This completes the proof.
== Sources ==
Category:Boolean Algebras
Category:Stone Spaces