# 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 $F \in X$:

 $\ds F$ $\in$ $\ds \map \phi {\neg a}$ $\ds \leadstoandfrom \ \$ $\ds \neg a$ $\in$ $\ds F$ $\ds \leadstoandfrom \ \$ $\ds a$ $\notin$ $\ds F$ $\ds \leadstoandfrom \ \$ $\ds F$ $\in$ $\ds \relcomp X {\map \phi a}$ $\ds \leadstoandfrom \ \$ $\ds F$ $\in$ $\ds X \setminus \map \phi a$ $\ds \leadstoandfrom \ \$ $\ds F$ $\in$ $\ds \map \phi {\neg \map \phi a}$

Therefore:

$\map \phi {\neg a} = X \setminus \map \phi a$

Similarly, for all $F \in X$:

 $\ds F$ $\in$ $\ds \map \phi {a \wedge b}$ $\ds \leadstoandfrom \ \$ $\ds a \wedge b$ $\in$ $\ds F$ $\ds \leadstoandfrom \ \$ $\ds a, b$ $\in$ $\ds F$ $\ds \leadstoandfrom \ \$ $\ds F$ $\in$ $\ds \map \phi a \cap \map \phi b$ $\ds \leadstoandfrom \ \$ $\ds F$ $\in$ $\ds \map \phi {a \wedge b}$

Therefore:

$\map \phi {a \wedge b} = \map \phi a \cap \map \phi b$

Similarly, for all $F \in X$:

 $\ds F$ $\in$ $\ds \map \phi {a \vee b}$ $\ds \leadstoandfrom \ \$ $\ds a \vee b$ $\in$ $\ds F$ $\ds \leadstoandfrom \ \$ $\ds a$ $\in$ $\ds F$ $\, \ds \text {or} \,$ $\ds b$ $\in$ $\ds F$ $\ds \leadstoandfrom \ \$ $\ds F$ $\in$ $\ds \map \phi a \cup \map \phi b$ $\ds \leadstoandfrom \ \$ $\ds F$ $\in$ $\ds \map \phi {a \vee b}$

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 $F \in X$:

 $\ds F$ $\in$ $\ds \map \phi a$ $\ds \leadstoandfrom \ \$ $\ds a$ $\in$ $\ds F$ $\ds \leadsto \ \$ $\ds a \wedge \neg b$ $\notin$ $\ds F$ $\ds \leadstoandfrom \ \$ $\ds b$ $\in$ $\ds F$ $\ds \leadstoandfrom \ \$ $\ds F$ $\in$ $\ds \map \phi b$ $\ds \leadstoandfrom \ \$ $\ds \map \phi a$ $\subseteq$ $\ds \map \phi b$

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

$\blacksquare$

## Source of Name

This entry was named for Marshall Harvey Stone.