# dual space of a Boolean algebra

Let $B$ be a Boolean algebra, and $B^{*}$ the set of all maximal ideals of $B$. In this entry, we will equip $B^{*}$ with a topology so it is a Boolean space.

Definition. For any $a\in B$, define $M(a):=\{M\in B^{*}\mid a\notin M\}$, and $\mathcal{B}:=\{M(a)\mid a\in B\}$.

It is know that in a Boolean algebra, maximal ideals and prime ideals coincide. From this entry (http://planetmath.org/RepresentingABooleanLatticeByFieldOfSets), we have the three following properties concerning $M(a)$:

 $M(a)\cap M(b)=M(a\wedge b),\qquad M(a)\cup M(b)=M(a\vee b),\qquad B^{*}-M(a)=M% (a^{\prime}).$

Furthermore, if $M(a)=M(b)$, then $a=b$.

From these properties, we see that $M(0)=\varnothing$ and $M(1)=B^{*}$. As a result, we see that

###### Proposition 1.

$B^{*}$ is a topological space, whose topology $\mathcal{T}$ is generated by the basis $\mathcal{B}$.

###### Proof.

$\varnothing$ and $B^{*}$ are both open, as they are $M(0)$ and $M(1)$ respectively. Also, the intersection of open sets $M(a)$ and $M(b)$ is again open, since it is $M(a\wedge b)$. ∎

We may in fact treat $\mathcal{B}$ as a subbasis for $\mathcal{T}$, since finite intersections of elements of $\mathcal{B}$ remain in $\mathcal{B}$.

###### Proposition 2.

Each member of $\mathcal{B}$ is closed, hence $\mathcal{T}$ is generated by a basis of clopen sets. In other words, $B^{*}$ is zero-dimensional.

###### Proof.

Each $M(a)$ is open, by definition, and closed, since it is the complement of the open set $M(a^{\prime})$. ∎

###### Proposition 3.

$B^{*}$ is Hausdorff.

###### Proof.

If $M,N\in B^{*}$ such that $M\neq N$, then there is some $a\in B$ such that $a\in M$ and $a\notin N$. This means that $N\in M(a)$ and $M\notin M(a)$, which means that $M\in B^{*}-M(a)=M(a^{\prime})$. Since $M(a)$ and $M(a^{\prime})$ are open and disjoint, with $N\in M(a)$ and $M\in M(a^{\prime})$, we see that $B^{*}$ is Hausdorff. ∎

Now, based on a topological fact, every zero-dimensional Hausdorff space is totally disconnected. Hence $B^{*}$ is totally disconnected.

###### Proposition 4.

$B^{*}$ is compact.

###### Proof.

Suppose $\{U_{i}\mid i\in I\}$ is a collection of open sets whose union is $B^{*}$. Since each $U_{i}$ is a union of elements of $\mathcal{B}$, we might as well assume that $B^{*}$ is covered by elements of $\mathcal{B}$. In other words, we may assume that each $U_{i}$ is some $M(a_{i})\in\mathcal{B}$.

Let $J$ be the ideal generated by the set $\{a_{i}\mid i\in I\}$. If $J\neq B$, then $J$ can be extended to a maximal ideal $M$. Since each $a_{i}\in M$, we see that $M\notin M(a_{i})$, so that $M\notin\bigcup\{M(a_{i})\mid i\in I\}=B^{*}$, which is a contradiction. Therefore, $J=B$. In particular, $1\in J$, which means that $1$ can be expressed as the join of a finite number of the $a_{i}$’s:

 $1=\bigvee\{a_{i}\mid i\in K\},$

where $K$ is a finite subset of $J$. As a result, we have

 $\bigcup\{M(a_{i})\mid i\in K\}=M(\bigvee\{a_{i}\mid i\in K\})=M(1)=B^{*}.$

So $B^{*}$ has a finite subcover, and hence is compact. ∎

Collecting the last three results, we see that $B^{*}$ is a Boolean space.

Remark. It can be shown that $B$ is isomorphic to the Boolean algebra of clopen sets in $B^{*}$. This is the famous Stone representation theorem.

 Title dual space of a Boolean algebra Canonical name DualSpaceOfABooleanAlgebra Date of creation 2013-03-22 19:08:35 Last modified on 2013-03-22 19:08:35 Owner CWoo (3771) Last modified by CWoo (3771) Numerical id 6 Author CWoo (3771) Entry type Definition Classification msc 06E05 Classification msc 03G05 Classification msc 06B20 Classification msc 03G10 Classification msc 06E20 Related topic StoneRepresentationTheorem Related topic MHStonesRepresentationTheorem