# 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 $M(a)$:

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

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

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

###### Proposition 1.

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

###### Proof.

$\mathrm{\varnothing}$ 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 $\mathrm{B}$ is closed, hence $\mathrm{T}$ is generated by a basis of clopen sets. In other words, ${B}^{\mathrm{*}}$ 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 $M\ne N$, 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 $J\ne B$, 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 |