representing a Boolean algebra by field of sets

In this entry, we show that every Boolean algebra is isomorphic to a field of sets, originally noted by Stone in 1936. The bulk of the proof has actually been carried out in this entry (http://planetmath.org/RepresentingADistributiveLatticeByRingOfSets), which we briefly state:

if $L$ is a distributive lattice, and $X$ the set of all prime ideals of $L$, then the map $F:L\to P(X)$ defined by $F(a)=\{P\mid a\notin P\}$ is an embedding.

Now, if $L$ is a Boolean lattice, then every element $a\in L$ has a complement $a^{\prime }\in L$. $a^{\prime }$ is in fact uniquely determined by $a$.

Proposition 1.

The embedding $F$ above preserves ${}^{\mathrm{\prime }}$ in the following sense:

$$F(a^{\prime })=X-F(a).$$

Proof.

$P\in F(a^{\prime })$ iff $a^{\prime }\notin P$ iff $a\in P$ iff $P\notin F(a)$ iff $P\in X-F(a)$. ∎

Theorem 1.

Every Boolean algebra is isomorphic to a field of sets.

Proof.

From what has been discussed so far, $F$ is a Boolean algebra isomorphism between $L$ and $F(L)$, which is a ring of sets first of all, and a field of sets, because $X-F(a)=F(a^{\prime })$. ∎

Remark. There are at least two other ways to characterize a Boolean algebra as a field of sets: let $L$ be a Boolean algebra:

• Every prime ideal is the kernel of a homomorphism into $\mathrm{𝟐}:=\{0,1\}$, and vice versa. So for an element $a$ to be not in a prime ideal $P$ is the same as saying that $\varphi (a)=1$ for some homomorphism $\varphi :L\to \mathrm{𝟐}$. If we take $Y$ to be the set of all homomorphisms from $L$ to $\mathrm{𝟐}$, and define $G:L\to P(Y)$ by $G(a)=\{\varphi \mid \varphi (a)=1\}$, then it is easy to see that $G$ is an embedding of $L$ into $P(Y)$.

• Every prime ideal is a maximal ideal, and vice versa. Furthermore, $P$ is maximal iff $P^{\prime }$ is an ultrafilter. So if we define $Z$ to be the set of all ultrafilters of $L$, and set $H:L\to P(Z)$ by $H(a)=\{U\mid a\in U\}$, then it is easy to see that $H$ is an embedding of $L$ into $P(Z)$.

If we appropriately topologize the sets $X,Y$, or $Z$, then we have the content of the Stone representation theorem.

Title	representing a Boolean algebra by field of sets
Canonical name	RepresentingABooleanAlgebraByFieldOfSets
Date of creation	2013-03-22 19:08:27
Last modified on	2013-03-22 19:08:27
Owner	CWoo (3771)
Last modified by	CWoo (3771)
Numerical id	13
Author	CWoo (3771)
Entry type	Theorem
Classification	msc 06E20
Classification	msc 06E05
Classification	msc 03G05
Classification	msc 06B20
Classification	msc 03G10
Related topic	FieldOfSets
Related topic	RepresentingADistributiveLatticeByRingOfSets
Related topic	LatticeHomomorphism
Related topic	RepresentingACompleteAtomicBooleanAlgebraByPowerSet
Related topic	StoneRepresentationTheorem
Related topic	MHStonesRepresentationTheorem