# 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 ${}^{\prime}$ 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 $\boldsymbol{2}:=\{0,1\}$, and vice versa. So for an element $a$ to be not in a prime ideal $P$ is the same as saying that $\phi(a)=1$ for some homomorphism $\phi:L\to\boldsymbol{2}$. If we take $Y$ to be the set of all homomorphisms from $L$ to $\boldsymbol{2}$, and define $G:L\to P(Y)$ by $G(a)=\{\phi\mid\phi(a)=1\}$, 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