representing a complete atomic Boolean algebra by power set

It is a known fact that every Boolean algebra is isomorphic to a field of sets (of some set) (proof here (http://planetmath.org/RepresentingABooleanLatticeByFieldOfSets)). In this entry, we show that, furthermore, if a Boolean algebra is atomic and complete, then it is isomorphic to the field of sets of some set, in other words, the powerset of some set, viewed as a Boolean algebra via the usual set-theoretic operations of union, intersection, and complement.

The proof is based on the following function, defined for any atomic Boolean algebra:

Definition. Let $B$ be an atomic Boolean algebra, and $X$ the set of its atoms. Define $f:B\to P(X)$ by

$$f(x):=\{a\mid a\le x\}.$$

It is easy to see that $f(x)=\{x\}$ iff $x$ is an atom of $B$.

Proposition 1.

$f(x)$ and $f\mathit{}\mathrm{(}{x}^{\mathrm{\prime }}\mathrm{)}$ are complement of one another in $X$.

Proof.

For any $a\in X$, $a\le 1=x\vee x^{\prime }$, so that $a\le x$ or $a\le x^{\prime }$, or $a\in f(x)$ or $a\in f(x^{\prime })$. This shows that $f(x)\cup f(x^{\prime })=X$. If $a\in f(x)\cap f(x^{\prime })$, then $a\le x$ and $a\le x^{\prime }$, so that $a\le x\wedge x^{\prime }=0$, which is impossible, since $a$ is an atom, and by definition, must be greater than $0$. ∎

Proposition 2.

$f$ is a Boolean algebra homomorphism.

Proof.

First, $f(x^{\prime })=X-f(x)$ by the last proposition.

Next, $f(x\vee y)=\{a\mid a\le x\vee y\}=\{a\mid a\le x\text{ or }a\le y\}$ since $a$ is an atom. But the right hand side equals $\{a\mid a\le x\}\cup \{a\mid a\le y\}=f(x)\cup f(y)$, we see that $f$ preserves $\vee$.

Finally, $f(0)=\{a\mid a\le 0\}=\mathrm{\varnothing }$ since any atom must be greater than $0$.

Hence, $f$ is a Boolean algebra homomorphism. ∎

Proposition 3.

$f$ is an injection.

Proof.

Suppose $f(x)=\mathrm{\varnothing }$. If $x\ne 0$, then there must be some atom $a$ such that $a\le x$. But this implies that $f(x)\ne \mathrm{\varnothing }$, a contradiction. Hence $x=0$ and $f$ is injective. ∎

Proposition 4.

$f$ is conditionally complete, in the sense that if $\mathrm{\bigvee }A$ is defined for any $A\mathrm{\subseteq }B$, then

$$f(\bigvee A)=\bigcup \{f(x)\mid x\in A\}.$$

Proof.

Suppose $y=\bigvee A$ and $Y=f(y)$. Let $Z=\bigcup \{f(x)\mid x\in A\}$. We want to show that $Y=Z$. If $a\in Y$, then $a\le y$, or $a\le x$ for some $x\in A$, since $a$ is an atom. So $a\in f(x)\subseteq Z$. Conversely, if $a\in Z$, then $a\in f(x)$, or $a\le x$ for some $x\in A$. This means that $a\le x\le \bigvee A=y$, and therefore $a\in f(y)=Y$. ∎

Proposition 5.

If $B$ is complete, so is $f$. Moreover, $f$ is surjective.

Proof.

The first sentence is a direct consequence of the previous proposition. For the second setnence, let $Y\in P(X)$. Let $y=\bigvee Y$. $x$ exists because $B$ is complete. So $f(y)=f(\bigvee Y)=\bigcup \{f(x)\mid x\in Y\}=\bigcup \{\{x\}\mid x\in Y\}=Y$, since each $x\in Y$ is an atom. ∎

Rewording the above proposition, we have

Theorem 1.

Any complete atomic Boolean algebra is isomorphic (as complete Boolean algebras) to the powerset of some set, namely, the set of all of its atoms.

A useful application of this representation theorem is the following:

Corollary 1.

The cardinality of a finite Boolean algebra is a power of $\mathrm{2}$.

Proof.

Every finite Boolean algebra is complete and atomic, and hence isomorphic to the powerset of a set, which is also finite, and the result follows. ∎

Remark. The proof does not depend on the representation of a Boolean algebra by a field of sets.

Title	representing a complete atomic Boolean algebra by power set
Canonical name	RepresentingACompleteAtomicBooleanAlgebraByPowerSet
Date of creation	2013-03-22 19:08:30
Last modified on	2013-03-22 19:08:30
Owner	CWoo (3771)
Last modified by	CWoo (3771)
Numerical id	8
Author	CWoo (3771)
Entry type	Definition
Classification	msc 06E10
Related topic	RepresentingABooleanLatticeByFieldOfSets