representing a complete atomic Boolean algebra by power set

It is a known fact that every Boolean algebraMathworldPlanetmath is isomorphicPlanetmathPlanetmathPlanetmath to a field of sets (of some set) (proof here ( In this entry, we show that, furthermore, if a Boolean algebra is atomic and completePlanetmathPlanetmathPlanetmathPlanetmath, 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 operationsMathworldPlanetmath of union, intersectionDlmfMathworldPlanetmath, and complementPlanetmathPlanetmath.

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

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


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

Proposition 1.

f(x) and f(x) are complement of one another in X.


For any aX, a1=xx, so that ax or ax, or af(x) or af(x). This shows that f(x)f(x)=X. If af(x)f(x), then ax and ax, so that axx=0, which is impossible, since a is an atom, and by definition, must be greater than 0. ∎

Proposition 2.

First, f(x)=X-f(x) by the last propositionPlanetmathPlanetmath.

Next, f(xy)={aaxy}={aax or ay} since a is an atom. But the right hand side equals {aax}{aay}=f(x)f(y), we see that f preserves .

Finally, f(0)={aa0}= since any atom must be greater than 0.

Hence, f is a Boolean algebra homomorphism. ∎

Proposition 3.

f is an injectionMathworldPlanetmath.


Suppose f(x)=. If x0, then there must be some atom a such that ax. But this implies that f(x), a contradictionMathworldPlanetmathPlanetmath. Hence x=0 and f is injective. ∎

Proposition 4.

f is conditionally complete, in the sense that if A is defined for any AB, then


Suppose y=A and Y=f(y). Let Z={f(x)xA}. We want to show that Y=Z. If aY, then ay, or ax for some xA, since a is an atom. So af(x)Z. Conversely, if aZ, then af(x), or ax for some xA. This means that axA=y, and therefore af(y)=Y. ∎

Proposition 5.

If B is complete, so is f. Moreover, f is surjectivePlanetmathPlanetmath.


The first sentenceMathworldPlanetmath is a direct consequence of the previous proposition. For the second setnence, let YP(X). Let y=Y. x exists because B is complete. So f(y)=f(Y)={f(x)xY}={{x}xY}=Y, since each xY 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 2.


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