# 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 |