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# 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). 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

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

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

###### Proposition 1.

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

###### Proof.

For any $a\in X$, $a\leq 1=x\vee x^{{\prime}}$, so that $a\leq x$ or $a\leq x^{{\prime}}$, 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 $a\leq x$ and $a\leq x^{{\prime}}$, so that $a\leq x\wedge x^{{\prime}}=0$, 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, $f(x\vee y)=\{a\mid a\leq x\vee y\}=\{a\mid a\leq x\mbox{ or }a\leq y\}$ since $a$ is an atom. But the right hand side equals $\{a\mid a\leq x\}\cup\{a\mid a\leq y\}=f(x)\cup f(y)$, we see that $f$ preserves $\vee$.

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

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

###### Proposition 3.

$f$ is an injection.

###### Proof.

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

###### Proposition 4.

$f$ is conditionally complete, in the sense that if $\bigvee A$ is defined for any $A\subseteq B$, 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 $a\leq y$, or $a\leq x$ 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 $a\leq x$ for some $x\in A$. This means that $a\leq x\leq\bigvee A=y$, 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 $2$.

###### 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.

## Mathematics Subject Classification

06E10*no label found*

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