Abstract Algebra: Theory and Applications

If $b$ is an atom, let $a=b\text{.}$ Otherwise, choose an element $b_1\text{,}$ not equal to $O$ or $b\text{,}$ such that $b_1⪯b\text{.}$ We are guaranteed that this is possible since $b$ is not an atom. If $b_1$ is an atom, then we are done. If not, choose $b_2\text{,}$ not equal to $O$ or $b_1\text{,}$ such that $b_2⪯b_1\text{.}$ Again, if $b_2$ is an atom, let $a=b_2\text{.}$ Continuing this process, we can obtain a chain

Since $B$ is a finite Boolean algebra, this chain must be finite. That is, for some $k\text{,}$ $b_k$ is an atom. Let $a=b_k\text{.}$

Since $a\wedge b$ is the greatest lower bound of $a$ and $b\text{,}$ we know that $a\wedge b⪯a\text{.}$ Hence, either $a\wedge b=a$ or $a\wedge b=O\text{.}$ However, if $a\wedge b=a\text{,}$ then either $a⪯b$ or $a=O\text{.}$ In either case we have a contradiction because $a$ and $b$ are both atoms; therefore, $a\wedge b=O\text{.}$

(1) $\Rightarrow$ (2). If $a⪯b\text{,}$ then $a\vee b=b\text{.}$ Therefore,

(2) $\Rightarrow$ (3). If $a\wedge b^{\prime }=O\text{,}$ then $a^{\prime }\vee b=(a\wedge b^{\prime }{)}^{\prime }=O^{\prime }=I\text{.}$

(3) $\Rightarrow$ (1). If $a^{\prime }\vee b=I\text{,}$ then

Thus, $a⪯b\text{.}$

By Lemma 19.20, $b\wedge c^{\prime }\ne O\text{.}$ Hence, there exists an atom $a$ such that $a⪯b\wedge c^{\prime }\text{.}$ Consequently, $a⪯b$ and $a⪯̸c\text{.}$

Let $b_1=a_1\vee \cdots \vee a_n\text{.}$ Since $a_i⪯b$ for each $i\text{,}$ we know that $b_1⪯b\text{.}$ If we can show that $b⪯b_1\text{,}$ then the lemma is true by antisymmetry. Assume $b⪯̸b_1\text{.}$ Then there exists an atom $a$ such that $a⪯b$ and $a⪯̸b_1\text{.}$ Since $a$ is an atom and $a⪯b\text{,}$ we can deduce that $a=a_i$ for some $a_i\text{.}$ However, this is impossible since $a⪯b_1\text{.}$ Therefore, $b⪯b_1\text{.}$

Now suppose that $b=a_1\vee \cdots \vee a_n\text{.}$ If $a$ is an atom less than $b\text{,}$

But each term is $O$ or $a$ with $a\wedge a_i$ occurring for only one $a_i\text{.}$ Hence, by Lemma 19.19, $a=a_i$ for some $i\text{.}$

We will show that $B$ is isomorphic to $\mathcal{P}(X)\text{,}$ where $X$ is the set of atoms of $B\text{.}$ Let $a\in B\text{.}$ By Lemma 19.22, we can write $a$ uniquely as $a=a_1\vee \cdots \vee a_n$ for $a_1,\dots ,a_n\in X\text{.}$ Consequently, we can define a map $\varphi :B\to \mathcal{P}(X)$ by

Clearly, $\varphi$ is onto.

Now let $a=a_1\vee \cdots \vee a_n$ and $b=b_1\vee \cdots \vee b_m$ be elements in $B\text{,}$ where each $a_i$ and each $b_i$ is an atom. If $\varphi (a)=\varphi (b)\text{,}$ then $\{a_1,\dots ,a_n\}=\{b_1,\dots ,b_m\}$ and $a=b\text{.}$ Consequently, $\varphi$ is injective.

The join of $a$ and $b$ is preserved by $\varphi$ since

Similarly, $\varphi (a\wedge b)=\varphi (a)\cap \varphi (b)\text{.}$