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\section{Boolean Algebra Definitions and Examples}

\begin{definition}
A \textbf{Boolean Algebra} is a tuple%
\[
\left\langle \delta ,\sqcap ,\sqcup ,^{\prime },\top ,\bot \right\rangle 
\]%
Where each element is defined as follows: 
\[
\begin{tabular}{ll}
$\delta $ & Non-empty set called the \textbf{domain} \\ 
$\sqcap $ & binary function $\sqcap :\delta \times \delta \rightarrow \delta 
$ \\ 
$\sqcup $ & binary function $\sqcup :\delta \times \delta \rightarrow \delta 
$ \\ 
$^{\prime }$ & unary function $^{\prime }:\delta \rightarrow \delta $ \\ 
$\bot $ & specific element called the \textbf{zero }element \\ 
$\top $ & specific element called the \textbf{one} element%
\end{tabular}%
\]
\end{definition}

\begin{definition}
A Boolean algebra of sets \emph{is any }\ Boolean algebra, where:

$\delta $ is a set of sets,

$\sqcup $ is interpreted as set union, denoted $\cup $.

$\sqcap $ is interpreted as set intersection denoted $\cap $

$^{\prime }$ is interpreted as set complement with repsect to $\top $,
denoted $\overline{}$, and $\sqsubseteq $ (or $\sqsupseteq $) is interpreted
as set containtment, denoted as $\subseteq $ (or $\supseteq $ ).
\end{definition}

\begin{definition}
An \emph{atom} of a Boolean algebra is an element $x\neq \bot $ such that
there is no other element $y\neq \bot $ with $y\sqsubseteq x$. I can happen
that there are no atoms at all in a Boolean algebra. In that case we call
the Boolean algebra \emph{atomless}; otherwise we call it \emph{atomic}.
\end{definition}

\begin{example}
\[
B_{Z}=\left\langle Powerset(\mathbb{Z}),\cap ,\cup ,~%
%TCIMACRO{\U{af}}%
%BeginExpansion
\bar{}%
%EndExpansion
~,\emptyset ,\mathbb{Z}\right\rangle 
\]%
is a Boolean algebra of sets. In this algebra for each $i\in \mathbb{Z}$,
the singleton $\left\{ i\right\} $ is an atom. Thus this is an \textbf{%
atomic Boolean algebra}. Clearly $x=\left\{ 1\right\} \neq \bot $ and $y$
can be either $\left\{ 1\right\} $ or $\emptyset $. We eliminate $\left\{
1\right\} $ from consideration because it is not an \textbf{other element}.
Since $y=\bot $, we conclude that there are no other elements such that $%
y\neq \bot $.
\end{example}

\begin{example}
Let $H$ be the set of all finite unions of half-open intervals of the form $%
[a,b)$ over the rational numbers, where $[a,b)$ means all rational numbers
that are greater than or equal to $a$ and less than $b$, where $a$ is a
rational or $-\infty $ and $b$ is a rational number or $\infty $.%
\[
B_{H}=\left\langle H,\cap ,\cup ,~%
%TCIMACRO{\U{af}}%
%BeginExpansion
\bar{}%
%EndExpansion
~,\emptyset ,\mathbb{Q}\right\rangle 
\]%
This is another Boolean algebra of sets, but this set is atomless. $\forall
x=[a,b)$, $a<b$ there exists a $c$ such that $a<c<b$. Thus $[a,c)\subseteq
\lbrack a,b)$, and $[a,c)\neq \emptyset $.
\end{example}

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