GeneralizedBooleanAlgebra 2007-05-23 17:21:44 2007-05-28 10:16:31 Definition \usepackage{amssymb,amscd} \usepackage{amsmath} \usepackage{amsfonts} \usepackage{mathrsfs} % used for TeXing text within eps files %\usepackage{psfrag} % need this for including graphics (\includegraphics) %\usepackage{graphicx} % for neatly defining theorems and propositions \usepackage{amsthm} % making logically defined graphics \usepackage{xypic} \usepackage{pst-plot} \usepackage{psfrag} % define commands here \newtheorem{prop}{Proposition} \newtheorem{thm}{Theorem} \newtheorem{ex}{Example} \newcommand{\real}{\mathbb{R}} \newcommand{\pdiff}[2]{\frac{\partial #1}{\partial #2}} \newcommand{\mpdiff}[3]{\frac{\partial^#1 #2}{\partial #3^#1}} A lattice $L$ is called a \emph{generalized Boolean algebra} if \begin{itemize} \item $L$ is distributive, \item $L$ is relatively complemented, and \item $L$ has $0$ as the bottom. \end{itemize} Clearly, a Boolean algebra is a generalized Boolean algebra. Conversely, a generalized Boolean algebra $L$ with a top $1$ is a Boolean algebra, since $L=[0,1]$ is a bounded distributive complemented lattice, so each element $a\in L$ has a unique complement $a'$ by distributivity. So $'$ is a unary operator on $L$ which makes $L$ into a de Morgan algebra. A complemented de Morgan algebra is, as a result, a Boolean algebra. As an example of a generalized Boolean algebra that is not Boolean, let $A$ be an infinite set and let $B$ be the set of all finite subsets of $A$. Then $B$ is generalized Boolean: order $B$ by inclusion, then $B$ is a distributive as the operation is inherited from $P(A)$, the powerset of $A$. It is also relatively complemented: if $C\in [X,Y]$ where $C,X,Y\in B$, then $(Y-C)\cup X$ is the relative complement of $C$ in $[X,Y]$. Finally, $\varnothing$ is, as usual, the bottom element in $B$. $B$ is not a Boolean algebra, because the union of all the singletons (all in $B$) is $A$, which is infinite, thus not in $B$. One property of a generalized Boolean algebra $L$ is the following: if $y$ and $z$ are complements of $x\in [a,b]$, then $y=z$; in other words, relative complements are uniquely determined. This is true because in any distributive lattice, complents are uniquely determined. As $L$ is distributive, so is each lattice interval $[a,b]$ in $L$. In fact, because of the existence of $0$, we can actually construct the relative complement. Let $b-x$ denote the unique complement of $x$ in $[0,b]$. Then $(b-x)\vee a$ is the unique complement of $x\in [a,b]$: $x\wedge ((b-x)\vee a)=(x\wedge (b-x))vee (x\wedge a)=0\vee a=a$ and $x\vee ((b-x)\vee a)=(x\vee (b-x))\vee a=b\vee a=b$. Conversely, if $L$ is a distributive lattice with $0$ such that any lattice interval $[0,a]$ is complemented, then $L$ is a generalized Boolean algebra. Again, $(b-x)\vee a$ provides the necessary complement of $x$ in $[a,b]$.