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'complemented lattice'
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| Title of object: |
complemented lattice |
| Canonical Name: |
ComplementedLattice |
| Type: |
Definition |
| Created on: |
2005-02-16 00:29:28 |
| Modified on: |
2006-03-08 20:52:32 |
| Classification: |
msc:06B05, msc:06C15 |
| Defines: |
related elements in lattice |
Preamble:
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\usepackage{amssymb,amscd}
\usepackage{amsmath}
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%\usepackage{psfrag}
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Content:
A \emph{complemented lattice} is a lattice $L$ such that there are
two elements $0$ and $1$ such that, for all $x\in L$,
\begin{enumerate}
\item $0\leq x$,
\item $x\leq 1$, and
\item there exists $y\in L$ such that $x\land y=0$ and $x\lor y=1$.
\end{enumerate}
$y$ is called a \emph{complement} of $x$.
In other words, a complemented lattice is a bounded lattice, such that the existence (not uniquness) of a complement of an element is guaranteed.
\textbf{Remarks}.
\begin{itemize}
\item Note that a complement of an element in a bounded lattice can be defined without the lattice being complemented. Given any element $a$ in a bounded lattice, two elements $b,c$ are said to be \emph{related} via $a$ if $b$ and $c$ are both complements of $a$.
\item If a complemented lattice $L$ is a distributive lattice, then every element of $L$ has a unique complement. For if
$y$ and $y^{\prime}$ are two complements of $x$, then
$$y^{\prime}=1\land y^{\prime}=(x\lor y)\land y^{\prime}=
(x\land y^{\prime})\lor(y\land y^{\prime})=0\lor(y\land
y^{\prime})=y\land y^{\prime}.$$ Similarly, $y=y^{\prime}\land y$.
So $y^{\prime}=y$.
\item In a complemented distributive lattice $L$, denote $\overline{x}$ to be the \emph{unique} complement of $x$. It is not hard to see
that $\overline{\overline{x}}=x$. Next, for any $x,y\in L$, define $x+y$ to be
$(x\land\overline{y})\lor(\overline{x}\land y)$, the $L$ together
with $\land$ and $+$ becomes a Boolean ring.
\end{itemize} |
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