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diff --git a/Master/texmf-dist/doc/latex/mil3/sampartb.tex b/Master/texmf-dist/doc/latex/mil3/sampartb.tex deleted file mode 100644 index 6515da73e05..00000000000 --- a/Master/texmf-dist/doc/latex/mil3/sampartb.tex +++ /dev/null @@ -1,236 +0,0 @@ -% Sample file: sampartb.tex -% The sample article for the amsart document class with BibTeX -% Typeset with LaTeX format - -\documentclass{amsart} -\usepackage{amssymb,latexsym} - -\theoremstyle{plain} -\newtheorem{theorem}{Theorem} -\newtheorem{corollary}{Corollary} -\newtheorem*{main}{Main~Theorem} -\newtheorem{lemma}{Lemma} -\newtheorem{proposition}{Proposition} - -\theoremstyle{definition} -\newtheorem{definition}{Definition} - -\theoremstyle{remark} -\newtheorem*{notation}{Notation} - -\numberwithin{equation}{section} - -\begin{document} -\title[Complete-simple distributive lattices] - {A construction of complete-simple\\ - distributive lattices} -\author{George~A. Menuhin} -\address{Computer Science Department\\ - University of Winnebago\\ - Winnebago, MN 53714} -\email{menuhin@ccw.uwinnebago.edu} -\urladdr{http://math.uwinnebago.edu/homepages/menuhin/} -\thanks{Research supported by the NSF under grant number -23466.} -\keywords{Complete lattice, distributive lattice, - complete congruence, congruence lattice} -\subjclass[2000]{Primary: 06B10; Secondary: 06D05} -\date{March 15, 1999} -\begin{abstract} - In this note we prove that there exist \emph{complete-simple distributive - lattices,} that is, complete distributive lattices in which there are - only two complete congruences. -\end{abstract} - -\maketitle - -\section{Introduction}\label{S:intro} -In this note we prove the following result: - -\begin{main} - There exists an infinite complete distributive lattice~$K$ with only - the two trivial complete congruence relations. -\end{main} - -\section{The $D^{\langle 2 \rangle}$ construction}\label{S:Ds} -For the basic notation in lattice theory and universal algebra, see Ferenc~R. -Richardson~\cite{fR82} and George~A. Menuhin~\cite{gM68}. We start with some -definitions: - -\begin{definition}\label{D:prime} - Let $V$ be a complete lattice, and let $\mathfrak{p} = [u, v]$ be - an interval of $V$. Then $\mathfrak{p}$ is called - \emph{complete-prime} if the following three conditions are satisfied: - \begin{enumerate} - \item $u$ is meet-irreducible but $u$ is \emph{not} - completely meet-irreducible; - \item $v$ is join-irreducible but $v$ is \emph{not} - completely join-irreducible; - \item $[u, v]$ is a complete-simple lattice. - \end{enumerate} -\end{definition} - -Now we prove the following result: - -\begin{lemma}\label{L:ds} - Let $D$ be a complete distributive lattice satisfying - conditions \textup{(1)} and~\textup{(2)}. Then - $D^{\langle 2 \rangle}$ is a sublattice of $D^{2}$; - hence $D^{\langle 2 \rangle}$ is a lattice, and - $D^{\langle 2 \rangle}$ is a complete distributive - lattice satisfying condition \textup{(1)} and~\textup{(2)}. -\end{lemma} - -\begin{proof} - By conditions (1) and~(2), $D^{\langle 2 \rangle}$ is a sublattice - of $D^{2}$. Hence, $D^{\langle 2 \rangle}$ is a lattice. - - Since $D^{\langle 2 \rangle}$ is a sublattice of a distributive - lattice, $D^{\langle 2 \rangle}$ is a distributive lattice. Using - the characterization of standard ideals in Ernest~T. Moynahan~\cite{eM57}, - $D^{\langle 2 \rangle}$ has a zero and a unit element, - namely, $\langle 0, 0 \rangle$ and $\langle 1, 1 \rangle$. - To show that $D^{\langle 2 \rangle}$ is complete, let - $\varnothing \ne A \subseteq D^{\langle 2 \rangle}$, and let - $a = \bigvee A$ in $D^{2}$. If - $a \in D^{\langle 2 \rangle}$, then - $a = \bigvee A$ in $D^{\langle 2 \rangle}$; otherwise, $a$ - is of the form $\langle b, 1 \rangle$ for some - $b \in D$ with $b < 1$. Now $\bigvee A = \langle 1, 1\rangle$ - in $D^{2}$ and the dual argument shows that $\bigwedge A$ also - exists in $D^{2}$. Hence $D$ is complete. Condition (1) - and~(2) are obvious for $D^{\langle 2 \rangle}$. -\end{proof} - -\begin{corollary}\label{C:prime} - If $D$ is complete-prime, then so is $D^{\langle 2 \rangle}$. -\end{corollary} - -The motivation for the following result comes from Soo-Key Foo~\cite{sF90}. - -\begin{lemma}\label{L:ccr} - Let $\Theta$ be a complete congruence relation of - $D^{\langle 2 \rangle}$ such that - \begin{equation}\label{E:rigid} - \langle 1, d \rangle \equiv \langle 1, 1 \rangle \pmod{\Theta}, - \end{equation} - for some $d \in D$ with $d < 1$. Then $\Theta = \iota$. -\end{lemma} - -\begin{proof} - Let $\Theta$ be a complete congruence relation of - $D^{\langle 2 \rangle}$ satisfying \eqref{E:rigid}. Then $\Theta = \iota$. -\end{proof} - -\section{The $\Pi^{*}$ construction}\label{S:P*} -The following construction is crucial to our proof of the Main Theorem: - -\begin{definition}\label{D:P*} - Let $D_{i}$, for $i \in I$, be complete distributive lattices - satisfying condition~\textup{(2)}. Their $\Pi^{*}$ product is defined as - follows: - \[ - \Pi^{*} ( D_{i} \mid i \in I ) = \Pi ( D_{i}^{-} \mid i \in I ) + 1; - \] - that is, $\Pi^{*} ( D_{i} \mid i \in I )$ is $\Pi ( D_{i}^{-} \mid - i \in I )$ with a new unit element. -\end{definition} - -\begin{notation} - If $i \in I$ and $d \in D_{i}^{-}$, then - \[ - \langle \dots, 0, \dots, \overset{i}{d}, \dots, 0, \dots \rangle - \] - is the element of $\Pi^{*} ( D_{i} \mid i \in I )$ whose $i$-th - component is $d$ and all the other components are $0$. -\end{notation} - -See also Ernest~T. Moynahan \cite{eM57a}. Next we verify: - -\begin{theorem}\label{T:P*} - Let $D_{i}$, for $i \in I$, be complete distributive lattices - satisfying condition~\textup{(2)}. Let $\Theta$ be a complete congruence - relation on $\Pi^{*} ( D_{i} \mid i \in I )$. If there exist - $i \in I$ and $d \in D_{i}$ with $d < 1_{i}$ such that for - all $d \leq c < 1_{i}$, - \begin{equation}\label{E:cong1} - \langle \dots, 0, \dots,\overset{i}{d}, - \dots, 0, \dots \rangle \equiv \langle \dots, 0, \dots, - \overset{i}{c}, \dots, 0, \dots \rangle \pmod{\Theta}, - \end{equation} - then $\Theta = \iota$. -\end{theorem} - -\begin{proof} - Since - \begin{equation}\label{E:cong2} - \langle \dots, 0, \dots, \overset{i}{d}, \dots, 0, - \dots \rangle \equiv \langle \dots, 0, \dots, - \overset{i}{c}, \dots, 0, \dots \rangle \pmod{\Theta}, - \end{equation} - and $\Theta$ is a complete congruence relation, it follows from - condition~(3) that - \begin{equation}\label{E:cong} - \begin{split} - &\langle \dots, \overset{i}{d}, \dots, 0, - \dots \rangle\\ - &\equiv \bigvee ( \langle \dots, 0, \dots, - \overset{i}{c}, \dots, 0, \dots \rangle \mid d \leq c < 1 ) - \equiv 1 \pmod{\Theta}. - \end{split} - \end{equation} - - Let $j \in I$ for $j \neq i$, and let $a \in D_{j}^{-}$. - Meeting both sides of the congruence \eqref{E:cong2} with - $\langle \dots, 0, \dots, \overset{j}{a}, \dots, 0, \dots \rangle$, - we obtain - \begin{equation}\label{E:comp} - \begin{split} - 0 &= \langle \dots, 0, \dots, \overset{i}{d}, \dots, 0, \dots - \rangle \wedge \langle \dots, 0, \dots, \overset{j}{a}, \dots, 0, - \dots \rangle\\ - &\equiv \langle \dots, 0, \dots, \overset{j}{a}, \dots, 0, \dots - \rangle \pmod{\Theta}. - \end{split} - \end{equation} - Using the completeness of $\Theta$ and \eqref{E:comp}, we get: - \[ - 0 \equiv \bigvee ( \langle \dots, 0, \dots, \overset{j}{a}, - \dots, 0, \dots \rangle \mid a \in D_{j}^{-} ) = 1 \pmod{\Theta}, - \] - hence $\Theta = \iota$. -\end{proof} - -\begin{theorem}\label{T:P*a} - Let $D_{i}$, for $i \in I$, be complete distributive lattices - satisfying conditions \textup{(2)} and~\textup{(3)}. Then - $\Pi^{*} ( D_{i} \mid i \in I )$ also satisfies conditions \textup{(2)} - and~\textup{(3)}. -\end{theorem} - -\begin{proof} - Let $\Theta$ be a complete congruence on - $\Pi^{*} ( D_{i} \mid i \in I )$. Let $i \in I$. Define - \[ - \widehat{D}_{i} = \{ \langle \dots, 0, \dots, \overset{i}{d}, - \dots, 0, \dots \rangle \mid d \in D_{i}^{-} \} \cup \{ 1 \}. - \] - Then $\widehat{D}_{i}$ is a complete sublattice of - $\Pi^{*} ( D_{i} \mid i \in I )$, and $\widehat{D}_{i}$ is - isomorphic to $D_{i}$. Let $\Theta_{i}$ be the restriction of - $\Theta$ to $\widehat{D}_{i}$. - - Since $D_{i}\) is complete-simple, so is $\widehat{D}_{i}$, and - hence $\Theta_{i}$ is $\omega$ or $\iota$. If - $\Theta_{i} = \rho$ for all $i \in I$, then - $\Theta = \omega$. If there is an $i \in I$, such that - $\Theta_{i} = \iota$, then $0 \equiv 1 \pmod{\Theta}$, hence - $\Theta = \iota$. -\end{proof} - -The Main Theorem follows easily from Theorems \ref{T:P*} and~\ref{T:P*a}. - -\bibliographystyle{amsplain} -\bibliography{sampartb} -\end{document} - |