SolidSet 2007-05-08 13:57:28 2007-05-12 23:56:41 Definition absolute value in a vector lattice vector lattice homomorphism solid closure \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}} Let $V$ be a vector lattice and $|\cdot|$ be the absolute value defined on $V$. A subset $A\subseteq V$ is said to be \emph{solid}, or \emph{absolutely convex}, if, $|v|\le |u|$ implies that $v\in A$, whenever $u\in A$ in the first place. From this definition, one deduces immediately that $0$ belongs to every non-empty solid set. Also, if $a$ is in a solid set, so is $a^+$, since $|a^+|=a^+\le a^++a^-=|a|$. Similarly $a^-\in S$, and $|a|\in S$, as $||a||=|a|$. Furthermore, we have \begin{prop} If $S$ is a solid subspace of $V$, then $S$ is a vector sublattice. \end{prop} \begin{proof} Suppose $a,b\in S$. We want to show that $a\wedge b\in S$, from which we see that $a\vee b= a+b-(a\wedge b)\in S$ also since $S$ is a vector subspace. Since both $a\wedge b, a\vee b\in S$, we have that $S$ is a sublattice. To show that $a\wedge b\in S$, we need to find $c\in S$ with $|a\wedge b|\le |c|$. Let $c=|a|+|b|$. Since $a,b\in S$, $|a|,|b|\in S$, and so $c\in S$ as well. We also have that $|c|=c$. So to show $a\wedge b\in S$, it is enough to show that $|a\wedge b|\le c$. To this end, note first that $a\le |a|$ and $b\le |b|$, so $a\wedge b\le |a|\wedge |b|\le |a|\vee |b|$. Also, since $-a\le |a|$ and $-b\le |b|$, $-(a\wedge b)=(-a)\vee (-b)\le |a|\vee |b|$. As a result, $|a\wedge b|=-(a\wedge b)\vee (a\wedge b)\le |a|\vee |b|$. But $|a|\vee |b|\le |a|\vee |b|+|a|\wedge |b|=|a|+|b|=c$, we have that $|a\wedge b|\le |a|\vee |b| \le c$. \end{proof} \textbf{Examples} Let $V$ be a vector lattice. \begin{itemize} \item $0$ and $V$ itself are solid subspaces. \item If $V$ is finite dimensional, the only solid subspaces are the improper ones. \item An example of a proper solid subspace of a vector lattice is found, when we take $V$ to be the countably infinite direct product of $\mathbb{R}$, and $S$ to be the countably infinite direct sum of $\mathbb{R}$. \item An example of a solid set that is not a subspace is the unit disk in $\mathbb{R}^2$, where the ordering is defined componentwise. \item Given any set $A$, the smallest solid set containing $A$ is called the \emph{solid closure} of $A$. For example, if $A=\lbrace a\rbrace$, then its solid closure is $\lbrace v\in V\mid |v|\le |a|\rbrace$. In $\mathbb{R}^2$, the solid closure of any point $p$ is the disk centered at $O$ whose radius is $|p|$. \item The solid closure of $V^+$, the positive cone, is $V$. \end{itemize} \begin{prop} If $V$ is a vector lattice and $S$ is a solid subspace of $V$, then $V/S$ is a vector lattice. \end{prop} \begin{proof} Since $S$ is a subspace $V/S$ has the structure of a vector space, whose vector space operations are inherited from the operations on $V$. Since $S$ is solid, it is a sublattice, so that $V/S$ has the structure of a lattice, whose lattice operations are inherited from those on $V$. It remains to show that the partial ordering is ``compatible'' with the vector operatons. We break this down into two steps: \begin{itemize} \item for any $u+S,v+S,w+S \in V/S$, if $(u+S)\le (v+S)$, then $(u+S)+(w+S)\le (v+S)+(w+S)$. This is a disguised form of the following: if $u-v\le a\in S$, then $(u+w)-(v+w)\le b\in S$ for some $b$. This is obvious: just pick $b=a$. \item if $0+S\le u+S\in V/S$, then for any $0<\lambda \in k$ ($k$ an ordered field), $0+S\le \lambda (u+S)$. This is the same as saying: if $c\le u$ for some $b\in S$, then $d\le \lambda u$ for some $d\in S$. This is also obvious: pick $d=\lambda c$. \end{itemize} The proof is now complete. \end{proof}