solid set
Let $V$ be a vector lattice and $\cdot $ be the absolute value^{} defined on $V$. A subset $A\subseteq V$ is said to be solid, or 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 nonempty 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
Proposition 1.
If $S$ is a solid subspace of $V$, then $S$ is a vector sublattice.
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 $a\wedge b\le a\vee b\le c$. ∎
Examples Let $V$ be a vector lattice.

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$0$ and $V$ itself are solid subspaces.

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If $V$ is finite dimensional, the only solid subspaces are the improper ones.

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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 $\mathbb{R}$.
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Given any set $A$, the smallest solid set containing $A$ is called the solid closure of $A$. For example, if $A=\{a\}$, then its solid closure is $\{v\in V\mid v\le a\}$. In ${\mathbb{R}}^{2}$, the solid closure of any point $p$ is the disk centered at $O$ whose radius is $p$.

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The solid closure of ${V}^{+}$, the positive cone^{}, is $V$.
Proposition 2.
If $V$ is a vector lattice and $S$ is a solid subspace of $V$, then $V\mathrm{/}S$ is a vector lattice.
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:

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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 $uv\le a\in S$, then $(u+w)(v+w)\le b\in S$ for some $b$. This is obvious: just pick $b=a$.

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if $0+S\le u+S\in V/S$, then for any $$ ($k$ 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 $d=\lambda c$.
The proof is now complete^{}. ∎
Title  solid set 

Canonical name  SolidSet 
Date of creation  20130322 17:03:19 
Last modified on  20130322 17:03:19 
Owner  CWoo (3771) 
Last modified by  CWoo (3771) 
Numerical id  6 
Author  CWoo (3771) 
Entry type  Definition 
Classification  msc 06F20 
Classification  msc 46A40 
Synonym  absolutely convex 
Defines  vector lattice homomorphism 
Defines  solid closure 