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Homesolid set
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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, $v\lequ$ 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 $a^{+}=a^{+}\leq a^{+}+a^{}=a$. 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 $a\wedge b\leqc$. 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 $a\wedge b\leq c$. To this end, note first that $a\leqa$ and $b\leqb$, so $a\wedge b\leqa\wedgeb\leqa\veeb$. Also, since $a\leqa$ and $b\leqb$, $(a\wedge b)=(a)\vee(b)\leqa\veeb$. As a result, $a\wedge b=(a\wedge b)\vee(a\wedge b)\leqa\veeb$. But $a\veeb\leqa\veeb+a\wedgeb=a+b=c$, we have that $a\wedge b\leqa\veeb\leq c$. ∎
Examples Let $V$ be a vector lattice.

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

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

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 $\mathbb{R}$.

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 $\{v\in V\midv\leqa\}$. In $\mathbb{R}^{2}$, the solid closure of any point $p$ is the disk centered at $O$ whose radius is $p$.

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 $V/S$ 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:

for any $u+S,v+S,w+S\in V/S$, if $(u+S)\leq(v+S)$, then $(u+S)+(w+S)\leq(v+S)+(w+S)$. This is a disguised form of the following: if $uv\leq a\in S$, then $(u+w)(v+w)\leq b\in S$ for some $b$. This is obvious: just pick $b=a$.

if $0+S\leq u+S\in V/S$, then for any $0<\lambda\in k$ ($k$ an ordered field), $0+S\leq\lambda(u+S)$. This is the same as saying: if $c\leq u$ for some $b\in S$, then $d\leq\lambda u$ for some $d\in S$. This is also obvious: pick $d=\lambda c$.
The proof is now complete. ∎
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