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 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

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.

• $0$ 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 $ℝ$, and $S$ to be the countably infinite direct sum of $ℝ$.

• An example of a solid set that is not a subspace is the unit disk in ${ℝ}^2$, where the ordering is defined componentwise.

• 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 ${ℝ}^2$, 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 $V\mathrm{/}S$ 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:

• 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 $b=a$.

• 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.∎