PlanetMath: solid set

Let

be a vector lattice and $\vert\cdot\vert$ be the absolute value defined on

. A subset $A\subseteq V$ is said to be solid, or absolutely convex, if, $\vert v\vert\le \vert u\vert$ 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

is in a solid set, so is

, since $\vert a^+\vert=a^+\le a^++a^-=\vert a\vert$ . Similarly $a^-\in S$ , and $\vert a\vert\in S$ , as $\vert\vert a\vert\vert=\vert a\vert$ . Furthermore, we have

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

is a vector subspace. Since both $a\wedge b, a\vee b\in S$ , we have that

is a sublattice.

To show that $a\wedge b\in S$ , we need to find $c\in S$ with $\vert a\wedge b\vert\le \vert c\vert$ . Let $c=\vert a\vert+\vert b\vert$ . Since $a,b\in S$ , $\vert a\vert,\vert b\vert\in S$ , and so $c\in S$ as well. We also have that $\vert c\vert=c$ . So to show $a\wedge b\in S$ , it is enough to show that $\vert a\wedge b\vert\le c$ . To this end, note first that $a\le \vert a\vert$ and $b\le \vert b\vert$ , so $a\wedge b\le \vert a\vert\wedge \vert b\vert\le \vert a\vert\vee \vert b\vert$ . Also, since $-a\le \vert a\vert$ and $-b\le \vert b\vert$ , $-(a\wedge b)=(-a)\vee (-b)\le \vert a\vert\vee \vert b\vert$ . As a result, $\vert a\wedge b\vert=-(a\wedge b)\vee (a\wedge b)\le \vert a\vert\vee \vert b\vert$ . But $\vert a\vert\vee \vert b\vert\le \vert a\vert\vee \vert b\vert+\vert a\vert\wedge \vert b\vert=\vert a\vert+\vert b\vert=c$ , we have that $\vert a\wedge b\vert\le \vert a\vert\vee \vert b\vert \le c$ . $\qedsymbol$

Proof. Since

is a subspace

has the structure of a vector space, whose vector space operations are inherited from the operations on

. Since

is solid, it is a sublattice, so that

has the structure of a lattice, whose lattice operations are inherited from those on

. 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 . This is obvious: just pick .
if $0+S\le u+S\in V/S$ , then for any $0<\lambda \in k$ ( 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. $\qedsymbol$