From this definition, one deduces immediately that belongs to every non-empty solid set. Also, if is in a solid set, so is , since . Similarly , and , as . Furthermore, we have
If is a solid subspace of , then is a vector sublattice.
Suppose . We want to show that , from which we see that also since is a vector subspace. Since both , we have that is a sublattice.
To show that , we need to find with . Let . Since , , and so as well. We also have that . So to show , it is enough to show that . To this end, note first that and , so . Also, since and , . As a result, . But , we have that . ∎
Examples Let be a vector lattice.
and itself are solid subspaces.
If is finite dimensional, the only solid subspaces are the improper ones.
Given any set , the smallest solid set containing is called the solid closure of . For example, if , then its solid closure is . In , the solid closure of any point is the disk centered at whose radius is .
The solid closure of , the positive cone, is .
If is a vector lattice and is a solid subspace of , then is a vector lattice.
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 , if , then . This is a disguised form of the following: if , then for some . This is obvious: just pick .
if , then for any ( an ordered field), . This is the same as saying: if for some , then for some . This is also obvious: pick .
The proof is now complete. ∎
|Date of creation||2013-03-22 17:03:19|
|Last modified on||2013-03-22 17:03:19|
|Last modified by||CWoo (3771)|
|Defines||vector lattice homomorphism|