solid set


Let V be a vector lattice and || be the absolute valueMathworldPlanetmathPlanetmathPlanetmathPlanetmath defined on V. A subset AV is said to be solid, or absolutely convex, if, |v||u| implies that vA, whenever uA 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+a++a-=|a|. Similarly a-S, and |a|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,bS. We want to show that abS, from which we see that ab=a+b-(ab)S also since S is a vector subspace. Since both ab,abS, we have that S is a sublattice.

To show that abS, we need to find cS with |ab||c|. Let c=|a|+|b|. Since a,bS, |a|,|b|S, and so cS as well. We also have that |c|=c. So to show abS, it is enough to show that |ab|c. To this end, note first that a|a| and b|b|, so ab|a||b||a||b|. Also, since -a|a| and -b|b|, -(ab)=(-a)(-b)|a||b|. As a result, |ab|=-(ab)(ab)|a||b|. But |a||b||a||b|+|a||b|=|a|+|b|=c, we have that |ab||a||b|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 infiniteMathworldPlanetmath direct product of , and S to be the countably infinite direct sum of .

  • An example of a solid set that is not a subspace is the unit disk in 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 {vV|v||a|}. In 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 conePlanetmathPlanetmathPlanetmathPlanetmathPlanetmath, 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 structureMathworldPlanetmath of a vector space, whose vector space operationsMathworldPlanetmath 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+SV/S, if (u+S)(v+S), then (u+S)+(w+S)(v+S)+(w+S). This is a disguised form of the following: if u-vaS, then (u+w)-(v+w)bS for some b. This is obvious: just pick b=a.

  • if 0+Su+SV/S, then for any 0<λk (k an ordered field), 0+Sλ(u+S). This is the same as saying: if cu for some bS, then dλu for some dS. This is also obvious: pick d=λc.

The proof is now completePlanetmathPlanetmathPlanetmathPlanetmath. ∎

Title solid set
Canonical name SolidSet
Date of creation 2013-03-22 17:03:19
Last modified on 2013-03-22 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