topological vector lattice

A topological vector lattice V over is

Proposition 1.

A topological vector lattice V is a topological lattice.

Before proving this, we show the following equivalence on the continuity of various operationsMathworldPlanetmath on a vector lattice V that is also a topological vector space.

Lemma 1.

Let V be a vector lattice and a topological vector space. The following are equivalentMathworldPlanetmathPlanetmathPlanetmathPlanetmathPlanetmath:

  1. 1.

    :V2V is continuousPlanetmathPlanetmath (simultaneously in both arguments)

  2. 2.

    :V2V is continuous (simultaneously in both arguments)

  3. 3.

    :+VV given by x+:=x0 is continuous

  4. 4.

    :-VV given by x-:=-x0 is continuous

  5. 5.

    ||:VV given by |x|:=-xx is continuous


(12). If is continuous, then xy=x+y-xy is continuous too, as + and - are both continuous under a topological vector space. This proof works in reverse too. (13), (14), and (34) are obvious. To see (45), we see that |x|=x++x-, since - is continuous, + is continuous also, so that || is continuous. To see (54), we use the identityPlanetmathPlanetmathPlanetmathPlanetmathPlanetmath x=x+-x-, so that |x|=(x+x-)+x-, which implies x-=12(|x|-x) is continuous. Finally, (31) is given by xy=(x-y+y)(0+y)=(x-y)0+y=(x-y)++y, which is continuous. ∎

In addition, we show an important inequality that is true on any vector lattice:

Lemma 2.

Let V be a vector lattice. Then |a+-b+||a-b| for any a,bV.


|a+-b+|=(b+-a+)(a+-b+)=(b0-a0)(a0-b0). Next, a0-b0=(b+(-a0)(-a0)=((b-a)b)(-a0) so that |a+-b+|=((b-a)b)(-a0)((a-b)a)(-b0)(b-a)(-a0)(a-b)(-b0). Since (b-a)(a-b)=|a-b| and a0 are both in the positive conePlanetmathPlanetmathPlanetmathPlanetmath of V, so is their sum, so that 0(b-a)(a-b)+(a0)=(b-a)(a-b)-(-a0), which means that (-a0)(b-a)(a-b). Similarly, (-b0)(b-a)(a-b). Combining these two inequalities, we see that |a+-b+|(b-a)(-a0)(a-b)(-b0)(b-a)(a-b)=|a-b|. ∎

We are now ready to prove the main assertion.


To show that V is a topological lattice, we need to show that the latticeMathworldPlanetmath operations meet and join are continuous, which, by Lemma 1, is equivalent in showing, say, that + is continuous. Suppose N is a neighborhood base of 0 consisting of solid sets. We prove that + is continuous. This amounts to showing that if x is close to x0, then x+ is close to x0+, which is the same as saying that if x-x0 is in a solid neighborhood U of 0 (UN), then so is x+-x0+ in U. Since x-x0U, |x-x0|U. But |x+-x0+||x-x0| by Lemma 2, and U is solid, x+-x0+U as well, and therefore + is continuous. ∎

As a corollary, we have

Proposition 2.

A topological vector lattice is an ordered topological vector space.


All we need to show is that the positive cone is a closed setPlanetmathPlanetmath. But the positive cone is defined as {x0x}={xx-=0}, which is closed since - is continuous, and the positive cone is the inverse imagePlanetmathPlanetmath of a singleton, a closed set in . ∎

Title topological vector lattice
Canonical name TopologicalVectorLattice
Date of creation 2013-03-22 17:03:51
Last modified on 2013-03-22 17:03:51
Owner CWoo (3771)
Last modified by CWoo (3771)
Numerical id 6
Author CWoo (3771)
Entry type Definition
Classification msc 06F20
Classification msc 46A40
Defines locally solid