# vector lattice

For example, given a topological space  $X$, its ring of continuous functions $C(X)$ is a vector lattice. In particular, any finite dimensional Euclidean space $\mathbb{R}^{n}$ is a vector lattice.

Below are some properties of the join ($\vee$) and meet ($\wedge$) operations  on a vector lattice $L$. Suppose $u,v,w\in L$, then

1. 1.

$(u+w)\vee(v+w)=(u\vee v)+w$

2. 2.

$u\wedge v=(u+v)-(u\vee v)$

3. 3.

If $\lambda\geq 0$, then $\lambda u\vee\lambda v=\lambda(u\vee v)$

4. 4.

If $\lambda\leq 0$, then $\lambda u\vee\lambda v=\lambda(u\wedge v)$

5. 5.
6. 6.

If $L$ is an ordered vector space, and if for any $u,v\in L$, either $u\vee v$ or $u\wedge v$ exists, then $L$ is a vector lattice. This is basically the result of property 2 above.

7. 7.

$(u\wedge v)+w=(u+w)\wedge(v+w)$ (dual of statement 1)

8. 8.

$u\wedge v=-(-u\vee-v)$ (a direct consequence of statement 4, with $\lambda=-1$)

9. 9.

$(-u)\wedge u\leq 0\leq(-u)\vee u$

###### Proof.

$(-u)\wedge u\leq u$ and $(-u)\wedge u\leq-u$ imply that $2((-u)\wedge u)\leq u+(-u)=0$, so $(-u)\wedge u\leq 0$, which means $0\leq-((-u)\wedge u)=u\vee(-u)$. ∎

10. 10.

$(a\vee b)+(c\vee d)=(a+c)\vee(a+d)\vee(b+c)\vee(b+d)$, by repeated application of 1 above.

Remark. The first five properties are also satisfied by an ordered vector space, with the assumptions  that the suprema exist for the appropriate pairs of elements (see the entry on ordered vector space for detail).

Title vector lattice VectorLattice 2013-03-22 17:03:13 2013-03-22 17:03:13 CWoo (3771) CWoo (3771) 7 CWoo (3771) Definition msc 06F20 msc 46A40 Riesz Space vector sublattice