In a Riesz space, does $(u+v)^+=u^+ + v^+$ hold?

Question

Assume that $E$ is a Riesz space (lattice ordered vector space). For $u\in E$, let $u^+ = u\vee 0$.

Then, for $u,v\in E$, does $(u+v)^+=u^+ + v^+$ hold?

That isn't even true on $\mathbb R$. Take $u = 1$ and $v=-1$. — user251257, Mar 26 '16 at 03:38
Then is there any relationship between $(u+v)^+$ and $u^+,v^+$? — seunglim, Mar 27 '16 at 03:15

score 3 · Answer 1 · answered Jun 10 '16 at 17:21

3

In general, you will only have $(u+v)^+ \le u^+ + v^+$ and this is easy to prove.

answered Jun 10 '16 at 17:21

gerw

1 Answers1