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?

Answer 1

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