Suppose that $A$ is a Riesz space (vector lattice). Recall the following terminology:
A is Archimedean if for any $x,y\geq 0$ such that $n x\leq y$ for $n=1,2,\ldots$ is follows that $x=0$;
A is Dedekind complete if every upper bounded subset of $A$ has a supremum.
As the title of my post suggest, my question is: Does Dedekind completeness imply the Archimeadean property?
I have the following proof, which I would like to check if it is right:
Suppose that $x,y\geq 0$ satisfied that $n x\leq y$ for $n=1,2,\ldots$ in a Dedekind complete Riesz space $A$. To reach a contradiction, suppose that $x\neq 0$. Then the set $$B:=\{ n x\colon n=1,2,\ldots\} $$ has an upper bound; namely $y$. Since $A$ is Dedekind complete, we can take $z:=\sup B$. Since $x> 0$ (i.e. $x\geq 0$ and $x\neq 0$), it follows that $z>z-x$. Hence $z-x$ is not an upper bound of $B$, hence there exists $m\in\mathbb{N}$ such that $z-x\ngeq m x$. But this means that $z\ngeq m x + x=(m+1)x$, a contradiction.