an ordered abelian group has no order units An element $e$ in $G^{+}$ is called an ordered unit in an ordered abelian group  $(G,G^{+})$ if for any $g\in G$,there exits a positive integer such that $-ne\leq g \leq ne$.
In Rordam's book,there is an example to show that not all ordered abelian groups have order units.
He takes $G$ as $c_0(\Bbb N,\Bbb Z)$,which is the group of all sequences of integers such that eventually converge to $0$. Let $G^{+}$ be the set of those sequences $(x_n)$ such that $x_n\geq 0$. Then $(G,G^{+})$ is an ordered abelian group without order units.
 Suppose $(G,G^{+})$ has an ordered unit $f\in G^{+}$,how to choose $g\in G$ such that there does not exist $n$ such that  $-nf\leq g \leq nf$.

Suppose $f\in c_0(\mathbb N,\mathbb Z)$ is an order unit.  Put $k_0=\max\{k\in\mathbb N:f(k)\neq0\}$.  Define $g\in c_0(\mathbb N,\mathbb Z)$ by $g(k_0+1)=1$ and $g(k)=0$ for $k\neq k_0+1$.  Then there is no $n\in\mathbb N$ such that $g\leq nf$, a contradiction.
