# 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.
• $\mathbb Z$ is discrete, so a sequence in $\mathbb Z$ which converges is eventually constant. Oct 2, 2019 at 20:45