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 $-nf\leq g \leq nf$.


1 Answer 1


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.

  • $\begingroup$ I have a quesion:you mentioned that k_0=\max\{k\in\mathbb N:f(k)\neq0\},how to ensure that maximum can be attained? $\endgroup$
    – math112358
    Oct 2, 2019 at 9:35
  • $\begingroup$ $\mathbb Z$ is discrete, so a sequence in $\mathbb Z$ which converges is eventually constant. $\endgroup$
    – Aweygan
    Oct 2, 2019 at 20:45

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