Given the additive group of the module $\mathbb{Z}^\mathbb{N}$ and a total ordering of the group that is compatible with addition and where $\chi\_{\{n\}} > 0$ for all $n \in \mathbb{N}$, can we say for sure that $\chi\_\mathbb{N} > 0$?


By "compatible" I mean that, for all $a, b, c \in \mathbb{Z}^\mathbb{N}$, if $a \le b$ then $a + c \le b + c$. By $\chi\_S$ for some subset $S$ of $\mathbb{N}$ I mean the characteristic function, i.e. $\chi\_S(n) = 1$ if $n \in S$ and $0$ otherwise.


In the context where this came up I need a proof without the axiom of choice; but I'm also interested in the answer in general.



Compatible total orderings of the group $\mathbb{Z}^\mathbb{N}$
So I think using the Axiom of Choice, you can see the answer is no.
Use Hamel's basis theorem to obtain a basis of $\mathbb Q^{\mathbb N}$ including $\chi\_{n}$ for each $n$ and $-\chi\_{\mathbb N}$. Well order the basis elements and then say $a<b$ if the expansion in the Hamel basis of $a$ is lexicographically smaller than the expansion of $b$. 


In this ordering $\chi\_{n}>0$ for each $n$, but $\chi\_{\mathbb N}<0$. 


