3
$\begingroup$

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.

Improve this question
$\endgroup$

1 Answer 1

Reset to default
5
$\begingroup$

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 $\chi_{\mathbb N}<0$.

Improve this answer
$\endgroup$
3
  • $\begingroup$ Welp, that's not the answer I wanted but I can in fact see it. Thanks for the prompt response. $\endgroup$
    – Anschel Schaffer-Cohen
    Commented Dec 3, 2015 at 0:59
  • 5
    $\begingroup$ I could see if I can come up with a proof of the opposite answer... $\endgroup$
    – Anthony Quas
    Commented Dec 3, 2015 at 1:17
  • 1
    $\begingroup$ That'd be very gentlepersonly of you. $\endgroup$
    – Anschel Schaffer-Cohen
    Commented Dec 3, 2015 at 1:22

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .