Convergence in ordered rings

Question

The question is, essentially, whether we can compute the limits of some sequences in ordered rings. Given an ordered ring $R$ with ordering $\le$, we can say that a sequence $(a_n) \to a$ if and only if for all $r \in R, r>0$ there exists $N\in\mathbb{N}$ such that $-r < a_n - a < r$ for $n > N$.

The question is: given an ordered ring $R$ with unity $1$, and for any $r \in R$ that satisfies $0 < r < 1$, is is true that the sequence $(r^n)$ converges to zero? It seems unlikely to me, as the proof of this fact in $\mathbb{R}$ relies on the Archimedean Property, but I am unable to construct a counterexample personally.

Answer 1

Your observation is correct. Proof of $r^n\to 0$ as $n\to\infty$ must require the Archimedian property. If an ordered ring contains infinitesimal, then $r^n$ does not converges to zero in general. For example, consider Levi-Civita field and the sequence $a_n=2^{-n}$. You can check that $\varepsilon<2^{-n}$ for all $n$ so it does not coverges to zero.

Answer 2

A perhaps high-brow counterexample is the field of hyperreals http://en.wikipedia.org/wiki/Hyperreal_number.

Answer 3

Take $k$ a totally ordered field ( like $\mathbb{Q}$ or $\mathbb{R}$). Let $K= k(t)$ the field of fractions with coefficients in $k$. Order $K$ as follows:

$$\frac{a_m t^m + \cdots + a_0}{b_n t^n + \cdots + b_0} > 0 \text{ in } K \iff a_m\cdot b_n > 0 \text{ in } k$$

( one can think of it as : the values of the fraction for $t$ very large are $>0$)

We have $ (1-\frac{1}{t})^n \in (\frac{1}{2},1)$ for all $n > 0$ integer. ( on can use any $x < 1$ \instead of $\frac{1}{2}$)