Consider an ordered field $(\mathbb{F}, +, \times, <)$ with additive identity $0$ and multiplicative identity $1$. It is not hard to show that $1\ne 0$ and then $1=1\times 1>0$. I am aware that the Archimedean property would imply the following function (or sequence, as we are talking about sequential limit) converges to $0$ as $n\to+\infty$. \begin{align*} f:\mathbb{N}&\to\mathbb{F}\\ n&\mapsto\frac{1}{\underbrace{1+\cdots+1}_n} \end{align*} However, the Archimedean property was just not takend as granted for any ordered field.
Please let me clarify my question to eliminate ambiguities. I am trying to define sequential limit in the following manner:
If $f:\mathbb{N}\to\mathbb{F}$ and $L\in\mathbb{F}$ satisfy $\forall \epsilon\in\mathbb{F}^+, \exists N\in\mathbb{N}:\forall n\in\mathbb{N}, \big(n>N\Rightarrow L-\epsilon<f(n)<L+\epsilon\big)$, we shall say that $f(n)\to L$ as $n\to+\infty$. $\mathbb{F}^+$ is defined as $\{x\in\mathbb{F}|x>0\}$.
My question is, whether any ordered field has a positive sequence ($f:\mathbb{N}\to\mathbb{F}^+$) converging to $0$?
The question occurs to me when I am trying to prove the equivalence of some forms of completeness axiom each of which characterizes $\mathbb{R}$, and I have decided to consider these axioms on some general ordered field $\mathbb{F}$ instead of on $\mathbb{Q}$. Of course, perhaps it's my naive way of proving that somehow encounters with the existence of such an auxiliary sequence (which may take place of common positive sequences in $\mathbb{Q}$ like $n^{-1}$ or $2^{-n}$), as a declaimer.
I am curious of such an answer. On top of that, I wish that it could be more detailed, as I am not totally familiar with the rich treasury of counter-examples in Mathematics. Anyway, any help will be appreciated :)
Some of my attempts (not necessary to read)
I am aware that we may make a copy of $\mathbb{Q}$ in $\mathbb{F}$ by considering ‘positive integers’ in the form of $\underbrace{1+\cdots+1}_n$, then ‘integers’, then ‘rational numbers’. The hypothetical setback is that an $\epsilon\in\mathbb{F}^+$ does not necessarily corresponds to some ‘positive rational number’. I am trying to consider some ordered field with no Archimedean property, such as the one mentioned here, and I am aware that $\frac{1}{\underbrace{1+\cdots+1}_n}$ does not converge to $0$ in this sense, as we may consider $\epsilon=\frac{1}{x}\in(\mathbb{R}(x))^+$. However, as I am trying to figure out a counter-example, I started to think of the diagonal argument, and so I am not able to construct a counter-example after all, though it may obviously due to the fact that I am not very experienced in dealing with $\mathbb{R}(x)$.