How could I show, that if $\lim_{n\rightarrow\infty}q^n=0$ for $0<q<1$ in an ordered field (not $\mathbb{R}$), that the ordered field is archimedean, meaning for every $x,y$ in this field with $x>0$ there exists an integer $n$, such that $xn>y$?
I tried to use many equivalent definitions of the archimedean property to show this, but got stuck everytime.
Key is probably, that you need the Archimedean Property to show that the sequence converges. So the other direction has to have to do something with this. I can't figure out how: Direct proof? Indirect? Any ideas?
Cheers.
I suggest an indirect proof. Assume the field is not Archimedean and conclude that $q^n$ doesn't converge to $0$ for some $0 < q < 1$. For example, $q = 1/2$ works well.
Now, the assumption that the field isn't Archimedian says that there is a $y$ such that $n < y$ for all $n \in \mathbb{N}$. [There's a little work to do to obtain this from the definition.]
Next, $2^n \in \mathbb{N}$ for all $n \in \mathbb{N}$, hence $2^n < y$ for all $n$. From this it follows easily that $\bigl(\frac{1}{2}\bigr)^n \not\to 0$.