TITLE: Cauchy sequence property
QUESTION [7 upvotes]: I do not see how this is even valid. Could someone point this out to me:
Assume that $x_n$ is a cauchy sequence of rational numbers satisfying $|x_n| \geq r$ for all $n\in\mathbb{N}$. Show that there is $N\in\mathbb{N}$ s.t. either $x_n > r$ for all $n \geq N$ or $x_n < -r$ for all $n\geq N$.
Here is my immediate thought. If $x_n$ is a constant sequence then it satisfies the conditions in the assumption. So, I was thinking perhaps that fact that it is Cauchy must guarantee what I need to prove. However, it must be Cauchy as $x_n-x_m=0$ for all $n\in\mathbb{N}$. So the definition: $\forall_{\epsilon>0} \exists_N \forall_{m,n}\implies |x_n - x_m| < \epsilon$ Holds.
This is contrary to what I am supposed to prove since $x_n \not > r$ as it is equal to $r$ for all $n\in\mathbb{N}$

Am I overlooking something?

Thanks, all feedback is welcomed and appreciated.

REPLY [3 votes]: First you need to assume that $r$ is positive and the condition needs to be $|x_n| > r$, not $\geq$.  Then, as you always have $|x_n| > r$ each term $x_n$ satisfies either $x_n > r$ or $x_n < -r$.  What you need to show is that after a point it stops jumping back and forth and settles on always being one or the other.
Start your proof by saying that the sequence is Cauchy so there exists an $n$ such that for all $n, m \geq N$ we have $|x_n - x_m| < r$, in particular $|x_N - x_m| < r$.  Now note that for $x_m$ to flip to the other side it needs to be a distance of at least $2r$ away from $x_N$.