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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 $n\in\mathbb{N}$

  1. Am I overlooking something?

Thanks, all feedback is welcomed and appreciated.

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    $\begingroup$ You’re not overlooking anything: the result is false as stated. Those two strict inequalities should be non-strict inequalities. $\endgroup$ – Brian M. Scott Feb 28 '13 at 7:58
  • $\begingroup$ As you pointed out, the result cannot be correct as stated. Perhaps you can modify it so that it becomes correct. $\endgroup$ – André Nicolas Feb 28 '13 at 7:59
  • $\begingroup$ Either that, or the non-strict inequality should be strict. $\endgroup$ – Cameron Buie Feb 28 '13 at 7:59
  • $\begingroup$ And even with non-strict inequalities, why can't the first few terms differ in sign from the rest? $\endgroup$ – Trevor Wilson Feb 28 '13 at 8:01
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    $\begingroup$ I see. I re-read, and as stated, it is obviously wrong - as you say. But you should assume that where it says "for all $n \in N$", is should read "for all $n > N$" (and simply remark on that too in your solution, or even point out a typo in an email to your professor so he can alert you co-students to the typo in what I assume to be a problem set). As formulated, it is claimed there is an $N$, but then this $N$ is not used; but the above is meant. $\endgroup$ – gnometorule Feb 28 '13 at 8:06
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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 $x_N$.

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Your objection is correct if you qualify "constant at $r$" (if it is constant at any other value, it holds vacuously). However, write this as part 1 of your answer; then show in part 2 that it is true if $x_n \neq r \,$ eventually.

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