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Suppose $x_n$ is a Cauchy sequence and eventually positive and it's not true that $x_n \rightarrow 0$ as $n \rightarrow \infty$. Prove that there is a real $\delta >0$, so that if $y_n = x_n - \delta$ then $y_n$ is eventually positive.

Not really sure about my proof, could use some opinions. Trying to do this with epsilon argument.

Since $x_n$ doesn't converge to $0$ then $\exists \varepsilon_o > 0$ s.t $\forall N \ \exists m \geq N$ so that:

$|x_m| > \varepsilon_o$

And since $x_n$ is eventually positive then $\exists N_1, \forall m \geq N_1$

$x_m > 0$

Also note that since $x_n$ is Cauchy for $\frac{\varepsilon_o}{2} > 0$ $\exists N_2$, s.t. $\forall \ m,n \geq N_2:$

$|x_m - x_n| < \frac{\varepsilon_o}{2}$.

Fix $m \geq \max\{N_1, N_2\}$, so that $x_m > \varepsilon_o$, then for $n \geq \max\{N_1, N_2\}$ we have:

$|x_m - x_n| < \frac{\varepsilon_o}{2}$ and $ x_n > x_m - \frac{\varepsilon_o}{2} > \frac{\varepsilon_o}{2}$.

Let $\delta = \frac{\varepsilon_o}{2} > 0$, then we have:

$y_n = x_n - \delta = x_n - \frac{\varepsilon_o}{2}$ but since $x_n > \frac{\varepsilon_o}{2}$, then $y_n > 0$ for $n \geq \max\{N_1, N_2\}$.

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  • $\begingroup$ Looks OK to me. But the title of the question is a badly garbled translation of the question you quote. Could you fix that? $\endgroup$ – Ethan Bolker Feb 25 '17 at 19:59
  • $\begingroup$ Yes haha sorry I really couldn't figure out a good title. Do you have any suggestions? $\endgroup$ – student_t Feb 25 '17 at 20:00
  • $\begingroup$ I edited the title. Your proof would be easier if you know that Cauchy sequences converge. Then you could use the positive limit instead of the given fact that $0$ is not the limit. But maybe that theorem is in your future.. $\endgroup$ – Ethan Bolker Feb 25 '17 at 20:27

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