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 $n \geq \max\{N_1, N_2\}$.