I'm trying to show that if $(x_n)_n∈N $ is a Cauchy sequence of rational numbers with $\lim_{n \to \infty} \neq 0$, that there exists $b \in \mathbb{Q}$ with $b> 0$ and $N \in \mathbb{N}$ so that $|x_n| > b$ for all $n \geq N$
So far, my proof is as follows:
Given that $x_n$ is a Cautchy sequence it is convergent. Hence, $x_n$ is a convergent sequence. Also, since $\lim_{n \to \infty} \neq 0 $, say $\lim_{n \to \infty} = L$, where $L \neq 0$. Then, given $\epsilon > 0$, there exists $N \in \mathbb{N}$ such that
$|x_n - L| < \epsilon \forall \; n \geq N$
$ - \epsilon < x_n - L < \epsilon \forall \; n \geq N$
$L - \epsilon < x_n < L + \epsilon \; \; \; \forall \; n \geq N$
Let $b = L - \epsilon$. It follows then that $|x_n| > b$.
However, I realize I have not shown that $b>0$. How could I do so to complete my proof? Also, are there any changes you would recommend to make my proof that it is better?
Choose $N$ such that $|x_n-L| <\frac {|L|} 2$ for $n \geq N$. Then $|x_n|=|L+(x_n-L)| \geq |L| -|x_n-L| >\frac {|L|} 2$ for $n \geq N$. Take $b$ to be any rational number in $(0, \frac {|L|} 2)$.