Definition: A sequence $(x_{n})$ is bounded away from zero if there exists some bound $b$ such that $\lvert x_{n}\rvert >b$ for all $n$; in particular $x_{n}\neq 0$ for all $n$.
I want to prove that if a Cauchy sequence does not tend to zero, then it is bounded away from zero. My attemp: If $(x_{n})$ does not tend to zero, there exists $\varepsilon_{0}>0$ such that, for all $M\geq 0$ then there exists $n>M$ such that $\lvert x_{n}\rvert >\varepsilon_{0}$. I have also that $(x_{n})$ is Cauchy, hence for all $\varepsilon > 0$ there exists $N\geq 0$ such that $\lvert x_{m}-x_{n}\rvert <\varepsilon$ whenever $m,n\geq N$. If I take $\varepsilon =\varepsilon_{0}$, $M = N$ $m>M$ such that $\lvert x_{m}\rvert >\varepsilon_{0}$ then if for some $n>M, x_{n}=0$ we have $\lvert x_{m}-x_{n}\rvert >\varepsilon_{0}$ which is a contraddiction with the Cauchy condition. The problem is that I can't see why $x_{n}\neq 0$ also when $n<M$ and so I can't conclude the proof. Can someone help me? Thanks before!