I'm trying to prove the following:
If $(a_n)$ is a cauchy sequence that does NOT tend to 0, then $\exists N$ s.t. $\forall n > N, a_n \neq 0$.
Here's my proposed proof (please excuse the poor writing):
$(a_n)$ does not tend to 0 implies that there exists a smallest $\epsilon > 0$ s.t. $\forall M \exists k > M$ s.t. $|a_k| > \epsilon$. This smallest $\epsilon$ must exist, else we could choose arbitrarily smaller $\epsilon$ s.t. $|a_n| < \epsilon$ and then $(a_n)$ would tend to 0, which is false.
Since $(a_n)$ is a cauchy sequence, choose $N$ s.t. $|a_n - a_m| < \epsilon$ (the $\epsilon$ above) $\forall n, m > N$.
Then $|a_n - a_k| < \epsilon$ (substituting $a_k$ for $a_m$ since one of the $a_m$ must satisfy the property that $a_k$ has)
$-\epsilon < a_n - a_k < \epsilon$
$a_k - \epsilon < a_n < a_k + \epsilon$
If $a_k > \epsilon$, then $0 < a_k - \epsilon < a_n$
If $a_k < -\epsilon$, then $a_n < a_k + \epsilon < 0$
In either case, $a_n \neq 0$
Is my proposed proof correct? Is there a simpler, more elegant way to do it? The part I'm most wary of is where I claim the existence of a smallest $\epsilon$ since the sequence doesn't tend to 0. It makes sense to me, but I'm not sure if I'm allowed to.
P.S. We haven't proved that all cauchy sequences in the reals converge, so I'm not sure if I'm allowed to use that fact.