Proof Check: Every Cauchy Sequence is Bounded Sorry if I keep asking for proof checks. I'll try to keep it to a minimum after this. I know this has a well-known proof. I understand that proof as well but I thought I'd do a proof that made sense to me and seemed, in some ways, simpler. Trouble is I'm not sure if it's totally correct. It's quite short though. I was just hoping someone could look it over and see if it is a valid proof. Thank you!
Lemma: Every Cauchy sequence is bounded.
Proof: Let $(a_{n})$ be Cauchy. We choose $ 0<\epsilon_{0}$. So $ \forall \; n>m\geq N_{0}$ we have that $\vert a_{n}-a_{m} \vert < \epsilon_{0}$. Therefore $(a_{n})$ is bounded for all $ m \geq N_{0} $ by $ \epsilon_{0} $. Since $ \mathbb{N}_{N_{0}}$ is finite, it is bounded. So, for all $ m<N_{0} $, $ (a_{n})$ is bounded. Therefore $(a_{n})$ is bounded.
I realize I haven't said what the bounds are but I think that's sort of irrelevant. So long as we know it's bounded. Any help is much appreciated!
 A: Yes, it is bounded, because (since the tag is Real-analysis):
1)The Reals are complete, so that the sequence converges to, say $a$, so that, for
any $\epsilon>0$, all-but-finitely many terms are in $(a-\epsilon, a+\epsilon)$.
2) The  terms that are (possibly) not in $(a-\epsilon, a+\epsilon)$ are finitely-many. A finite collection of Real numbers has an actual maximum, say $M$, and an actual minimum, say $m$. 
3) All the terms of the sequence are contained in the interval $(c,d)$, where :
$c$=Min{m, $a-\epsilon$} ; $d$=Max{M, $a+\epsilon$}
A: It looks like you skipped some steps and used some weird notation ($\mathbb{N}_{N_0}$). 
Choose $\epsilon > 0$. Then, there exists $N$ such that for $m,n \geq N$, $|a_m - a_n| < \epsilon$. By the triangle inequality, $|a_m| - |a_n| \leq |a_m - a_n | < \epsilon$. Take $n = N$ and we see $|a_m| - |a_N| < \epsilon$ for all $m  \geq N$. 
Rearranging, we have $|a_m| < \epsilon + |a_N|$ for all $m \geq N$. Thus, $|a_m| \leq \max \left\{ |a_0|, |a_1|, \ldots, |a_{N-1}|, |a_{N}|, \epsilon + |a_N|\right\}$ for all $m$. Thus, $a_m$ is bounded (it is sandwiched in $\pm \max \left\{ |a_0|, |a_1|, \ldots, |a_{N-1}|, |a_{N}|, \epsilon + |a_N|\right\}$). 
