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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!

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  • $\begingroup$ Yes, this looks correct; basically, you have infinitely-many terms very close to the (finite) limit, and only finitely-many that are not, and these finitely-many are all Real(finite) numbers. $\endgroup$
    – user99680
    Commented Jun 16, 2014 at 6:34
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    $\begingroup$ Mild error. We can say that for all $n\ge N_0$, we have $|a_n|\lt |a_{N_0}|+\epsilon_0$. $\endgroup$ Commented Jun 16, 2014 at 6:37
  • $\begingroup$ @AndréNicolas Ah! I see what you mean. I was thinking to myself that since every $a_{m}$ could only be so far away from $a_{n}$ it would be bounded. I can make the correction you suggested or could I possibly also just remove the " by $\epsilon_{0}$" from "is bounded for... by $\epsilon_{0}$". I see now that it isn't necessarily bounded by $\epsilon_{0}$. $\endgroup$ Commented Jun 16, 2014 at 6:46
  • $\begingroup$ One might as well be explicit, as in the answers given. $\endgroup$ Commented Jun 16, 2014 at 6:54
  • $\begingroup$ @user99680 Why are there finitely-many terms not close to the limit? I understand why, as Andre Nicolas put it, for all $n \geq N_{0}$, we have $|a_{n}| < |a_{N_{0}}| + \epsilon_{0}$. Can't there be infinitely indexes, $n$, less than $N_{0}$, however? Or, more convincingly, can't the chosen $N_{0}$ be infinitely far into the sequence, meaning that the remaining indexes before $N_{0}$ are also infinite and, thus, have no bound? $\endgroup$
    – Muno
    Commented Feb 28, 2016 at 23:12

2 Answers 2

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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 $\pm \max \left\{ |a_0|, |a_1|, \ldots, |a_{N-1}|, |a_{N}|, \epsilon + |a_N|\right\}$).

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  • $\begingroup$ Yeah, admittedly my notation is probably a bit funky. I was going for $\mathbb{N}_{k}$ but the 'naught' made it look weird.I actually like this proof as it is still a slight variation to the popular proof. That said, I was hoping to verify this ascetic approach to the proof, not necessarily describing the bounds but establishing that they're there. $\endgroup$ Commented Jun 16, 2014 at 6:54
  • $\begingroup$ It is unclear what $N_{1}$ is. Downvoting. $\endgroup$ Commented Oct 29, 2016 at 8:40
  • $\begingroup$ sorry if this question is a bit silly but doesnt the triangle inequality state that $ | |a_m| - |a_n|| \leq |a_m -a_n | $ in your second line? how did we get rid of the outer brackets? we don't know if the sequence is increasing or decreasing or if $a_m$ is less or greater than $a_n$ so i'm a bit confused and stuck in my proof $\endgroup$
    – bluemuse
    Commented Feb 19, 2019 at 1:16
  • $\begingroup$ $|a_m|-|a_n| \leq ||a_m| - |a_n||$; the left hand side is equal to the right hand side when non-negative, and is smaller when it is negative (since the right hand side is positive. $\endgroup$
    – Batman
    Commented Feb 19, 2019 at 4:34
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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 $(a-\epsilon, a+\epsilon)$.

  1. 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 $m$.

  2. All the terms of the sequence are contained in the interval $(c,d)$, where :

$c=\min\{m, a-\epsilon\}$ ; $d=\max\{M, a+\epsilon\}$

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  • $\begingroup$ I hope it was not overkill. $\endgroup$
    – user99680
    Commented Jun 16, 2014 at 7:00
  • $\begingroup$ This is what I was looking for actually! Thank you! $\endgroup$ Commented Jun 16, 2014 at 7:05
  • $\begingroup$ Glad you liked it. $\endgroup$
    – user99680
    Commented Jun 16, 2014 at 7:09

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