0
$\begingroup$

I would like to know if this is an acceptable proof. I have the following statement

Show directly that a bounded, monotone increasing sequence is a Cauchy sequence.

Let $\{x_n\}$ be a sequence that is bounded and monotone increasing. Then, by the Monotone Convergence Theorem, $\{x_n\}$ converges. Furthermore, by the Cauchy Criterion, we know that every convergent sequence is a Cauchy sequence and so we are done. QED

Does this work?

$\endgroup$
  • 3
    $\begingroup$ This is not what "show directly" means. $\endgroup$ – Qiaochu Yuan Nov 23 '17 at 20:35
  • $\begingroup$ In general, when asked to prove “directly” it means using the basic definitions. $\endgroup$ – Anurag A Nov 23 '17 at 20:35
  • $\begingroup$ $L=\sup_n x_n$ is an upper bound, and you can find $n$ such that $L-x_n$ is as small as you like. $\endgroup$ – copper.hat Nov 23 '17 at 20:36
  • $\begingroup$ Thank you very much! But does my argument make sense if they were not asking me to ''show directly''? $\endgroup$ – Michelle Drolet Nov 23 '17 at 22:21
  • $\begingroup$ For a direct argument, see math.stackexchange.com/questions/566635/… $\endgroup$ – Arnaud D. Nov 24 '17 at 9:10
1
$\begingroup$

Let $L=\sup_{n}x_{n}$, for $\epsilon>0$, find some $N$ such that $L-\epsilon<x_{N}\leq L$. As $\{x_{n}\}$ is increasing, $L-\epsilon<x_{N}\leq x_{n}\leq L$ for all $n\geq N$. In particular, $|x_{n+p}-x_{n}|=x_{n+p}-x_{n}=x_{n+p}-L+L-x_{n}\leq L-x_{n}<\epsilon$ for all $n\geq N$ and $p=1,2,...$

$\endgroup$
0
$\begingroup$

Start here. Put $s = \sup_n x_n$; show that $x_n \to s$.

$\endgroup$
  • $\begingroup$ This does not provide an answer to the question. To critique or request clarification from an author, leave a comment below their post. - From Review $\endgroup$ – Arnaud D. Nov 23 '17 at 21:44
  • $\begingroup$ He is looking for a hint, not for someone to give away the shop. $\endgroup$ – ncmathsadist Nov 24 '17 at 1:12
  • $\begingroup$ But the question says "Prove directly (...)". I think this means "without using the fact that it is convergent." $\endgroup$ – Arnaud D. Nov 24 '17 at 9:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.