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?

  • 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

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 $p=1,2,...$


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

  • $\begingroup$ 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

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