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Is my argument correct?

Proposition. There is no divergent monotone sequence with a Cauchy subsequence.

Proof. Assume that we have a divergent monotone sequence $(a_n)$ with a Cauchy sequence $(a_{n_k})$, from theorem $\textbf{2.6.3}$ we know that there exists an $M>0$ such that $|a_{n_k}|<M,\forall k\in\mathbf{N}$.

We now examine the case where $(a_n)$ is increasing, the case where $(a_n)$ is decreasing will be handled similarly. Let $r\in\mathbf{N}$, evidently $r\leq n_r$ but then $a_r\leq a_{n_r}$ and since $a_{n_r}\leq |a_{n_r}|\leq M$ consequently $a_r\leq M$, implying that $(a_n)$ is bounded, but $(a_n)$ cannot be bounded since this together with $(a_n)$ being increasing would imply $(a_n)$ is convergent by theorem $\textbf{2.4.1}$.

$\blacksquare$

Note:

  • $\textbf{2.6.3}$ All Cauchy sequences are bounded.

  • $\textbf{2.4.1}$ All bounded monotone sequeces are convergent.

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  • $\begingroup$ Correct! Good job. And yes, you showed that there is no such thing. $\endgroup$ – A. Pongrácz Jul 31 '18 at 18:13
  • $\begingroup$ @DavidC.Ullrich that was indeed my intention what would you say is wrong? $\endgroup$ – Atif Farooq Jul 31 '18 at 18:14
  • $\begingroup$ The statement. Say "there is no"... $\endgroup$ – A. Pongrácz Jul 31 '18 at 18:15
  • $\begingroup$ @A.Pongrácz Thanks corrections made $\endgroup$ – Atif Farooq Jul 31 '18 at 18:16

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