Concerning the existence of a divergent monotone sequence with a Cauchy subsequence.

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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 $textbf2.6.3$ we know that there exists an $M>0$ such that $|a_n_k|<M,forall kinmathbfN$.



We now examine the case where $(a_n)$ is increasing, the case where $(a_n)$ is decreasing will be handled similarly. Let $rinmathbfN$, evidently $rleq n_r$ but then $a_rleq a_n_r$ and since $a_n_rleq |a_n_r|leq M$ consequently $a_rleq M$, 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 $textbf2.4.1$.



$blacksquare$



Note:



  • $textbf2.6.3$ All Cauchy sequences are bounded.


  • $textbf2.4.1$ All bounded monotone sequeces are convergent.




real-analysis sequences-and-series proof-verification cauchy-sequences




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  • Correct! Good job. And yes, you showed that there is no such thing.
    – A. Pongrácz
    Jul 31 at 18:13











  • @DavidC.Ullrich that was indeed my intention what would you say is wrong?
    – Atif Farooq
    Jul 31 at 18:14










  • The statement. Say "there is no"...
    – A. Pongrácz
    Jul 31 at 18:15










  • @A.Pongrácz Thanks corrections made
    – Atif Farooq
    Jul 31 at 18:16















up vote
1
down vote

favorite












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 $textbf2.6.3$ we know that there exists an $M>0$ such that $|a_n_k|<M,forall kinmathbfN$.



We now examine the case where $(a_n)$ is increasing, the case where $(a_n)$ is decreasing will be handled similarly. Let $rinmathbfN$, evidently $rleq n_r$ but then $a_rleq a_n_r$ and since $a_n_rleq |a_n_r|leq M$ consequently $a_rleq M$, 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 $textbf2.4.1$.



$blacksquare$



Note:



  • $textbf2.6.3$ All Cauchy sequences are bounded.


  • $textbf2.4.1$ All bounded monotone sequeces are convergent.




real-analysis sequences-and-series proof-verification cauchy-sequences




share|cite|improve this question





















  • Correct! Good job. And yes, you showed that there is no such thing.
    – A. Pongrácz
    Jul 31 at 18:13











  • @DavidC.Ullrich that was indeed my intention what would you say is wrong?
    – Atif Farooq
    Jul 31 at 18:14










  • The statement. Say "there is no"...
    – A. Pongrácz
    Jul 31 at 18:15










  • @A.Pongrácz Thanks corrections made
    – Atif Farooq
    Jul 31 at 18:16













up vote
1
down vote

favorite









up vote
1
down vote

favorite











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 $textbf2.6.3$ we know that there exists an $M>0$ such that $|a_n_k|<M,forall kinmathbfN$.



We now examine the case where $(a_n)$ is increasing, the case where $(a_n)$ is decreasing will be handled similarly. Let $rinmathbfN$, evidently $rleq n_r$ but then $a_rleq a_n_r$ and since $a_n_rleq |a_n_r|leq M$ consequently $a_rleq M$, 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 $textbf2.4.1$.



$blacksquare$



Note:



  • $textbf2.6.3$ All Cauchy sequences are bounded.


  • $textbf2.4.1$ All bounded monotone sequeces are convergent.




real-analysis sequences-and-series proof-verification cauchy-sequences




share|cite|improve this question













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 $textbf2.6.3$ we know that there exists an $M>0$ such that $|a_n_k|<M,forall kinmathbfN$.



We now examine the case where $(a_n)$ is increasing, the case where $(a_n)$ is decreasing will be handled similarly. Let $rinmathbfN$, evidently $rleq n_r$ but then $a_rleq a_n_r$ and since $a_n_rleq |a_n_r|leq M$ consequently $a_rleq M$, 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 $textbf2.4.1$.



$blacksquare$



Note:



  • $textbf2.6.3$ All Cauchy sequences are bounded.


  • $textbf2.4.1$ All bounded monotone sequeces are convergent.





real-analysis sequences-and-series proof-verification cauchy-sequences





share|cite|improve this question












share|cite|improve this question




share|cite|improve this question








edited Jul 31 at 18:16
























asked Jul 31 at 18:00









Atif Farooq

2,7352824




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  • Correct! Good job. And yes, you showed that there is no such thing.
    – A. Pongrácz
    Jul 31 at 18:13











  • @DavidC.Ullrich that was indeed my intention what would you say is wrong?
    – Atif Farooq
    Jul 31 at 18:14










  • The statement. Say "there is no"...
    – A. Pongrácz
    Jul 31 at 18:15










  • @A.Pongrácz Thanks corrections made
    – Atif Farooq
    Jul 31 at 18:16

















  • Correct! Good job. And yes, you showed that there is no such thing.
    – A. Pongrácz
    Jul 31 at 18:13











  • @DavidC.Ullrich that was indeed my intention what would you say is wrong?
    – Atif Farooq
    Jul 31 at 18:14










  • The statement. Say "there is no"...
    – A. Pongrácz
    Jul 31 at 18:15










  • @A.Pongrácz Thanks corrections made
    – Atif Farooq
    Jul 31 at 18:16
















Correct! Good job. And yes, you showed that there is no such thing.
– A. Pongrácz
Jul 31 at 18:13





Correct! Good job. And yes, you showed that there is no such thing.
– A. Pongrácz
Jul 31 at 18:13













@DavidC.Ullrich that was indeed my intention what would you say is wrong?
– Atif Farooq
Jul 31 at 18:14




@DavidC.Ullrich that was indeed my intention what would you say is wrong?
– Atif Farooq
Jul 31 at 18:14












The statement. Say "there is no"...
– A. Pongrácz
Jul 31 at 18:15




The statement. Say "there is no"...
– A. Pongrácz
Jul 31 at 18:15












@A.Pongrácz Thanks corrections made
– Atif Farooq
Jul 31 at 18:16





@A.Pongrácz Thanks corrections made
– Atif Farooq
Jul 31 at 18:16
















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