A sequence in $mathbb{R}$ that has no Cauchy subsequence












0












$begingroup$



Give an example of a sequence in $mathbb{R}$ which has no subsequence which is a Cauchy sequence.




I can find out a sequence that is not a Cauchy sequence such as ${ln(n)}$ once $|ln(n)-ln(n+1)|=0$ but $|ln(n)-ln(2n)|=|ln(frac{1}{2})|>epsilon$
$forall epsilon<ln(frac{1}{2})$



I can still find a subsequence of the type ${ln(2n)}_{2ninmathbb{N}}$ such that $|ln(2n)-ln(2n+1)|=0$



Question:



What should I do to get a sequence that has no Cauchy subsequence?



Thanks in advance!






real-analysis sequences-and-series general-topology metric-spaces cauchy-sequences







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$endgroup$








  • 2




    $begingroup$
    In the first place, you should only be looking at sequences which are not convergent, right?
    $endgroup$
    – MPW
    Nov 22 '18 at 22:57






  • 4




    $begingroup$
    I do not understand your question. What do you mean by $|ln(2n)-ln(2n+1)|=0$?
    $endgroup$
    – Carsten S
    Nov 23 '18 at 10:29
















0












$begingroup$



Give an example of a sequence in $mathbb{R}$ which has no subsequence which is a Cauchy sequence.




I can find out a sequence that is not a Cauchy sequence such as ${ln(n)}$ once $|ln(n)-ln(n+1)|=0$ but $|ln(n)-ln(2n)|=|ln(frac{1}{2})|>epsilon$
$forall epsilon<ln(frac{1}{2})$



I can still find a subsequence of the type ${ln(2n)}_{2ninmathbb{N}}$ such that $|ln(2n)-ln(2n+1)|=0$



Question:



What should I do to get a sequence that has no Cauchy subsequence?



Thanks in advance!






real-analysis sequences-and-series general-topology metric-spaces cauchy-sequences







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    In the first place, you should only be looking at sequences which are not convergent, right?
    $endgroup$
    – MPW
    Nov 22 '18 at 22:57






  • 4




    $begingroup$
    I do not understand your question. What do you mean by $|ln(2n)-ln(2n+1)|=0$?
    $endgroup$
    – Carsten S
    Nov 23 '18 at 10:29














0












0








0





$begingroup$



Give an example of a sequence in $mathbb{R}$ which has no subsequence which is a Cauchy sequence.




I can find out a sequence that is not a Cauchy sequence such as ${ln(n)}$ once $|ln(n)-ln(n+1)|=0$ but $|ln(n)-ln(2n)|=|ln(frac{1}{2})|>epsilon$
$forall epsilon<ln(frac{1}{2})$



I can still find a subsequence of the type ${ln(2n)}_{2ninmathbb{N}}$ such that $|ln(2n)-ln(2n+1)|=0$



Question:



What should I do to get a sequence that has no Cauchy subsequence?



Thanks in advance!






real-analysis sequences-and-series general-topology metric-spaces cauchy-sequences







share|cite|improve this question











$endgroup$





Give an example of a sequence in $mathbb{R}$ which has no subsequence which is a Cauchy sequence.




I can find out a sequence that is not a Cauchy sequence such as ${ln(n)}$ once $|ln(n)-ln(n+1)|=0$ but $|ln(n)-ln(2n)|=|ln(frac{1}{2})|>epsilon$
$forall epsilon<ln(frac{1}{2})$



I can still find a subsequence of the type ${ln(2n)}_{2ninmathbb{N}}$ such that $|ln(2n)-ln(2n+1)|=0$



Question:



What should I do to get a sequence that has no Cauchy subsequence?



Thanks in advance!






real-analysis sequences-and-series general-topology metric-spaces cauchy-sequences




real-analysis sequences-and-series general-topology metric-spaces cauchy-sequences






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share|cite|improve this question













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edited Nov 24 '18 at 7:17









user21820

39.4k543155




39.4k543155










asked Nov 22 '18 at 22:36









Pedro GomesPedro Gomes

1,9252721




1,9252721








  • 2




    $begingroup$
    In the first place, you should only be looking at sequences which are not convergent, right?
    $endgroup$
    – MPW
    Nov 22 '18 at 22:57






  • 4




    $begingroup$
    I do not understand your question. What do you mean by $|ln(2n)-ln(2n+1)|=0$?
    $endgroup$
    – Carsten S
    Nov 23 '18 at 10:29














  • 2




    $begingroup$
    In the first place, you should only be looking at sequences which are not convergent, right?
    $endgroup$
    – MPW
    Nov 22 '18 at 22:57






  • 4




    $begingroup$
    I do not understand your question. What do you mean by $|ln(2n)-ln(2n+1)|=0$?
    $endgroup$
    – Carsten S
    Nov 23 '18 at 10:29








2




2




$begingroup$
In the first place, you should only be looking at sequences which are not convergent, right?
$endgroup$
– MPW
Nov 22 '18 at 22:57




$begingroup$
In the first place, you should only be looking at sequences which are not convergent, right?
$endgroup$
– MPW
Nov 22 '18 at 22:57




4




4




$begingroup$
I do not understand your question. What do you mean by $|ln(2n)-ln(2n+1)|=0$?
$endgroup$
– Carsten S
Nov 23 '18 at 10:29




$begingroup$
I do not understand your question. What do you mean by $|ln(2n)-ln(2n+1)|=0$?
$endgroup$
– Carsten S
Nov 23 '18 at 10:29










6 Answers
6






active

oldest

votes


















27












$begingroup$

Take $a_n=n$. Then for any subsequence $n_k$, $|n_k-n_{k-1}|geq 1$. So, it has no Cauchy sub-sequence.






share|cite|improve this answer









$endgroup$





















    12












    $begingroup$

    Take a sequence $(a_n)$ such that $a_nto infty$. Then each subsequence $(a_{n_k})$ is such that $a_{n_k}to infty$ and thus can't be Cauchy since necessarily Cauchy sequences are bounded.






    share|cite|improve this answer









    $endgroup$





















      10












      $begingroup$

      Take the sequence $1,2,3,4,ldots$ It has no Cauchy subsequence since the distance between any two distinct terms is at least $1$.






      share|cite|improve this answer









      $endgroup$





















        7












        $begingroup$

        I didn't understand your analysis but sequence $a_n = ln n$ doesn't have Cauchy subsequence since its limit is $infty$






        share|cite|improve this answer









        $endgroup$













        • $begingroup$
          I guess this one uses the theorem that every Cauchy sequence converges
          $endgroup$
          – user334732
          Nov 23 '18 at 21:15



















        7












        $begingroup$

        As Foobaz John says, any sequence with $a_ntoinfty$ works, and it's easy to see that $|a_n|toinfty$ is also sufficient. In fact this latter condition is necessary and sufficient. If $|a_n|nottoinfty$, then by definition there is some $c>0$ for which there are infinitely many values $n_i$ such that $|a_{n_i}|<c$ for each $i$. Now by Bolzano-Weierstrass the subsequence $a_{n_i}$ has an infinite subsequence $a_{n_{i_j}}$ which is convergent, and therefore Cauchy.






        share|cite|improve this answer









        $endgroup$













        • $begingroup$
          Ah, Bolzano-Weierstrass, also known as the theorem about subsubsubscripts.
          $endgroup$
          – Misha Lavrov
          Nov 23 '18 at 18:32



















        1












        $begingroup$

        Bolzano-Weierstrass: Every bounded sequence has a convergent subsequence.



        Fact: Every convergent sequence is bounded.



        Strategy: Try an unbounded sequence.



        Guess: $a_n=n$



        Conclusion: (I leave it to you)






        share|cite|improve this answer









        $endgroup$













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          6 Answers
          6






          active

          oldest

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          6 Answers
          6






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          27












          $begingroup$

          Take $a_n=n$. Then for any subsequence $n_k$, $|n_k-n_{k-1}|geq 1$. So, it has no Cauchy sub-sequence.






          share|cite|improve this answer









          $endgroup$


















            27












            $begingroup$

            Take $a_n=n$. Then for any subsequence $n_k$, $|n_k-n_{k-1}|geq 1$. So, it has no Cauchy sub-sequence.






            share|cite|improve this answer









            $endgroup$
















              27












              27








              27





              $begingroup$

              Take $a_n=n$. Then for any subsequence $n_k$, $|n_k-n_{k-1}|geq 1$. So, it has no Cauchy sub-sequence.






              share|cite|improve this answer









              $endgroup$



              Take $a_n=n$. Then for any subsequence $n_k$, $|n_k-n_{k-1}|geq 1$. So, it has no Cauchy sub-sequence.







              share|cite|improve this answer












              share|cite|improve this answer



              share|cite|improve this answer










              answered Nov 22 '18 at 22:40









              John_WickJohn_Wick

              1,616111




              1,616111























                  12












                  $begingroup$

                  Take a sequence $(a_n)$ such that $a_nto infty$. Then each subsequence $(a_{n_k})$ is such that $a_{n_k}to infty$ and thus can't be Cauchy since necessarily Cauchy sequences are bounded.






                  share|cite|improve this answer









                  $endgroup$


















                    12












                    $begingroup$

                    Take a sequence $(a_n)$ such that $a_nto infty$. Then each subsequence $(a_{n_k})$ is such that $a_{n_k}to infty$ and thus can't be Cauchy since necessarily Cauchy sequences are bounded.






                    share|cite|improve this answer









                    $endgroup$
















                      12












                      12








                      12





                      $begingroup$

                      Take a sequence $(a_n)$ such that $a_nto infty$. Then each subsequence $(a_{n_k})$ is such that $a_{n_k}to infty$ and thus can't be Cauchy since necessarily Cauchy sequences are bounded.






                      share|cite|improve this answer









                      $endgroup$



                      Take a sequence $(a_n)$ such that $a_nto infty$. Then each subsequence $(a_{n_k})$ is such that $a_{n_k}to infty$ and thus can't be Cauchy since necessarily Cauchy sequences are bounded.







                      share|cite|improve this answer












                      share|cite|improve this answer



                      share|cite|improve this answer










                      answered Nov 22 '18 at 22:56









                      Foobaz JohnFoobaz John

                      22.7k41452




                      22.7k41452























                          10












                          $begingroup$

                          Take the sequence $1,2,3,4,ldots$ It has no Cauchy subsequence since the distance between any two distinct terms is at least $1$.






                          share|cite|improve this answer









                          $endgroup$


















                            10












                            $begingroup$

                            Take the sequence $1,2,3,4,ldots$ It has no Cauchy subsequence since the distance between any two distinct terms is at least $1$.






                            share|cite|improve this answer









                            $endgroup$
















                              10












                              10








                              10





                              $begingroup$

                              Take the sequence $1,2,3,4,ldots$ It has no Cauchy subsequence since the distance between any two distinct terms is at least $1$.






                              share|cite|improve this answer









                              $endgroup$



                              Take the sequence $1,2,3,4,ldots$ It has no Cauchy subsequence since the distance between any two distinct terms is at least $1$.







                              share|cite|improve this answer












                              share|cite|improve this answer



                              share|cite|improve this answer










                              answered Nov 22 '18 at 22:39









                              José Carlos SantosJosé Carlos Santos

                              168k22132236




                              168k22132236























                                  7












                                  $begingroup$

                                  I didn't understand your analysis but sequence $a_n = ln n$ doesn't have Cauchy subsequence since its limit is $infty$






                                  share|cite|improve this answer









                                  $endgroup$













                                  • $begingroup$
                                    I guess this one uses the theorem that every Cauchy sequence converges
                                    $endgroup$
                                    – user334732
                                    Nov 23 '18 at 21:15
















                                  7












                                  $begingroup$

                                  I didn't understand your analysis but sequence $a_n = ln n$ doesn't have Cauchy subsequence since its limit is $infty$






                                  share|cite|improve this answer









                                  $endgroup$













                                  • $begingroup$
                                    I guess this one uses the theorem that every Cauchy sequence converges
                                    $endgroup$
                                    – user334732
                                    Nov 23 '18 at 21:15














                                  7












                                  7








                                  7





                                  $begingroup$

                                  I didn't understand your analysis but sequence $a_n = ln n$ doesn't have Cauchy subsequence since its limit is $infty$






                                  share|cite|improve this answer









                                  $endgroup$



                                  I didn't understand your analysis but sequence $a_n = ln n$ doesn't have Cauchy subsequence since its limit is $infty$







                                  share|cite|improve this answer












                                  share|cite|improve this answer



                                  share|cite|improve this answer










                                  answered Nov 23 '18 at 8:09









                                  RiaDRiaD

                                  725718




                                  725718












                                  • $begingroup$
                                    I guess this one uses the theorem that every Cauchy sequence converges
                                    $endgroup$
                                    – user334732
                                    Nov 23 '18 at 21:15


















                                  • $begingroup$
                                    I guess this one uses the theorem that every Cauchy sequence converges
                                    $endgroup$
                                    – user334732
                                    Nov 23 '18 at 21:15
















                                  $begingroup$
                                  I guess this one uses the theorem that every Cauchy sequence converges
                                  $endgroup$
                                  – user334732
                                  Nov 23 '18 at 21:15




                                  $begingroup$
                                  I guess this one uses the theorem that every Cauchy sequence converges
                                  $endgroup$
                                  – user334732
                                  Nov 23 '18 at 21:15











                                  7












                                  $begingroup$

                                  As Foobaz John says, any sequence with $a_ntoinfty$ works, and it's easy to see that $|a_n|toinfty$ is also sufficient. In fact this latter condition is necessary and sufficient. If $|a_n|nottoinfty$, then by definition there is some $c>0$ for which there are infinitely many values $n_i$ such that $|a_{n_i}|<c$ for each $i$. Now by Bolzano-Weierstrass the subsequence $a_{n_i}$ has an infinite subsequence $a_{n_{i_j}}$ which is convergent, and therefore Cauchy.






                                  share|cite|improve this answer









                                  $endgroup$













                                  • $begingroup$
                                    Ah, Bolzano-Weierstrass, also known as the theorem about subsubsubscripts.
                                    $endgroup$
                                    – Misha Lavrov
                                    Nov 23 '18 at 18:32
















                                  7












                                  $begingroup$

                                  As Foobaz John says, any sequence with $a_ntoinfty$ works, and it's easy to see that $|a_n|toinfty$ is also sufficient. In fact this latter condition is necessary and sufficient. If $|a_n|nottoinfty$, then by definition there is some $c>0$ for which there are infinitely many values $n_i$ such that $|a_{n_i}|<c$ for each $i$. Now by Bolzano-Weierstrass the subsequence $a_{n_i}$ has an infinite subsequence $a_{n_{i_j}}$ which is convergent, and therefore Cauchy.






                                  share|cite|improve this answer









                                  $endgroup$













                                  • $begingroup$
                                    Ah, Bolzano-Weierstrass, also known as the theorem about subsubsubscripts.
                                    $endgroup$
                                    – Misha Lavrov
                                    Nov 23 '18 at 18:32














                                  7












                                  7








                                  7





                                  $begingroup$

                                  As Foobaz John says, any sequence with $a_ntoinfty$ works, and it's easy to see that $|a_n|toinfty$ is also sufficient. In fact this latter condition is necessary and sufficient. If $|a_n|nottoinfty$, then by definition there is some $c>0$ for which there are infinitely many values $n_i$ such that $|a_{n_i}|<c$ for each $i$. Now by Bolzano-Weierstrass the subsequence $a_{n_i}$ has an infinite subsequence $a_{n_{i_j}}$ which is convergent, and therefore Cauchy.






                                  share|cite|improve this answer









                                  $endgroup$



                                  As Foobaz John says, any sequence with $a_ntoinfty$ works, and it's easy to see that $|a_n|toinfty$ is also sufficient. In fact this latter condition is necessary and sufficient. If $|a_n|nottoinfty$, then by definition there is some $c>0$ for which there are infinitely many values $n_i$ such that $|a_{n_i}|<c$ for each $i$. Now by Bolzano-Weierstrass the subsequence $a_{n_i}$ has an infinite subsequence $a_{n_{i_j}}$ which is convergent, and therefore Cauchy.







                                  share|cite|improve this answer












                                  share|cite|improve this answer



                                  share|cite|improve this answer










                                  answered Nov 23 '18 at 13:02









                                  Especially LimeEspecially Lime

                                  22.4k22858




                                  22.4k22858












                                  • $begingroup$
                                    Ah, Bolzano-Weierstrass, also known as the theorem about subsubsubscripts.
                                    $endgroup$
                                    – Misha Lavrov
                                    Nov 23 '18 at 18:32


















                                  • $begingroup$
                                    Ah, Bolzano-Weierstrass, also known as the theorem about subsubsubscripts.
                                    $endgroup$
                                    – Misha Lavrov
                                    Nov 23 '18 at 18:32
















                                  $begingroup$
                                  Ah, Bolzano-Weierstrass, also known as the theorem about subsubsubscripts.
                                  $endgroup$
                                  – Misha Lavrov
                                  Nov 23 '18 at 18:32




                                  $begingroup$
                                  Ah, Bolzano-Weierstrass, also known as the theorem about subsubsubscripts.
                                  $endgroup$
                                  – Misha Lavrov
                                  Nov 23 '18 at 18:32











                                  1












                                  $begingroup$

                                  Bolzano-Weierstrass: Every bounded sequence has a convergent subsequence.



                                  Fact: Every convergent sequence is bounded.



                                  Strategy: Try an unbounded sequence.



                                  Guess: $a_n=n$



                                  Conclusion: (I leave it to you)






                                  share|cite|improve this answer









                                  $endgroup$


















                                    1












                                    $begingroup$

                                    Bolzano-Weierstrass: Every bounded sequence has a convergent subsequence.



                                    Fact: Every convergent sequence is bounded.



                                    Strategy: Try an unbounded sequence.



                                    Guess: $a_n=n$



                                    Conclusion: (I leave it to you)






                                    share|cite|improve this answer









                                    $endgroup$
















                                      1












                                      1








                                      1





                                      $begingroup$

                                      Bolzano-Weierstrass: Every bounded sequence has a convergent subsequence.



                                      Fact: Every convergent sequence is bounded.



                                      Strategy: Try an unbounded sequence.



                                      Guess: $a_n=n$



                                      Conclusion: (I leave it to you)






                                      share|cite|improve this answer









                                      $endgroup$



                                      Bolzano-Weierstrass: Every bounded sequence has a convergent subsequence.



                                      Fact: Every convergent sequence is bounded.



                                      Strategy: Try an unbounded sequence.



                                      Guess: $a_n=n$



                                      Conclusion: (I leave it to you)







                                      share|cite|improve this answer












                                      share|cite|improve this answer



                                      share|cite|improve this answer










                                      answered Nov 24 '18 at 3:26







                                      user198044





































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