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












0















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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  • 2




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






  • 4




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
















0















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




















  • 2




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






  • 4




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














0












0








0








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
















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

38.7k543153




38.7k543153










asked Nov 22 '18 at 22:36









Pedro Gomes

1,6982720




1,6982720








  • 2




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






  • 4




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














  • 2




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






  • 4




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








2




2




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




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




4




4




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




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










6 Answers
6






active

oldest

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27














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





























    12














    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





























      10














      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





























        7














        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





















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



















        7














        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





















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



















        1














        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





















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          6 Answers
          6






          active

          oldest

          votes








          6 Answers
          6






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          27














          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


























            27














            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
























              27












              27








              27






              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












              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_Wick

              1,366111




              1,366111























                  12














                  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


























                    12














                    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
























                      12












                      12








                      12






                      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












                      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 John

                      21k41250




                      21k41250























                          10














                          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


























                            10














                            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
























                              10












                              10








                              10






                              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












                              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 Santos

                              150k22121221




                              150k22121221























                                  7














                                  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





















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
















                                  7














                                  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





















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














                                  7












                                  7








                                  7






                                  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












                                  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









                                  RiaD

                                  720717




                                  720717












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


















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
















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




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











                                  7














                                  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





















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
















                                  7














                                  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





















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














                                  7












                                  7








                                  7






                                  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












                                  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 Lime

                                  21.6k22858




                                  21.6k22858












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


















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
















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




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











                                  1














                                  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


























                                    1














                                    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
























                                      1












                                      1








                                      1






                                      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












                                      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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