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







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















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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








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

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





























    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





















          Your Answer





          StackExchange.ifUsing("editor", function () {
          return StackExchange.using("mathjaxEditing", function () {
          StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
          StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
          });
          });
          }, "mathjax-editing");

          StackExchange.ready(function() {
          var channelOptions = {
          tags: "".split(" "),
          id: "69"
          };
          initTagRenderer("".split(" "), "".split(" "), channelOptions);

          StackExchange.using("externalEditor", function() {
          // Have to fire editor after snippets, if snippets enabled
          if (StackExchange.settings.snippets.snippetsEnabled) {
          StackExchange.using("snippets", function() {
          createEditor();
          });
          }
          else {
          createEditor();
          }
          });

          function createEditor() {
          StackExchange.prepareEditor({
          heartbeatType: 'answer',
          autoActivateHeartbeat: false,
          convertImagesToLinks: true,
          noModals: true,
          showLowRepImageUploadWarning: true,
          reputationToPostImages: 10,
          bindNavPrevention: true,
          postfix: "",
          imageUploader: {
          brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
          contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
          allowUrls: true
          },
          noCode: true, onDemand: true,
          discardSelector: ".discard-answer"
          ,immediatelyShowMarkdownHelp:true
          });


          }
          });














          draft saved

          draft discarded


















          StackExchange.ready(
          function () {
          StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3009766%2fa-sequence-in-mathbbr-that-has-no-cauchy-subsequence%23new-answer', 'question_page');
          }
          );

          Post as a guest















          Required, but never shown

























          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





































                                          draft saved

                                          draft discarded




















































                                          Thanks for contributing an answer to Mathematics Stack Exchange!


                                          • Please be sure to answer the question. Provide details and share your research!

                                          But avoid …



                                          • Asking for help, clarification, or responding to other answers.

                                          • Making statements based on opinion; back them up with references or personal experience.


                                          Use MathJax to format equations. MathJax reference.


                                          To learn more, see our tips on writing great answers.





                                          Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


                                          Please pay close attention to the following guidance:


                                          • Please be sure to answer the question. Provide details and share your research!

                                          But avoid …



                                          • Asking for help, clarification, or responding to other answers.

                                          • Making statements based on opinion; back them up with references or personal experience.


                                          To learn more, see our tips on writing great answers.




                                          draft saved


                                          draft discarded














                                          StackExchange.ready(
                                          function () {
                                          StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3009766%2fa-sequence-in-mathbbr-that-has-no-cauchy-subsequence%23new-answer', 'question_page');
                                          }
                                          );

                                          Post as a guest















                                          Required, but never shown





















































                                          Required, but never shown














                                          Required, but never shown












                                          Required, but never shown







                                          Required, but never shown

































                                          Required, but never shown














                                          Required, but never shown












                                          Required, but never shown







                                          Required, but never shown







                                          -

                                          Popular posts from this blog

                                          AnyDesk - Fatal Program Failure

                                          Image
                                          up vote 0 down vote favorite Good day all. I have a window that pops up saying Fatal Program Failure (AnyDesk). This is what I have done so far: Uninstalled AnyDesk and reinstalled it. ( The window still appeared ) I disabled AnyDesk start up in Msconfig and the window seems to have disapeared. Lastly I made sure that AnyDesk worked by remoting into it, and it worked. Today the same window appeared. How do I get rid of this window? remote-desktop share | improve this question edited 10 hours ago Tomasz Jakub Rup 552 3 13 ...
                                          Read more

                                          How to calibrate 16:9 built-in touch-screen to a 4:3 resolution?

                                          Image
                                          up vote 0 down vote favorite I need help with an issue we're facing in a project where we are moving from a Windows-based POS environment to one running on CentOS7. The problem consists of a number of factors, please find the details below: The supplier of our POS application will only deliver an application that is hard-coded to run in a 1024:768 4:3 format. The hardware we are using is a POS unit from HP that has a built-in 16:9 touch-screen. The employees using this solution is not supposed to see nor have access to anything but the POS application. Our sales organization demands that touch is an available feature when going live with the new POS environment. Going further into this, allow me to explain what we have encountered on the way trying to achieve the goal of having a touch-screen: We've simply change...
                                          Read more

                                          QoS: MAC-Priority for clients behind a repeater

                                          Image
                                          1 1 I've setup a wireless network using a Linksys-Router (DD-WRT) as an AP and a TP-Link repeater for range extension. To manage bandwith for specific users, I've setup QoS-Rules on the AP (DD-WRT) using MAC-Priority: However the AP shows only the Repeater's MAC and the MACs of users inside the AP-Range. The MAC addresses of users behind the Repeater are not listed : The purple MAC is not listed in the AP-Clientlist. Its related to a device connected via the TP-Link repeater. Q: How can I setup MAC-Priority for repeater-clients? Do I simply add the purple MAC in AP-MAC priority list although it's not listed as a client? I want to setup specific priority instead of giving one priority for all clients behind the repeater by adding the repeater's MAC in the MAC priority list in the AP as ...
                                          Read more