An example of a sequence which does not have any subsequence with a finite limit.












2















Find and example of a sequence ${x_n}$ which does not contain any subsequences having a finite limit.




I've been thinking of the following seqeunce:
$$
x_n = sin(n)cdotsin(sqrt{3}n)
$$



But is that true? Also what are other examples of such sequences?






calculus limits examples-counterexamples







share|cite|improve this question


















  • 10




    Natural numbers $1,2,3,...$? No subsequence is Cauchy, hence convergent?
    – астон вілла олоф мэллбэрг
    Dec 10 at 15:46








  • 1




    @астонвіллаолофмэллбэрг oh my, shame on me
    – roman
    Dec 10 at 15:48






  • 4




    Remember, every bounded sequence of real numbers has a convergent subsequence. Therefore, your sequence being bounded does not have a convergent subsequence. The best bet is to take an unbounded sequence where the distance between any two terms is greater than some fixed number.
    – астон вілла олоф мэллбэрг
    Dec 10 at 15:50










  • @астонвіллаолофмэллбэрг thank you! it's crystal clear now
    – roman
    Dec 10 at 15:52






  • 1




    Great. You can now answer this question yourself and close it.
    – астон вілла олоф мэллбэрг
    Dec 10 at 15:52


















2















Find and example of a sequence ${x_n}$ which does not contain any subsequences having a finite limit.




I've been thinking of the following seqeunce:
$$
x_n = sin(n)cdotsin(sqrt{3}n)
$$



But is that true? Also what are other examples of such sequences?






calculus limits examples-counterexamples







share|cite|improve this question


















  • 10




    Natural numbers $1,2,3,...$? No subsequence is Cauchy, hence convergent?
    – астон вілла олоф мэллбэрг
    Dec 10 at 15:46








  • 1




    @астонвіллаолофмэллбэрг oh my, shame on me
    – roman
    Dec 10 at 15:48






  • 4




    Remember, every bounded sequence of real numbers has a convergent subsequence. Therefore, your sequence being bounded does not have a convergent subsequence. The best bet is to take an unbounded sequence where the distance between any two terms is greater than some fixed number.
    – астон вілла олоф мэллбэрг
    Dec 10 at 15:50










  • @астонвіллаолофмэллбэрг thank you! it's crystal clear now
    – roman
    Dec 10 at 15:52






  • 1




    Great. You can now answer this question yourself and close it.
    – астон вілла олоф мэллбэрг
    Dec 10 at 15:52
















2












2








2








Find and example of a sequence ${x_n}$ which does not contain any subsequences having a finite limit.




I've been thinking of the following seqeunce:
$$
x_n = sin(n)cdotsin(sqrt{3}n)
$$



But is that true? Also what are other examples of such sequences?






calculus limits examples-counterexamples







share|cite|improve this question














Find and example of a sequence ${x_n}$ which does not contain any subsequences having a finite limit.




I've been thinking of the following seqeunce:
$$
x_n = sin(n)cdotsin(sqrt{3}n)
$$



But is that true? Also what are other examples of such sequences?






calculus limits examples-counterexamples




calculus limits examples-counterexamples






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 10 at 15:45









roman

1,93421221




1,93421221








  • 10




    Natural numbers $1,2,3,...$? No subsequence is Cauchy, hence convergent?
    – астон вілла олоф мэллбэрг
    Dec 10 at 15:46








  • 1




    @астонвіллаолофмэллбэрг oh my, shame on me
    – roman
    Dec 10 at 15:48






  • 4




    Remember, every bounded sequence of real numbers has a convergent subsequence. Therefore, your sequence being bounded does not have a convergent subsequence. The best bet is to take an unbounded sequence where the distance between any two terms is greater than some fixed number.
    – астон вілла олоф мэллбэрг
    Dec 10 at 15:50










  • @астонвіллаолофмэллбэрг thank you! it's crystal clear now
    – roman
    Dec 10 at 15:52






  • 1




    Great. You can now answer this question yourself and close it.
    – астон вілла олоф мэллбэрг
    Dec 10 at 15:52
















  • 10




    Natural numbers $1,2,3,...$? No subsequence is Cauchy, hence convergent?
    – астон вілла олоф мэллбэрг
    Dec 10 at 15:46








  • 1




    @астонвіллаолофмэллбэрг oh my, shame on me
    – roman
    Dec 10 at 15:48






  • 4




    Remember, every bounded sequence of real numbers has a convergent subsequence. Therefore, your sequence being bounded does not have a convergent subsequence. The best bet is to take an unbounded sequence where the distance between any two terms is greater than some fixed number.
    – астон вілла олоф мэллбэрг
    Dec 10 at 15:50










  • @астонвіллаолофмэллбэрг thank you! it's crystal clear now
    – roman
    Dec 10 at 15:52






  • 1




    Great. You can now answer this question yourself and close it.
    – астон вілла олоф мэллбэрг
    Dec 10 at 15:52










10




10




Natural numbers $1,2,3,...$? No subsequence is Cauchy, hence convergent?
– астон вілла олоф мэллбэрг
Dec 10 at 15:46






Natural numbers $1,2,3,...$? No subsequence is Cauchy, hence convergent?
– астон вілла олоф мэллбэрг
Dec 10 at 15:46






1




1




@астонвіллаолофмэллбэрг oh my, shame on me
– roman
Dec 10 at 15:48




@астонвіллаолофмэллбэрг oh my, shame on me
– roman
Dec 10 at 15:48




4




4




Remember, every bounded sequence of real numbers has a convergent subsequence. Therefore, your sequence being bounded does not have a convergent subsequence. The best bet is to take an unbounded sequence where the distance between any two terms is greater than some fixed number.
– астон вілла олоф мэллбэрг
Dec 10 at 15:50




Remember, every bounded sequence of real numbers has a convergent subsequence. Therefore, your sequence being bounded does not have a convergent subsequence. The best bet is to take an unbounded sequence where the distance between any two terms is greater than some fixed number.
– астон вілла олоф мэллбэрг
Dec 10 at 15:50












@астонвіллаолофмэллбэрг thank you! it's crystal clear now
– roman
Dec 10 at 15:52




@астонвіллаолофмэллбэрг thank you! it's crystal clear now
– roman
Dec 10 at 15:52




1




1




Great. You can now answer this question yourself and close it.
– астон вілла олоф мэллбэрг
Dec 10 at 15:52






Great. You can now answer this question yourself and close it.
– астон вілла олоф мэллбэрг
Dec 10 at 15:52












2 Answers
2






active

oldest

votes


















7














Any sequence ${x_n}$ with $|x_n| to infty$ will do.



By Bolzano-Weierstrass Theorem others wont: if $|x_n| not to infty$, there is $M$, such that you can find a subsequence of ${x_n}$ that is bounded by $M$. Then you can find a converging subsequence of that subsequence (and hence of the original sequence).




EDIT: By the way, if we think of the one point compactification of $mathbb{R}$ where we add $infty$ in, then every sequence has a converging subsequence in this space (which is homeomorphic to circle). So this also shows that the only way to have no subsequence with finite limit is to make that converging subsequence that is quaranteed to exists in the compactified space to have limit $infty$. Well I guess you still have to make the above argument to show that we must have $|x_n|to infty$ (i.e $x to infty$ in the compactified space).






share|cite|improve this answer































    8














    The example in the OP is wrong. This can be shown by Bolzano-Weierstrass theorem which states that every bounded sequence contains a convergent subsequence.



    Now if we consider:
    $$
    x_n = sin(n)sin(sqrt{3}n)
    $$
    its clear that the sequence is bounded by $[-1, 1]$. Hence it contains a convergent subsequence.






    share|cite|improve this answer























    • Upvoted. Good job.
      – астон вілла олоф мэллбэрг
      Dec 10 at 16:00






    • 1




      Take for example $n=1, 8,77, 146, 721, 5270, 19492 , 33714 ,148357, 263000, 674357$ and you get $ 0.8305543, 0.9506199, 0.9883301, 0.9962933, 0.9996635, 0.9997120, 0.9998930, 0.9999708, 0.9999954, 0.9999987, 0.9999999$
      – Henry
      Dec 10 at 23:00













    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%2f3034077%2fan-example-of-a-sequence-which-does-not-have-any-subsequence-with-a-finite-limit%23new-answer', 'question_page');
    }
    );

    Post as a guest















    Required, but never shown

























    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    7














    Any sequence ${x_n}$ with $|x_n| to infty$ will do.



    By Bolzano-Weierstrass Theorem others wont: if $|x_n| not to infty$, there is $M$, such that you can find a subsequence of ${x_n}$ that is bounded by $M$. Then you can find a converging subsequence of that subsequence (and hence of the original sequence).




    EDIT: By the way, if we think of the one point compactification of $mathbb{R}$ where we add $infty$ in, then every sequence has a converging subsequence in this space (which is homeomorphic to circle). So this also shows that the only way to have no subsequence with finite limit is to make that converging subsequence that is quaranteed to exists in the compactified space to have limit $infty$. Well I guess you still have to make the above argument to show that we must have $|x_n|to infty$ (i.e $x to infty$ in the compactified space).






    share|cite|improve this answer




























      7














      Any sequence ${x_n}$ with $|x_n| to infty$ will do.



      By Bolzano-Weierstrass Theorem others wont: if $|x_n| not to infty$, there is $M$, such that you can find a subsequence of ${x_n}$ that is bounded by $M$. Then you can find a converging subsequence of that subsequence (and hence of the original sequence).




      EDIT: By the way, if we think of the one point compactification of $mathbb{R}$ where we add $infty$ in, then every sequence has a converging subsequence in this space (which is homeomorphic to circle). So this also shows that the only way to have no subsequence with finite limit is to make that converging subsequence that is quaranteed to exists in the compactified space to have limit $infty$. Well I guess you still have to make the above argument to show that we must have $|x_n|to infty$ (i.e $x to infty$ in the compactified space).






      share|cite|improve this answer


























        7












        7








        7






        Any sequence ${x_n}$ with $|x_n| to infty$ will do.



        By Bolzano-Weierstrass Theorem others wont: if $|x_n| not to infty$, there is $M$, such that you can find a subsequence of ${x_n}$ that is bounded by $M$. Then you can find a converging subsequence of that subsequence (and hence of the original sequence).




        EDIT: By the way, if we think of the one point compactification of $mathbb{R}$ where we add $infty$ in, then every sequence has a converging subsequence in this space (which is homeomorphic to circle). So this also shows that the only way to have no subsequence with finite limit is to make that converging subsequence that is quaranteed to exists in the compactified space to have limit $infty$. Well I guess you still have to make the above argument to show that we must have $|x_n|to infty$ (i.e $x to infty$ in the compactified space).






        share|cite|improve this answer














        Any sequence ${x_n}$ with $|x_n| to infty$ will do.



        By Bolzano-Weierstrass Theorem others wont: if $|x_n| not to infty$, there is $M$, such that you can find a subsequence of ${x_n}$ that is bounded by $M$. Then you can find a converging subsequence of that subsequence (and hence of the original sequence).




        EDIT: By the way, if we think of the one point compactification of $mathbb{R}$ where we add $infty$ in, then every sequence has a converging subsequence in this space (which is homeomorphic to circle). So this also shows that the only way to have no subsequence with finite limit is to make that converging subsequence that is quaranteed to exists in the compactified space to have limit $infty$. Well I guess you still have to make the above argument to show that we must have $|x_n|to infty$ (i.e $x to infty$ in the compactified space).







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Dec 11 at 9:01

























        answered Dec 10 at 15:58









        ploosu2

        4,5771023




        4,5771023























            8














            The example in the OP is wrong. This can be shown by Bolzano-Weierstrass theorem which states that every bounded sequence contains a convergent subsequence.



            Now if we consider:
            $$
            x_n = sin(n)sin(sqrt{3}n)
            $$
            its clear that the sequence is bounded by $[-1, 1]$. Hence it contains a convergent subsequence.






            share|cite|improve this answer























            • Upvoted. Good job.
              – астон вілла олоф мэллбэрг
              Dec 10 at 16:00






            • 1




              Take for example $n=1, 8,77, 146, 721, 5270, 19492 , 33714 ,148357, 263000, 674357$ and you get $ 0.8305543, 0.9506199, 0.9883301, 0.9962933, 0.9996635, 0.9997120, 0.9998930, 0.9999708, 0.9999954, 0.9999987, 0.9999999$
              – Henry
              Dec 10 at 23:00


















            8














            The example in the OP is wrong. This can be shown by Bolzano-Weierstrass theorem which states that every bounded sequence contains a convergent subsequence.



            Now if we consider:
            $$
            x_n = sin(n)sin(sqrt{3}n)
            $$
            its clear that the sequence is bounded by $[-1, 1]$. Hence it contains a convergent subsequence.






            share|cite|improve this answer























            • Upvoted. Good job.
              – астон вілла олоф мэллбэрг
              Dec 10 at 16:00






            • 1




              Take for example $n=1, 8,77, 146, 721, 5270, 19492 , 33714 ,148357, 263000, 674357$ and you get $ 0.8305543, 0.9506199, 0.9883301, 0.9962933, 0.9996635, 0.9997120, 0.9998930, 0.9999708, 0.9999954, 0.9999987, 0.9999999$
              – Henry
              Dec 10 at 23:00
















            8












            8








            8






            The example in the OP is wrong. This can be shown by Bolzano-Weierstrass theorem which states that every bounded sequence contains a convergent subsequence.



            Now if we consider:
            $$
            x_n = sin(n)sin(sqrt{3}n)
            $$
            its clear that the sequence is bounded by $[-1, 1]$. Hence it contains a convergent subsequence.






            share|cite|improve this answer














            The example in the OP is wrong. This can be shown by Bolzano-Weierstrass theorem which states that every bounded sequence contains a convergent subsequence.



            Now if we consider:
            $$
            x_n = sin(n)sin(sqrt{3}n)
            $$
            its clear that the sequence is bounded by $[-1, 1]$. Hence it contains a convergent subsequence.







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Dec 10 at 16:00

























            answered Dec 10 at 15:58









            roman

            1,93421221




            1,93421221












            • Upvoted. Good job.
              – астон вілла олоф мэллбэрг
              Dec 10 at 16:00






            • 1




              Take for example $n=1, 8,77, 146, 721, 5270, 19492 , 33714 ,148357, 263000, 674357$ and you get $ 0.8305543, 0.9506199, 0.9883301, 0.9962933, 0.9996635, 0.9997120, 0.9998930, 0.9999708, 0.9999954, 0.9999987, 0.9999999$
              – Henry
              Dec 10 at 23:00




















            • Upvoted. Good job.
              – астон вілла олоф мэллбэрг
              Dec 10 at 16:00






            • 1




              Take for example $n=1, 8,77, 146, 721, 5270, 19492 , 33714 ,148357, 263000, 674357$ and you get $ 0.8305543, 0.9506199, 0.9883301, 0.9962933, 0.9996635, 0.9997120, 0.9998930, 0.9999708, 0.9999954, 0.9999987, 0.9999999$
              – Henry
              Dec 10 at 23:00


















            Upvoted. Good job.
            – астон вілла олоф мэллбэрг
            Dec 10 at 16:00




            Upvoted. Good job.
            – астон вілла олоф мэллбэрг
            Dec 10 at 16:00




            1




            1




            Take for example $n=1, 8,77, 146, 721, 5270, 19492 , 33714 ,148357, 263000, 674357$ and you get $ 0.8305543, 0.9506199, 0.9883301, 0.9962933, 0.9996635, 0.9997120, 0.9998930, 0.9999708, 0.9999954, 0.9999987, 0.9999999$
            – Henry
            Dec 10 at 23:00






            Take for example $n=1, 8,77, 146, 721, 5270, 19492 , 33714 ,148357, 263000, 674357$ and you get $ 0.8305543, 0.9506199, 0.9883301, 0.9962933, 0.9996635, 0.9997120, 0.9998930, 0.9999708, 0.9999954, 0.9999987, 0.9999999$
            – Henry
            Dec 10 at 23:00




















            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%2f3034077%2fan-example-of-a-sequence-which-does-not-have-any-subsequence-with-a-finite-limit%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

            How to change which sound is reproduced for terminal bell?

            Image
            3 Recently upgrade to 18.10 from 18.04, and using the terminal is becoming annoying: the new alert sound is a drip ( /usr/share/sounds/gnome/default/alerts/drip.ogg ), and the previous one is still located at ( /usr/share/sounds/ubuntu/stereo/bell.ogg ). My issue is that at the gnome-terminal, when autocompleting or pressing left BOTH sounds are reproduced, at the same time. I want to just have ubuntu/stereo/bell.ogg active, not both . At the moment I have been only able to reduce the volume of the alerts (but still both are played), or turn the alerts down (but I dislike both "solutions"). gnome-terminal 18.10 teminal-bell share | improve this question
            Read more

            Title Spacing in Bjornstrup Chapter, Removing Chapter Number From Contents

            Image
            up vote 5 down vote favorite 2 I would like to make it so that the vertical space below the title of each chapter is the same when using the Bjornstrup package (i.e. so that the descenders do not affect the spacing - as shown by the red arrow in the picture below). I think I'm fairly happy with the spacing around "A chapter", so I'd really like titles without descenders (e.g. "Contents") to have the same vertical spacing as "A chapter" below the title. I would also like to remove the large "0" in the contents title. Any help appreciated, as always. documentclass[11pt,a4paper]{book} % Packages[![enter image description here][1]][1] usepackage[left=3.5cm, right=3.5cm, top=4.3cm, bottom=4.25cm]{geometry} usepackage[utf8]{inputenc} usepackage{libertine} usepackage[libertine]{newtx
            Read more

            Can I use Tabulator js library in my java Spring + Thymeleaf project?

            Image
            0 I want to add Tabulator js to my java project. I added tabulator.min.js, tabulator.min.css to resources folder. After that added to my html page these dependencies as recommending in readme.md. <link rel="stylesheet" th:href="@{/css/tabulator.min.css}"> <script type="text/javascript" th:src="@{/js/tabulator.min.js}"></script> Now my tables view all elements, that I have in the html page, but not correct. In the documentation of Tabulator is example how to create header and table, var table = new Tabulator("#example-table", { columns:[ {title:"Name", field:"name", sorter:"string", width:200, editor:true}, {title:"Age", field:"age", sorter:"
            Read more