Can a Cauchy sequence converge for one metric while not converging for another?












10












$begingroup$


Is there an easy example of one and the same space $X$ with two different metrics $d$ and $e$ such that one and the same sequence ${x_n}$ is a Cauchy sequence for both metrics, but converges only for one of them?






convergence metric-spaces cauchy-sequences







share|cite|improve this question









New contributor




rplantiko is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$








  • 1




    $begingroup$
    You should define some additional properties you want your metric to have. Otherwise something like "every two different points have distance one" will also work as an example.
    $endgroup$
    – Dirk
    2 days ago










  • $begingroup$
    @Dirk In a discrete space a Cauchy sequence is eventually constant, hence if a Cauchy sequence converges in one discrete metric, it converges in all of them to the same limit. Thus, it will not work as an example.
    $endgroup$
    – uniquesolution
    2 days ago








  • 1




    $begingroup$
    @uniquesolution Yeah, but on the other hand, every converging sequence (in whatever metric you like) that doesn't get constant will not converge in a discrete metric.
    $endgroup$
    – Dirk
    2 days ago










  • $begingroup$
    @Dirk, Every convergent sequence is Cauchy in any metric space. Hence in a discrete space every convergent sequence will also get eventually constant. Therefore there is no converging sequence that does not get eventually constant in a discrete space.
    $endgroup$
    – uniquesolution
    2 days ago






  • 3




    $begingroup$
    Goblin has an excellent point. You say the "same space $X$". If by "space" you are referring to the topology, not just the set, then the answer is no.
    $endgroup$
    – Paul Sinclair
    2 days ago
















10












$begingroup$


Is there an easy example of one and the same space $X$ with two different metrics $d$ and $e$ such that one and the same sequence ${x_n}$ is a Cauchy sequence for both metrics, but converges only for one of them?






convergence metric-spaces cauchy-sequences







share|cite|improve this question









New contributor




rplantiko is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$








  • 1




    $begingroup$
    You should define some additional properties you want your metric to have. Otherwise something like "every two different points have distance one" will also work as an example.
    $endgroup$
    – Dirk
    2 days ago










  • $begingroup$
    @Dirk In a discrete space a Cauchy sequence is eventually constant, hence if a Cauchy sequence converges in one discrete metric, it converges in all of them to the same limit. Thus, it will not work as an example.
    $endgroup$
    – uniquesolution
    2 days ago








  • 1




    $begingroup$
    @uniquesolution Yeah, but on the other hand, every converging sequence (in whatever metric you like) that doesn't get constant will not converge in a discrete metric.
    $endgroup$
    – Dirk
    2 days ago










  • $begingroup$
    @Dirk, Every convergent sequence is Cauchy in any metric space. Hence in a discrete space every convergent sequence will also get eventually constant. Therefore there is no converging sequence that does not get eventually constant in a discrete space.
    $endgroup$
    – uniquesolution
    2 days ago






  • 3




    $begingroup$
    Goblin has an excellent point. You say the "same space $X$". If by "space" you are referring to the topology, not just the set, then the answer is no.
    $endgroup$
    – Paul Sinclair
    2 days ago














10












10








10


1



$begingroup$


Is there an easy example of one and the same space $X$ with two different metrics $d$ and $e$ such that one and the same sequence ${x_n}$ is a Cauchy sequence for both metrics, but converges only for one of them?






convergence metric-spaces cauchy-sequences







share|cite|improve this question









New contributor




rplantiko is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$




Is there an easy example of one and the same space $X$ with two different metrics $d$ and $e$ such that one and the same sequence ${x_n}$ is a Cauchy sequence for both metrics, but converges only for one of them?






convergence metric-spaces cauchy-sequences




convergence metric-spaces cauchy-sequences






share|cite|improve this question









New contributor




rplantiko is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











share|cite|improve this question









New contributor




rplantiko is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









share|cite|improve this question




share|cite|improve this question








edited 2 days ago









José Carlos Santos

173k23133241




173k23133241






New contributor




rplantiko is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









asked 2 days ago









rplantikorplantiko

1625




1625




New contributor




rplantiko is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.





New contributor





rplantiko is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






rplantiko is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.








  • 1




    $begingroup$
    You should define some additional properties you want your metric to have. Otherwise something like "every two different points have distance one" will also work as an example.
    $endgroup$
    – Dirk
    2 days ago










  • $begingroup$
    @Dirk In a discrete space a Cauchy sequence is eventually constant, hence if a Cauchy sequence converges in one discrete metric, it converges in all of them to the same limit. Thus, it will not work as an example.
    $endgroup$
    – uniquesolution
    2 days ago








  • 1




    $begingroup$
    @uniquesolution Yeah, but on the other hand, every converging sequence (in whatever metric you like) that doesn't get constant will not converge in a discrete metric.
    $endgroup$
    – Dirk
    2 days ago










  • $begingroup$
    @Dirk, Every convergent sequence is Cauchy in any metric space. Hence in a discrete space every convergent sequence will also get eventually constant. Therefore there is no converging sequence that does not get eventually constant in a discrete space.
    $endgroup$
    – uniquesolution
    2 days ago






  • 3




    $begingroup$
    Goblin has an excellent point. You say the "same space $X$". If by "space" you are referring to the topology, not just the set, then the answer is no.
    $endgroup$
    – Paul Sinclair
    2 days ago














  • 1




    $begingroup$
    You should define some additional properties you want your metric to have. Otherwise something like "every two different points have distance one" will also work as an example.
    $endgroup$
    – Dirk
    2 days ago










  • $begingroup$
    @Dirk In a discrete space a Cauchy sequence is eventually constant, hence if a Cauchy sequence converges in one discrete metric, it converges in all of them to the same limit. Thus, it will not work as an example.
    $endgroup$
    – uniquesolution
    2 days ago








  • 1




    $begingroup$
    @uniquesolution Yeah, but on the other hand, every converging sequence (in whatever metric you like) that doesn't get constant will not converge in a discrete metric.
    $endgroup$
    – Dirk
    2 days ago










  • $begingroup$
    @Dirk, Every convergent sequence is Cauchy in any metric space. Hence in a discrete space every convergent sequence will also get eventually constant. Therefore there is no converging sequence that does not get eventually constant in a discrete space.
    $endgroup$
    – uniquesolution
    2 days ago






  • 3




    $begingroup$
    Goblin has an excellent point. You say the "same space $X$". If by "space" you are referring to the topology, not just the set, then the answer is no.
    $endgroup$
    – Paul Sinclair
    2 days ago








1




1




$begingroup$
You should define some additional properties you want your metric to have. Otherwise something like "every two different points have distance one" will also work as an example.
$endgroup$
– Dirk
2 days ago




$begingroup$
You should define some additional properties you want your metric to have. Otherwise something like "every two different points have distance one" will also work as an example.
$endgroup$
– Dirk
2 days ago












$begingroup$
@Dirk In a discrete space a Cauchy sequence is eventually constant, hence if a Cauchy sequence converges in one discrete metric, it converges in all of them to the same limit. Thus, it will not work as an example.
$endgroup$
– uniquesolution
2 days ago






$begingroup$
@Dirk In a discrete space a Cauchy sequence is eventually constant, hence if a Cauchy sequence converges in one discrete metric, it converges in all of them to the same limit. Thus, it will not work as an example.
$endgroup$
– uniquesolution
2 days ago






1




1




$begingroup$
@uniquesolution Yeah, but on the other hand, every converging sequence (in whatever metric you like) that doesn't get constant will not converge in a discrete metric.
$endgroup$
– Dirk
2 days ago




$begingroup$
@uniquesolution Yeah, but on the other hand, every converging sequence (in whatever metric you like) that doesn't get constant will not converge in a discrete metric.
$endgroup$
– Dirk
2 days ago












$begingroup$
@Dirk, Every convergent sequence is Cauchy in any metric space. Hence in a discrete space every convergent sequence will also get eventually constant. Therefore there is no converging sequence that does not get eventually constant in a discrete space.
$endgroup$
– uniquesolution
2 days ago




$begingroup$
@Dirk, Every convergent sequence is Cauchy in any metric space. Hence in a discrete space every convergent sequence will also get eventually constant. Therefore there is no converging sequence that does not get eventually constant in a discrete space.
$endgroup$
– uniquesolution
2 days ago




3




3




$begingroup$
Goblin has an excellent point. You say the "same space $X$". If by "space" you are referring to the topology, not just the set, then the answer is no.
$endgroup$
– Paul Sinclair
2 days ago




$begingroup$
Goblin has an excellent point. You say the "same space $X$". If by "space" you are referring to the topology, not just the set, then the answer is no.
$endgroup$
– Paul Sinclair
2 days ago










4 Answers
4






active

oldest

votes


















21












$begingroup$

Take $X=[0,infty)$, let $d(x,y)=lvert x-yrvert$ and let$$e(x,y)=begin{cases}lvert x-yrvert&text{ if }x,yneq0\lvert x+1rvert&text{ if }xneq0text{ and }y=0\lvert y+1rvert&text{ if }x=0text{ and }yneq0\0&text{ if }x,y=0.end{cases}$$Then $left(frac1nright)_{ninmathbb N}$ is a Cauchy sequence with respect to both metrics, but it is convergent only in $(X,d)$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Is there a reason this example needs to use $X = [0, infty)$ rather than $X = mathbb{R}$?
    $endgroup$
    – Michael Seifert
    2 days ago






  • 1




    $begingroup$
    If, in my answer, you work with $mathbb R$ instead of $[0,infty)$, then $e$ will not be a distance, because then $e(0,-1)=0$.
    $endgroup$
    – José Carlos Santos
    2 days ago










  • $begingroup$
    The second metric basically makes it ${-1}cup (0,infty)$, which can't really be said to be a metric for the same space.
    $endgroup$
    – Matt Samuel
    2 days ago






  • 3




    $begingroup$
    @MattSamuel That seems like an easier way to define this function. Let $f(0)=-1$ and $f(x)=x$ for $x>0$, and define $e(x,y)=|f(x)-f(y)|$. This is a metric because $f$ is injective.
    $endgroup$
    – Mario Carneiro
    yesterday










  • $begingroup$
    For $X = mathbb{R}$, one could use $e(0,x)=e(x,0)=|x|+1$ when $xneq 0$.
    $endgroup$
    – Litho
    yesterday



















16












$begingroup$

You can always have some artificial example where you just "move the limit elsewhere". For example, let $X=mathbb{R_{ge 0}}$, $d_1$ be the Euclidean metric and $d_2(x, y)=|hat{x}-hat{y}|$, where $ hat{x}=-1$ if $x=0$ and $ hat{x}=x$ otherwise. Then the sequence $x_n=frac 1n$ converges in $(X, d_1)$ but not in $(X, d_2)$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Is this somehow a standard example? Since you other poster also jsed it? Which book is it from
    $endgroup$
    – lalala
    2 days ago






  • 1




    $begingroup$
    AFAIK, this isn't from some standard reference (at least I didn't refer to one). I guess that's just a coincidence. After all, $x_n=frac 1n$ is often the first (non-trivial) convergent sequence when you are asked to think of one and moving $0$ to $-1$ just seems like a natural thing to do if you have to move $0$ away
    $endgroup$
    – Poon Levi
    2 days ago










  • $begingroup$
    But how did you see immediately that this metric is a metric? (Maybe its obvious if one has a geometric interpretation like you seem to have). Triangular inequality doesnt seem obvious to me.
    $endgroup$
    – lalala
    yesterday








  • 1




    $begingroup$
    $d_2$ is essentially the Euclidean metric on a new set ${-1}cup(0,infty)$. But since we can't change the base set, we instead use a map $xmapstohat{x}$ to rename $0$ to $-1$.
    $endgroup$
    – Poon Levi
    yesterday










  • $begingroup$
    Thanks! Yes. It is just a renameing. (I am not so used to metrics which do not derive from a norm)
    $endgroup$
    – lalala
    yesterday



















9












$begingroup$

Depending on what you mean, the answer is "no". In particular, if by "two metrics on the one space" you mean two metrics on the same set that induce the same topology, then the answer is no because convergence is a topological property.



However the other answers are correct if the induced topologies are allowed to be different.






share|cite|improve this answer









$endgroup$





















    8












    $begingroup$

    Having "the same X" is somewhat meaningless, since there's no shared structure other than cardinality. Given any other set Y with metric $d_Y$ and a injective map $phi: X rightarrow Y$, you can define $e(x_1,x_2) = d_Y(phi(x_1),phi(x_2))$. So all you have to do is find two spaces with the same cardinality where one has a convergent Cauchy sequence and the other has a non-convergent one. Or find a convergent Cauchy sequence and a non-convergent Cauchy sequence in the same space, then map one sequence to the other.






    share|cite|improve this answer









    $endgroup$














      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
      });


      }
      });






      rplantiko is a new contributor. Be nice, and check out our Code of Conduct.










      draft saved

      draft discarded


















      StackExchange.ready(
      function () {
      StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3175604%2fcan-a-cauchy-sequence-converge-for-one-metric-while-not-converging-for-another%23new-answer', 'question_page');
      }
      );

      Post as a guest















      Required, but never shown

























      4 Answers
      4






      active

      oldest

      votes








      4 Answers
      4






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      21












      $begingroup$

      Take $X=[0,infty)$, let $d(x,y)=lvert x-yrvert$ and let$$e(x,y)=begin{cases}lvert x-yrvert&text{ if }x,yneq0\lvert x+1rvert&text{ if }xneq0text{ and }y=0\lvert y+1rvert&text{ if }x=0text{ and }yneq0\0&text{ if }x,y=0.end{cases}$$Then $left(frac1nright)_{ninmathbb N}$ is a Cauchy sequence with respect to both metrics, but it is convergent only in $(X,d)$.






      share|cite|improve this answer









      $endgroup$













      • $begingroup$
        Is there a reason this example needs to use $X = [0, infty)$ rather than $X = mathbb{R}$?
        $endgroup$
        – Michael Seifert
        2 days ago






      • 1




        $begingroup$
        If, in my answer, you work with $mathbb R$ instead of $[0,infty)$, then $e$ will not be a distance, because then $e(0,-1)=0$.
        $endgroup$
        – José Carlos Santos
        2 days ago










      • $begingroup$
        The second metric basically makes it ${-1}cup (0,infty)$, which can't really be said to be a metric for the same space.
        $endgroup$
        – Matt Samuel
        2 days ago






      • 3




        $begingroup$
        @MattSamuel That seems like an easier way to define this function. Let $f(0)=-1$ and $f(x)=x$ for $x>0$, and define $e(x,y)=|f(x)-f(y)|$. This is a metric because $f$ is injective.
        $endgroup$
        – Mario Carneiro
        yesterday










      • $begingroup$
        For $X = mathbb{R}$, one could use $e(0,x)=e(x,0)=|x|+1$ when $xneq 0$.
        $endgroup$
        – Litho
        yesterday
















      21












      $begingroup$

      Take $X=[0,infty)$, let $d(x,y)=lvert x-yrvert$ and let$$e(x,y)=begin{cases}lvert x-yrvert&text{ if }x,yneq0\lvert x+1rvert&text{ if }xneq0text{ and }y=0\lvert y+1rvert&text{ if }x=0text{ and }yneq0\0&text{ if }x,y=0.end{cases}$$Then $left(frac1nright)_{ninmathbb N}$ is a Cauchy sequence with respect to both metrics, but it is convergent only in $(X,d)$.






      share|cite|improve this answer









      $endgroup$













      • $begingroup$
        Is there a reason this example needs to use $X = [0, infty)$ rather than $X = mathbb{R}$?
        $endgroup$
        – Michael Seifert
        2 days ago






      • 1




        $begingroup$
        If, in my answer, you work with $mathbb R$ instead of $[0,infty)$, then $e$ will not be a distance, because then $e(0,-1)=0$.
        $endgroup$
        – José Carlos Santos
        2 days ago










      • $begingroup$
        The second metric basically makes it ${-1}cup (0,infty)$, which can't really be said to be a metric for the same space.
        $endgroup$
        – Matt Samuel
        2 days ago






      • 3




        $begingroup$
        @MattSamuel That seems like an easier way to define this function. Let $f(0)=-1$ and $f(x)=x$ for $x>0$, and define $e(x,y)=|f(x)-f(y)|$. This is a metric because $f$ is injective.
        $endgroup$
        – Mario Carneiro
        yesterday










      • $begingroup$
        For $X = mathbb{R}$, one could use $e(0,x)=e(x,0)=|x|+1$ when $xneq 0$.
        $endgroup$
        – Litho
        yesterday














      21












      21








      21





      $begingroup$

      Take $X=[0,infty)$, let $d(x,y)=lvert x-yrvert$ and let$$e(x,y)=begin{cases}lvert x-yrvert&text{ if }x,yneq0\lvert x+1rvert&text{ if }xneq0text{ and }y=0\lvert y+1rvert&text{ if }x=0text{ and }yneq0\0&text{ if }x,y=0.end{cases}$$Then $left(frac1nright)_{ninmathbb N}$ is a Cauchy sequence with respect to both metrics, but it is convergent only in $(X,d)$.






      share|cite|improve this answer









      $endgroup$



      Take $X=[0,infty)$, let $d(x,y)=lvert x-yrvert$ and let$$e(x,y)=begin{cases}lvert x-yrvert&text{ if }x,yneq0\lvert x+1rvert&text{ if }xneq0text{ and }y=0\lvert y+1rvert&text{ if }x=0text{ and }yneq0\0&text{ if }x,y=0.end{cases}$$Then $left(frac1nright)_{ninmathbb N}$ is a Cauchy sequence with respect to both metrics, but it is convergent only in $(X,d)$.







      share|cite|improve this answer












      share|cite|improve this answer



      share|cite|improve this answer










      answered 2 days ago









      José Carlos SantosJosé Carlos Santos

      173k23133241




      173k23133241












      • $begingroup$
        Is there a reason this example needs to use $X = [0, infty)$ rather than $X = mathbb{R}$?
        $endgroup$
        – Michael Seifert
        2 days ago






      • 1




        $begingroup$
        If, in my answer, you work with $mathbb R$ instead of $[0,infty)$, then $e$ will not be a distance, because then $e(0,-1)=0$.
        $endgroup$
        – José Carlos Santos
        2 days ago










      • $begingroup$
        The second metric basically makes it ${-1}cup (0,infty)$, which can't really be said to be a metric for the same space.
        $endgroup$
        – Matt Samuel
        2 days ago






      • 3




        $begingroup$
        @MattSamuel That seems like an easier way to define this function. Let $f(0)=-1$ and $f(x)=x$ for $x>0$, and define $e(x,y)=|f(x)-f(y)|$. This is a metric because $f$ is injective.
        $endgroup$
        – Mario Carneiro
        yesterday










      • $begingroup$
        For $X = mathbb{R}$, one could use $e(0,x)=e(x,0)=|x|+1$ when $xneq 0$.
        $endgroup$
        – Litho
        yesterday


















      • $begingroup$
        Is there a reason this example needs to use $X = [0, infty)$ rather than $X = mathbb{R}$?
        $endgroup$
        – Michael Seifert
        2 days ago






      • 1




        $begingroup$
        If, in my answer, you work with $mathbb R$ instead of $[0,infty)$, then $e$ will not be a distance, because then $e(0,-1)=0$.
        $endgroup$
        – José Carlos Santos
        2 days ago










      • $begingroup$
        The second metric basically makes it ${-1}cup (0,infty)$, which can't really be said to be a metric for the same space.
        $endgroup$
        – Matt Samuel
        2 days ago






      • 3




        $begingroup$
        @MattSamuel That seems like an easier way to define this function. Let $f(0)=-1$ and $f(x)=x$ for $x>0$, and define $e(x,y)=|f(x)-f(y)|$. This is a metric because $f$ is injective.
        $endgroup$
        – Mario Carneiro
        yesterday










      • $begingroup$
        For $X = mathbb{R}$, one could use $e(0,x)=e(x,0)=|x|+1$ when $xneq 0$.
        $endgroup$
        – Litho
        yesterday
















      $begingroup$
      Is there a reason this example needs to use $X = [0, infty)$ rather than $X = mathbb{R}$?
      $endgroup$
      – Michael Seifert
      2 days ago




      $begingroup$
      Is there a reason this example needs to use $X = [0, infty)$ rather than $X = mathbb{R}$?
      $endgroup$
      – Michael Seifert
      2 days ago




      1




      1




      $begingroup$
      If, in my answer, you work with $mathbb R$ instead of $[0,infty)$, then $e$ will not be a distance, because then $e(0,-1)=0$.
      $endgroup$
      – José Carlos Santos
      2 days ago




      $begingroup$
      If, in my answer, you work with $mathbb R$ instead of $[0,infty)$, then $e$ will not be a distance, because then $e(0,-1)=0$.
      $endgroup$
      – José Carlos Santos
      2 days ago












      $begingroup$
      The second metric basically makes it ${-1}cup (0,infty)$, which can't really be said to be a metric for the same space.
      $endgroup$
      – Matt Samuel
      2 days ago




      $begingroup$
      The second metric basically makes it ${-1}cup (0,infty)$, which can't really be said to be a metric for the same space.
      $endgroup$
      – Matt Samuel
      2 days ago




      3




      3




      $begingroup$
      @MattSamuel That seems like an easier way to define this function. Let $f(0)=-1$ and $f(x)=x$ for $x>0$, and define $e(x,y)=|f(x)-f(y)|$. This is a metric because $f$ is injective.
      $endgroup$
      – Mario Carneiro
      yesterday




      $begingroup$
      @MattSamuel That seems like an easier way to define this function. Let $f(0)=-1$ and $f(x)=x$ for $x>0$, and define $e(x,y)=|f(x)-f(y)|$. This is a metric because $f$ is injective.
      $endgroup$
      – Mario Carneiro
      yesterday












      $begingroup$
      For $X = mathbb{R}$, one could use $e(0,x)=e(x,0)=|x|+1$ when $xneq 0$.
      $endgroup$
      – Litho
      yesterday




      $begingroup$
      For $X = mathbb{R}$, one could use $e(0,x)=e(x,0)=|x|+1$ when $xneq 0$.
      $endgroup$
      – Litho
      yesterday











      16












      $begingroup$

      You can always have some artificial example where you just "move the limit elsewhere". For example, let $X=mathbb{R_{ge 0}}$, $d_1$ be the Euclidean metric and $d_2(x, y)=|hat{x}-hat{y}|$, where $ hat{x}=-1$ if $x=0$ and $ hat{x}=x$ otherwise. Then the sequence $x_n=frac 1n$ converges in $(X, d_1)$ but not in $(X, d_2)$.






      share|cite|improve this answer









      $endgroup$













      • $begingroup$
        Is this somehow a standard example? Since you other poster also jsed it? Which book is it from
        $endgroup$
        – lalala
        2 days ago






      • 1




        $begingroup$
        AFAIK, this isn't from some standard reference (at least I didn't refer to one). I guess that's just a coincidence. After all, $x_n=frac 1n$ is often the first (non-trivial) convergent sequence when you are asked to think of one and moving $0$ to $-1$ just seems like a natural thing to do if you have to move $0$ away
        $endgroup$
        – Poon Levi
        2 days ago










      • $begingroup$
        But how did you see immediately that this metric is a metric? (Maybe its obvious if one has a geometric interpretation like you seem to have). Triangular inequality doesnt seem obvious to me.
        $endgroup$
        – lalala
        yesterday








      • 1




        $begingroup$
        $d_2$ is essentially the Euclidean metric on a new set ${-1}cup(0,infty)$. But since we can't change the base set, we instead use a map $xmapstohat{x}$ to rename $0$ to $-1$.
        $endgroup$
        – Poon Levi
        yesterday










      • $begingroup$
        Thanks! Yes. It is just a renameing. (I am not so used to metrics which do not derive from a norm)
        $endgroup$
        – lalala
        yesterday
















      16












      $begingroup$

      You can always have some artificial example where you just "move the limit elsewhere". For example, let $X=mathbb{R_{ge 0}}$, $d_1$ be the Euclidean metric and $d_2(x, y)=|hat{x}-hat{y}|$, where $ hat{x}=-1$ if $x=0$ and $ hat{x}=x$ otherwise. Then the sequence $x_n=frac 1n$ converges in $(X, d_1)$ but not in $(X, d_2)$.






      share|cite|improve this answer









      $endgroup$













      • $begingroup$
        Is this somehow a standard example? Since you other poster also jsed it? Which book is it from
        $endgroup$
        – lalala
        2 days ago






      • 1




        $begingroup$
        AFAIK, this isn't from some standard reference (at least I didn't refer to one). I guess that's just a coincidence. After all, $x_n=frac 1n$ is often the first (non-trivial) convergent sequence when you are asked to think of one and moving $0$ to $-1$ just seems like a natural thing to do if you have to move $0$ away
        $endgroup$
        – Poon Levi
        2 days ago










      • $begingroup$
        But how did you see immediately that this metric is a metric? (Maybe its obvious if one has a geometric interpretation like you seem to have). Triangular inequality doesnt seem obvious to me.
        $endgroup$
        – lalala
        yesterday








      • 1




        $begingroup$
        $d_2$ is essentially the Euclidean metric on a new set ${-1}cup(0,infty)$. But since we can't change the base set, we instead use a map $xmapstohat{x}$ to rename $0$ to $-1$.
        $endgroup$
        – Poon Levi
        yesterday










      • $begingroup$
        Thanks! Yes. It is just a renameing. (I am not so used to metrics which do not derive from a norm)
        $endgroup$
        – lalala
        yesterday














      16












      16








      16





      $begingroup$

      You can always have some artificial example where you just "move the limit elsewhere". For example, let $X=mathbb{R_{ge 0}}$, $d_1$ be the Euclidean metric and $d_2(x, y)=|hat{x}-hat{y}|$, where $ hat{x}=-1$ if $x=0$ and $ hat{x}=x$ otherwise. Then the sequence $x_n=frac 1n$ converges in $(X, d_1)$ but not in $(X, d_2)$.






      share|cite|improve this answer









      $endgroup$



      You can always have some artificial example where you just "move the limit elsewhere". For example, let $X=mathbb{R_{ge 0}}$, $d_1$ be the Euclidean metric and $d_2(x, y)=|hat{x}-hat{y}|$, where $ hat{x}=-1$ if $x=0$ and $ hat{x}=x$ otherwise. Then the sequence $x_n=frac 1n$ converges in $(X, d_1)$ but not in $(X, d_2)$.







      share|cite|improve this answer












      share|cite|improve this answer



      share|cite|improve this answer










      answered 2 days ago









      Poon LeviPoon Levi

      67639




      67639












      • $begingroup$
        Is this somehow a standard example? Since you other poster also jsed it? Which book is it from
        $endgroup$
        – lalala
        2 days ago






      • 1




        $begingroup$
        AFAIK, this isn't from some standard reference (at least I didn't refer to one). I guess that's just a coincidence. After all, $x_n=frac 1n$ is often the first (non-trivial) convergent sequence when you are asked to think of one and moving $0$ to $-1$ just seems like a natural thing to do if you have to move $0$ away
        $endgroup$
        – Poon Levi
        2 days ago










      • $begingroup$
        But how did you see immediately that this metric is a metric? (Maybe its obvious if one has a geometric interpretation like you seem to have). Triangular inequality doesnt seem obvious to me.
        $endgroup$
        – lalala
        yesterday








      • 1




        $begingroup$
        $d_2$ is essentially the Euclidean metric on a new set ${-1}cup(0,infty)$. But since we can't change the base set, we instead use a map $xmapstohat{x}$ to rename $0$ to $-1$.
        $endgroup$
        – Poon Levi
        yesterday










      • $begingroup$
        Thanks! Yes. It is just a renameing. (I am not so used to metrics which do not derive from a norm)
        $endgroup$
        – lalala
        yesterday


















      • $begingroup$
        Is this somehow a standard example? Since you other poster also jsed it? Which book is it from
        $endgroup$
        – lalala
        2 days ago






      • 1




        $begingroup$
        AFAIK, this isn't from some standard reference (at least I didn't refer to one). I guess that's just a coincidence. After all, $x_n=frac 1n$ is often the first (non-trivial) convergent sequence when you are asked to think of one and moving $0$ to $-1$ just seems like a natural thing to do if you have to move $0$ away
        $endgroup$
        – Poon Levi
        2 days ago










      • $begingroup$
        But how did you see immediately that this metric is a metric? (Maybe its obvious if one has a geometric interpretation like you seem to have). Triangular inequality doesnt seem obvious to me.
        $endgroup$
        – lalala
        yesterday








      • 1




        $begingroup$
        $d_2$ is essentially the Euclidean metric on a new set ${-1}cup(0,infty)$. But since we can't change the base set, we instead use a map $xmapstohat{x}$ to rename $0$ to $-1$.
        $endgroup$
        – Poon Levi
        yesterday










      • $begingroup$
        Thanks! Yes. It is just a renameing. (I am not so used to metrics which do not derive from a norm)
        $endgroup$
        – lalala
        yesterday
















      $begingroup$
      Is this somehow a standard example? Since you other poster also jsed it? Which book is it from
      $endgroup$
      – lalala
      2 days ago




      $begingroup$
      Is this somehow a standard example? Since you other poster also jsed it? Which book is it from
      $endgroup$
      – lalala
      2 days ago




      1




      1




      $begingroup$
      AFAIK, this isn't from some standard reference (at least I didn't refer to one). I guess that's just a coincidence. After all, $x_n=frac 1n$ is often the first (non-trivial) convergent sequence when you are asked to think of one and moving $0$ to $-1$ just seems like a natural thing to do if you have to move $0$ away
      $endgroup$
      – Poon Levi
      2 days ago




      $begingroup$
      AFAIK, this isn't from some standard reference (at least I didn't refer to one). I guess that's just a coincidence. After all, $x_n=frac 1n$ is often the first (non-trivial) convergent sequence when you are asked to think of one and moving $0$ to $-1$ just seems like a natural thing to do if you have to move $0$ away
      $endgroup$
      – Poon Levi
      2 days ago












      $begingroup$
      But how did you see immediately that this metric is a metric? (Maybe its obvious if one has a geometric interpretation like you seem to have). Triangular inequality doesnt seem obvious to me.
      $endgroup$
      – lalala
      yesterday






      $begingroup$
      But how did you see immediately that this metric is a metric? (Maybe its obvious if one has a geometric interpretation like you seem to have). Triangular inequality doesnt seem obvious to me.
      $endgroup$
      – lalala
      yesterday






      1




      1




      $begingroup$
      $d_2$ is essentially the Euclidean metric on a new set ${-1}cup(0,infty)$. But since we can't change the base set, we instead use a map $xmapstohat{x}$ to rename $0$ to $-1$.
      $endgroup$
      – Poon Levi
      yesterday




      $begingroup$
      $d_2$ is essentially the Euclidean metric on a new set ${-1}cup(0,infty)$. But since we can't change the base set, we instead use a map $xmapstohat{x}$ to rename $0$ to $-1$.
      $endgroup$
      – Poon Levi
      yesterday












      $begingroup$
      Thanks! Yes. It is just a renameing. (I am not so used to metrics which do not derive from a norm)
      $endgroup$
      – lalala
      yesterday




      $begingroup$
      Thanks! Yes. It is just a renameing. (I am not so used to metrics which do not derive from a norm)
      $endgroup$
      – lalala
      yesterday











      9












      $begingroup$

      Depending on what you mean, the answer is "no". In particular, if by "two metrics on the one space" you mean two metrics on the same set that induce the same topology, then the answer is no because convergence is a topological property.



      However the other answers are correct if the induced topologies are allowed to be different.






      share|cite|improve this answer









      $endgroup$


















        9












        $begingroup$

        Depending on what you mean, the answer is "no". In particular, if by "two metrics on the one space" you mean two metrics on the same set that induce the same topology, then the answer is no because convergence is a topological property.



        However the other answers are correct if the induced topologies are allowed to be different.






        share|cite|improve this answer









        $endgroup$
















          9












          9








          9





          $begingroup$

          Depending on what you mean, the answer is "no". In particular, if by "two metrics on the one space" you mean two metrics on the same set that induce the same topology, then the answer is no because convergence is a topological property.



          However the other answers are correct if the induced topologies are allowed to be different.






          share|cite|improve this answer









          $endgroup$



          Depending on what you mean, the answer is "no". In particular, if by "two metrics on the one space" you mean two metrics on the same set that induce the same topology, then the answer is no because convergence is a topological property.



          However the other answers are correct if the induced topologies are allowed to be different.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered 2 days ago









          goblingoblin

          37.1k1159194




          37.1k1159194























              8












              $begingroup$

              Having "the same X" is somewhat meaningless, since there's no shared structure other than cardinality. Given any other set Y with metric $d_Y$ and a injective map $phi: X rightarrow Y$, you can define $e(x_1,x_2) = d_Y(phi(x_1),phi(x_2))$. So all you have to do is find two spaces with the same cardinality where one has a convergent Cauchy sequence and the other has a non-convergent one. Or find a convergent Cauchy sequence and a non-convergent Cauchy sequence in the same space, then map one sequence to the other.






              share|cite|improve this answer









              $endgroup$


















                8












                $begingroup$

                Having "the same X" is somewhat meaningless, since there's no shared structure other than cardinality. Given any other set Y with metric $d_Y$ and a injective map $phi: X rightarrow Y$, you can define $e(x_1,x_2) = d_Y(phi(x_1),phi(x_2))$. So all you have to do is find two spaces with the same cardinality where one has a convergent Cauchy sequence and the other has a non-convergent one. Or find a convergent Cauchy sequence and a non-convergent Cauchy sequence in the same space, then map one sequence to the other.






                share|cite|improve this answer









                $endgroup$
















                  8












                  8








                  8





                  $begingroup$

                  Having "the same X" is somewhat meaningless, since there's no shared structure other than cardinality. Given any other set Y with metric $d_Y$ and a injective map $phi: X rightarrow Y$, you can define $e(x_1,x_2) = d_Y(phi(x_1),phi(x_2))$. So all you have to do is find two spaces with the same cardinality where one has a convergent Cauchy sequence and the other has a non-convergent one. Or find a convergent Cauchy sequence and a non-convergent Cauchy sequence in the same space, then map one sequence to the other.






                  share|cite|improve this answer









                  $endgroup$



                  Having "the same X" is somewhat meaningless, since there's no shared structure other than cardinality. Given any other set Y with metric $d_Y$ and a injective map $phi: X rightarrow Y$, you can define $e(x_1,x_2) = d_Y(phi(x_1),phi(x_2))$. So all you have to do is find two spaces with the same cardinality where one has a convergent Cauchy sequence and the other has a non-convergent one. Or find a convergent Cauchy sequence and a non-convergent Cauchy sequence in the same space, then map one sequence to the other.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered 2 days ago









                  AcccumulationAcccumulation

                  7,3052619




                  7,3052619






















                      rplantiko is a new contributor. Be nice, and check out our Code of Conduct.










                      draft saved

                      draft discarded


















                      rplantiko is a new contributor. Be nice, and check out our Code of Conduct.













                      rplantiko is a new contributor. Be nice, and check out our Code of Conduct.












                      rplantiko is a new contributor. Be nice, and check out our Code of Conduct.
















                      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.




                      draft saved


                      draft discarded














                      StackExchange.ready(
                      function () {
                      StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3175604%2fcan-a-cauchy-sequence-converge-for-one-metric-while-not-converging-for-another%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

                      數位音樂下載

                      Image
                      數位音樂下載 為非以「實體」方式來販賣音樂的相關產品，而使用數位格式如mp3、AAC等文件格式來進行銷售販賣的一种音乐产业形式。著名的數位音樂下載服務商有iTunes Store、RecoChoku、Google Play音乐和微软Zune等。数字音乐下载在二十一世纪初一度风靡全球， [1] [2] 使得传统唱片产业走向衰落，但其自身也在2010年代流媒体音乐服务如Apple Music、Spotify等的挤压下迅速失去市场份额。 [3] [4] 数字音乐下载有便于携带、传播迅速、制作简单成本低等优点，与流媒体音乐相比也能更好保障音乐人权益； [5] 但它同时很可能带来侵犯版权和滥用的情形， [6] 也给音乐产业带来了诸多不利影响。数字音乐下载为音乐人和乐迷提供了更加便捷直接的交流渠道，也一定程度上推动了网络音乐人的出现和独立音乐的迅速发展。 [7] 目录 1 历史 1.1 技术积淀（1987-2000） 1.2 出现与兴起（2000-2004） 1.3 唱片市场的颠覆者（2004-2007） 1.4 巅峰和衰落（2005-2013） 1.5 摇摇欲坠的现状与不可预知的明天（2014至今） 2 主要文件格式 2.1 PCM编码格式 2.1.1 WAV 2.1.2 APE 2.1.3 FLAC 2.2 苹果推出的无损文件格式 2.2.1 AIFF 2.2.2 ALAC 2.3 有损压缩格式 2.3.1 MP3 2.3.2 AAC 3 主要市场 3.1 iTunes Store 3.2 Google Play音乐商店 3.3 亚马逊音乐商店 3.4 部分其他音乐商店 4 影响与评价 4.1 对音乐产业的影响 4.2 对科技行业的影响 4.3 正面评价 4.4 负面评价 4.4.1 侵权问题 4.4.2 经营模式不明晰 4.4.3 不适应移动时代 4.5 文化现象 5 延伸阅读 6 参考资料 历史 苹果iTunes+iPod的软硬件搭配让数字音...
                      Read more

                      When can things happen in Etherscan, such as the picture below?

                      Image
                      2 I don't know why balance of the address is 0. when can this situation? etherscan balances share | improve this question asked yesterday Ru Hi Ru Hi 11 1 New contributor Ru Hi is a new contributor to this site. Take care in asking for clarification, commenting, and answering. Check out our Code of Conduct. add a comment  |  ...
                      Read more

                      格利澤436b

                      Image
                      格利泽436b 太陽系外行星 太陽系外行星列表 艺术家笔下的格利泽436b 母恆星 母恆星 Gliese 436 星座 獅子座 赤經 ( α ) 11 h  42 m  11.0941 s [1] 赤緯 ( δ ) +26° 42′ 23.652″ [1] 距離 33.4 ly (10.2 pc) 光譜類型 M2.5 V [1] 軌道參數 半长轴 ( a ) 0.0291±0.0004 [2] AU 軌道離心率 ( e ) 0.150±0.012 [2] 公轉週期 ( P ) 2.643904±0.000005 [3] d (0.00723849 y) 軌道傾角 ( i ) 85.8 +0.21 −0.25 [3] ° 角距 ( θ ) 2.794 mas 近星點時間 ( T 0 ) 2,451,551.716 ±0.01 JD 半振幅 ( K ) 18.68±0.8 m/s 物理性质 质量 ( m ) 22.2±1.0 [2] M ⊕ 半徑 ( r ) 4.327±0.183 [2] [4] R ⊕ 密度 ( ρ ) 1.51 g cm -3 表面重力 ( g ) 1.18 g 溫度 ( T ) 712±36 [2] K 發現 發現時間 2004 發現者 巴特勒、沃格特、 馬西 et al. 發現方法 徑向速度、凌日法 發現地點 美國加利福尼亞州 發表論文 出版 其他名稱 Ross 905 b, GJ 436 b, [5] LTT 13213 b, GCTP 2704.10 b, LHS 310, AC+27:28217 b, Vyssotsky 616 b, HIP 57087 b, GEN# +9.80120068 b, LP 319-75 b, G 121-7 b, LSPM J1142+2642 b, 1RXS J114211.9+264328 b, ASCC 683818 b, G 147-68 b, UCAC2 41198281 b, BP...
                      Read more