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












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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







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  • 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.
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  • 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









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rplantiko is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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share|cite|improve this question









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share|cite|improve this question




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edited 2 days ago









José Carlos Santos

173k23133241




173k23133241






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asked 2 days ago









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  • 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

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21












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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



















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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 $(X, d_2)$.






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$endgroup$













  • $begingroup$
    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



















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$begingroup$

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.






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    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.






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      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






















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