Can a Cauchy sequence converge for one metric while not converging for another?A “non-trivial” example of...

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


A “non-trivial” example of a Cauchy sequence that does not converge?How can one know every Cauchy sequence in a complete metric space converges?Does every Cauchy sequence converge in $mathbb{C}$?Cauchy sequence in vector topological and metric spaceAny Cauchy sequence in the metric space $(Bbb N, d)$ is either “converging” to infinity or ultimately constant.Incomplete metric space or normed space with only one non-convergent Cauchy sequenceConvergent but not Cauchy under two equivalent metricsSequence that is Cauchy, bounded, in a complete metric space but does not converge.Showing a sequence in a metric space is CauchyWhy can't completeness be defined on topological spaces without using metrics?













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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?










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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
    22 hours 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
    22 hours 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
    22 hours 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
    22 hours ago






  • 1




    $begingroup$
    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
    12 hours ago
















12












$begingroup$


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?










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
    22 hours 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
    22 hours 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
    22 hours 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
    22 hours ago






  • 1




    $begingroup$
    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
    12 hours ago














12












12








12


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?










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






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share|cite|improve this question









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share|cite|improve this question




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edited 22 hours ago









José Carlos Santos

172k23133241




172k23133241






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asked 22 hours ago









rplantikorplantiko

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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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  • 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
    22 hours 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
    22 hours 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
    22 hours 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
    22 hours ago






  • 1




    $begingroup$
    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
    12 hours 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
    22 hours 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
    22 hours 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
    22 hours 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
    22 hours ago






  • 1




    $begingroup$
    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
    12 hours 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
22 hours 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
22 hours 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
22 hours 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
22 hours 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
22 hours 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
22 hours 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
22 hours 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
22 hours ago




1




1




$begingroup$
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
12 hours 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
12 hours ago










4 Answers
4






active

oldest

votes


















18












$begingroup$

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
    17 hours 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
    16 hours 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
    5 hours ago






  • 1




    $begingroup$
    @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
    1 hour ago



















11












$begingroup$

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
    10 hours 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
    8 hours ago



















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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.






share|cite|improve this answer









$endgroup$





















    2












    $begingroup$

    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









    $endgroup$














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      4 Answers
      4






      active

      oldest

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      4 Answers
      4






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      18












      $begingroup$

      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
        17 hours 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
        16 hours 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
        5 hours ago






      • 1




        $begingroup$
        @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
        1 hour ago
















      18












      $begingroup$

      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
        17 hours 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
        16 hours 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
        5 hours ago






      • 1




        $begingroup$
        @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
        1 hour ago














      18












      18








      18





      $begingroup$

      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 22 hours ago









      José Carlos SantosJosé Carlos Santos

      172k23133241




      172k23133241












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






      • 1




        $begingroup$
        @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
        1 hour ago


















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






      • 1




        $begingroup$
        @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
        1 hour ago
















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




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




      1




      1




      $begingroup$
      @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
      1 hour ago




      $begingroup$
      @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
      1 hour ago











      11












      $begingroup$

      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
        10 hours 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
        8 hours ago
















      11












      $begingroup$

      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
        10 hours 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
        8 hours ago














      11












      11








      11





      $begingroup$

      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 21 hours ago









      Poon LeviPoon Levi

      62639




      62639












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


















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
















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




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











      7












      $begingroup$

      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









      $endgroup$


















        7












        $begingroup$

        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









        $endgroup$
















          7












          7








          7





          $begingroup$

          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 16 hours ago









          goblingoblin

          37.1k1159194




          37.1k1159194























              2












              $begingroup$

              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









              $endgroup$


















                2












                $begingroup$

                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









                $endgroup$
















                  2












                  2








                  2





                  $begingroup$

                  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 12 hours ago









                  AcccumulationAcccumulation

                  7,2452619




                  7,2452619






















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