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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?

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?

convergence metric-spaces cauchy-sequences

share|cite|improve this question

edited 22 hours ago

José Carlos Santos

172k23133241

asked 22 hours ago

rplantikorplantiko

1635

New contributor

rplantiko is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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$endgroup$

• 
  
  
  

  
  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
  

  
  
  
  


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  @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
  
  
  

  
  
  
  


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  @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
  

  
  
  
  


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  @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
  

  
  
  
  


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  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
  

  
  
  
  

add a comment | 

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?

convergence metric-spaces cauchy-sequences

share|cite|improve this question

edited 22 hours ago

José Carlos Santos

172k23133241

asked 22 hours ago

rplantikorplantiko

1635

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
  
  
  

  
  
  
  


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  @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
  

  
  
  
  


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  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
  

  
  
  
  

add a comment | 

$begingroup$

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

edited 22 hours ago

José Carlos Santos

172k23133241

asked 22 hours ago

rplantikorplantiko

1635

New contributor

rplantiko is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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$endgroup$

share|cite|improve this question

edited 22 hours ago

José Carlos Santos

172k23133241

asked 22 hours ago

rplantikorplantiko

1635

New contributor

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

edited 22 hours ago

José Carlos Santos

172k23133241

asked 22 hours ago

rplantikorplantiko

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edited 22 hours ago

José Carlos Santos

172k23133241

edited 22 hours ago

José Carlos Santos

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asked 22 hours ago

rplantikorplantiko

1635

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asked 22 hours ago

rplantikorplantiko

1635

4 Answers
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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

answered 22 hours ago

José Carlos SantosJosé Carlos Santos

172k23133241

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

  
  
  
  


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  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
  

  
  
  
  


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  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
  

  
  
  
  


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  1
  

  
  
  
  
  
  

  
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  @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
  

  
  
  
  

add a comment | 

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

answered 21 hours ago

Poon LeviPoon Levi

62639

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  

  
  
  
  

add a comment | 

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

answered 16 hours ago

goblingoblin

37.1k1159194

$endgroup$

add a comment | 

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

answered 12 hours ago

AcccumulationAcccumulation

7,2452619

$endgroup$

add a comment | 

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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

answered 22 hours ago

José Carlos SantosJosé Carlos Santos

172k23133241

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  

  
  
  
  

add a comment | 

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

answered 22 hours ago

José Carlos SantosJosé Carlos Santos

172k23133241

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  

  
  
  
  

add a comment | 

$begingroup$

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

answered 22 hours ago

José Carlos SantosJosé Carlos Santos

172k23133241

$endgroup$

share|cite|improve this answer

answered 22 hours ago

José Carlos SantosJosé Carlos Santos

172k23133241

answered 22 hours ago

José Carlos SantosJosé Carlos Santos

172k23133241

answered 22 hours ago

José Carlos SantosJosé Carlos Santos

172k23133241

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

answered 21 hours ago

Poon LeviPoon Levi

62639

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  

  
  
  
  

add a comment | 

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

answered 21 hours ago

Poon LeviPoon Levi

62639

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  

  
  
  
  

add a comment | 

$begingroup$

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

answered 21 hours ago

Poon LeviPoon Levi

62639

$endgroup$

share|cite|improve this answer

answered 21 hours ago

Poon LeviPoon Levi

62639

answered 21 hours ago

Poon LeviPoon Levi

62639

answered 21 hours ago

Poon LeviPoon Levi

62639

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

answered 16 hours ago

goblingoblin

37.1k1159194

$endgroup$

add a comment | 

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

answered 16 hours ago

goblingoblin

37.1k1159194

$endgroup$

add a comment | 

$begingroup$

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

answered 16 hours ago

goblingoblin

37.1k1159194

$endgroup$

share|cite|improve this answer

answered 16 hours ago

goblingoblin

37.1k1159194

answered 16 hours ago

goblingoblin

37.1k1159194

answered 16 hours ago

goblingoblin

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

answered 12 hours ago

AcccumulationAcccumulation

7,2452619

$endgroup$

add a comment | 

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

answered 12 hours ago

AcccumulationAcccumulation

7,2452619

$endgroup$

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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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answered 12 hours ago

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share|cite|improve this answer

answered 12 hours ago

AcccumulationAcccumulation

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answered 12 hours ago

AcccumulationAcccumulation

7,2452619

answered 12 hours ago

AcccumulationAcccumulation

7,2452619