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Can a Cauchy sequence converge for one metric while not converging for another?
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$begingroup$
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
edited Apr 5 at 7:23
José Carlos Santos
173k23133241
asked Apr 5 at 6:56
rplantikorplantiko
1605
New contributor
rplantiko is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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$endgroup$
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
edited Apr 5 at 7:23
José Carlos Santos
173k23133241
asked Apr 5 at 6:56
rplantikorplantiko
1605
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
Apr 5 at 7:12
$begingroup$
@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
Apr 5 at 7:14
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
Apr 5 at 7:18
$begingroup$
@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
Apr 5 at 7:19
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
Apr 5 at 16:44
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
edited Apr 5 at 7:23
José Carlos Santos
173k23133241
asked Apr 5 at 6:56
rplantikorplantiko
1605
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
edited Apr 5 at 7:23
José Carlos Santos
173k23133241
asked Apr 5 at 6:56
rplantikorplantiko
1605
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.
edited Apr 5 at 7:23
José Carlos Santos
173k23133241
asked Apr 5 at 6:56
rplantikorplantiko
1605
New contributor
rplantiko is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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edited Apr 5 at 7:23
José Carlos Santos
173k23133241
edited Apr 5 at 7:23
José Carlos Santos
173k23133241
edited Apr 5 at 7:23
José Carlos Santos
173k23133241
173k23133241
asked Apr 5 at 6:56
rplantikorplantiko
1605
New contributor
rplantiko is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Apr 5 at 6:56
rplantikorplantiko
1605
asked Apr 5 at 6:56
rplantikorplantiko
1605
1605
New contributor
rplantiko is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
rplantiko is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
rplantiko is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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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
Apr 5 at 7:12
$begingroup$
@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
Apr 5 at 7:14
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
Apr 5 at 7:18
$begingroup$
@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
Apr 5 at 7:19
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
Apr 5 at 16:44
add a comment |
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
Apr 5 at 7:12
$begingroup$
@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
Apr 5 at 7:14
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
Apr 5 at 7:18
$begingroup$
@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
Apr 5 at 7:19
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
Apr 5 at 16:44
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
Apr 5 at 7:12
$begingroup$
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
Apr 5 at 7:12
$begingroup$
@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
Apr 5 at 7:14
$begingroup$
@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
Apr 5 at 7:14
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
Apr 5 at 7:18
$begingroup$
@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
Apr 5 at 7:18
$begingroup$
@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
Apr 5 at 7:19
$begingroup$
@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
Apr 5 at 7:19
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
Apr 5 at 16:44
$begingroup$
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
Apr 5 at 16:44
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
Take $X=[0,infty)$, let $d(x,y)=lvert x-yrvert$ and let$$e(x,y)=begincaseslvert x-yrvert&text if x,yneq0\lvert x+1rvert&text if xneq0text and y=0\lvert y+1rvert&text if x=0text and yneq0\0&text if x,y=0.endcases$$Then $left(frac1nright)_ninmathbb N$ is a Cauchy sequence with respect to both metrics, but it is convergent only in $(X,d)$.
answered Apr 5 at 7:11
José Carlos SantosJosé Carlos Santos
173k23133241
$endgroup$
$begingroup$
Is there a reason this example needs to use $X = [0, infty)$ rather than $X = mathbbR$?
$endgroup$
– Michael Seifert
Apr 5 at 12:13
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
Apr 5 at 12:47
$begingroup$
The second metric basically makes it $-1cup (0,infty)$, which can't really be said to be a metric for the same space.
$endgroup$
– Matt Samuel
Apr 5 at 23:57
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
2 days ago
$begingroup$
For $X = mathbbR$, one could use $e(0,x)=e(x,0)=|x|+1$ when $xneq 0$.
$endgroup$
– Litho
2 days ago
add a comment |
$begingroup$
You can always have some artificial example where you just "move the limit elsewhere". For example, let $X=mathbbR_ge 0$, $d_1$ be the Euclidean metric and $d_2(x, y)=|hatx-haty|$, where $ hatx=-1$ if $x=0$ and $ hatx=x$ otherwise. Then the sequence $x_n=frac 1n$ converges in $(X, d_1)$ but not in $(X, d_2)$.
answered Apr 5 at 7:32
Poon LeviPoon Levi
70139
$endgroup$
$begingroup$
Is this somehow a standard example? Since you other poster also jsed it? Which book is it from
$endgroup$
– lalala
Apr 5 at 18:50
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
Apr 5 at 21:25
$begingroup$
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
2 days ago
1
$begingroup$
$d_2$ is essentially the Euclidean metric on a new set $-1cup(0,infty)$. But since we can't change the base set, we instead use a map $xmapstohatx$ to rename $0$ to $-1$.
$endgroup$
– Poon Levi
2 days ago
$begingroup$
Thanks! Yes. It is just a renameing. (I am not so used to metrics which do not derive from a norm)
$endgroup$
– lalala
2 days ago
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.
answered Apr 5 at 13:26
goblingoblin
37.1k1159194
$endgroup$
add a comment |
$begingroup$
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.
answered Apr 5 at 16:50
AcccumulationAcccumulation
7,3052619
$endgroup$
add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Take $X=[0,infty)$, let $d(x,y)=lvert x-yrvert$ and let$$e(x,y)=begincaseslvert x-yrvert&text if x,yneq0\lvert x+1rvert&text if xneq0text and y=0\lvert y+1rvert&text if x=0text and yneq0\0&text if x,y=0.endcases$$Then $left(frac1nright)_ninmathbb N$ is a Cauchy sequence with respect to both metrics, but it is convergent only in $(X,d)$.
answered Apr 5 at 7:11
José Carlos SantosJosé Carlos Santos
173k23133241
$endgroup$
$begingroup$
Is there a reason this example needs to use $X = [0, infty)$ rather than $X = mathbbR$?
$endgroup$
– Michael Seifert
Apr 5 at 12:13
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
Apr 5 at 12:47
$begingroup$
The second metric basically makes it $-1cup (0,infty)$, which can't really be said to be a metric for the same space.
$endgroup$
– Matt Samuel
Apr 5 at 23:57
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
2 days ago
$begingroup$
For $X = mathbbR$, one could use $e(0,x)=e(x,0)=|x|+1$ when $xneq 0$.
$endgroup$
– Litho
2 days ago
add a comment |
$begingroup$
Take $X=[0,infty)$, let $d(x,y)=lvert x-yrvert$ and let$$e(x,y)=begincaseslvert x-yrvert&text if x,yneq0\lvert x+1rvert&text if xneq0text and y=0\lvert y+1rvert&text if x=0text and yneq0\0&text if x,y=0.endcases$$Then $left(frac1nright)_ninmathbb N$ is a Cauchy sequence with respect to both metrics, but it is convergent only in $(X,d)$.
answered Apr 5 at 7:11
José Carlos SantosJosé Carlos Santos
173k23133241
$endgroup$
$begingroup$
Is there a reason this example needs to use $X = [0, infty)$ rather than $X = mathbbR$?
$endgroup$
– Michael Seifert
Apr 5 at 12:13
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
Apr 5 at 12:47
$begingroup$
The second metric basically makes it $-1cup (0,infty)$, which can't really be said to be a metric for the same space.
$endgroup$
– Matt Samuel
Apr 5 at 23:57
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
2 days ago
$begingroup$
For $X = mathbbR$, one could use $e(0,x)=e(x,0)=|x|+1$ when $xneq 0$.
$endgroup$
– Litho
2 days ago
add a comment |
$begingroup$
Take $X=[0,infty)$, let $d(x,y)=lvert x-yrvert$ and let$$e(x,y)=begincaseslvert x-yrvert&text if x,yneq0\lvert x+1rvert&text if xneq0text and y=0\lvert y+1rvert&text if x=0text and yneq0\0&text if x,y=0.endcases$$Then $left(frac1nright)_ninmathbb N$ is a Cauchy sequence with respect to both metrics, but it is convergent only in $(X,d)$.
answered Apr 5 at 7:11
José Carlos SantosJosé Carlos Santos
173k23133241
$endgroup$
Take $X=[0,infty)$, let $d(x,y)=lvert x-yrvert$ and let$$e(x,y)=begincaseslvert x-yrvert&text if x,yneq0\lvert x+1rvert&text if xneq0text and y=0\lvert y+1rvert&text if x=0text and yneq0\0&text if x,y=0.endcases$$Then $left(frac1nright)_ninmathbb N$ is a Cauchy sequence with respect to both metrics, but it is convergent only in $(X,d)$.
answered Apr 5 at 7:11
José Carlos SantosJosé Carlos Santos
173k23133241
answered Apr 5 at 7:11
José Carlos SantosJosé Carlos Santos
173k23133241
answered Apr 5 at 7:11
José Carlos SantosJosé Carlos Santos
173k23133241
answered Apr 5 at 7:11
José Carlos SantosJosé Carlos Santos
173k23133241
173k23133241
$begingroup$
Is there a reason this example needs to use $X = [0, infty)$ rather than $X = mathbbR$?
$endgroup$
– Michael Seifert
Apr 5 at 12:13
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
Apr 5 at 12:47
$begingroup$
The second metric basically makes it $-1cup (0,infty)$, which can't really be said to be a metric for the same space.
$endgroup$
– Matt Samuel
Apr 5 at 23:57
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
2 days ago
$begingroup$
For $X = mathbbR$, one could use $e(0,x)=e(x,0)=|x|+1$ when $xneq 0$.
$endgroup$
– Litho
2 days ago
add a comment |
$begingroup$
Is there a reason this example needs to use $X = [0, infty)$ rather than $X = mathbbR$?
$endgroup$
– Michael Seifert
Apr 5 at 12:13
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
Apr 5 at 12:47
$begingroup$
The second metric basically makes it $-1cup (0,infty)$, which can't really be said to be a metric for the same space.
$endgroup$
– Matt Samuel
Apr 5 at 23:57
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
2 days ago
$begingroup$
For $X = mathbbR$, one could use $e(0,x)=e(x,0)=|x|+1$ when $xneq 0$.
$endgroup$
– Litho
2 days ago
$begingroup$
Is there a reason this example needs to use $X = [0, infty)$ rather than $X = mathbbR$?
$endgroup$
– Michael Seifert
Apr 5 at 12:13
$begingroup$
Is there a reason this example needs to use $X = [0, infty)$ rather than $X = mathbbR$?
$endgroup$
– Michael Seifert
Apr 5 at 12:13
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
Apr 5 at 12:47
$begingroup$
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
Apr 5 at 12:47
$begingroup$
The second metric basically makes it $-1cup (0,infty)$, which can't really be said to be a metric for the same space.
$endgroup$
– Matt Samuel
Apr 5 at 23:57
$begingroup$
The second metric basically makes it $-1cup (0,infty)$, which can't really be said to be a metric for the same space.
$endgroup$
– Matt Samuel
Apr 5 at 23:57
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
2 days 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
2 days ago
$begingroup$
For $X = mathbbR$, one could use $e(0,x)=e(x,0)=|x|+1$ when $xneq 0$.
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– Litho
2 days ago
$begingroup$
For $X = mathbbR$, one could use $e(0,x)=e(x,0)=|x|+1$ when $xneq 0$.
$endgroup$
– Litho
2 days ago
add a comment |
$begingroup$
You can always have some artificial example where you just "move the limit elsewhere". For example, let $X=mathbbR_ge 0$, $d_1$ be the Euclidean metric and $d_2(x, y)=|hatx-haty|$, where $ hatx=-1$ if $x=0$ and $ hatx=x$ otherwise. Then the sequence $x_n=frac 1n$ converges in $(X, d_1)$ but not in $(X, d_2)$.
answered Apr 5 at 7:32
Poon LeviPoon Levi
70139
$endgroup$
$begingroup$
Is this somehow a standard example? Since you other poster also jsed it? Which book is it from
$endgroup$
– lalala
Apr 5 at 18:50
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
Apr 5 at 21:25
$begingroup$
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
2 days ago
1
$begingroup$
$d_2$ is essentially the Euclidean metric on a new set $-1cup(0,infty)$. But since we can't change the base set, we instead use a map $xmapstohatx$ to rename $0$ to $-1$.
$endgroup$
– Poon Levi
2 days ago
$begingroup$
Thanks! Yes. It is just a renameing. (I am not so used to metrics which do not derive from a norm)
$endgroup$
– lalala
2 days ago
add a comment |
$begingroup$
You can always have some artificial example where you just "move the limit elsewhere". For example, let $X=mathbbR_ge 0$, $d_1$ be the Euclidean metric and $d_2(x, y)=|hatx-haty|$, where $ hatx=-1$ if $x=0$ and $ hatx=x$ otherwise. Then the sequence $x_n=frac 1n$ converges in $(X, d_1)$ but not in $(X, d_2)$.
answered Apr 5 at 7:32
Poon LeviPoon Levi
70139
$endgroup$
$begingroup$
Is this somehow a standard example? Since you other poster also jsed it? Which book is it from
$endgroup$
– lalala
Apr 5 at 18:50
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
Apr 5 at 21:25
$begingroup$
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
2 days ago
1
$begingroup$
$d_2$ is essentially the Euclidean metric on a new set $-1cup(0,infty)$. But since we can't change the base set, we instead use a map $xmapstohatx$ to rename $0$ to $-1$.
$endgroup$
– Poon Levi
2 days ago
$begingroup$
Thanks! Yes. It is just a renameing. (I am not so used to metrics which do not derive from a norm)
$endgroup$
– lalala
2 days ago
add a comment |
$begingroup$
You can always have some artificial example where you just "move the limit elsewhere". For example, let $X=mathbbR_ge 0$, $d_1$ be the Euclidean metric and $d_2(x, y)=|hatx-haty|$, where $ hatx=-1$ if $x=0$ and $ hatx=x$ otherwise. Then the sequence $x_n=frac 1n$ converges in $(X, d_1)$ but not in $(X, d_2)$.
answered Apr 5 at 7:32
Poon LeviPoon Levi
70139
$endgroup$
You can always have some artificial example where you just "move the limit elsewhere". For example, let $X=mathbbR_ge 0$, $d_1$ be the Euclidean metric and $d_2(x, y)=|hatx-haty|$, where $ hatx=-1$ if $x=0$ and $ hatx=x$ otherwise. Then the sequence $x_n=frac 1n$ converges in $(X, d_1)$ but not in $(X, d_2)$.
answered Apr 5 at 7:32
Poon LeviPoon Levi
70139
answered Apr 5 at 7:32
Poon LeviPoon Levi
70139
answered Apr 5 at 7:32
Poon LeviPoon Levi
70139
answered Apr 5 at 7:32
Poon LeviPoon Levi
70139
70139
$begingroup$
Is this somehow a standard example? Since you other poster also jsed it? Which book is it from
$endgroup$
– lalala
Apr 5 at 18:50
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
Apr 5 at 21:25
$begingroup$
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
2 days ago
1
$begingroup$
$d_2$ is essentially the Euclidean metric on a new set $-1cup(0,infty)$. But since we can't change the base set, we instead use a map $xmapstohatx$ to rename $0$ to $-1$.
$endgroup$
– Poon Levi
2 days ago
$begingroup$
Thanks! Yes. It is just a renameing. (I am not so used to metrics which do not derive from a norm)
$endgroup$
– lalala
2 days ago
add a comment |
$begingroup$
Is this somehow a standard example? Since you other poster also jsed it? Which book is it from
$endgroup$
– lalala
Apr 5 at 18:50
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
Apr 5 at 21:25
$begingroup$
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
2 days ago
1
$begingroup$
$d_2$ is essentially the Euclidean metric on a new set $-1cup(0,infty)$. But since we can't change the base set, we instead use a map $xmapstohatx$ to rename $0$ to $-1$.
$endgroup$
– Poon Levi
2 days ago
$begingroup$
Thanks! Yes. It is just a renameing. (I am not so used to metrics which do not derive from a norm)
$endgroup$
– 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
Apr 5 at 18:50
$begingroup$
Is this somehow a standard example? Since you other poster also jsed it? Which book is it from
$endgroup$
– lalala
Apr 5 at 18:50
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
Apr 5 at 21:25
$begingroup$
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
Apr 5 at 21:25
$begingroup$
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
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
2 days ago
1
1
$begingroup$
$d_2$ is essentially the Euclidean metric on a new set $-1cup(0,infty)$. But since we can't change the base set, we instead use a map $xmapstohatx$ to rename $0$ to $-1$.
$endgroup$
– Poon Levi
2 days ago
$begingroup$
$d_2$ is essentially the Euclidean metric on a new set $-1cup(0,infty)$. But since we can't change the base set, we instead use a map $xmapstohatx$ to rename $0$ to $-1$.
$endgroup$
– Poon Levi
2 days ago
$begingroup$
Thanks! Yes. It is just a renameing. (I am not so used to metrics which do not derive from a norm)
$endgroup$
– lalala
2 days ago
$begingroup$
Thanks! Yes. It is just a renameing. (I am not so used to metrics which do not derive from a norm)
$endgroup$
– lalala
2 days ago
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.
answered Apr 5 at 13:26
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.
answered Apr 5 at 13:26
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.
answered Apr 5 at 13:26
goblingoblin
37.1k1159194
$endgroup$
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.
answered Apr 5 at 13:26
goblingoblin
37.1k1159194
answered Apr 5 at 13:26
goblingoblin
37.1k1159194
answered Apr 5 at 13:26
goblingoblin
37.1k1159194
answered Apr 5 at 13:26
goblingoblin
37.1k1159194
37.1k1159194
add a comment |
add a comment |
$begingroup$
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.
answered Apr 5 at 16:50
AcccumulationAcccumulation
7,3052619
$endgroup$
add a comment |
$begingroup$
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.
answered Apr 5 at 16:50
AcccumulationAcccumulation
7,3052619
$endgroup$
add a comment |
$begingroup$
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.
answered Apr 5 at 16:50
AcccumulationAcccumulation
7,3052619
$endgroup$
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.
answered Apr 5 at 16:50
AcccumulationAcccumulation
7,3052619
answered Apr 5 at 16:50
AcccumulationAcccumulation
7,3052619
answered Apr 5 at 16:50
AcccumulationAcccumulation
7,3052619
answered Apr 5 at 16:50
AcccumulationAcccumulation
7,3052619
7,3052619
add a comment |
add a comment |
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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
Apr 5 at 7:12
$begingroup$
@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
Apr 5 at 7:14
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
Apr 5 at 7:18
$begingroup$
@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
Apr 5 at 7:19
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
Apr 5 at 16:44