Why the limit $rho_2(i,-i)=0$? [closed]

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Why the limit $rho_2(i,-i)=0$?

general-topology

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asked Jul 17 at 12:21

Shen Chong

386

closed as off-topic by amWhy, Jyrki Lahtonen, José Carlos Santos, abiessu, Shailesh Jul 19 at 0:27

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, Jyrki Lahtonen, José Carlos Santos, abiessu, Shailesh

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I suppose the book is giving an example of two equivalent metrics (even complete metrics) that do not have the same set of Cauchy sequences. (Although equivalent metrics must have the same set of $convergent$ sequences, because in a metric space $S,$ the set of convergent sequences, and the closure operator on subsets of $S,$ completely determine each other.)
  – DanielWainfleet
  Jul 17 at 19:54
  
  

  
  
  
  

add a comment | 

up vote
0
down vote

favorite

Why the limit $rho_2(i,-i)=0$?

general-topology

share|cite|improve this question

asked Jul 17 at 12:21

Shen Chong

386

closed as off-topic by amWhy, Jyrki Lahtonen, José Carlos Santos, abiessu, Shailesh Jul 19 at 0:27

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, Jyrki Lahtonen, José Carlos Santos, abiessu, Shailesh

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I suppose the book is giving an example of two equivalent metrics (even complete metrics) that do not have the same set of Cauchy sequences. (Although equivalent metrics must have the same set of $convergent$ sequences, because in a metric space $S,$ the set of convergent sequences, and the closure operator on subsets of $S,$ completely determine each other.)
  – DanielWainfleet
  Jul 17 at 19:54
  
  

  
  
  
  

add a comment | 

up vote
0
down vote

favorite

up vote
0
down vote

favorite

Why the limit $rho_2(i,-i)=0$?

general-topology

share|cite|improve this question

asked Jul 17 at 12:21

Shen Chong

386

Why the limit $rho_2(i,-i)=0$?

general-topology

share|cite|improve this question

asked Jul 17 at 12:21

Shen Chong

386

share|cite|improve this question

share|cite|improve this question

asked Jul 17 at 12:21

Shen Chong

386

asked Jul 17 at 12:21

Shen Chong

386

asked Jul 17 at 12:21

Shen Chong

386

386

closed as off-topic by amWhy, Jyrki Lahtonen, José Carlos Santos, abiessu, Shailesh Jul 19 at 0:27

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, Jyrki Lahtonen, José Carlos Santos, abiessu, Shailesh

If this question can be reworded to fit the rules in the help center, please edit the question.

closed as off-topic by amWhy, Jyrki Lahtonen, José Carlos Santos, abiessu, Shailesh Jul 19 at 0:27

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, Jyrki Lahtonen, José Carlos Santos, abiessu, Shailesh

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I suppose the book is giving an example of two equivalent metrics (even complete metrics) that do not have the same set of Cauchy sequences. (Although equivalent metrics must have the same set of $convergent$ sequences, because in a metric space $S,$ the set of convergent sequences, and the closure operator on subsets of $S,$ completely determine each other.)
  – DanielWainfleet
  Jul 17 at 19:54
  
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I suppose the book is giving an example of two equivalent metrics (even complete metrics) that do not have the same set of Cauchy sequences. (Although equivalent metrics must have the same set of $convergent$ sequences, because in a metric space $S,$ the set of convergent sequences, and the closure operator on subsets of $S,$ completely determine each other.)
  – DanielWainfleet
  Jul 17 at 19:54
  
  

  
  
  
  

I suppose the book is giving an example of two equivalent metrics (even complete metrics) that do not have the same set of Cauchy sequences. (Although equivalent metrics must have the same set of $convergent$ sequences, because in a metric space $S,$ the set of convergent sequences, and the closure operator on subsets of $S,$ completely determine each other.)
– DanielWainfleet
Jul 17 at 19:54

I suppose the book is giving an example of two equivalent metrics (even complete metrics) that do not have the same set of Cauchy sequences. (Although equivalent metrics must have the same set of $convergent$ sequences, because in a metric space $S,$ the set of convergent sequences, and the closure operator on subsets of $S,$ completely determine each other.)
– DanielWainfleet
Jul 17 at 19:54

add a comment | 

2 Answers
2

active

oldest

votes

up vote
1
down vote



accepted

For any $yin S^1$ $(0,1)$ there is a unique $T(y)in (-3pi/2, pi/2)$ such that $y=(cos T(y),sin T(y)).$

And $T: S^1$ $(0,1)to (-3pi /2, pi /2)$ is a homeomorphism. So the composite function $(Th):Bbb R to (-3pi/2,pi /2)$ is a homeomorphism.

Consider any homeomorphism $H: Bbb R to (a,b)$ to a bounded real interval $(a,b).$ Now $H$ must be monotonic because it is continuous and injective. We must have either

(I). $ lim_ito infty H(i)=a$ and $lim_ito -infty H(i)=b,$... or

(II). $ lim_ito inftyH(i)=b$ and $lim_ito -inftyH(i)=a.$

If neither (I) nor (II) held then the monotonicity of $H$ would imply that $H$ is not surjective onto $(a,b)$.

Apply this with $H=Tf$ and $(a,b)=(-3pi /2,pi /2).$

share|cite|improve this answer

answered Jul 17 at 19:40

DanielWainfleet

31.7k31644

• 
  
  
  

  
  

  
  
  
  
  
  

  
  every good explanation! thanks very much
  – Shen Chong
  Jul 18 at 8:09
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  In your answer, is the topology of $(-3pi/2,pi/2)$ equipped with the subspace of $R$?
  – Shen Chong
  Jul 18 at 8:11
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Yes. The "usual", or "standard" topology on $(-3pi /2,pi /2).$
  – DanielWainfleet
  Jul 19 at 16:48
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Yes....................
  – DanielWainfleet
  Jul 19 at 17:03
  

  
  
  
  

add a comment | 

up vote
4
down vote

Note that since $h:mathbbRto S^1backslash(0,1)$ is a homeomorphism then

$$lim_xtoinfty h(x)=(0,1)=lim_xtoinfty h(-x)$$

And since $rho$ is continuous with respect to the Euclidean (natural) metric then the thesis easily follows by simple evaluation:

$$limlimits_itoinfty rho_2(-i, i)=limrhobig(h(-i), h(i)big)=$$
$$=rhobigg(lim big(h(-i), h(i)big)bigg)=$$
$$=rhobig(lim h(-i), lim h(i)big)=$$
$$=rhobig((0,1), (0,1)big)=0.$$

share|cite|improve this answer

edited Jul 17 at 13:13

answered Jul 17 at 12:27

freakish

8,5771524

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I still don't understand the first equation that why $lim h(x)$ and $h(-x)$ must be $(0,1)$? Is there a specific proof for this? thanks
  – Shen Chong
  Jul 18 at 7:58
  

  
  
  
  

add a comment | 

2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
1
down vote



accepted

For any $yin S^1$ $(0,1)$ there is a unique $T(y)in (-3pi/2, pi/2)$ such that $y=(cos T(y),sin T(y)).$

And $T: S^1$ $(0,1)to (-3pi /2, pi /2)$ is a homeomorphism. So the composite function $(Th):Bbb R to (-3pi/2,pi /2)$ is a homeomorphism.

Consider any homeomorphism $H: Bbb R to (a,b)$ to a bounded real interval $(a,b).$ Now $H$ must be monotonic because it is continuous and injective. We must have either

(I). $ lim_ito infty H(i)=a$ and $lim_ito -infty H(i)=b,$... or

(II). $ lim_ito inftyH(i)=b$ and $lim_ito -inftyH(i)=a.$

If neither (I) nor (II) held then the monotonicity of $H$ would imply that $H$ is not surjective onto $(a,b)$.

Apply this with $H=Tf$ and $(a,b)=(-3pi /2,pi /2).$

share|cite|improve this answer

answered Jul 17 at 19:40

DanielWainfleet

31.7k31644

• 
  
  
  

  
  

  
  
  
  
  
  

  
  every good explanation! thanks very much
  – Shen Chong
  Jul 18 at 8:09
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  In your answer, is the topology of $(-3pi/2,pi/2)$ equipped with the subspace of $R$?
  – Shen Chong
  Jul 18 at 8:11
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Yes. The "usual", or "standard" topology on $(-3pi /2,pi /2).$
  – DanielWainfleet
  Jul 19 at 16:48
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Yes....................
  – DanielWainfleet
  Jul 19 at 17:03
  

  
  
  
  

add a comment | 

up vote
1
down vote



accepted

For any $yin S^1$ $(0,1)$ there is a unique $T(y)in (-3pi/2, pi/2)$ such that $y=(cos T(y),sin T(y)).$

And $T: S^1$ $(0,1)to (-3pi /2, pi /2)$ is a homeomorphism. So the composite function $(Th):Bbb R to (-3pi/2,pi /2)$ is a homeomorphism.

Consider any homeomorphism $H: Bbb R to (a,b)$ to a bounded real interval $(a,b).$ Now $H$ must be monotonic because it is continuous and injective. We must have either

(I). $ lim_ito infty H(i)=a$ and $lim_ito -infty H(i)=b,$... or

(II). $ lim_ito inftyH(i)=b$ and $lim_ito -inftyH(i)=a.$

If neither (I) nor (II) held then the monotonicity of $H$ would imply that $H$ is not surjective onto $(a,b)$.

Apply this with $H=Tf$ and $(a,b)=(-3pi /2,pi /2).$

share|cite|improve this answer

answered Jul 17 at 19:40

DanielWainfleet

31.7k31644

• 
  
  
  

  
  

  
  
  
  
  
  

  
  every good explanation! thanks very much
  – Shen Chong
  Jul 18 at 8:09
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  In your answer, is the topology of $(-3pi/2,pi/2)$ equipped with the subspace of $R$?
  – Shen Chong
  Jul 18 at 8:11
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Yes. The "usual", or "standard" topology on $(-3pi /2,pi /2).$
  – DanielWainfleet
  Jul 19 at 16:48
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Yes....................
  – DanielWainfleet
  Jul 19 at 17:03
  

  
  
  
  

add a comment | 

up vote
1
down vote



accepted

up vote
1
down vote



accepted

For any $yin S^1$ $(0,1)$ there is a unique $T(y)in (-3pi/2, pi/2)$ such that $y=(cos T(y),sin T(y)).$

And $T: S^1$ $(0,1)to (-3pi /2, pi /2)$ is a homeomorphism. So the composite function $(Th):Bbb R to (-3pi/2,pi /2)$ is a homeomorphism.

Consider any homeomorphism $H: Bbb R to (a,b)$ to a bounded real interval $(a,b).$ Now $H$ must be monotonic because it is continuous and injective. We must have either

(I). $ lim_ito infty H(i)=a$ and $lim_ito -infty H(i)=b,$... or

(II). $ lim_ito inftyH(i)=b$ and $lim_ito -inftyH(i)=a.$

If neither (I) nor (II) held then the monotonicity of $H$ would imply that $H$ is not surjective onto $(a,b)$.

Apply this with $H=Tf$ and $(a,b)=(-3pi /2,pi /2).$

share|cite|improve this answer

answered Jul 17 at 19:40

DanielWainfleet

31.7k31644

For any $yin S^1$ $(0,1)$ there is a unique $T(y)in (-3pi/2, pi/2)$ such that $y=(cos T(y),sin T(y)).$

And $T: S^1$ $(0,1)to (-3pi /2, pi /2)$ is a homeomorphism. So the composite function $(Th):Bbb R to (-3pi/2,pi /2)$ is a homeomorphism.

Consider any homeomorphism $H: Bbb R to (a,b)$ to a bounded real interval $(a,b).$ Now $H$ must be monotonic because it is continuous and injective. We must have either

(I). $ lim_ito infty H(i)=a$ and $lim_ito -infty H(i)=b,$... or

(II). $ lim_ito inftyH(i)=b$ and $lim_ito -inftyH(i)=a.$

If neither (I) nor (II) held then the monotonicity of $H$ would imply that $H$ is not surjective onto $(a,b)$.

Apply this with $H=Tf$ and $(a,b)=(-3pi /2,pi /2).$

share|cite|improve this answer

answered Jul 17 at 19:40

DanielWainfleet

31.7k31644

share|cite|improve this answer

share|cite|improve this answer

answered Jul 17 at 19:40

DanielWainfleet

31.7k31644

answered Jul 17 at 19:40

DanielWainfleet

31.7k31644

answered Jul 17 at 19:40

DanielWainfleet

31.7k31644

31.7k31644

• 
  
  
  

  
  

  
  
  
  
  
  

  
  every good explanation! thanks very much
  – Shen Chong
  Jul 18 at 8:09
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  In your answer, is the topology of $(-3pi/2,pi/2)$ equipped with the subspace of $R$?
  – Shen Chong
  Jul 18 at 8:11
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Yes. The "usual", or "standard" topology on $(-3pi /2,pi /2).$
  – DanielWainfleet
  Jul 19 at 16:48
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Yes....................
  – DanielWainfleet
  Jul 19 at 17:03
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  every good explanation! thanks very much
  – Shen Chong
  Jul 18 at 8:09
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  In your answer, is the topology of $(-3pi/2,pi/2)$ equipped with the subspace of $R$?
  – Shen Chong
  Jul 18 at 8:11
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Yes. The "usual", or "standard" topology on $(-3pi /2,pi /2).$
  – DanielWainfleet
  Jul 19 at 16:48
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Yes....................
  – DanielWainfleet
  Jul 19 at 17:03
  

  
  
  
  

every good explanation! thanks very much
– Shen Chong
Jul 18 at 8:09

every good explanation! thanks very much
– Shen Chong
Jul 18 at 8:09

In your answer, is the topology of $(-3pi/2,pi/2)$ equipped with the subspace of $R$?
– Shen Chong
Jul 18 at 8:11

In your answer, is the topology of $(-3pi/2,pi/2)$ equipped with the subspace of $R$?
– Shen Chong
Jul 18 at 8:11

Yes. The "usual", or "standard" topology on $(-3pi /2,pi /2).$
– DanielWainfleet
Jul 19 at 16:48

Yes. The "usual", or "standard" topology on $(-3pi /2,pi /2).$
– DanielWainfleet
Jul 19 at 16:48

Yes....................
– DanielWainfleet
Jul 19 at 17:03

Yes....................
– DanielWainfleet
Jul 19 at 17:03

add a comment | 

up vote
4
down vote

Note that since $h:mathbbRto S^1backslash(0,1)$ is a homeomorphism then

$$lim_xtoinfty h(x)=(0,1)=lim_xtoinfty h(-x)$$

And since $rho$ is continuous with respect to the Euclidean (natural) metric then the thesis easily follows by simple evaluation:

$$limlimits_itoinfty rho_2(-i, i)=limrhobig(h(-i), h(i)big)=$$
$$=rhobigg(lim big(h(-i), h(i)big)bigg)=$$
$$=rhobig(lim h(-i), lim h(i)big)=$$
$$=rhobig((0,1), (0,1)big)=0.$$

share|cite|improve this answer

edited Jul 17 at 13:13

answered Jul 17 at 12:27

freakish

8,5771524

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I still don't understand the first equation that why $lim h(x)$ and $h(-x)$ must be $(0,1)$? Is there a specific proof for this? thanks
  – Shen Chong
  Jul 18 at 7:58
  

  
  
  
  

add a comment | 

up vote
4
down vote

Note that since $h:mathbbRto S^1backslash(0,1)$ is a homeomorphism then

$$lim_xtoinfty h(x)=(0,1)=lim_xtoinfty h(-x)$$

And since $rho$ is continuous with respect to the Euclidean (natural) metric then the thesis easily follows by simple evaluation:

$$limlimits_itoinfty rho_2(-i, i)=limrhobig(h(-i), h(i)big)=$$
$$=rhobigg(lim big(h(-i), h(i)big)bigg)=$$
$$=rhobig(lim h(-i), lim h(i)big)=$$
$$=rhobig((0,1), (0,1)big)=0.$$

share|cite|improve this answer

edited Jul 17 at 13:13

answered Jul 17 at 12:27

freakish

8,5771524

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I still don't understand the first equation that why $lim h(x)$ and $h(-x)$ must be $(0,1)$? Is there a specific proof for this? thanks
  – Shen Chong
  Jul 18 at 7:58
  

  
  
  
  

add a comment | 

up vote
4
down vote

up vote
4
down vote

Note that since $h:mathbbRto S^1backslash(0,1)$ is a homeomorphism then

$$lim_xtoinfty h(x)=(0,1)=lim_xtoinfty h(-x)$$

And since $rho$ is continuous with respect to the Euclidean (natural) metric then the thesis easily follows by simple evaluation:

$$limlimits_itoinfty rho_2(-i, i)=limrhobig(h(-i), h(i)big)=$$
$$=rhobigg(lim big(h(-i), h(i)big)bigg)=$$
$$=rhobig(lim h(-i), lim h(i)big)=$$
$$=rhobig((0,1), (0,1)big)=0.$$

share|cite|improve this answer

edited Jul 17 at 13:13

answered Jul 17 at 12:27

freakish

8,5771524

Note that since $h:mathbbRto S^1backslash(0,1)$ is a homeomorphism then

$$lim_xtoinfty h(x)=(0,1)=lim_xtoinfty h(-x)$$

And since $rho$ is continuous with respect to the Euclidean (natural) metric then the thesis easily follows by simple evaluation:

$$limlimits_itoinfty rho_2(-i, i)=limrhobig(h(-i), h(i)big)=$$
$$=rhobigg(lim big(h(-i), h(i)big)bigg)=$$
$$=rhobig(lim h(-i), lim h(i)big)=$$
$$=rhobig((0,1), (0,1)big)=0.$$

share|cite|improve this answer

edited Jul 17 at 13:13

answered Jul 17 at 12:27

freakish

8,5771524

share|cite|improve this answer

share|cite|improve this answer

edited Jul 17 at 13:13

edited Jul 17 at 13:13

edited Jul 17 at 13:13

answered Jul 17 at 12:27

freakish

8,5771524

answered Jul 17 at 12:27

freakish

8,5771524

answered Jul 17 at 12:27

freakish

8,5771524

8,5771524

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I still don't understand the first equation that why $lim h(x)$ and $h(-x)$ must be $(0,1)$? Is there a specific proof for this? thanks
  – Shen Chong
  Jul 18 at 7:58
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I still don't understand the first equation that why $lim h(x)$ and $h(-x)$ must be $(0,1)$? Is there a specific proof for this? thanks
  – Shen Chong
  Jul 18 at 7:58
  

  
  
  
  

I still don't understand the first equation that why $lim h(x)$ and $h(-x)$ must be $(0,1)$? Is there a specific proof for this? thanks
– Shen Chong
Jul 18 at 7:58

I still don't understand the first equation that why $lim h(x)$ and $h(-x)$ must be $(0,1)$? Is there a specific proof for this? thanks
– Shen Chong
Jul 18 at 7:58

add a comment |