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If two metric spaces are topologically equivalent (homeomorphic) imply that they are complete?

Real numbers equipped with the metric $ d (x,y) = | arctan(x) - arctan(y)| $ is an incomplete metric spaceWhich of the following metric spaces are complete?Showing that a metric space is completeUnderstanding Complete Metric Spaces and Cauchy SequencesCompact homeomorphic non bilipschitz homeomorphic metric spacesA metric space is complete when every closed and bounded subset of it is compactTopologically equivalent metrics? Ceiling function of metric $d$Limit Points of closure of $A$ is subset of limit points of $A$If $d_1,d_2$ are not equivalent metrics, is it true $(X,d_1)$ is not homeomorphic to $(X,d_2)$?Under what metric spaces are pointwise and uniform convergence equivalent?How to *disprove* topological equivalence

2

$begingroup$

My definition of topologically equivalent is: every convergent sequence in $(X, d_A)$ converges at $(X, d_B)$ to the same limit.

I know that

Let $(X,d_A)$ and $(X, d_B)$ be two strongly equivalent metric spaces, if $(X, d_A)$ is complete then $(X, d_B)$ is complete.

I was thinking that if these spaces were just topologically equivalent yet would the statement be true?

My answer is no.

But I can't think of a counter example.

general-topology convergence metric-spaces

share|cite|improve this question

edited 1 hour ago

YuiTo Cheng

2,3084937

asked 3 hours ago

Juliana de SouzaJuliana de Souza

876

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  I think there is missing a part :definition is every convergent sequence in $(X, d_A)$ converges at $(X, d_B)$ to the same limit - and vice versa
  $endgroup$
  – Maksim
  3 hours ago
  

  
  
  
  

add a comment | 

2

$begingroup$

My definition of topologically equivalent is: every convergent sequence in $(X, d_A)$ converges at $(X, d_B)$ to the same limit.

I know that

Let $(X,d_A)$ and $(X, d_B)$ be two strongly equivalent metric spaces, if $(X, d_A)$ is complete then $(X, d_B)$ is complete.

I was thinking that if these spaces were just topologically equivalent yet would the statement be true?

My answer is no.

But I can't think of a counter example.

general-topology convergence metric-spaces

share|cite|improve this question

edited 1 hour ago

YuiTo Cheng

2,3084937

asked 3 hours ago

Juliana de SouzaJuliana de Souza

876

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  I think there is missing a part :definition is every convergent sequence in $(X, d_A)$ converges at $(X, d_B)$ to the same limit - and vice versa
  $endgroup$
  – Maksim
  3 hours ago
  

  
  
  
  

add a comment | 

$begingroup$

My definition of topologically equivalent is: every convergent sequence in $(X, d_A)$ converges at $(X, d_B)$ to the same limit.

I know that

Let $(X,d_A)$ and $(X, d_B)$ be two strongly equivalent metric spaces, if $(X, d_A)$ is complete then $(X, d_B)$ is complete.

I was thinking that if these spaces were just topologically equivalent yet would the statement be true?

My answer is no.

But I can't think of a counter example.

general-topology convergence metric-spaces

share|cite|improve this question

edited 1 hour ago

YuiTo Cheng

2,3084937

asked 3 hours ago

Juliana de SouzaJuliana de Souza

876

$endgroup$

share|cite|improve this question

edited 1 hour ago

YuiTo Cheng

2,3084937

asked 3 hours ago

Juliana de SouzaJuliana de Souza

876

share|cite|improve this question

edited 1 hour ago

YuiTo Cheng

2,3084937

asked 3 hours ago

Juliana de SouzaJuliana de Souza

876

edited 1 hour ago

YuiTo Cheng

2,3084937

edited 1 hour ago

YuiTo Cheng

2,3084937

asked 3 hours ago

Juliana de SouzaJuliana de Souza

876

asked 3 hours ago

Juliana de SouzaJuliana de Souza

876

2 Answers
2

active

oldest

votes

3

$begingroup$

Consider $(Bbb R,d)$ , where $d$ is the usual metric and $(Bbb R,rho)$ , where $rho$ is the metric defined by $$rho(x,y)=vert arctan x-arctan y vert$$ Then $(Bbb R,d)$ is topologically equivalent to $(Bbb R,rho)$(How ?), but one is complete while the other is not (How?)

Try to fill the gaps!

share|cite|improve this answer

answered 3 hours ago

Chinnapparaj RChinnapparaj R

5,9382928

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  See math.stackexchange.com/questions/152243/… for the proof of non-completeness
  $endgroup$
  – Chinnapparaj R
  3 hours ago
  

  
  
  
  

add a comment | 

5

$begingroup$

$(-1,1)$ is topologically equivalent to $mathbb R$ via the homeomorphism $x to frac x x$. $(-1,1)$ is not complete but $mathbb R$ is complete. [Usual metric on both spaces].

share|cite|improve this answer

answered 3 hours ago

Kavi Rama MurthyKavi Rama Murthy

72.7k53170

$endgroup$

add a comment | 

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3

$begingroup$

Consider $(Bbb R,d)$ , where $d$ is the usual metric and $(Bbb R,rho)$ , where $rho$ is the metric defined by $$rho(x,y)=vert arctan x-arctan y vert$$ Then $(Bbb R,d)$ is topologically equivalent to $(Bbb R,rho)$(How ?), but one is complete while the other is not (How?)

Try to fill the gaps!

share|cite|improve this answer

answered 3 hours ago

Chinnapparaj RChinnapparaj R

5,9382928

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  See math.stackexchange.com/questions/152243/… for the proof of non-completeness
  $endgroup$
  – Chinnapparaj R
  3 hours ago
  

  
  
  
  

add a comment | 

3

$begingroup$

Consider $(Bbb R,d)$ , where $d$ is the usual metric and $(Bbb R,rho)$ , where $rho$ is the metric defined by $$rho(x,y)=vert arctan x-arctan y vert$$ Then $(Bbb R,d)$ is topologically equivalent to $(Bbb R,rho)$(How ?), but one is complete while the other is not (How?)

Try to fill the gaps!

share|cite|improve this answer

answered 3 hours ago

Chinnapparaj RChinnapparaj R

5,9382928

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  See math.stackexchange.com/questions/152243/… for the proof of non-completeness
  $endgroup$
  – Chinnapparaj R
  3 hours ago
  

  
  
  
  

add a comment | 

$begingroup$

Consider $(Bbb R,d)$ , where $d$ is the usual metric and $(Bbb R,rho)$ , where $rho$ is the metric defined by $$rho(x,y)=vert arctan x-arctan y vert$$ Then $(Bbb R,d)$ is topologically equivalent to $(Bbb R,rho)$(How ?), but one is complete while the other is not (How?)

Try to fill the gaps!

share|cite|improve this answer

answered 3 hours ago

Chinnapparaj RChinnapparaj R

5,9382928

$endgroup$

share|cite|improve this answer

answered 3 hours ago

Chinnapparaj RChinnapparaj R

5,9382928

answered 3 hours ago

Chinnapparaj RChinnapparaj R

5,9382928

answered 3 hours ago

Chinnapparaj RChinnapparaj R

5,9382928

5

$begingroup$

$(-1,1)$ is topologically equivalent to $mathbb R$ via the homeomorphism $x to frac x x$. $(-1,1)$ is not complete but $mathbb R$ is complete. [Usual metric on both spaces].

share|cite|improve this answer

answered 3 hours ago

Kavi Rama MurthyKavi Rama Murthy

72.7k53170

$endgroup$

add a comment | 

5

$begingroup$

$(-1,1)$ is topologically equivalent to $mathbb R$ via the homeomorphism $x to frac x x$. $(-1,1)$ is not complete but $mathbb R$ is complete. [Usual metric on both spaces].

share|cite|improve this answer

answered 3 hours ago

Kavi Rama MurthyKavi Rama Murthy

72.7k53170

$endgroup$

add a comment | 

$begingroup$

$(-1,1)$ is topologically equivalent to $mathbb R$ via the homeomorphism $x to frac x x$. $(-1,1)$ is not complete but $mathbb R$ is complete. [Usual metric on both spaces].

share|cite|improve this answer

answered 3 hours ago

Kavi Rama MurthyKavi Rama Murthy

72.7k53170

$endgroup$

share|cite|improve this answer

answered 3 hours ago

Kavi Rama MurthyKavi Rama Murthy

72.7k53170

answered 3 hours ago

Kavi Rama MurthyKavi Rama Murthy

72.7k53170

answered 3 hours ago

Kavi Rama MurthyKavi Rama Murthy

72.7k53170