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

2

$begingroup$

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

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 Apr 7 at 18:22

Juliana de Souza

asked Apr 7 at 11:05

Juliana de SouzaJuliana de Souza

877

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  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
  Apr 7 at 11:56
  

  
  
  
  

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 and vice versa.

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 Apr 7 at 18:22

Juliana de Souza

asked Apr 7 at 11:05

Juliana de SouzaJuliana de Souza

877

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  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
  Apr 7 at 11:56
  

  
  
  
  

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 and vice versa.

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 Apr 7 at 18:22

Juliana de Souza

asked Apr 7 at 11:05

Juliana de SouzaJuliana de Souza

877

$endgroup$

share|cite|improve this question

edited Apr 7 at 18:22

Juliana de Souza

asked Apr 7 at 11:05

Juliana de SouzaJuliana de Souza

877

share|cite|improve this question

edited Apr 7 at 18:22

Juliana de Souza

asked Apr 7 at 11:05

Juliana de SouzaJuliana de Souza

877

asked Apr 7 at 11:05

Juliana de SouzaJuliana de Souza

877

asked Apr 7 at 11:05

Juliana de SouzaJuliana de Souza

877

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 Apr 7 at 11:31

Chinnapparaj RChinnapparaj R

6,2672929

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  See math.stackexchange.com/questions/152243/… for the proof of non-completeness
  $endgroup$
  – Chinnapparaj R
  Apr 7 at 11:38
  

  
  
  
  

add a comment | 

5

$begingroup$

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

share|cite|improve this answer

answered Apr 7 at 11:32

Kavi Rama MurthyKavi Rama Murthy

73.9k53170

$endgroup$

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 Apr 7 at 11:31

Chinnapparaj RChinnapparaj R

6,2672929

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  See math.stackexchange.com/questions/152243/… for the proof of non-completeness
  $endgroup$
  – Chinnapparaj R
  Apr 7 at 11:38
  

  
  
  
  

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 Apr 7 at 11:31

Chinnapparaj RChinnapparaj R

6,2672929

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  See math.stackexchange.com/questions/152243/… for the proof of non-completeness
  $endgroup$
  – Chinnapparaj R
  Apr 7 at 11:38
  

  
  
  
  

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 Apr 7 at 11:31

Chinnapparaj RChinnapparaj R

6,2672929

$endgroup$

share|cite|improve this answer

answered Apr 7 at 11:31

Chinnapparaj RChinnapparaj R

6,2672929

answered Apr 7 at 11:31

Chinnapparaj RChinnapparaj R

6,2672929

answered Apr 7 at 11:31

Chinnapparaj RChinnapparaj R

6,2672929

5

$begingroup$

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

share|cite|improve this answer

answered Apr 7 at 11:32

Kavi Rama MurthyKavi Rama Murthy

73.9k53170

$endgroup$

add a comment | 

5

$begingroup$

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

share|cite|improve this answer

answered Apr 7 at 11:32

Kavi Rama MurthyKavi Rama Murthy

73.9k53170

$endgroup$

add a comment | 

$begingroup$

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

share|cite|improve this answer

answered Apr 7 at 11:32

Kavi Rama MurthyKavi Rama Murthy

73.9k53170

$endgroup$

share|cite|improve this answer

answered Apr 7 at 11:32

Kavi Rama MurthyKavi Rama Murthy

73.9k53170

answered Apr 7 at 11:32

Kavi Rama MurthyKavi Rama Murthy

73.9k53170

answered Apr 7 at 11:32

Kavi Rama MurthyKavi Rama Murthy

73.9k53170