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







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  • 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
    2 days ago
















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











$endgroup$








  • 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
    2 days ago














2












2








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











$endgroup$




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




general-topology convergence metric-spaces






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 2 days ago







Juliana de Souza

















asked 2 days ago









Juliana de SouzaJuliana de Souza

877




877








  • 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
    2 days ago














  • 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
    2 days ago








2




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
2 days ago




$begingroup$
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
2 days ago










2 Answers
2






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$begingroup$

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









$endgroup$









  • 1




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



















5












$begingroup$

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






share|cite|improve this answer









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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    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









    $endgroup$









    • 1




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
















    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









    $endgroup$









    • 1




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














    3












    3








    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









    $endgroup$



    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












    share|cite|improve this answer



    share|cite|improve this answer










    answered 2 days ago









    Chinnapparaj RChinnapparaj R

    6,1712929




    6,1712929








    • 1




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














    • 1




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








    1




    1




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




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











    5












    $begingroup$

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






    share|cite|improve this answer









    $endgroup$


















      5












      $begingroup$

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






      share|cite|improve this answer









      $endgroup$
















        5












        5








        5





        $begingroup$

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






        share|cite|improve this answer









        $endgroup$



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







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered 2 days ago









        Kavi Rama MurthyKavi Rama Murthy

        73.6k53170




        73.6k53170






























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