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

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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 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.










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
    16 hours 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.










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
    16 hours 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.










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






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share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 9 hours ago







Juliana de Souza

















asked 17 hours 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
    16 hours 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
    16 hours 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
16 hours 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
16 hours ago










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









$endgroup$









  • 1




    $begingroup$
    See math.stackexchange.com/questions/152243/… for the proof of non-completeness
    $endgroup$
    – Chinnapparaj R
    16 hours 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
      16 hours 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
      16 hours 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 16 hours ago









    Chinnapparaj RChinnapparaj R

    5,9382928




    5,9382928








    • 1




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














    • 1




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








    1




    1




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




    $begingroup$
    See math.stackexchange.com/questions/152243/… for the proof of non-completeness
    $endgroup$
    – Chinnapparaj R
    16 hours 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 16 hours ago









        Kavi Rama MurthyKavi Rama Murthy

        72.9k53170




        72.9k53170






























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