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Is there a metric for which the open unit interval is complete?



The Next CEO of Stack OverflowWhich of the following metric spaces are complete?which of the following metric spaces are separable?Metric on half-open interval s.t. subset is open w.r.t. $d$ iff open w.r.t. Euclidean metricQuestion on complete metric spaces and whether the following is a complete metric space:Open interval $(0,1)$ with the usual topology admits a metric spaceComplete Metric Space examplechoose the complete metric space?Spaces of (complete) separable metric spacesLocally Complete Metric SpaceIs the family of all continuous functions from a compact metric space to a separable complete metric space separable?










0












$begingroup$


Let, $I= (0,1)$ It is well known that $I$ is not a complete with respect to the Euclidean metric $(x,y)mapsto |x-y|$.



However, $(I,|cdot|)$ is separable.




Question: Can we find a metric $d: Itimes I to(0,infty)$ for which, $(I,d)$ is separable and complete?











share|cite|improve this question









$endgroup$







  • 2




    $begingroup$
    Pull back the metric from $mathbbR$ according to some homeomorphism between them.
    $endgroup$
    – user647486
    Mar 19 at 20:52















0












$begingroup$


Let, $I= (0,1)$ It is well known that $I$ is not a complete with respect to the Euclidean metric $(x,y)mapsto |x-y|$.



However, $(I,|cdot|)$ is separable.




Question: Can we find a metric $d: Itimes I to(0,infty)$ for which, $(I,d)$ is separable and complete?











share|cite|improve this question









$endgroup$







  • 2




    $begingroup$
    Pull back the metric from $mathbbR$ according to some homeomorphism between them.
    $endgroup$
    – user647486
    Mar 19 at 20:52













0












0








0





$begingroup$


Let, $I= (0,1)$ It is well known that $I$ is not a complete with respect to the Euclidean metric $(x,y)mapsto |x-y|$.



However, $(I,|cdot|)$ is separable.




Question: Can we find a metric $d: Itimes I to(0,infty)$ for which, $(I,d)$ is separable and complete?











share|cite|improve this question









$endgroup$




Let, $I= (0,1)$ It is well known that $I$ is not a complete with respect to the Euclidean metric $(x,y)mapsto |x-y|$.



However, $(I,|cdot|)$ is separable.




Question: Can we find a metric $d: Itimes I to(0,infty)$ for which, $(I,d)$ is separable and complete?








real-analysis analysis metric-spaces complete-spaces separable-spaces






share|cite|improve this question













share|cite|improve this question











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share|cite|improve this question










asked Mar 19 at 20:50









Guy FsoneGuy Fsone

17.3k43074




17.3k43074







  • 2




    $begingroup$
    Pull back the metric from $mathbbR$ according to some homeomorphism between them.
    $endgroup$
    – user647486
    Mar 19 at 20:52












  • 2




    $begingroup$
    Pull back the metric from $mathbbR$ according to some homeomorphism between them.
    $endgroup$
    – user647486
    Mar 19 at 20:52







2




2




$begingroup$
Pull back the metric from $mathbbR$ according to some homeomorphism between them.
$endgroup$
– user647486
Mar 19 at 20:52




$begingroup$
Pull back the metric from $mathbbR$ according to some homeomorphism between them.
$endgroup$
– user647486
Mar 19 at 20:52










2 Answers
2






active

oldest

votes


















2












$begingroup$

Try $J = (-pi/2,pi/2)$ instead. How about $d(x,y) = |tan x - tan y|$?






share|cite|improve this answer









$endgroup$




















    1












    $begingroup$

    Take a homeomorphism from the interval to $Bbb R$ and pull back the usual Euclidean metric.






    share|cite|improve this answer









    $endgroup$













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      2 Answers
      2






      active

      oldest

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      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      2












      $begingroup$

      Try $J = (-pi/2,pi/2)$ instead. How about $d(x,y) = |tan x - tan y|$?






      share|cite|improve this answer









      $endgroup$

















        2












        $begingroup$

        Try $J = (-pi/2,pi/2)$ instead. How about $d(x,y) = |tan x - tan y|$?






        share|cite|improve this answer









        $endgroup$















          2












          2








          2





          $begingroup$

          Try $J = (-pi/2,pi/2)$ instead. How about $d(x,y) = |tan x - tan y|$?






          share|cite|improve this answer









          $endgroup$



          Try $J = (-pi/2,pi/2)$ instead. How about $d(x,y) = |tan x - tan y|$?







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Mar 19 at 20:52









          Umberto P.Umberto P.

          40.2k13370




          40.2k13370





















              1












              $begingroup$

              Take a homeomorphism from the interval to $Bbb R$ and pull back the usual Euclidean metric.






              share|cite|improve this answer









              $endgroup$

















                1












                $begingroup$

                Take a homeomorphism from the interval to $Bbb R$ and pull back the usual Euclidean metric.






                share|cite|improve this answer









                $endgroup$















                  1












                  1








                  1





                  $begingroup$

                  Take a homeomorphism from the interval to $Bbb R$ and pull back the usual Euclidean metric.






                  share|cite|improve this answer









                  $endgroup$



                  Take a homeomorphism from the interval to $Bbb R$ and pull back the usual Euclidean metric.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Mar 19 at 20:52









                  Ted ShifrinTed Shifrin

                  64.7k44692




                  64.7k44692



























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