Complete+bounded homeomorphic to incomplete+unbounded












3














I'm aware that completeness and (total) boundedness are not preserved under homeomorphism, with $(0,1) cong mathbb R$ being a counterexample to both simultaneously. I'm curious if there exists a "double counterexample" in another way:



Are there homeomorphic metric spaces $M$ and $N$ such that
$M$ is both complete and bounded,
but $N$ is neither complete nor bounded?



Of course in that case, $M$ could not be totally bounded, else it would be compact, which is certainly a topological property!






general-topology metric-spaces







share|cite|improve this question



























    3














    I'm aware that completeness and (total) boundedness are not preserved under homeomorphism, with $(0,1) cong mathbb R$ being a counterexample to both simultaneously. I'm curious if there exists a "double counterexample" in another way:



    Are there homeomorphic metric spaces $M$ and $N$ such that
    $M$ is both complete and bounded,
    but $N$ is neither complete nor bounded?



    Of course in that case, $M$ could not be totally bounded, else it would be compact, which is certainly a topological property!






    general-topology metric-spaces







    share|cite|improve this question

























      3












      3








      3







      I'm aware that completeness and (total) boundedness are not preserved under homeomorphism, with $(0,1) cong mathbb R$ being a counterexample to both simultaneously. I'm curious if there exists a "double counterexample" in another way:



      Are there homeomorphic metric spaces $M$ and $N$ such that
      $M$ is both complete and bounded,
      but $N$ is neither complete nor bounded?



      Of course in that case, $M$ could not be totally bounded, else it would be compact, which is certainly a topological property!






      general-topology metric-spaces







      share|cite|improve this question













      I'm aware that completeness and (total) boundedness are not preserved under homeomorphism, with $(0,1) cong mathbb R$ being a counterexample to both simultaneously. I'm curious if there exists a "double counterexample" in another way:



      Are there homeomorphic metric spaces $M$ and $N$ such that
      $M$ is both complete and bounded,
      but $N$ is neither complete nor bounded?



      Of course in that case, $M$ could not be totally bounded, else it would be compact, which is certainly a topological property!






      general-topology metric-spaces




      general-topology metric-spaces






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked 4 hours ago









      Evan Chen

      1,145516




      1,145516






















          2 Answers
          2






          active

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          3














          You could let $M$ be $mathbb R$ with the metric $d(x,y)=min(1,|x-y|)$ and $N$ be $(0,infty)$ with the usual metric.






          share|cite|improve this answer





















          • Thanks! Do you mind if I include this example in my metric topology notes? I would credit you and link here.
            – Evan Chen
            4 hours ago










          • @EvanChen: Sure, go ahead.
            – Henning Makholm
            4 hours ago



















          2














          Yes. Let $M$ be $mathbb N$ with the discrete metric, and let $N= {1/n:nin mathbb N}cup {2,3,4,dots}$ with the usual $mathbb R$ metric.






          share|cite|improve this answer





















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            2 Answers
            2






            active

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            2 Answers
            2






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            3














            You could let $M$ be $mathbb R$ with the metric $d(x,y)=min(1,|x-y|)$ and $N$ be $(0,infty)$ with the usual metric.






            share|cite|improve this answer





















            • Thanks! Do you mind if I include this example in my metric topology notes? I would credit you and link here.
              – Evan Chen
              4 hours ago










            • @EvanChen: Sure, go ahead.
              – Henning Makholm
              4 hours ago
















            3














            You could let $M$ be $mathbb R$ with the metric $d(x,y)=min(1,|x-y|)$ and $N$ be $(0,infty)$ with the usual metric.






            share|cite|improve this answer





















            • Thanks! Do you mind if I include this example in my metric topology notes? I would credit you and link here.
              – Evan Chen
              4 hours ago










            • @EvanChen: Sure, go ahead.
              – Henning Makholm
              4 hours ago














            3












            3








            3






            You could let $M$ be $mathbb R$ with the metric $d(x,y)=min(1,|x-y|)$ and $N$ be $(0,infty)$ with the usual metric.






            share|cite|improve this answer












            You could let $M$ be $mathbb R$ with the metric $d(x,y)=min(1,|x-y|)$ and $N$ be $(0,infty)$ with the usual metric.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered 4 hours ago









            Henning Makholm

            237k16301536




            237k16301536












            • Thanks! Do you mind if I include this example in my metric topology notes? I would credit you and link here.
              – Evan Chen
              4 hours ago










            • @EvanChen: Sure, go ahead.
              – Henning Makholm
              4 hours ago


















            • Thanks! Do you mind if I include this example in my metric topology notes? I would credit you and link here.
              – Evan Chen
              4 hours ago










            • @EvanChen: Sure, go ahead.
              – Henning Makholm
              4 hours ago
















            Thanks! Do you mind if I include this example in my metric topology notes? I would credit you and link here.
            – Evan Chen
            4 hours ago




            Thanks! Do you mind if I include this example in my metric topology notes? I would credit you and link here.
            – Evan Chen
            4 hours ago












            @EvanChen: Sure, go ahead.
            – Henning Makholm
            4 hours ago




            @EvanChen: Sure, go ahead.
            – Henning Makholm
            4 hours ago











            2














            Yes. Let $M$ be $mathbb N$ with the discrete metric, and let $N= {1/n:nin mathbb N}cup {2,3,4,dots}$ with the usual $mathbb R$ metric.






            share|cite|improve this answer


























              2














              Yes. Let $M$ be $mathbb N$ with the discrete metric, and let $N= {1/n:nin mathbb N}cup {2,3,4,dots}$ with the usual $mathbb R$ metric.






              share|cite|improve this answer
























                2












                2








                2






                Yes. Let $M$ be $mathbb N$ with the discrete metric, and let $N= {1/n:nin mathbb N}cup {2,3,4,dots}$ with the usual $mathbb R$ metric.






                share|cite|improve this answer












                Yes. Let $M$ be $mathbb N$ with the discrete metric, and let $N= {1/n:nin mathbb N}cup {2,3,4,dots}$ with the usual $mathbb R$ metric.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered 4 hours ago









                zhw.

                71.5k43075




                71.5k43075






























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