Does there exist a metric on $X$ such that $X^{∗} − X$ is uncountable?












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Let $X^{∗}$ denote the completion of the metric space $X$. Is there a metric on the open interval $X = (0, 1)$ generating the Euclidean topology for which $X^{∗} − X$ is uncountable? Any idea and hint would be helpful. Thanks!






general-topology







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    $begingroup$


    Let $X^{∗}$ denote the completion of the metric space $X$. Is there a metric on the open interval $X = (0, 1)$ generating the Euclidean topology for which $X^{∗} − X$ is uncountable? Any idea and hint would be helpful. Thanks!






    general-topology







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    $endgroup$















      2












      2








      2





      $begingroup$


      Let $X^{∗}$ denote the completion of the metric space $X$. Is there a metric on the open interval $X = (0, 1)$ generating the Euclidean topology for which $X^{∗} − X$ is uncountable? Any idea and hint would be helpful. Thanks!






      general-topology







      share|cite|improve this question









      $endgroup$




      Let $X^{∗}$ denote the completion of the metric space $X$. Is there a metric on the open interval $X = (0, 1)$ generating the Euclidean topology for which $X^{∗} − X$ is uncountable? Any idea and hint would be helpful. Thanks!






      general-topology




      general-topology






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      asked Jan 25 at 18:19









      Ergin SuerErgin Suer

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          $begingroup$

          Consider the "topologist's sine curve" in the form
          $$C={(x,sin(1/x):0<x<1}.$$
          This is a subspace of $Bbb R^2$ homeomorphic to $(0,1)$, so we can think
          of $(0,1)$ as a metric space by considering the metric of $C$ inside $Bbb R^2$.
          Thus
          $$d(x,y)=sqrt{(x-y)^2+left(sinfrac1x-sinfrac1yright)^2}.$$
          The closure of $C$ in $Bbb R^2$ is the completion of $C$ with respect
          to this metric, and contains the uncountable set ${(0,y):-1<y<1}$.
          Thus $C^*-C$ is uncountable.






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            $begingroup$

            Consider the "topologist's sine curve" in the form
            $$C={(x,sin(1/x):0<x<1}.$$
            This is a subspace of $Bbb R^2$ homeomorphic to $(0,1)$, so we can think
            of $(0,1)$ as a metric space by considering the metric of $C$ inside $Bbb R^2$.
            Thus
            $$d(x,y)=sqrt{(x-y)^2+left(sinfrac1x-sinfrac1yright)^2}.$$
            The closure of $C$ in $Bbb R^2$ is the completion of $C$ with respect
            to this metric, and contains the uncountable set ${(0,y):-1<y<1}$.
            Thus $C^*-C$ is uncountable.






            share|cite|improve this answer









            $endgroup$


















              4












              $begingroup$

              Consider the "topologist's sine curve" in the form
              $$C={(x,sin(1/x):0<x<1}.$$
              This is a subspace of $Bbb R^2$ homeomorphic to $(0,1)$, so we can think
              of $(0,1)$ as a metric space by considering the metric of $C$ inside $Bbb R^2$.
              Thus
              $$d(x,y)=sqrt{(x-y)^2+left(sinfrac1x-sinfrac1yright)^2}.$$
              The closure of $C$ in $Bbb R^2$ is the completion of $C$ with respect
              to this metric, and contains the uncountable set ${(0,y):-1<y<1}$.
              Thus $C^*-C$ is uncountable.






              share|cite|improve this answer









              $endgroup$
















                4












                4








                4





                $begingroup$

                Consider the "topologist's sine curve" in the form
                $$C={(x,sin(1/x):0<x<1}.$$
                This is a subspace of $Bbb R^2$ homeomorphic to $(0,1)$, so we can think
                of $(0,1)$ as a metric space by considering the metric of $C$ inside $Bbb R^2$.
                Thus
                $$d(x,y)=sqrt{(x-y)^2+left(sinfrac1x-sinfrac1yright)^2}.$$
                The closure of $C$ in $Bbb R^2$ is the completion of $C$ with respect
                to this metric, and contains the uncountable set ${(0,y):-1<y<1}$.
                Thus $C^*-C$ is uncountable.






                share|cite|improve this answer









                $endgroup$



                Consider the "topologist's sine curve" in the form
                $$C={(x,sin(1/x):0<x<1}.$$
                This is a subspace of $Bbb R^2$ homeomorphic to $(0,1)$, so we can think
                of $(0,1)$ as a metric space by considering the metric of $C$ inside $Bbb R^2$.
                Thus
                $$d(x,y)=sqrt{(x-y)^2+left(sinfrac1x-sinfrac1yright)^2}.$$
                The closure of $C$ in $Bbb R^2$ is the completion of $C$ with respect
                to this metric, and contains the uncountable set ${(0,y):-1<y<1}$.
                Thus $C^*-C$ is uncountable.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Jan 25 at 18:26









                Lord Shark the UnknownLord Shark the Unknown

                106k1161133




                106k1161133






























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