Showing this metric space is complete












1












$begingroup$


Let $X=(0,1]$ and $d(x,y)=left|frac{1}{x}-frac{1}{y}right|$. I've proven $(X,d)$ is a metric space but I don't know how to show its completeness. How can I do that?






metric-spaces







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$endgroup$












  • $begingroup$
    Start with a Cauchy Sequence in $(0,1]$ and show that it has a limit in $(0,1]$
    $endgroup$
    – crskhr
    Oct 12 '14 at 8:59










  • $begingroup$
    {1/x_n} is a Cauchy sequence but its limit 0 isn't in (0,1]
    $endgroup$
    – GiulyB
    Oct 12 '14 at 9:01










  • $begingroup$
    what is $1/x_n$???
    $endgroup$
    – Ittay Weiss
    Oct 12 '14 at 9:03






  • 2




    $begingroup$
    Assuming that your $frac1{x_n}$ is simply $frac1n$, it’s not a Cauchy sequence with respect to the metric $d$: $$dleft(frac1n,frac1mright)=|n-m|;.$$
    $endgroup$
    – Brian M. Scott
    Oct 12 '14 at 9:03


















1












$begingroup$


Let $X=(0,1]$ and $d(x,y)=left|frac{1}{x}-frac{1}{y}right|$. I've proven $(X,d)$ is a metric space but I don't know how to show its completeness. How can I do that?






metric-spaces







share|cite|improve this question











$endgroup$












  • $begingroup$
    Start with a Cauchy Sequence in $(0,1]$ and show that it has a limit in $(0,1]$
    $endgroup$
    – crskhr
    Oct 12 '14 at 8:59










  • $begingroup$
    {1/x_n} is a Cauchy sequence but its limit 0 isn't in (0,1]
    $endgroup$
    – GiulyB
    Oct 12 '14 at 9:01










  • $begingroup$
    what is $1/x_n$???
    $endgroup$
    – Ittay Weiss
    Oct 12 '14 at 9:03






  • 2




    $begingroup$
    Assuming that your $frac1{x_n}$ is simply $frac1n$, it’s not a Cauchy sequence with respect to the metric $d$: $$dleft(frac1n,frac1mright)=|n-m|;.$$
    $endgroup$
    – Brian M. Scott
    Oct 12 '14 at 9:03
















1












1








1


0



$begingroup$


Let $X=(0,1]$ and $d(x,y)=left|frac{1}{x}-frac{1}{y}right|$. I've proven $(X,d)$ is a metric space but I don't know how to show its completeness. How can I do that?






metric-spaces







share|cite|improve this question











$endgroup$




Let $X=(0,1]$ and $d(x,y)=left|frac{1}{x}-frac{1}{y}right|$. I've proven $(X,d)$ is a metric space but I don't know how to show its completeness. How can I do that?






metric-spaces




metric-spaces






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Oct 12 '14 at 11:01









MickG

4,36431957




4,36431957










asked Oct 12 '14 at 8:57









GiulyBGiulyB

436




436












  • $begingroup$
    Start with a Cauchy Sequence in $(0,1]$ and show that it has a limit in $(0,1]$
    $endgroup$
    – crskhr
    Oct 12 '14 at 8:59










  • $begingroup$
    {1/x_n} is a Cauchy sequence but its limit 0 isn't in (0,1]
    $endgroup$
    – GiulyB
    Oct 12 '14 at 9:01










  • $begingroup$
    what is $1/x_n$???
    $endgroup$
    – Ittay Weiss
    Oct 12 '14 at 9:03






  • 2




    $begingroup$
    Assuming that your $frac1{x_n}$ is simply $frac1n$, it’s not a Cauchy sequence with respect to the metric $d$: $$dleft(frac1n,frac1mright)=|n-m|;.$$
    $endgroup$
    – Brian M. Scott
    Oct 12 '14 at 9:03




















  • $begingroup$
    Start with a Cauchy Sequence in $(0,1]$ and show that it has a limit in $(0,1]$
    $endgroup$
    – crskhr
    Oct 12 '14 at 8:59










  • $begingroup$
    {1/x_n} is a Cauchy sequence but its limit 0 isn't in (0,1]
    $endgroup$
    – GiulyB
    Oct 12 '14 at 9:01










  • $begingroup$
    what is $1/x_n$???
    $endgroup$
    – Ittay Weiss
    Oct 12 '14 at 9:03






  • 2




    $begingroup$
    Assuming that your $frac1{x_n}$ is simply $frac1n$, it’s not a Cauchy sequence with respect to the metric $d$: $$dleft(frac1n,frac1mright)=|n-m|;.$$
    $endgroup$
    – Brian M. Scott
    Oct 12 '14 at 9:03


















$begingroup$
Start with a Cauchy Sequence in $(0,1]$ and show that it has a limit in $(0,1]$
$endgroup$
– crskhr
Oct 12 '14 at 8:59




$begingroup$
Start with a Cauchy Sequence in $(0,1]$ and show that it has a limit in $(0,1]$
$endgroup$
– crskhr
Oct 12 '14 at 8:59












$begingroup$
{1/x_n} is a Cauchy sequence but its limit 0 isn't in (0,1]
$endgroup$
– GiulyB
Oct 12 '14 at 9:01




$begingroup$
{1/x_n} is a Cauchy sequence but its limit 0 isn't in (0,1]
$endgroup$
– GiulyB
Oct 12 '14 at 9:01












$begingroup$
what is $1/x_n$???
$endgroup$
– Ittay Weiss
Oct 12 '14 at 9:03




$begingroup$
what is $1/x_n$???
$endgroup$
– Ittay Weiss
Oct 12 '14 at 9:03




2




2




$begingroup$
Assuming that your $frac1{x_n}$ is simply $frac1n$, it’s not a Cauchy sequence with respect to the metric $d$: $$dleft(frac1n,frac1mright)=|n-m|;.$$
$endgroup$
– Brian M. Scott
Oct 12 '14 at 9:03






$begingroup$
Assuming that your $frac1{x_n}$ is simply $frac1n$, it’s not a Cauchy sequence with respect to the metric $d$: $$dleft(frac1n,frac1mright)=|n-m|;.$$
$endgroup$
– Brian M. Scott
Oct 12 '14 at 9:03












3 Answers
3






active

oldest

votes


















3












$begingroup$

HINT:



The map $x mapsto frac{1}{x}$ from your space to $[1, infty)$ with the usual metric is an isometry.






share|cite|improve this answer









$endgroup$





















    1












    $begingroup$

    Expanding on orangeskid's hint-answer, "isometry" means that if we call that map $f$, the distance on $X$ $d$ and the usual metric $d_e$, then $d(x,y)=d_e(f(x),f(y))$. This can be used to turn $x_n$ into a Cauchy sequence in $[1,infty)$, thus finding a limit there and getting back to $X$.



    $frac{1}{n}$ is a Cauchy sequence in $[0,1]$ or any subset of $mathbb{R}$ containing all its terms with the Euclid metric, not with your metric, where, as Brian M. Scott pointed out, the distances are never less than one.



    Finally, a comment on format: it is better to provide a generic title like "Proving a metric space is complete" or "Proving a certain metric space is complete" and then putting the definition into the question rather than placing the definition of the space and distance in the title and then simply stating you can't find a proof in the question. I'll edit the question accordingly, but keep this in mind for the future :).






    share|cite|improve this answer









    $endgroup$





















      1












      $begingroup$

      Let $x_n$ be a Cauchy sequence in $(X,d)$. This is equivalent to
      $$forall_{epsilon>0}exists_{N in mathbb{N}}forall_{n,m>N} left|frac{1}{x_n}-frac{1}{x_m}right| < epsilon$$
      But then $frac{1}{x_n} =:y_n$ is a Cauchy sequence in $(mathbb{R},p)$ with $p(x,y) = |x-y|$. Because $(mathbb{R},p)$ is complete there exists a number $x$ such that $lim_{n to infty}y_n = x$. Note that $x geq 1$, because $0<x_nleq 1$. $lim_{n to infty}y_n = x$ is again equivalent by the definition of convergence to $$forall_{epsilon>0}exists_{N in mathbb{N}}forall_{n,m>N} left|frac{1}{x_n}-xright| < epsilon$$
      But this is the same as
      $$forall_{epsilon>0}exists_{N in mathbb{N}}forall_{n,m>N} d(x_n,frac{1}{x}) < epsilon$$ which means that $lim_{n to infty}x_n = frac{1}{x} in X$.






      share|cite|improve this answer











      $endgroup$













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        3 Answers
        3






        active

        oldest

        votes








        3 Answers
        3






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        3












        $begingroup$

        HINT:



        The map $x mapsto frac{1}{x}$ from your space to $[1, infty)$ with the usual metric is an isometry.






        share|cite|improve this answer









        $endgroup$


















          3












          $begingroup$

          HINT:



          The map $x mapsto frac{1}{x}$ from your space to $[1, infty)$ with the usual metric is an isometry.






          share|cite|improve this answer









          $endgroup$
















            3












            3








            3





            $begingroup$

            HINT:



            The map $x mapsto frac{1}{x}$ from your space to $[1, infty)$ with the usual metric is an isometry.






            share|cite|improve this answer









            $endgroup$



            HINT:



            The map $x mapsto frac{1}{x}$ from your space to $[1, infty)$ with the usual metric is an isometry.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Oct 12 '14 at 9:06









            Orest BucicovschiOrest Bucicovschi

            28.6k31748




            28.6k31748























                1












                $begingroup$

                Expanding on orangeskid's hint-answer, "isometry" means that if we call that map $f$, the distance on $X$ $d$ and the usual metric $d_e$, then $d(x,y)=d_e(f(x),f(y))$. This can be used to turn $x_n$ into a Cauchy sequence in $[1,infty)$, thus finding a limit there and getting back to $X$.



                $frac{1}{n}$ is a Cauchy sequence in $[0,1]$ or any subset of $mathbb{R}$ containing all its terms with the Euclid metric, not with your metric, where, as Brian M. Scott pointed out, the distances are never less than one.



                Finally, a comment on format: it is better to provide a generic title like "Proving a metric space is complete" or "Proving a certain metric space is complete" and then putting the definition into the question rather than placing the definition of the space and distance in the title and then simply stating you can't find a proof in the question. I'll edit the question accordingly, but keep this in mind for the future :).






                share|cite|improve this answer









                $endgroup$


















                  1












                  $begingroup$

                  Expanding on orangeskid's hint-answer, "isometry" means that if we call that map $f$, the distance on $X$ $d$ and the usual metric $d_e$, then $d(x,y)=d_e(f(x),f(y))$. This can be used to turn $x_n$ into a Cauchy sequence in $[1,infty)$, thus finding a limit there and getting back to $X$.



                  $frac{1}{n}$ is a Cauchy sequence in $[0,1]$ or any subset of $mathbb{R}$ containing all its terms with the Euclid metric, not with your metric, where, as Brian M. Scott pointed out, the distances are never less than one.



                  Finally, a comment on format: it is better to provide a generic title like "Proving a metric space is complete" or "Proving a certain metric space is complete" and then putting the definition into the question rather than placing the definition of the space and distance in the title and then simply stating you can't find a proof in the question. I'll edit the question accordingly, but keep this in mind for the future :).






                  share|cite|improve this answer









                  $endgroup$
















                    1












                    1








                    1





                    $begingroup$

                    Expanding on orangeskid's hint-answer, "isometry" means that if we call that map $f$, the distance on $X$ $d$ and the usual metric $d_e$, then $d(x,y)=d_e(f(x),f(y))$. This can be used to turn $x_n$ into a Cauchy sequence in $[1,infty)$, thus finding a limit there and getting back to $X$.



                    $frac{1}{n}$ is a Cauchy sequence in $[0,1]$ or any subset of $mathbb{R}$ containing all its terms with the Euclid metric, not with your metric, where, as Brian M. Scott pointed out, the distances are never less than one.



                    Finally, a comment on format: it is better to provide a generic title like "Proving a metric space is complete" or "Proving a certain metric space is complete" and then putting the definition into the question rather than placing the definition of the space and distance in the title and then simply stating you can't find a proof in the question. I'll edit the question accordingly, but keep this in mind for the future :).






                    share|cite|improve this answer









                    $endgroup$



                    Expanding on orangeskid's hint-answer, "isometry" means that if we call that map $f$, the distance on $X$ $d$ and the usual metric $d_e$, then $d(x,y)=d_e(f(x),f(y))$. This can be used to turn $x_n$ into a Cauchy sequence in $[1,infty)$, thus finding a limit there and getting back to $X$.



                    $frac{1}{n}$ is a Cauchy sequence in $[0,1]$ or any subset of $mathbb{R}$ containing all its terms with the Euclid metric, not with your metric, where, as Brian M. Scott pointed out, the distances are never less than one.



                    Finally, a comment on format: it is better to provide a generic title like "Proving a metric space is complete" or "Proving a certain metric space is complete" and then putting the definition into the question rather than placing the definition of the space and distance in the title and then simply stating you can't find a proof in the question. I'll edit the question accordingly, but keep this in mind for the future :).







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Oct 12 '14 at 10:47









                    MickGMickG

                    4,36431957




                    4,36431957























                        1












                        $begingroup$

                        Let $x_n$ be a Cauchy sequence in $(X,d)$. This is equivalent to
                        $$forall_{epsilon>0}exists_{N in mathbb{N}}forall_{n,m>N} left|frac{1}{x_n}-frac{1}{x_m}right| < epsilon$$
                        But then $frac{1}{x_n} =:y_n$ is a Cauchy sequence in $(mathbb{R},p)$ with $p(x,y) = |x-y|$. Because $(mathbb{R},p)$ is complete there exists a number $x$ such that $lim_{n to infty}y_n = x$. Note that $x geq 1$, because $0<x_nleq 1$. $lim_{n to infty}y_n = x$ is again equivalent by the definition of convergence to $$forall_{epsilon>0}exists_{N in mathbb{N}}forall_{n,m>N} left|frac{1}{x_n}-xright| < epsilon$$
                        But this is the same as
                        $$forall_{epsilon>0}exists_{N in mathbb{N}}forall_{n,m>N} d(x_n,frac{1}{x}) < epsilon$$ which means that $lim_{n to infty}x_n = frac{1}{x} in X$.






                        share|cite|improve this answer











                        $endgroup$


















                          1












                          $begingroup$

                          Let $x_n$ be a Cauchy sequence in $(X,d)$. This is equivalent to
                          $$forall_{epsilon>0}exists_{N in mathbb{N}}forall_{n,m>N} left|frac{1}{x_n}-frac{1}{x_m}right| < epsilon$$
                          But then $frac{1}{x_n} =:y_n$ is a Cauchy sequence in $(mathbb{R},p)$ with $p(x,y) = |x-y|$. Because $(mathbb{R},p)$ is complete there exists a number $x$ such that $lim_{n to infty}y_n = x$. Note that $x geq 1$, because $0<x_nleq 1$. $lim_{n to infty}y_n = x$ is again equivalent by the definition of convergence to $$forall_{epsilon>0}exists_{N in mathbb{N}}forall_{n,m>N} left|frac{1}{x_n}-xright| < epsilon$$
                          But this is the same as
                          $$forall_{epsilon>0}exists_{N in mathbb{N}}forall_{n,m>N} d(x_n,frac{1}{x}) < epsilon$$ which means that $lim_{n to infty}x_n = frac{1}{x} in X$.






                          share|cite|improve this answer











                          $endgroup$
















                            1












                            1








                            1





                            $begingroup$

                            Let $x_n$ be a Cauchy sequence in $(X,d)$. This is equivalent to
                            $$forall_{epsilon>0}exists_{N in mathbb{N}}forall_{n,m>N} left|frac{1}{x_n}-frac{1}{x_m}right| < epsilon$$
                            But then $frac{1}{x_n} =:y_n$ is a Cauchy sequence in $(mathbb{R},p)$ with $p(x,y) = |x-y|$. Because $(mathbb{R},p)$ is complete there exists a number $x$ such that $lim_{n to infty}y_n = x$. Note that $x geq 1$, because $0<x_nleq 1$. $lim_{n to infty}y_n = x$ is again equivalent by the definition of convergence to $$forall_{epsilon>0}exists_{N in mathbb{N}}forall_{n,m>N} left|frac{1}{x_n}-xright| < epsilon$$
                            But this is the same as
                            $$forall_{epsilon>0}exists_{N in mathbb{N}}forall_{n,m>N} d(x_n,frac{1}{x}) < epsilon$$ which means that $lim_{n to infty}x_n = frac{1}{x} in X$.






                            share|cite|improve this answer











                            $endgroup$



                            Let $x_n$ be a Cauchy sequence in $(X,d)$. This is equivalent to
                            $$forall_{epsilon>0}exists_{N in mathbb{N}}forall_{n,m>N} left|frac{1}{x_n}-frac{1}{x_m}right| < epsilon$$
                            But then $frac{1}{x_n} =:y_n$ is a Cauchy sequence in $(mathbb{R},p)$ with $p(x,y) = |x-y|$. Because $(mathbb{R},p)$ is complete there exists a number $x$ such that $lim_{n to infty}y_n = x$. Note that $x geq 1$, because $0<x_nleq 1$. $lim_{n to infty}y_n = x$ is again equivalent by the definition of convergence to $$forall_{epsilon>0}exists_{N in mathbb{N}}forall_{n,m>N} left|frac{1}{x_n}-xright| < epsilon$$
                            But this is the same as
                            $$forall_{epsilon>0}exists_{N in mathbb{N}}forall_{n,m>N} d(x_n,frac{1}{x}) < epsilon$$ which means that $lim_{n to infty}x_n = frac{1}{x} in X$.







                            share|cite|improve this answer














                            share|cite|improve this answer



                            share|cite|improve this answer








                            edited Jul 22 '15 at 9:48

























                            answered Jul 21 '15 at 21:50









                            Andrei KhAndrei Kh

                            1,135818




                            1,135818






























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