Showing this metric space is complete












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$begingroup$


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

4,35731856




4,35731856










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

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        3 Answers
        3






        active

        oldest

        votes









        active

        oldest

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        active

        oldest

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        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

            28.5k31747




            28.5k31747























                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

                    4,35731856




                    4,35731856























                        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

                            1,100818




                            1,100818






























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                                張江高科駅

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                                張江高科駅 ちょうこうこうか Zhangjiang Hi-Tech Park 所在地 中国上海市浦東新区 所属事業者 上海地鉄運営有限公司（上海軌道交通・駅詳細） 上海浦東現代有軌交通有限公司（駅詳細） テンプレートを表示 張江高科駅 各種表記 繁体字： 張江高科站 簡体字： 张江高科站 拼音 ： Zhàngjiāng Gāokē Zhàn 発音： ヂャンジャン ガオクー ヂャン 英文： Zhangjiang Hi-Tech Park Station テンプレートを表示 張江高科駅 （ちょうこうこうか-えき）は中華人民共和国上海市浦東新区に位置する、上海軌道交通2号線の駅である。 なお、張江高科駅前にある張江軌道電車の 張江地鉄駅駅 についてもここで述べる。 目次 1 利用可能な鉄道路線 2 沿革 3 駅構造 3.1 上海軌道交通 3.2 張江軌道電車 4 駅周辺 5 隣の駅 6 注釈 7 関連項目 利用可能な鉄道路線 上海軌道交通 ■ 2号線 上海浦東現代有軌交通有限公司 ■ 張江軌道電車 沿革 高架駅時代の姿 2000年12月27日 - 2号線の駅が高架駅として開業。 2010年 1月1日 - 張江軌道電車の開業に伴い、張江地鉄駅駅（停留所）を設置。 2月13日 - 駅地下化に伴い高架駅の運用停止。 2月24日 - 10日間の閉鎖を経て地下新駅の開業。 2017年2月9日 - リニューアル工事を経て再開業。乗降客の増加に対応し、ホームへのエレベーター、エスカレーターを増設、券売機エリアを拡張した [1] 。 駅構造 上海軌道交通 上海軌道交通 張江高科駅 5号口 ちょうこうこうか Zhangjiang Hi-Tech Park ◄ 竜陽路 (2.60km) (1.61km) 金科路 ► 所在地 中国上海市浦東新区松涛路 祖冲之路 （中国語版） 北緯31度12分11秒 東経121度34分50秒  /  北緯31.2...
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