Identical Cauchy sequences and continuity.












1












$begingroup$


Find a set $X$ and two metrics $d$ and $m$ on $X$ such that the Cauchy sequences of $(X,d)$ and $(X,m)$ are identical and the identity map from $(X,d)$ to $(X,m)$ is continuous but not uniformly continuous.



My attempt: if we take $X=mathbb{R}$ and the identity map $f:mathbb{R} rightarrow mathbb{R}$ and the metrics are equivalent, then we get our result.(?)



A question: Uniform continuity means that the cauchy sequences of the first metric space maps to the second. How can it not be uniformly continuous?






metric-spaces cauchy-sequences uniform-continuity







share|cite|improve this question











$endgroup$












  • $begingroup$
    Uniform continuity implies that the function maps Cauchy to Cauchy. But is not equivalent.
    $endgroup$
    – freakish
    Dec 18 '18 at 11:56
















1












$begingroup$


Find a set $X$ and two metrics $d$ and $m$ on $X$ such that the Cauchy sequences of $(X,d)$ and $(X,m)$ are identical and the identity map from $(X,d)$ to $(X,m)$ is continuous but not uniformly continuous.



My attempt: if we take $X=mathbb{R}$ and the identity map $f:mathbb{R} rightarrow mathbb{R}$ and the metrics are equivalent, then we get our result.(?)



A question: Uniform continuity means that the cauchy sequences of the first metric space maps to the second. How can it not be uniformly continuous?






metric-spaces cauchy-sequences uniform-continuity







share|cite|improve this question











$endgroup$












  • $begingroup$
    Uniform continuity implies that the function maps Cauchy to Cauchy. But is not equivalent.
    $endgroup$
    – freakish
    Dec 18 '18 at 11:56














1












1








1





$begingroup$


Find a set $X$ and two metrics $d$ and $m$ on $X$ such that the Cauchy sequences of $(X,d)$ and $(X,m)$ are identical and the identity map from $(X,d)$ to $(X,m)$ is continuous but not uniformly continuous.



My attempt: if we take $X=mathbb{R}$ and the identity map $f:mathbb{R} rightarrow mathbb{R}$ and the metrics are equivalent, then we get our result.(?)



A question: Uniform continuity means that the cauchy sequences of the first metric space maps to the second. How can it not be uniformly continuous?






metric-spaces cauchy-sequences uniform-continuity







share|cite|improve this question











$endgroup$




Find a set $X$ and two metrics $d$ and $m$ on $X$ such that the Cauchy sequences of $(X,d)$ and $(X,m)$ are identical and the identity map from $(X,d)$ to $(X,m)$ is continuous but not uniformly continuous.



My attempt: if we take $X=mathbb{R}$ and the identity map $f:mathbb{R} rightarrow mathbb{R}$ and the metrics are equivalent, then we get our result.(?)



A question: Uniform continuity means that the cauchy sequences of the first metric space maps to the second. How can it not be uniformly continuous?






metric-spaces cauchy-sequences uniform-continuity




metric-spaces cauchy-sequences uniform-continuity






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 18 '18 at 11:57









José Carlos Santos

168k22132236




168k22132236










asked Dec 18 '18 at 11:29









argiriskarargiriskar

1409




1409












  • $begingroup$
    Uniform continuity implies that the function maps Cauchy to Cauchy. But is not equivalent.
    $endgroup$
    – freakish
    Dec 18 '18 at 11:56


















  • $begingroup$
    Uniform continuity implies that the function maps Cauchy to Cauchy. But is not equivalent.
    $endgroup$
    – freakish
    Dec 18 '18 at 11:56
















$begingroup$
Uniform continuity implies that the function maps Cauchy to Cauchy. But is not equivalent.
$endgroup$
– freakish
Dec 18 '18 at 11:56




$begingroup$
Uniform continuity implies that the function maps Cauchy to Cauchy. But is not equivalent.
$endgroup$
– freakish
Dec 18 '18 at 11:56










2 Answers
2






active

oldest

votes


















2












$begingroup$

Let $X=mathbb R$, $d(x,y)=|x-y|$ and $m(x,y)=|x^{2}-y^{2}|+|x-y|$. Use the fact that d-Cauchy sequences are bounded to show that Cauchy sequences under $d$ and $m$ are the same. If the identity map from $(X,d)$ to $(X.m)$ is uniformly continuous then, given $epsilon >0$ there exists $delta >0$ such that $|x^{2}-y^{2}|+|x-y| <epsilon$ whenever $|x-y| < delta$ $,,,$ (1). But this would imply that $x to x^{2}$ is uniformly continuous function (when both domain and range are given the metric $d$) which you know is false. [ You can take $x=n,y=n+frac 1 n$ with large $n$ to get a contradiction to (1)].






share|cite|improve this answer









$endgroup$





















    2












    $begingroup$

    Take $X=mathbb R$, let $d$ be the usual distance and define$$m(x,y)=bigllvert x^3-y^3bigrrvert.$$Then $(X,d)$ and $(X,m)$ have the same Cauchy sequences, but $operatorname{Id}colon(X,d)longrightarrow(X,m)$ is not uniformly continuous (but it is continuous).






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      WLOG assume $|x| le |y|$, $|x^3-y^3|=|x-y||x^2+xy+y^2| le d(x,y)3|y|^2 le 3Md(x,y)$. Is this why these different metrics have the same Cauchy sequences ?
      $endgroup$
      – argiriskar
      Dec 19 '18 at 15:15












    • $begingroup$
      It seems to me that, yes, the fact that the Cauchy sequences are the same in both cases can be deduced from this.
      $endgroup$
      – José Carlos Santos
      Dec 19 '18 at 17:07










    • $begingroup$
      Thank you very much !
      $endgroup$
      – argiriskar
      Dec 19 '18 at 17:07











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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    2












    $begingroup$

    Let $X=mathbb R$, $d(x,y)=|x-y|$ and $m(x,y)=|x^{2}-y^{2}|+|x-y|$. Use the fact that d-Cauchy sequences are bounded to show that Cauchy sequences under $d$ and $m$ are the same. If the identity map from $(X,d)$ to $(X.m)$ is uniformly continuous then, given $epsilon >0$ there exists $delta >0$ such that $|x^{2}-y^{2}|+|x-y| <epsilon$ whenever $|x-y| < delta$ $,,,$ (1). But this would imply that $x to x^{2}$ is uniformly continuous function (when both domain and range are given the metric $d$) which you know is false. [ You can take $x=n,y=n+frac 1 n$ with large $n$ to get a contradiction to (1)].






    share|cite|improve this answer









    $endgroup$


















      2












      $begingroup$

      Let $X=mathbb R$, $d(x,y)=|x-y|$ and $m(x,y)=|x^{2}-y^{2}|+|x-y|$. Use the fact that d-Cauchy sequences are bounded to show that Cauchy sequences under $d$ and $m$ are the same. If the identity map from $(X,d)$ to $(X.m)$ is uniformly continuous then, given $epsilon >0$ there exists $delta >0$ such that $|x^{2}-y^{2}|+|x-y| <epsilon$ whenever $|x-y| < delta$ $,,,$ (1). But this would imply that $x to x^{2}$ is uniformly continuous function (when both domain and range are given the metric $d$) which you know is false. [ You can take $x=n,y=n+frac 1 n$ with large $n$ to get a contradiction to (1)].






      share|cite|improve this answer









      $endgroup$
















        2












        2








        2





        $begingroup$

        Let $X=mathbb R$, $d(x,y)=|x-y|$ and $m(x,y)=|x^{2}-y^{2}|+|x-y|$. Use the fact that d-Cauchy sequences are bounded to show that Cauchy sequences under $d$ and $m$ are the same. If the identity map from $(X,d)$ to $(X.m)$ is uniformly continuous then, given $epsilon >0$ there exists $delta >0$ such that $|x^{2}-y^{2}|+|x-y| <epsilon$ whenever $|x-y| < delta$ $,,,$ (1). But this would imply that $x to x^{2}$ is uniformly continuous function (when both domain and range are given the metric $d$) which you know is false. [ You can take $x=n,y=n+frac 1 n$ with large $n$ to get a contradiction to (1)].






        share|cite|improve this answer









        $endgroup$



        Let $X=mathbb R$, $d(x,y)=|x-y|$ and $m(x,y)=|x^{2}-y^{2}|+|x-y|$. Use the fact that d-Cauchy sequences are bounded to show that Cauchy sequences under $d$ and $m$ are the same. If the identity map from $(X,d)$ to $(X.m)$ is uniformly continuous then, given $epsilon >0$ there exists $delta >0$ such that $|x^{2}-y^{2}|+|x-y| <epsilon$ whenever $|x-y| < delta$ $,,,$ (1). But this would imply that $x to x^{2}$ is uniformly continuous function (when both domain and range are given the metric $d$) which you know is false. [ You can take $x=n,y=n+frac 1 n$ with large $n$ to get a contradiction to (1)].







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 18 '18 at 11:55









        Kavi Rama MurthyKavi Rama Murthy

        67.6k53067




        67.6k53067























            2












            $begingroup$

            Take $X=mathbb R$, let $d$ be the usual distance and define$$m(x,y)=bigllvert x^3-y^3bigrrvert.$$Then $(X,d)$ and $(X,m)$ have the same Cauchy sequences, but $operatorname{Id}colon(X,d)longrightarrow(X,m)$ is not uniformly continuous (but it is continuous).






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              WLOG assume $|x| le |y|$, $|x^3-y^3|=|x-y||x^2+xy+y^2| le d(x,y)3|y|^2 le 3Md(x,y)$. Is this why these different metrics have the same Cauchy sequences ?
              $endgroup$
              – argiriskar
              Dec 19 '18 at 15:15












            • $begingroup$
              It seems to me that, yes, the fact that the Cauchy sequences are the same in both cases can be deduced from this.
              $endgroup$
              – José Carlos Santos
              Dec 19 '18 at 17:07










            • $begingroup$
              Thank you very much !
              $endgroup$
              – argiriskar
              Dec 19 '18 at 17:07
















            2












            $begingroup$

            Take $X=mathbb R$, let $d$ be the usual distance and define$$m(x,y)=bigllvert x^3-y^3bigrrvert.$$Then $(X,d)$ and $(X,m)$ have the same Cauchy sequences, but $operatorname{Id}colon(X,d)longrightarrow(X,m)$ is not uniformly continuous (but it is continuous).






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              WLOG assume $|x| le |y|$, $|x^3-y^3|=|x-y||x^2+xy+y^2| le d(x,y)3|y|^2 le 3Md(x,y)$. Is this why these different metrics have the same Cauchy sequences ?
              $endgroup$
              – argiriskar
              Dec 19 '18 at 15:15












            • $begingroup$
              It seems to me that, yes, the fact that the Cauchy sequences are the same in both cases can be deduced from this.
              $endgroup$
              – José Carlos Santos
              Dec 19 '18 at 17:07










            • $begingroup$
              Thank you very much !
              $endgroup$
              – argiriskar
              Dec 19 '18 at 17:07














            2












            2








            2





            $begingroup$

            Take $X=mathbb R$, let $d$ be the usual distance and define$$m(x,y)=bigllvert x^3-y^3bigrrvert.$$Then $(X,d)$ and $(X,m)$ have the same Cauchy sequences, but $operatorname{Id}colon(X,d)longrightarrow(X,m)$ is not uniformly continuous (but it is continuous).






            share|cite|improve this answer









            $endgroup$



            Take $X=mathbb R$, let $d$ be the usual distance and define$$m(x,y)=bigllvert x^3-y^3bigrrvert.$$Then $(X,d)$ and $(X,m)$ have the same Cauchy sequences, but $operatorname{Id}colon(X,d)longrightarrow(X,m)$ is not uniformly continuous (but it is continuous).







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Dec 18 '18 at 11:56









            José Carlos SantosJosé Carlos Santos

            168k22132236




            168k22132236












            • $begingroup$
              WLOG assume $|x| le |y|$, $|x^3-y^3|=|x-y||x^2+xy+y^2| le d(x,y)3|y|^2 le 3Md(x,y)$. Is this why these different metrics have the same Cauchy sequences ?
              $endgroup$
              – argiriskar
              Dec 19 '18 at 15:15












            • $begingroup$
              It seems to me that, yes, the fact that the Cauchy sequences are the same in both cases can be deduced from this.
              $endgroup$
              – José Carlos Santos
              Dec 19 '18 at 17:07










            • $begingroup$
              Thank you very much !
              $endgroup$
              – argiriskar
              Dec 19 '18 at 17:07


















            • $begingroup$
              WLOG assume $|x| le |y|$, $|x^3-y^3|=|x-y||x^2+xy+y^2| le d(x,y)3|y|^2 le 3Md(x,y)$. Is this why these different metrics have the same Cauchy sequences ?
              $endgroup$
              – argiriskar
              Dec 19 '18 at 15:15












            • $begingroup$
              It seems to me that, yes, the fact that the Cauchy sequences are the same in both cases can be deduced from this.
              $endgroup$
              – José Carlos Santos
              Dec 19 '18 at 17:07










            • $begingroup$
              Thank you very much !
              $endgroup$
              – argiriskar
              Dec 19 '18 at 17:07
















            $begingroup$
            WLOG assume $|x| le |y|$, $|x^3-y^3|=|x-y||x^2+xy+y^2| le d(x,y)3|y|^2 le 3Md(x,y)$. Is this why these different metrics have the same Cauchy sequences ?
            $endgroup$
            – argiriskar
            Dec 19 '18 at 15:15






            $begingroup$
            WLOG assume $|x| le |y|$, $|x^3-y^3|=|x-y||x^2+xy+y^2| le d(x,y)3|y|^2 le 3Md(x,y)$. Is this why these different metrics have the same Cauchy sequences ?
            $endgroup$
            – argiriskar
            Dec 19 '18 at 15:15














            $begingroup$
            It seems to me that, yes, the fact that the Cauchy sequences are the same in both cases can be deduced from this.
            $endgroup$
            – José Carlos Santos
            Dec 19 '18 at 17:07




            $begingroup$
            It seems to me that, yes, the fact that the Cauchy sequences are the same in both cases can be deduced from this.
            $endgroup$
            – José Carlos Santos
            Dec 19 '18 at 17:07












            $begingroup$
            Thank you very much !
            $endgroup$
            – argiriskar
            Dec 19 '18 at 17:07




            $begingroup$
            Thank you very much !
            $endgroup$
            – argiriskar
            Dec 19 '18 at 17:07


















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