Convergent sequences in metric space $(mathbb{R}^n,d)$












1












$begingroup$


Given a metric $d : mathbb{R}^n times mathbb{R}^n mapsto mathbb{R}$ defined as
$$ d(x,y) = begin{cases}
lVert x rVert + lVert y rVert & xneq y \
0 & x = y
end{cases}
$$



How can one describe the convergent sequences in metric space $(mathbb{R}^n,d)$?



My work:



This metric will return a longer distance between any two points $x, y in mathbb{R}^n$ So by the triangle inequality if a sequence converges in $(mathbb{R}^n,d)$ it must also converge in $mathbb{R}^n$ with the standard metric $lVert x - y rVert$ but the reverse is not necessarily true, the standard metric on $mathbb{R}^n$ is finer.



What about Cauchy sequences? It's not very clear to me whether $(mathbb{R}^n,d)$ is a complete metric space.






real-analysis metric-spaces







share|cite|improve this question









$endgroup$












  • $begingroup$
    Hints: What is in a small neighbourhood of $x$ when (a) $xneq 0$ and (b) $x=0$?
    $endgroup$
    – Peter Franek
    Dec 16 '18 at 22:07










  • $begingroup$
    In case $x = 0$, it's the same epsilon neighbourhood around zero with the standard metric. If $x neq 0$ then it's an epsilon neighbourhood that completely contains the epsilon neighbourhood with the standard metric..
    $endgroup$
    – Sptmp
    Dec 16 '18 at 22:18










  • $begingroup$
    @Sptmp And note that if $xne 0$, there are lots of $epsilon>0$ for which you cant be within an $epsilon$ ball of the point, indeed $epsilon<||x||$.
    $endgroup$
    – qbert
    Dec 16 '18 at 22:23








  • 1




    $begingroup$
    (@Sptmp Indeed, everywhere apart of zero, the space is discrete.)
    $endgroup$
    – Peter Franek
    Dec 16 '18 at 22:51
















1












$begingroup$


Given a metric $d : mathbb{R}^n times mathbb{R}^n mapsto mathbb{R}$ defined as
$$ d(x,y) = begin{cases}
lVert x rVert + lVert y rVert & xneq y \
0 & x = y
end{cases}
$$



How can one describe the convergent sequences in metric space $(mathbb{R}^n,d)$?



My work:



This metric will return a longer distance between any two points $x, y in mathbb{R}^n$ So by the triangle inequality if a sequence converges in $(mathbb{R}^n,d)$ it must also converge in $mathbb{R}^n$ with the standard metric $lVert x - y rVert$ but the reverse is not necessarily true, the standard metric on $mathbb{R}^n$ is finer.



What about Cauchy sequences? It's not very clear to me whether $(mathbb{R}^n,d)$ is a complete metric space.






real-analysis metric-spaces







share|cite|improve this question









$endgroup$












  • $begingroup$
    Hints: What is in a small neighbourhood of $x$ when (a) $xneq 0$ and (b) $x=0$?
    $endgroup$
    – Peter Franek
    Dec 16 '18 at 22:07










  • $begingroup$
    In case $x = 0$, it's the same epsilon neighbourhood around zero with the standard metric. If $x neq 0$ then it's an epsilon neighbourhood that completely contains the epsilon neighbourhood with the standard metric..
    $endgroup$
    – Sptmp
    Dec 16 '18 at 22:18










  • $begingroup$
    @Sptmp And note that if $xne 0$, there are lots of $epsilon>0$ for which you cant be within an $epsilon$ ball of the point, indeed $epsilon<||x||$.
    $endgroup$
    – qbert
    Dec 16 '18 at 22:23








  • 1




    $begingroup$
    (@Sptmp Indeed, everywhere apart of zero, the space is discrete.)
    $endgroup$
    – Peter Franek
    Dec 16 '18 at 22:51














1












1








1





$begingroup$


Given a metric $d : mathbb{R}^n times mathbb{R}^n mapsto mathbb{R}$ defined as
$$ d(x,y) = begin{cases}
lVert x rVert + lVert y rVert & xneq y \
0 & x = y
end{cases}
$$



How can one describe the convergent sequences in metric space $(mathbb{R}^n,d)$?



My work:



This metric will return a longer distance between any two points $x, y in mathbb{R}^n$ So by the triangle inequality if a sequence converges in $(mathbb{R}^n,d)$ it must also converge in $mathbb{R}^n$ with the standard metric $lVert x - y rVert$ but the reverse is not necessarily true, the standard metric on $mathbb{R}^n$ is finer.



What about Cauchy sequences? It's not very clear to me whether $(mathbb{R}^n,d)$ is a complete metric space.






real-analysis metric-spaces







share|cite|improve this question









$endgroup$




Given a metric $d : mathbb{R}^n times mathbb{R}^n mapsto mathbb{R}$ defined as
$$ d(x,y) = begin{cases}
lVert x rVert + lVert y rVert & xneq y \
0 & x = y
end{cases}
$$



How can one describe the convergent sequences in metric space $(mathbb{R}^n,d)$?



My work:



This metric will return a longer distance between any two points $x, y in mathbb{R}^n$ So by the triangle inequality if a sequence converges in $(mathbb{R}^n,d)$ it must also converge in $mathbb{R}^n$ with the standard metric $lVert x - y rVert$ but the reverse is not necessarily true, the standard metric on $mathbb{R}^n$ is finer.



What about Cauchy sequences? It's not very clear to me whether $(mathbb{R}^n,d)$ is a complete metric space.






real-analysis metric-spaces




real-analysis metric-spaces






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 16 '18 at 22:04









SptmpSptmp

7914




7914












  • $begingroup$
    Hints: What is in a small neighbourhood of $x$ when (a) $xneq 0$ and (b) $x=0$?
    $endgroup$
    – Peter Franek
    Dec 16 '18 at 22:07










  • $begingroup$
    In case $x = 0$, it's the same epsilon neighbourhood around zero with the standard metric. If $x neq 0$ then it's an epsilon neighbourhood that completely contains the epsilon neighbourhood with the standard metric..
    $endgroup$
    – Sptmp
    Dec 16 '18 at 22:18










  • $begingroup$
    @Sptmp And note that if $xne 0$, there are lots of $epsilon>0$ for which you cant be within an $epsilon$ ball of the point, indeed $epsilon<||x||$.
    $endgroup$
    – qbert
    Dec 16 '18 at 22:23








  • 1




    $begingroup$
    (@Sptmp Indeed, everywhere apart of zero, the space is discrete.)
    $endgroup$
    – Peter Franek
    Dec 16 '18 at 22:51


















  • $begingroup$
    Hints: What is in a small neighbourhood of $x$ when (a) $xneq 0$ and (b) $x=0$?
    $endgroup$
    – Peter Franek
    Dec 16 '18 at 22:07










  • $begingroup$
    In case $x = 0$, it's the same epsilon neighbourhood around zero with the standard metric. If $x neq 0$ then it's an epsilon neighbourhood that completely contains the epsilon neighbourhood with the standard metric..
    $endgroup$
    – Sptmp
    Dec 16 '18 at 22:18










  • $begingroup$
    @Sptmp And note that if $xne 0$, there are lots of $epsilon>0$ for which you cant be within an $epsilon$ ball of the point, indeed $epsilon<||x||$.
    $endgroup$
    – qbert
    Dec 16 '18 at 22:23








  • 1




    $begingroup$
    (@Sptmp Indeed, everywhere apart of zero, the space is discrete.)
    $endgroup$
    – Peter Franek
    Dec 16 '18 at 22:51
















$begingroup$
Hints: What is in a small neighbourhood of $x$ when (a) $xneq 0$ and (b) $x=0$?
$endgroup$
– Peter Franek
Dec 16 '18 at 22:07




$begingroup$
Hints: What is in a small neighbourhood of $x$ when (a) $xneq 0$ and (b) $x=0$?
$endgroup$
– Peter Franek
Dec 16 '18 at 22:07












$begingroup$
In case $x = 0$, it's the same epsilon neighbourhood around zero with the standard metric. If $x neq 0$ then it's an epsilon neighbourhood that completely contains the epsilon neighbourhood with the standard metric..
$endgroup$
– Sptmp
Dec 16 '18 at 22:18




$begingroup$
In case $x = 0$, it's the same epsilon neighbourhood around zero with the standard metric. If $x neq 0$ then it's an epsilon neighbourhood that completely contains the epsilon neighbourhood with the standard metric..
$endgroup$
– Sptmp
Dec 16 '18 at 22:18












$begingroup$
@Sptmp And note that if $xne 0$, there are lots of $epsilon>0$ for which you cant be within an $epsilon$ ball of the point, indeed $epsilon<||x||$.
$endgroup$
– qbert
Dec 16 '18 at 22:23






$begingroup$
@Sptmp And note that if $xne 0$, there are lots of $epsilon>0$ for which you cant be within an $epsilon$ ball of the point, indeed $epsilon<||x||$.
$endgroup$
– qbert
Dec 16 '18 at 22:23






1




1




$begingroup$
(@Sptmp Indeed, everywhere apart of zero, the space is discrete.)
$endgroup$
– Peter Franek
Dec 16 '18 at 22:51




$begingroup$
(@Sptmp Indeed, everywhere apart of zero, the space is discrete.)
$endgroup$
– Peter Franek
Dec 16 '18 at 22:51










2 Answers
2






active

oldest

votes


















1












$begingroup$

So assume that $(x_n)$ converges to $x$. It follows that $d(x_n, x)$ is arbitrarly small. Now assume that $(x_n)$ is not eventually constant, i.e. $x_nneq x$ for infinitely many $n$. Then $d(x_n,x)=lVert x_nrVert+lVert xrVertgeq lVert xrVert$. If $xneq 0$ then $d(x_n,x)geq lVert xrVert>0$ and thus $x_n$ cannot be convergent to $x$.



It follows that if $xneq 0$ then $(x_n)to x$ if and only if $(x_n)$ is eventually constant.



On the other hand following similar reasoning we conclude that $(x_n)to 0$ in $d$ if and only if $(x_n)to 0$ in $lVertcdotrVert$.



Similarly if $(x_n)$ is Cauchy then $d(x_n, x_m)=lVert x_nrVert +lVert x_mrVert$ assuming $(x_n)$ is not eventually constant (which is the trivial case). So $d(x_n,x_m)$ is arbitrarly small if and only if $lVert x_nrVert$ is arbitrarly small. Meaning that $(x_n)$ is Cauchy if and only if it converges to $0$ (I omit the case when $(x_n)$ is eventually constant). In particular $(mathbb{R}^n,d)$ is complete.






share|cite|improve this answer









$endgroup$





















    2












    $begingroup$

    Note that if for any $kgeq 0$,
    $$
    d(x_{n},x_{n+k})to 0iff ||x_n||+||x_{n+k}||to 0implies ||x_n||to 0
    $$
    So, for any Cauchy sequence, we have
    $$
    d(x_n,0)=||x_{n}||to 0
    $$
    and $x_nto 0$ in this metric.



    Since convergent sequences are always Cauchy, this goes a long way towards describing the convergent sequences, even in terms of the usual Euclidean metric.






    share|cite|improve this answer









    $endgroup$













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      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      1












      $begingroup$

      So assume that $(x_n)$ converges to $x$. It follows that $d(x_n, x)$ is arbitrarly small. Now assume that $(x_n)$ is not eventually constant, i.e. $x_nneq x$ for infinitely many $n$. Then $d(x_n,x)=lVert x_nrVert+lVert xrVertgeq lVert xrVert$. If $xneq 0$ then $d(x_n,x)geq lVert xrVert>0$ and thus $x_n$ cannot be convergent to $x$.



      It follows that if $xneq 0$ then $(x_n)to x$ if and only if $(x_n)$ is eventually constant.



      On the other hand following similar reasoning we conclude that $(x_n)to 0$ in $d$ if and only if $(x_n)to 0$ in $lVertcdotrVert$.



      Similarly if $(x_n)$ is Cauchy then $d(x_n, x_m)=lVert x_nrVert +lVert x_mrVert$ assuming $(x_n)$ is not eventually constant (which is the trivial case). So $d(x_n,x_m)$ is arbitrarly small if and only if $lVert x_nrVert$ is arbitrarly small. Meaning that $(x_n)$ is Cauchy if and only if it converges to $0$ (I omit the case when $(x_n)$ is eventually constant). In particular $(mathbb{R}^n,d)$ is complete.






      share|cite|improve this answer









      $endgroup$


















        1












        $begingroup$

        So assume that $(x_n)$ converges to $x$. It follows that $d(x_n, x)$ is arbitrarly small. Now assume that $(x_n)$ is not eventually constant, i.e. $x_nneq x$ for infinitely many $n$. Then $d(x_n,x)=lVert x_nrVert+lVert xrVertgeq lVert xrVert$. If $xneq 0$ then $d(x_n,x)geq lVert xrVert>0$ and thus $x_n$ cannot be convergent to $x$.



        It follows that if $xneq 0$ then $(x_n)to x$ if and only if $(x_n)$ is eventually constant.



        On the other hand following similar reasoning we conclude that $(x_n)to 0$ in $d$ if and only if $(x_n)to 0$ in $lVertcdotrVert$.



        Similarly if $(x_n)$ is Cauchy then $d(x_n, x_m)=lVert x_nrVert +lVert x_mrVert$ assuming $(x_n)$ is not eventually constant (which is the trivial case). So $d(x_n,x_m)$ is arbitrarly small if and only if $lVert x_nrVert$ is arbitrarly small. Meaning that $(x_n)$ is Cauchy if and only if it converges to $0$ (I omit the case when $(x_n)$ is eventually constant). In particular $(mathbb{R}^n,d)$ is complete.






        share|cite|improve this answer









        $endgroup$
















          1












          1








          1





          $begingroup$

          So assume that $(x_n)$ converges to $x$. It follows that $d(x_n, x)$ is arbitrarly small. Now assume that $(x_n)$ is not eventually constant, i.e. $x_nneq x$ for infinitely many $n$. Then $d(x_n,x)=lVert x_nrVert+lVert xrVertgeq lVert xrVert$. If $xneq 0$ then $d(x_n,x)geq lVert xrVert>0$ and thus $x_n$ cannot be convergent to $x$.



          It follows that if $xneq 0$ then $(x_n)to x$ if and only if $(x_n)$ is eventually constant.



          On the other hand following similar reasoning we conclude that $(x_n)to 0$ in $d$ if and only if $(x_n)to 0$ in $lVertcdotrVert$.



          Similarly if $(x_n)$ is Cauchy then $d(x_n, x_m)=lVert x_nrVert +lVert x_mrVert$ assuming $(x_n)$ is not eventually constant (which is the trivial case). So $d(x_n,x_m)$ is arbitrarly small if and only if $lVert x_nrVert$ is arbitrarly small. Meaning that $(x_n)$ is Cauchy if and only if it converges to $0$ (I omit the case when $(x_n)$ is eventually constant). In particular $(mathbb{R}^n,d)$ is complete.






          share|cite|improve this answer









          $endgroup$



          So assume that $(x_n)$ converges to $x$. It follows that $d(x_n, x)$ is arbitrarly small. Now assume that $(x_n)$ is not eventually constant, i.e. $x_nneq x$ for infinitely many $n$. Then $d(x_n,x)=lVert x_nrVert+lVert xrVertgeq lVert xrVert$. If $xneq 0$ then $d(x_n,x)geq lVert xrVert>0$ and thus $x_n$ cannot be convergent to $x$.



          It follows that if $xneq 0$ then $(x_n)to x$ if and only if $(x_n)$ is eventually constant.



          On the other hand following similar reasoning we conclude that $(x_n)to 0$ in $d$ if and only if $(x_n)to 0$ in $lVertcdotrVert$.



          Similarly if $(x_n)$ is Cauchy then $d(x_n, x_m)=lVert x_nrVert +lVert x_mrVert$ assuming $(x_n)$ is not eventually constant (which is the trivial case). So $d(x_n,x_m)$ is arbitrarly small if and only if $lVert x_nrVert$ is arbitrarly small. Meaning that $(x_n)$ is Cauchy if and only if it converges to $0$ (I omit the case when $(x_n)$ is eventually constant). In particular $(mathbb{R}^n,d)$ is complete.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 16 '18 at 22:23









          freakishfreakish

          12.8k1631




          12.8k1631























              2












              $begingroup$

              Note that if for any $kgeq 0$,
              $$
              d(x_{n},x_{n+k})to 0iff ||x_n||+||x_{n+k}||to 0implies ||x_n||to 0
              $$
              So, for any Cauchy sequence, we have
              $$
              d(x_n,0)=||x_{n}||to 0
              $$
              and $x_nto 0$ in this metric.



              Since convergent sequences are always Cauchy, this goes a long way towards describing the convergent sequences, even in terms of the usual Euclidean metric.






              share|cite|improve this answer









              $endgroup$


















                2












                $begingroup$

                Note that if for any $kgeq 0$,
                $$
                d(x_{n},x_{n+k})to 0iff ||x_n||+||x_{n+k}||to 0implies ||x_n||to 0
                $$
                So, for any Cauchy sequence, we have
                $$
                d(x_n,0)=||x_{n}||to 0
                $$
                and $x_nto 0$ in this metric.



                Since convergent sequences are always Cauchy, this goes a long way towards describing the convergent sequences, even in terms of the usual Euclidean metric.






                share|cite|improve this answer









                $endgroup$
















                  2












                  2








                  2





                  $begingroup$

                  Note that if for any $kgeq 0$,
                  $$
                  d(x_{n},x_{n+k})to 0iff ||x_n||+||x_{n+k}||to 0implies ||x_n||to 0
                  $$
                  So, for any Cauchy sequence, we have
                  $$
                  d(x_n,0)=||x_{n}||to 0
                  $$
                  and $x_nto 0$ in this metric.



                  Since convergent sequences are always Cauchy, this goes a long way towards describing the convergent sequences, even in terms of the usual Euclidean metric.






                  share|cite|improve this answer









                  $endgroup$



                  Note that if for any $kgeq 0$,
                  $$
                  d(x_{n},x_{n+k})to 0iff ||x_n||+||x_{n+k}||to 0implies ||x_n||to 0
                  $$
                  So, for any Cauchy sequence, we have
                  $$
                  d(x_n,0)=||x_{n}||to 0
                  $$
                  and $x_nto 0$ in this metric.



                  Since convergent sequences are always Cauchy, this goes a long way towards describing the convergent sequences, even in terms of the usual Euclidean metric.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 16 '18 at 22:17









                  qbertqbert

                  22.2k32561




                  22.2k32561






























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