A $delta$- chain is $delta$ shadowed by identity map












0












$begingroup$


Let $f$ be the identity map on the space $X={x_i}_{i=1}^{infty} $, where $x_i=sum_{n=1}^ifrac{1}{n}$, given the metric inherited from $mathbb{R}$. Also, let for finite sequence ${y_n}_{n=0}^msubseteq X$, we have $|y_0-y_m|<delta$ and $|y_n-y_{n+1}|<delta$ for $n=0, 1, ldots m-1$.



Question. Is it true there is $zin{y_n}_{n=0}^m$ such that $|z-y_n|<delta$ for all $n=0, 1, ldots, m$?






real-analysis calculus analysis







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$endgroup$












  • $begingroup$
    No, it is not true. You can go as far away as you want and then come back. Why do you think this should be true? Maybe you want to impose further conditions?
    $endgroup$
    – Severin Schraven
    Dec 2 '18 at 18:44










  • $begingroup$
    @S.Schraven. There is no further conditions. give an example please , if $delta>0$ is sufficiently small
    $endgroup$
    – user479859
    Dec 3 '18 at 0:00










  • $begingroup$
    Only I can choice $zin X$, hence
    $endgroup$
    – user479859
    Dec 3 '18 at 0:34










  • $begingroup$
    Is it true there is $zin X$ such that $|z-y_n|<delta$ for all $n=0, 1, ldots, m$?
    $endgroup$
    – user479859
    Dec 3 '18 at 0:34
















0












$begingroup$


Let $f$ be the identity map on the space $X={x_i}_{i=1}^{infty} $, where $x_i=sum_{n=1}^ifrac{1}{n}$, given the metric inherited from $mathbb{R}$. Also, let for finite sequence ${y_n}_{n=0}^msubseteq X$, we have $|y_0-y_m|<delta$ and $|y_n-y_{n+1}|<delta$ for $n=0, 1, ldots m-1$.



Question. Is it true there is $zin{y_n}_{n=0}^m$ such that $|z-y_n|<delta$ for all $n=0, 1, ldots, m$?






real-analysis calculus analysis







share|cite|improve this question









$endgroup$












  • $begingroup$
    No, it is not true. You can go as far away as you want and then come back. Why do you think this should be true? Maybe you want to impose further conditions?
    $endgroup$
    – Severin Schraven
    Dec 2 '18 at 18:44










  • $begingroup$
    @S.Schraven. There is no further conditions. give an example please , if $delta>0$ is sufficiently small
    $endgroup$
    – user479859
    Dec 3 '18 at 0:00










  • $begingroup$
    Only I can choice $zin X$, hence
    $endgroup$
    – user479859
    Dec 3 '18 at 0:34










  • $begingroup$
    Is it true there is $zin X$ such that $|z-y_n|<delta$ for all $n=0, 1, ldots, m$?
    $endgroup$
    – user479859
    Dec 3 '18 at 0:34














0












0








0





$begingroup$


Let $f$ be the identity map on the space $X={x_i}_{i=1}^{infty} $, where $x_i=sum_{n=1}^ifrac{1}{n}$, given the metric inherited from $mathbb{R}$. Also, let for finite sequence ${y_n}_{n=0}^msubseteq X$, we have $|y_0-y_m|<delta$ and $|y_n-y_{n+1}|<delta$ for $n=0, 1, ldots m-1$.



Question. Is it true there is $zin{y_n}_{n=0}^m$ such that $|z-y_n|<delta$ for all $n=0, 1, ldots, m$?






real-analysis calculus analysis







share|cite|improve this question









$endgroup$




Let $f$ be the identity map on the space $X={x_i}_{i=1}^{infty} $, where $x_i=sum_{n=1}^ifrac{1}{n}$, given the metric inherited from $mathbb{R}$. Also, let for finite sequence ${y_n}_{n=0}^msubseteq X$, we have $|y_0-y_m|<delta$ and $|y_n-y_{n+1}|<delta$ for $n=0, 1, ldots m-1$.



Question. Is it true there is $zin{y_n}_{n=0}^m$ such that $|z-y_n|<delta$ for all $n=0, 1, ldots, m$?






real-analysis calculus analysis




real-analysis calculus analysis






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 2 '18 at 18:27









user479859user479859

756




756












  • $begingroup$
    No, it is not true. You can go as far away as you want and then come back. Why do you think this should be true? Maybe you want to impose further conditions?
    $endgroup$
    – Severin Schraven
    Dec 2 '18 at 18:44










  • $begingroup$
    @S.Schraven. There is no further conditions. give an example please , if $delta>0$ is sufficiently small
    $endgroup$
    – user479859
    Dec 3 '18 at 0:00










  • $begingroup$
    Only I can choice $zin X$, hence
    $endgroup$
    – user479859
    Dec 3 '18 at 0:34










  • $begingroup$
    Is it true there is $zin X$ such that $|z-y_n|<delta$ for all $n=0, 1, ldots, m$?
    $endgroup$
    – user479859
    Dec 3 '18 at 0:34


















  • $begingroup$
    No, it is not true. You can go as far away as you want and then come back. Why do you think this should be true? Maybe you want to impose further conditions?
    $endgroup$
    – Severin Schraven
    Dec 2 '18 at 18:44










  • $begingroup$
    @S.Schraven. There is no further conditions. give an example please , if $delta>0$ is sufficiently small
    $endgroup$
    – user479859
    Dec 3 '18 at 0:00










  • $begingroup$
    Only I can choice $zin X$, hence
    $endgroup$
    – user479859
    Dec 3 '18 at 0:34










  • $begingroup$
    Is it true there is $zin X$ such that $|z-y_n|<delta$ for all $n=0, 1, ldots, m$?
    $endgroup$
    – user479859
    Dec 3 '18 at 0:34
















$begingroup$
No, it is not true. You can go as far away as you want and then come back. Why do you think this should be true? Maybe you want to impose further conditions?
$endgroup$
– Severin Schraven
Dec 2 '18 at 18:44




$begingroup$
No, it is not true. You can go as far away as you want and then come back. Why do you think this should be true? Maybe you want to impose further conditions?
$endgroup$
– Severin Schraven
Dec 2 '18 at 18:44












$begingroup$
@S.Schraven. There is no further conditions. give an example please , if $delta>0$ is sufficiently small
$endgroup$
– user479859
Dec 3 '18 at 0:00




$begingroup$
@S.Schraven. There is no further conditions. give an example please , if $delta>0$ is sufficiently small
$endgroup$
– user479859
Dec 3 '18 at 0:00












$begingroup$
Only I can choice $zin X$, hence
$endgroup$
– user479859
Dec 3 '18 at 0:34




$begingroup$
Only I can choice $zin X$, hence
$endgroup$
– user479859
Dec 3 '18 at 0:34












$begingroup$
Is it true there is $zin X$ such that $|z-y_n|<delta$ for all $n=0, 1, ldots, m$?
$endgroup$
– user479859
Dec 3 '18 at 0:34




$begingroup$
Is it true there is $zin X$ such that $|z-y_n|<delta$ for all $n=0, 1, ldots, m$?
$endgroup$
– user479859
Dec 3 '18 at 0:34










1 Answer
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$begingroup$

Choose $delta >0$ in any way you like. Pick $N>0$ sufficiently large s.t. $1/N<delta$. Then consider the sequence $y_0=x_N=y_{2l}, y_1=x_{N+1}=y_{2l-1}, dots, y_{l}=x_{N+l}$. Clearly $vert y_0 -y_{2l} vert =0<delta$ and $vert y_n - y_{n+1} vert leq 1/N <delta $, but $vert y_0 - y_l vert = sum_{k=N+1}^{N+l} 1/k$ which we can make as large as we want by making $l$ large.



Note that in fact there does not even exist $zin X$ (i.e. if we are not forced to pick $zin (y_m)_{m=0}^{2l})$ such that $vert z - y_m vert<delta$ for all $m=0, dots , 2l$.






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    $begingroup$

    Choose $delta >0$ in any way you like. Pick $N>0$ sufficiently large s.t. $1/N<delta$. Then consider the sequence $y_0=x_N=y_{2l}, y_1=x_{N+1}=y_{2l-1}, dots, y_{l}=x_{N+l}$. Clearly $vert y_0 -y_{2l} vert =0<delta$ and $vert y_n - y_{n+1} vert leq 1/N <delta $, but $vert y_0 - y_l vert = sum_{k=N+1}^{N+l} 1/k$ which we can make as large as we want by making $l$ large.



    Note that in fact there does not even exist $zin X$ (i.e. if we are not forced to pick $zin (y_m)_{m=0}^{2l})$ such that $vert z - y_m vert<delta$ for all $m=0, dots , 2l$.






    share|cite|improve this answer











    $endgroup$


















      1












      $begingroup$

      Choose $delta >0$ in any way you like. Pick $N>0$ sufficiently large s.t. $1/N<delta$. Then consider the sequence $y_0=x_N=y_{2l}, y_1=x_{N+1}=y_{2l-1}, dots, y_{l}=x_{N+l}$. Clearly $vert y_0 -y_{2l} vert =0<delta$ and $vert y_n - y_{n+1} vert leq 1/N <delta $, but $vert y_0 - y_l vert = sum_{k=N+1}^{N+l} 1/k$ which we can make as large as we want by making $l$ large.



      Note that in fact there does not even exist $zin X$ (i.e. if we are not forced to pick $zin (y_m)_{m=0}^{2l})$ such that $vert z - y_m vert<delta$ for all $m=0, dots , 2l$.






      share|cite|improve this answer











      $endgroup$
















        1












        1








        1





        $begingroup$

        Choose $delta >0$ in any way you like. Pick $N>0$ sufficiently large s.t. $1/N<delta$. Then consider the sequence $y_0=x_N=y_{2l}, y_1=x_{N+1}=y_{2l-1}, dots, y_{l}=x_{N+l}$. Clearly $vert y_0 -y_{2l} vert =0<delta$ and $vert y_n - y_{n+1} vert leq 1/N <delta $, but $vert y_0 - y_l vert = sum_{k=N+1}^{N+l} 1/k$ which we can make as large as we want by making $l$ large.



        Note that in fact there does not even exist $zin X$ (i.e. if we are not forced to pick $zin (y_m)_{m=0}^{2l})$ such that $vert z - y_m vert<delta$ for all $m=0, dots , 2l$.






        share|cite|improve this answer











        $endgroup$



        Choose $delta >0$ in any way you like. Pick $N>0$ sufficiently large s.t. $1/N<delta$. Then consider the sequence $y_0=x_N=y_{2l}, y_1=x_{N+1}=y_{2l-1}, dots, y_{l}=x_{N+l}$. Clearly $vert y_0 -y_{2l} vert =0<delta$ and $vert y_n - y_{n+1} vert leq 1/N <delta $, but $vert y_0 - y_l vert = sum_{k=N+1}^{N+l} 1/k$ which we can make as large as we want by making $l$ large.



        Note that in fact there does not even exist $zin X$ (i.e. if we are not forced to pick $zin (y_m)_{m=0}^{2l})$ such that $vert z - y_m vert<delta$ for all $m=0, dots , 2l$.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Dec 4 '18 at 7:20

























        answered Dec 3 '18 at 14:23









        Severin SchravenSeverin Schraven

        6,0231934




        6,0231934






























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