Prove $(x_n)$ has a subsequence convergent to $x$.












1












$begingroup$


I’m trying to do this one but I don’t know how to start.




Let $(x_n)$ be a sequence on a metric space $(X,d)$, and $x$ a limit point of the image set of the sequence. Prove $(x_n)$ has a subsequence convergent to $x$.




My definition of limit point is that a point $x$ is a limit point on a space $Ssubset X$ if every neighbourhood of $x$ contains a point of $S$ different of $x$.



I don’t know how to build the convergent subsequence with this assumptions.



Thanks for your help.






calculus general-topology analysis







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  • 1




    $begingroup$
    Hint: consider a sequence of shrinking neighbourhoods of $x$.
    $endgroup$
    – SmileyCraft
    Dec 14 '18 at 12:52
















1












$begingroup$


I’m trying to do this one but I don’t know how to start.




Let $(x_n)$ be a sequence on a metric space $(X,d)$, and $x$ a limit point of the image set of the sequence. Prove $(x_n)$ has a subsequence convergent to $x$.




My definition of limit point is that a point $x$ is a limit point on a space $Ssubset X$ if every neighbourhood of $x$ contains a point of $S$ different of $x$.



I don’t know how to build the convergent subsequence with this assumptions.



Thanks for your help.






calculus general-topology analysis







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Hint: consider a sequence of shrinking neighbourhoods of $x$.
    $endgroup$
    – SmileyCraft
    Dec 14 '18 at 12:52














1












1








1





$begingroup$


I’m trying to do this one but I don’t know how to start.




Let $(x_n)$ be a sequence on a metric space $(X,d)$, and $x$ a limit point of the image set of the sequence. Prove $(x_n)$ has a subsequence convergent to $x$.




My definition of limit point is that a point $x$ is a limit point on a space $Ssubset X$ if every neighbourhood of $x$ contains a point of $S$ different of $x$.



I don’t know how to build the convergent subsequence with this assumptions.



Thanks for your help.






calculus general-topology analysis







share|cite|improve this question











$endgroup$




I’m trying to do this one but I don’t know how to start.




Let $(x_n)$ be a sequence on a metric space $(X,d)$, and $x$ a limit point of the image set of the sequence. Prove $(x_n)$ has a subsequence convergent to $x$.




My definition of limit point is that a point $x$ is a limit point on a space $Ssubset X$ if every neighbourhood of $x$ contains a point of $S$ different of $x$.



I don’t know how to build the convergent subsequence with this assumptions.



Thanks for your help.






calculus general-topology analysis




calculus general-topology analysis






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 14 '18 at 12:49









H. R.

9,48093263




9,48093263










asked Dec 14 '18 at 12:41









RelureRelure

2,1671035




2,1671035








  • 1




    $begingroup$
    Hint: consider a sequence of shrinking neighbourhoods of $x$.
    $endgroup$
    – SmileyCraft
    Dec 14 '18 at 12:52














  • 1




    $begingroup$
    Hint: consider a sequence of shrinking neighbourhoods of $x$.
    $endgroup$
    – SmileyCraft
    Dec 14 '18 at 12:52








1




1




$begingroup$
Hint: consider a sequence of shrinking neighbourhoods of $x$.
$endgroup$
– SmileyCraft
Dec 14 '18 at 12:52




$begingroup$
Hint: consider a sequence of shrinking neighbourhoods of $x$.
$endgroup$
– SmileyCraft
Dec 14 '18 at 12:52










2 Answers
2






active

oldest

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1












$begingroup$

For every positive integer $k$ the set ${ninmathbb Nmid d(x_n,x)<frac1k}$ is infinite.



(If not then it can be shown that $x$ is not a limit point of ${x_nmid ninmathbb N}$)



So for $k=1$ choose some $n_1$ with $d(x_{n_1},x)<frac11=1$.



Then for $k=2$ choose some $n_2>n_1$ with $d(x_{n_2},x)<frac12$.



Et cetera.






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    Let denote : $B_epsilon = {y in X mid d(y-x) < epsilon }$ this is an open ball with center $x$ and radius $epsilon$.




    Since $x$ is a limit point of the $(x_n)$ it means that for all $epsilon$ we can find an $N_{epsilon}$ such that : $x_{N_{epsilon}} in B_epsilon backslash {x}$.




    Now let's construct a striclty increasing function $phi : mathbb{N} to mathbb{N}$ such that : $x_{phi(n)} to x$.



    Let's choose an $epsilon > 0$, and let $phi(0) = N_{epsilon}$. Now suppose that the function $phi$ has been construct for all $n leq N$. Then we can take : $0 < epsilon'= frac{1}{K} < min {d(x_{phi(n)}-x) mid n leq N }$. And we can let : $phi(N+1) = N_{epsilon'}$.






    share|cite|improve this answer









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      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      1












      $begingroup$

      For every positive integer $k$ the set ${ninmathbb Nmid d(x_n,x)<frac1k}$ is infinite.



      (If not then it can be shown that $x$ is not a limit point of ${x_nmid ninmathbb N}$)



      So for $k=1$ choose some $n_1$ with $d(x_{n_1},x)<frac11=1$.



      Then for $k=2$ choose some $n_2>n_1$ with $d(x_{n_2},x)<frac12$.



      Et cetera.






      share|cite|improve this answer









      $endgroup$


















        1












        $begingroup$

        For every positive integer $k$ the set ${ninmathbb Nmid d(x_n,x)<frac1k}$ is infinite.



        (If not then it can be shown that $x$ is not a limit point of ${x_nmid ninmathbb N}$)



        So for $k=1$ choose some $n_1$ with $d(x_{n_1},x)<frac11=1$.



        Then for $k=2$ choose some $n_2>n_1$ with $d(x_{n_2},x)<frac12$.



        Et cetera.






        share|cite|improve this answer









        $endgroup$
















          1












          1








          1





          $begingroup$

          For every positive integer $k$ the set ${ninmathbb Nmid d(x_n,x)<frac1k}$ is infinite.



          (If not then it can be shown that $x$ is not a limit point of ${x_nmid ninmathbb N}$)



          So for $k=1$ choose some $n_1$ with $d(x_{n_1},x)<frac11=1$.



          Then for $k=2$ choose some $n_2>n_1$ with $d(x_{n_2},x)<frac12$.



          Et cetera.






          share|cite|improve this answer









          $endgroup$



          For every positive integer $k$ the set ${ninmathbb Nmid d(x_n,x)<frac1k}$ is infinite.



          (If not then it can be shown that $x$ is not a limit point of ${x_nmid ninmathbb N}$)



          So for $k=1$ choose some $n_1$ with $d(x_{n_1},x)<frac11=1$.



          Then for $k=2$ choose some $n_2>n_1$ with $d(x_{n_2},x)<frac12$.



          Et cetera.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 14 '18 at 12:56









          drhabdrhab

          101k545136




          101k545136























              0












              $begingroup$

              Let denote : $B_epsilon = {y in X mid d(y-x) < epsilon }$ this is an open ball with center $x$ and radius $epsilon$.




              Since $x$ is a limit point of the $(x_n)$ it means that for all $epsilon$ we can find an $N_{epsilon}$ such that : $x_{N_{epsilon}} in B_epsilon backslash {x}$.




              Now let's construct a striclty increasing function $phi : mathbb{N} to mathbb{N}$ such that : $x_{phi(n)} to x$.



              Let's choose an $epsilon > 0$, and let $phi(0) = N_{epsilon}$. Now suppose that the function $phi$ has been construct for all $n leq N$. Then we can take : $0 < epsilon'= frac{1}{K} < min {d(x_{phi(n)}-x) mid n leq N }$. And we can let : $phi(N+1) = N_{epsilon'}$.






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                Let denote : $B_epsilon = {y in X mid d(y-x) < epsilon }$ this is an open ball with center $x$ and radius $epsilon$.




                Since $x$ is a limit point of the $(x_n)$ it means that for all $epsilon$ we can find an $N_{epsilon}$ such that : $x_{N_{epsilon}} in B_epsilon backslash {x}$.




                Now let's construct a striclty increasing function $phi : mathbb{N} to mathbb{N}$ such that : $x_{phi(n)} to x$.



                Let's choose an $epsilon > 0$, and let $phi(0) = N_{epsilon}$. Now suppose that the function $phi$ has been construct for all $n leq N$. Then we can take : $0 < epsilon'= frac{1}{K} < min {d(x_{phi(n)}-x) mid n leq N }$. And we can let : $phi(N+1) = N_{epsilon'}$.






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  Let denote : $B_epsilon = {y in X mid d(y-x) < epsilon }$ this is an open ball with center $x$ and radius $epsilon$.




                  Since $x$ is a limit point of the $(x_n)$ it means that for all $epsilon$ we can find an $N_{epsilon}$ such that : $x_{N_{epsilon}} in B_epsilon backslash {x}$.




                  Now let's construct a striclty increasing function $phi : mathbb{N} to mathbb{N}$ such that : $x_{phi(n)} to x$.



                  Let's choose an $epsilon > 0$, and let $phi(0) = N_{epsilon}$. Now suppose that the function $phi$ has been construct for all $n leq N$. Then we can take : $0 < epsilon'= frac{1}{K} < min {d(x_{phi(n)}-x) mid n leq N }$. And we can let : $phi(N+1) = N_{epsilon'}$.






                  share|cite|improve this answer









                  $endgroup$



                  Let denote : $B_epsilon = {y in X mid d(y-x) < epsilon }$ this is an open ball with center $x$ and radius $epsilon$.




                  Since $x$ is a limit point of the $(x_n)$ it means that for all $epsilon$ we can find an $N_{epsilon}$ such that : $x_{N_{epsilon}} in B_epsilon backslash {x}$.




                  Now let's construct a striclty increasing function $phi : mathbb{N} to mathbb{N}$ such that : $x_{phi(n)} to x$.



                  Let's choose an $epsilon > 0$, and let $phi(0) = N_{epsilon}$. Now suppose that the function $phi$ has been construct for all $n leq N$. Then we can take : $0 < epsilon'= frac{1}{K} < min {d(x_{phi(n)}-x) mid n leq N }$. And we can let : $phi(N+1) = N_{epsilon'}$.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 14 '18 at 12:59









                  ThinkingThinking

                  1,11416




                  1,11416






























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