Facebook Page
Twitter
RSS
Thanks Thanks:  0
+ Reply to Thread
Results 1 to 4 of 4
  1. MHB Master
    MHB Site Helper
    mathmari's Avatar
    Status
    Online
    Join Date
    Apr 2013
    Posts
    2,694
    Thanks
    2,148 times
    Thanked
    682 times
    Awards
    MHB Chat Room Award (2015)  

MHB Model User Award (2015)  

MHB LaTeX Award (2015)
    #1
    Hey!!

    Let $(a_n)_{n=1}^{\infty}$ be a real sequence and let $(b_n)_{n=1}^{\infty}$ a sequence in the set of limit points of $(a_n)_{n=1}^{\infty}$, $L(a_n)$.

    There is also a $b_0\in \mathbb{R}$ with $b_n\rightarrow b_0$ for $n\rightarrow \infty$.

    I want to show that then $b_0\in L(a_n)$.

    How could we show this?


    I want to show also that a sequence $(a_n)_{n=1}^{\infty}$ converges to $a\in \mathbb{R}$ iff each subsequence $(a_{n_k})_{k=1}^{\infty}$ of $(a_n)_{n=1}^{\infty}$ has a subsequence $(a_{n_{k_j}})_{j=1}^{\infty}$ that converges to $a$.

    We have that a sequence converges to a limit in $\mathbb{R}$ iff each subsequence converges to that limit, or not?
    But how could we show the above condition?
    Last edited by mathmari; November 20th, 2016 at 13:54.

  2. MHB Master
    MHB Site Helper
    mathmari's Avatar
    Status
    Online
    Join Date
    Apr 2013
    Posts
    2,694
    Thanks
    2,148 times
    Thanked
    682 times
    Awards
    MHB Chat Room Award (2015)  

MHB Model User Award (2015)  

MHB LaTeX Award (2015)
    #2 Thread Author
    For the second question is it maybe as follows?

    Proposition:
    A sequence converges to a limit in $\mathbb{R}$ iff each subsequence converges to that limit


    $\Leftarrow$ :
    We suppose that each subsequence $(a_{n_k})_{k=1}^{\infty}$ of $(a_n)_{n=1}^{\infty}$ has a subsequence $(a_{n_{k_j}})_{j=1}^{\infty}$ that converges to $a$.
    From the proposition we get that each subsequence has to converge to $a$.
    Again from the proposition we get that the sequence has to converge to $a$.

    $\Rightarrow$ :
    We suppose that not each subsequence $(a_{n_k})_{k=1}^{\infty}$ of $(a_n)_{n=1}^{\infty}$ has a subsequence $(a_{n_{k_j}})_{j=1}^{\infty}$ that converges to $a$. Then there are some subseuqneces that vonverge to an other point, say $b\neq a$. So, it cannit be that the sequnec converges to $a$.


    Is this correct?

  3. MHB Master
    MHB Site Helper
    mathmari's Avatar
    Status
    Online
    Join Date
    Apr 2013
    Posts
    2,694
    Thanks
    2,148 times
    Thanked
    682 times
    Awards
    MHB Chat Room Award (2015)  

MHB Model User Award (2015)  

MHB LaTeX Award (2015)
    #3 Thread Author
    For the second question I changed it..


    $\Leftarrow$ :
    We suppose that $(a_n)$ does not converfe to $a$, so there is at least two limit points, so there are two different subsequences that converge to different points, say $a$ and $b$.
    From the proposition we have that the subsequence that converges to $b$ has no subsequence that converges to $a$.

    $\Rightarrow$ :
    We suppose that $(a_n)$ converges to $a$. From the proposition we have that every subsequence converges to $a$. If we apply the proposition to each subsequence we get that each subsequence of esch subsequence converges to $a$.


    Is this correct?

  4. MHB Master
    MHB Site Helper
    mathmari's Avatar
    Status
    Online
    Join Date
    Apr 2013
    Posts
    2,694
    Thanks
    2,148 times
    Thanked
    682 times
    Awards
    MHB Chat Room Award (2015)  

MHB Model User Award (2015)  

MHB LaTeX Award (2015)
    #4 Thread Author
    Quote Originally Posted by mathmari View Post
    Let $(a_n)_{n=1}^{\infty}$ be a real sequence and let $(b_n)_{n=1}^{\infty}$ a sequence in the set of limit points of $(a_n)_{n=1}^{\infty}$, $L(a_n)$.

    There is also a $b_0\in \mathbb{R}$ with $b_n\rightarrow b_0$ for $n\rightarrow \infty$.

    I want to show that then $b_0\in L(a_n)$.

    How could we show this?

    We have that $b_n\in H(a_n)$ is a limit point of $(a_n)$ if every neighbourhood of $b_n$ contains at least one point of $(a_n)$ different from $b_n$ itself, right?

    Let $y_n$ be that point.
    So we have then that $|y_n-b_n|>0$ ?

    If $b_n=b_0$ for some $n$ then we have that $b_0\in L(a_n)$.
    If $b_n\neq b_0, \forall n$ then from the above defintion $0<|y_n-b_n|<|b_n-b_0|$.

    Is this correct? Does this help?
    Last edited by mathmari; November 20th, 2016 at 15:39.

Similar Threads

  1. Limit of a ln sequence
    By tmt in forum Calculus
    Replies: 1
    Last Post: July 9th, 2016, 04:40
  2. Limit of a sequence
    By Alexmahone in forum Linear and Abstract Algebra
    Replies: 1
    Last Post: July 27th, 2015, 02:38
  3. [SOLVED] Limit of sequence
    By wishmaster in forum Pre-Calculus
    Replies: 35
    Last Post: October 25th, 2013, 09:24
  4. Proving limit of a given sequence
    By issacnewton in forum Analysis
    Replies: 2
    Last Post: October 15th, 2013, 03:44
  5. Limit of a sequence
    By Fernando Revilla in forum Calculus
    Replies: 2
    Last Post: January 21st, 2013, 08:37

Tags for this Thread

Posting Permissions

  • You may not post new threads
  • You may not post replies
  • You may not post attachments
  • You may not edit your posts
  •  
Math Help Boards