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  1. MHB Master
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    #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
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    #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
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    #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
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    #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.

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