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Thread: Sequence and accumulation point

  1. #1
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    Sequence and accumulation point

    Suppose $\displaystyle \{a_n\}_{n\in\mathbb{N}}\to A$ and $\displaystyle \{a_n: \ n\in\mathbb{N}\}$ is an infinite set. Show that A is an accumulation point of $\displaystyle \{a_n: \ n\in\mathbb{N}\}$.

    Let $\displaystyle A=\text{sup} \ \{a_n\}$ and Q be a neighborhood of A. There is an $\displaystyle \epsilon>0$ such that $\displaystyle (A-\epsilon, A+\epsilon)\subset Q$.
    Since A = sup and if c is an upper bound, $\displaystyle A\leq c$
    Let $\displaystyle a_i\in\{a_n\}, \ i=1,2,....$
    Assume there are finitely many $\displaystyle a_i\in Q$.
    Now, let $\displaystyle A'=\text{max}\{a_i\}$
    Since x' is the max, x' is an upper bound of $\displaystyle \{a_n\}$. Therefore, $\displaystyle A'<A$, but $\displaystyle A = \text{sup} \ \{a_n\}$ which is a contradiction. Thus, A is an accumulation point.

    Correct?
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  2. #2
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    Re: Sequence and accumulation point

    Quote Originally Posted by dwsmith View Post
    Suppose $\displaystyle \{a_n\}_{n\in\mathbb{N}}\to A$ and $\displaystyle \{a_n: \ n\in\mathbb{N}\}$ is an infinite set. Show that A is an accumulation point of $\displaystyle \{a_n: \ n\in\mathbb{N}\}$.

    Let $\displaystyle A=\text{sup} \ \{a_n\}$ and Q be a neighborhood of A. There is an $\displaystyle \epsilon>0$ such that $\displaystyle (A-\epsilon, A+\epsilon)\subset Q$.
    Since A = sup and if c is an upper bound, $\displaystyle A\leq c$
    Let $\displaystyle a_i\in\{a_n\}, \ i=1,2,....$
    Assume there are finitely many $\displaystyle a_i\in Q$.
    Now, let $\displaystyle A'=\text{max}\{a_i\}$
    Since x' is the max, x' is an upper bound of $\displaystyle \{a_n\}$. Therefore, $\displaystyle A'<A$, but $\displaystyle A = \text{sup} \ \{a_n\}$ which is a contradiction. Thus, A is an accumulation point.

    Correct?
    No, that's not correct. You don't get to just assume that $\displaystyle A=\sup\{a_n\}$. For example $\displaystyle \langle (-1)^n/n\rangle$ is a sequence converging to $\displaystyle 0$, which is neither an upper nor lower bound of the sequence.

    Recall that by definition $\displaystyle \langle a_n\rangle\to A$ iff for each $\displaystyle \epsilon>0$ there is $\displaystyle N\in\mathbb{N}$ such that if $\displaystyle n\geq N$ then $\displaystyle a_n\in(A\pm\epsilon)$. So just notice that there are infinitely many elements in $\displaystyle \{a_n:n\geq N\}\subseteq(A\pm\epsilon)$, and the proof is complete.
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