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Math Help - Sequence and accumulation point

  1. #1
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    Sequence and accumulation point

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

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

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

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

    Let A=\text{sup} \ \{a_n\} and Q be a neighborhood of A. There is an \epsilon>0 such that (A-\epsilon, A+\epsilon)\subset Q.
    Since A = sup and if c is an upper bound, A\leq c
    Let a_i\in\{a_n\}, \ i=1,2,....
    Assume there are finitely many a_i\in Q.
    Now, let A'=\text{max}\{a_i\}
    Since x' is the max, x' is an upper bound of \{a_n\}. Therefore, A'<A, but 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 A=\sup\{a_n\}. For example \langle (-1)^n/n\rangle is a sequence converging to 0, which is neither an upper nor lower bound of the sequence.

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