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Math Help - Accumulation point

  1. #1
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    Accumulation point

    Prove that a real number s is an accumulation point of a set S if and only if there exists some sequence {a} in S such that a is not equal to s for every natural number n where the limit of a as n approaches infinity is equal to s.

    I'm not really sure how to start. I know since it's an if and only if statement we have to prove it both ways.
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  2. #2
    Senior Member Dinkydoe's Avatar
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    s\in S is a accumulation point if for every \epsilon > 0 there exists a x_0\in S with a-\epsilon < x_0 < a+\epsilon and x_0\neq a

    Choose a \epsilon > 0.

    For every n\geq 1 you can choose a x_n\in S with x_n\neq a such that: a-\frac{\epsilon}{n} <  x_n < a+\frac{\epsilon}{n}

    Now x_n\to a as n\to \infty

    I'll leave the other implication to you.
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  3. #3
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    Quote Originally Posted by bethh View Post
    Prove that a real number s is an accumulation point of a set S if and only if there exists some sequence {a} in S such that a is not equal to s for every natural number n where the limit of a as n approaches infinity is equal to s.
    You did not tell us what sort of top-space we are considering.
    I will assume that you are working in a metric space (so the statement is true).
    If r>0 then \mathcal{B}(a;r)=\{x:d(a,x)<r\} is a ball.
    Using the definition of accumulation point we get:
    \left( {\exists x_1 } \right)\left[ {s \ne x_1 \,\& \,d\left( {x_1 ,s} \right) < 1} \right] .
     \left( {\exists x_2 } \right)\left[ {s \ne x_2 \,\& \,d\left( {x_2 ,s} \right) < \min(d\left( {x_1 ,s} \right),\frac{1}{2}} )\right].

    Now you have a way to generate your sequence by induction.
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  4. #4
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by bethh View Post
    Prove that a real number s is an accumulation point of a set S if and only if there exists some sequence {a} in S such that a is not equal to s for every natural number n where the limit of a as n approaches infinity is equal to s.

    I'm not really sure how to start. I know since it's an if and only if statement we have to prove it both ways.
    Let \left\{x_n\right\}_{n\in\mathbb{N}} be a sequence in S such that x_n\ne x,\text{ }\forall n and x_n\to x. Now, let B_{\varepsilon}(x) be arbitrary. Since x_n\to x there exists an N\in\mathbb{N} such that N\leqslant n\implies x_n\in B_{\delta}(x) and since x_N\ne x and x_N\in B_{\delta}(x) we see there exists a point of S distinct from S in this ball. The conclusion follows from the arbitrariness of the open ball.


    Conversely, choosing any arbitrary point in B_{\frac{1}{n}}(x)-\{x\} finishes the job. There is no need for them to be all distinct.
    Last edited by Drexel28; January 27th 2010 at 04:54 PM.
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