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Homework Help: Accumulation points of a set

  1. Aug 31, 2008 #1
    1. The problem statement, all variables and given/known data

    Determine the accumulation points of the following set:

    z(sub n)=i^n, (n=1,2,....);

    2. Relevant equations

    3. The attempt at a solution

    My book says z(sub n) does not have any accumulation points. When mapped onto a complex plane, z(sub n) forms a circle. For any set to contained each of its accumulation points, the set has to be closed. And the definition of a closed set is a set containing all of its boundary points. z(sub n) is a close set since the only points of z(sub n) are: i,-1,-i and 1. How can any of those four points not be accumulation points?
  2. jcsd
  3. Aug 31, 2008 #2


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    Homework Helper

    I think you're there. Non-mathematically speaking, an accumulation point for your sequence would be any number that is approached by infinitely many terms of the sequence.
  4. Sep 1, 2008 #3


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    Science Advisor

    No it does not. It is four distinct points as you say below.

    The definition of accumulation point is "p is an accumulation point of A if and only if every neighborhood contains as least one point of A other than p itself".

    i, -1, -i, and i are boundary points; they are not accumulation points.

    Notice the "direction" of your first statement: "For any set to contained each of its accumulation points, the set has to be closed." More formally "if a set contains all of its accumulation points then it is closed". If a set has NO accumulation points then it trivially includes "all of them".
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