Accumulation points of a set

  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. statdad

    statdad 1,455
    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. HallsofIvy

    HallsofIvy 40,310
    Staff Emeritus
    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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