
#1
Aug3108, 02:47 PM

P: 569

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
Aug3108, 04:36 PM

HW Helper
P: 1,344

I think you're there. Nonmathematically speaking, an accumulation point for your sequence would be any number that is approached by infinitely many terms of the sequence.




#3
Sep108, 02:56 AM

Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 38,881

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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