# Accumulation points of a set

## Homework Statement

Determine the accumulation points of the following set:

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

## 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?

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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.

HallsofIvy
Homework Helper

## Homework Statement

Determine the accumulation points of the following set:

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

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

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?
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".