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I am stuck on the last problem of my complex variable homework.

The problem is given the set $S=\{\frac{i}{n} \mid n \text{ is an integer}\}$,

a.) List the points in $S$.

b.) What are the accumulation points of $S$?

So far I have the points being $i$, $i/2$, $i/3,\ldots , i/n$ but that seems like it is too easy for this class.

Am I missing something?

The second part, I would think there are not any accumulation points because none of the neighborhoods of the points will include any other points for a small neighborhood.

Any help would be appreciated.

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    I believe the only accumulation point would be 0 since every neighborhood around 0 will contain a point in S(as n gets very large)2011-10-11
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    An accumulation point does not have to be in the set itself; it can be in the "larger" space we are considering, which in this case is presumably the set of all complex numbers.2011-10-11
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    I don't like the problem statement. 1. Zero is an integer, so your "set" contains $i/0$, which is a bit of a worry. 2. In a), you ask about $S$, which you haven't defined. I take it $S$ is that set from your previous sentence? As to what you are missing, you seem to be missing that the integers do not stop at $n$, but go on forever, so there are infinitely many points in $S$, not just $n$ points, as you have written.2011-10-11
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    Is the number $i$ fixed?2011-10-11
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    @GerryMyerson I agree that this would make more sense for the set S to only include positive integers or exclude 0. The answer I have so far for a.) +-i, +-i/2, +-i/3, ... b.) The only accumulation point would be zero as every neighborhood around zero would include a point in the set if we let n approach infinity.2011-10-11
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    Joe, that looks good to me.2011-10-11

1 Answers 1

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For every nonzero complex number $z$, let $A_z=\left\{\frac{z}n; n\geqslant1\right\}$. Then, for every positive $\varepsilon$:

  1. The set $A_z\cap D_\varepsilon$ is infinite,
  2. The set $A_z\setminus D_\varepsilon$ is finite.

Here, $D_\varepsilon=\{u\in\mathbb C;|u|<\varepsilon\}$. This shows that $A_z$ has one and only one accumulation point, which is $0$, for every $z\ne0$ and for example $z=i$.

To prove 1. and 2., note that for every nonzero $z$ and every positive $\varepsilon$, there exists a finite nonnegative integer $k$ such that $k\varepsilon\leqslant |z|<(k+1)\varepsilon$. Hence, $A_z\cap D_\varepsilon=\left\{\frac{z}n;n\geqslant k+1\right\}$ is always infinite and $A_z\setminus D_\varepsilon=\left\{\frac{z}n;1\leqslant n\leqslant k\right\}$ is always finite.

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    From the title and the tag, I assumed OP was using $i$ to mean $\sqrt{-1}$.2011-10-11
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    Yes I was using i to mean sqrt(-1)2011-10-11
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    Right. Then the proof is valid.2011-10-11