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I need to determine whether the following set has accumulation points:

$0 \le \arg z<\pi/2 (z\ne 0)$

Would the accumulation point be z=0, as the set does not include 0? If not, does it not have any accumulation points as the set fans out to $\infty$ in the complex plane?

I think that I am just having a bit of trouble understanding the concept of an accumulation point.

Thank you for your help in advance!

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    $\begingroup$ There are several accumulation points in this example. It's true that $z=0$ is one of the accumulation points, but the reason is not that $0$ is not in the set. What is your definition of an accumulation point? $\endgroup$
    – user108903
    Jan 31, 2013 at 21:57
  • $\begingroup$ Any point you can 'approach' from within the set is an accumulation point. Since $\frac{1}{n}$ is in the set and converges to $0$, then $0$ is an accumulation point. $\endgroup$
    – copper.hat
    Jan 31, 2013 at 22:00
  • $\begingroup$ The definition I have for an accumulation point is: point $z_0$is an accumulation point of S if each deleted neighborhood of $z_0$ contains at least one point of S. i.e. $\forall \epsilon >0 (D(z_0,\epsilon)\setminus {z_0})\cap S \neq \emptyset$ $\endgroup$
    – Jess
    Jan 31, 2013 at 23:58
  • $\begingroup$ Another question I have regarding accumulation points: can they be within the set themselves? $\endgroup$
    – Jess
    Feb 1, 2013 at 0:01

3 Answers 3

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An accumulation point of a set $A$ does not have to belong to $A$ itself. At least every interior point of $A$ and every non-isolated boundary point of $A$ is an accumulation point.

In your example, the set of accumulation points is the same as the closure of your set, i.e. the entire closed first quadrant $\{ z = x+iy | x \ge 0, y \ge 0 \}$. (In general, the set of accumulation points can be smaller than the closure: Take $A = \{ 0 \}$; then the set of accumulation points is empty.)

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In the text the definition is: “point z0 is said to be an accumulation point of a set S if each deleted neighborhood of z0 contains at least one point of S.” So this is independent whether it is in or outside S. Therefore the accumulation points of this unbounded region are the points of the positive parts of the real and imaginary axis of the z plane, including the origin.

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Be careful here: notice the inclusive and exclusive inequalities.

Since $\operatorname{arg}(z) \geq 0$, it includes the $x$-axis on the complex plane (except $z=0$). By contrast, $\operatorname{arg}(z) < \frac{\pi}{2}$, so the $y$-axis on the complex plane is excluded.

Hence, the set of accumulation points is $\{ z = x + iy\, |\, x = 0, y \geq 0 \}$.

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