State whether the following set is open or closed. State whether the following set is open or closed. And determine the accumulation points:
$$A=\{n\left(\sqrt[n]{2}\left(\sin\left(\frac{\pi}{2n}\right)+i\cos\left (\dfrac{\pi}{2n} \right)\right)-1\right): n\in\Bbb N\}$$
Try:
I can say that this set is not open, since it is a set of discrete points, and these sets are not open in $\Bbb C$. But I don't know how to find if it's closed and accumulation points.
 A: You are correct to say that $A$ is not open since any nonempty open set must contain an open rectangle/disk, and all such sets are uncountable.
One strategy to show that it's closed is to prove that every accumulation point of the set belongs to the set. Let's show the contrapositive instead, that is: if $z\notin A$ then $z$ is not an accumulation point of $A$.
Since the imaginary parts of the points in $A$ are of the form $n\sqrt[n]{2}\cos(2\pi/n)$, and we have
$$\begin{align}
n&\to\infty \\
\sqrt[n]{2}&\to 1 \\
\cos(2\pi/n)&\to 1
\end{align}
$$
as $n$ gets larger, then $n\sqrt[n]{2}\cos(2\pi/n)\to\infty$.
If $z\notin A$, then since the imaginary parts of points in $A$ grow arbitrarily large and positive, there are only finitely many points in $A$ whose imaginary parts are less than, say, $\Im(z) + 1$. Call this finite set $F$. Since finite sets are closed in $\mathbb{C}$ and $z$ does not belong to $F$, we can find an open neighborhood of $z$ which is disjoint from $F$. Intersecting this open set with the set of complex numbers with imaginary part less than $\Im(z) + 1$ gives an open neighborhood of $z$ which does not intersect $A$ at any point. Thus $z$ is not an accumulation point of $A$, and that is what needed to be shown.
