Is the set $\{z \in \mathbb{C}: z^n=1$ for some integer $n \geq 1\}$ open and/or closed?

Question

Is the set $\{z \in \mathbb{C}: z^n=1$ for some integer $n \geq 1\}$ open and/or closed?

I think the set is closed but not open as every point is a boundary point.

Answer 1

Let us call $$A=\{z\in \mathbb C: z^n =1 \text{ for some integer } n\geq1\}$$ A is neither closed nor open.

It is not open, because if $z\in A$, then $z^n=1$ for some $n$, and that means $|z|=1$. Now, every neighbourhood of a point $z$ in $\mathbb C$ would contain points of module not $1$.

But it is not closed. In fact, its closure is $S_1=\{z\in\mathbb C: |z|=1\}$. If you take a number $z$ that satisfies $|z|=1$ and $\theta=arg(z)\in \mathbb R\setminus\mathbb Q$, you can find a sequence of points in $A$ that converges to $z$.

Answer 2

The set consists of the union of the vertices of all regular $n$ gons inscribed in the unit circle, each having $z=1$ as the "start" vertex, then proceeding around the circle. This union is not open, since a little ball around $1$ will contain points not on the unit circle. It isn't closed either, since one can find a sequence of points in the union which approach for example the point $e^{i \sqrt{2}},$ for which no positive integer power is $1.$