Let [$\mathbb{Z}, \tau $] be a topological space, with $\mathbb{Z}$ being the set of all integer numbers and $\tau$ being the Discrete Topology. With $\alpha \in \mathbb{R} - \mathbb{Q}$ and $S^1$ being the unit sphere, define $$ f_\alpha:\mathbb{Z} >\rightarrow S^1\subset\mathbb{C} ~,~ n \mapsto exp(2\pi i n \alpha). $$
$\underline{Question:}$ How can the closure $\overline{f_\alpha(\mathbb{Z})} \subset S^1$ be characterized? I have made an approach to understand this by myself. Is my conclusion right? Thanks for your help!
$\underline{My~Approach:}$
$\overline{f_\alpha(\mathbb{Z})}$ can be written as the union of the boundary of $f_\alpha(\mathbb{Z})$ with the set of all its inner points $$ \overline{f_\alpha(\mathbb{Z})} =\partial f_\alpha(\mathbb{Z}) ~\cup~\mathring{f_\alpha(\mathbb{Z})},$$ the first of which is empty and the second of which is the set $f_\alpha(\mathbb{Z})$ itself. Thus $$ \overline{f_\alpha(\mathbb{Z})} = f_\alpha(\mathbb{Z}),$$ so that $f_\alpha(\mathbb{Z})$ is a closed subset of $S^1$. I got to this conclusion by thinking about the possible neighbourhoods of an element in the image of $f_\alpha$ in the unit sphere.
The conclusion is not right, I'm afraid.
The boundary $\partial f_\alpha(\mathbb{Z})$ is not empty. To see this, note that because $\alpha$ is not rational the function $f_\alpha$ is injective. If you divide $S^1$ into intervals then you can use the pigeonhole principle to show that some interval contains infinitely many $f_\alpha(n)$, and you can use these to construct a sequence without a limit in $f_\alpha(\mathbb{Z})$.
Hint to find the boundary: use the pigeonhole principle as above and the fact that $f_\alpha$ is a homomorphism to show that every open set in $S^1$ contains an element of $f_\alpha(\mathbb{Z})$.
The interior is not $f_\alpha(\mathbb{Z})$. You have to be careful, because after applying $f_\alpha$ you are inside $S^1$, so you are working with its topology, not the discrete topology. This means that singleton sets are not open.