Sequence of Imaginary Reciprocals/Boundary Points

Theorem
Every point of $S$, along with the point $z = 0$, is a boundary point of $S$.

Proof
Consider the point $z = \dfrac i n \in S$.

Let $\delta \in \R_{>0}$.

Let $\map {N_\delta} z$ be the $\delta$-neighborhood of $z$.

Then $\map {N_\delta} z$ contains at least one point of $S$ ($i / n$ itself) as well as points which are not in $S$.

Hence, by definition, $z$ is a boundary point of $S$.

Let $z = 0$.

Similarly, let $\map {N_\delta} z$,

Let $\delta \in \R_{>0}$.

Let $\map {N_\delta} z$ be the $\delta$-neighborhood of $z$.

Let $n \in \N$ such that $n > \dfrac 1 \delta$.

Then $\cmod {\dfrac i n} < \delta$ and so:
 * $\dfrac i n \in \map {N_\delta} z$

Thus every $\delta$-neighborhood of $z = 0$ also contains points of $S$ and points not in $S$.

Hence, by definition, $z = 0$ is also a boundary point of $S$.