Sequence of Imaginary Reciprocals/Limit Points

Theorem
The set $S$ has exactly one limit point, and that is $z = 0$.

Proof
We have that $S$ is countably infinite and also bounded.

Hence by the Bolzano-Weierstrass Theorem $S$ has at least one limit point.

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

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

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

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

where $\map {N_\epsilon} {z_0}$ denotes the $\epsilon$-neighborhood of $z$.

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

Now let $z_1 \in \C$ such that $z_1 \ne 0$.

Let $z \in S$ such that $z \ne z_1$.

Then:
 * $\cmod {z_1 - z} > 0$

Let $\epsilon \in \R_{>0}$ such that $\epsilon < \cmod {z_1 - z}$.

Then:
 * $z \notin \map {N_\epsilon} {z_1}$

By making $\epsilon$ smaller than the minimum $\cmod {z_1 - z}$ for $z \in S$, it is seen that:
 * $\forall z \in S: z \ne z_1 \implies z \notin \map {N_\epsilon} {z_1}$

and so $z_1$ is not a limit point of $S$.