Sequence of Imaginary Reciprocals/Closedness

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Theorem

Consider the subset $S$ of the complex plane defined as:

$S := \set {\dfrac i n : n \in \Z_{>0} }$

That is:

$S := \set {i, \dfrac i 2, \dfrac i 3, \dfrac i 4, \ldots}$

where $i$ is the imaginary unit.


The set $S$ is not closed.


Proof

From Sequence of Imaginary Reciprocals: Limit Points, $S$ has one limit point $z = 0$.

But:

$\nexists n \in \N: \dfrac i n = 0$

so $0 \notin S$.

As $S$ does not contain (all) its limit point(s), it follows by definition that $S$ is not closed.

$\blacksquare$


Sources