Sequence of Imaginary Reciprocals/Closure is Compact

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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 closure $S^-$ of the set $S$ is compact.


From Topological Closure is Closed, $S^-$ is closed.

From Sequence of Imaginary Reciprocals: Boundedness, $S$ is bounded in $\C$.

It follows trivially that $S^-$ is also bounded in $\C$.

Hence the result by definition of compact.