Sequence of Imaginary Reciprocals/Closure is Compact
Jump to navigation
Jump to search
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 closure $S^-$ of the set $S$ is compact.
Proof
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.
$\blacksquare$
Sources
- 1981: Murray R. Spiegel: Theory and Problems of Complex Variables (SI ed.) ... (previous) ... (next): $1$: Complex Numbers: Solved Problems: Point Sets: $45 \ \text {(l)}$