Limit Point of Set in Complex Plane not Element is Boundary Point
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Theorem
Let $S \subseteq \C$ be a subset of the complex plane.
Let $z \in \C$ be a limit point of $S$ such that $z \notin S$.
Then $z$ is a boundary point of $S$.
Proof
Let $z \in \C$ be a limit point of $S$ such that $z \notin S$.
Suppose $z$ is not a boundary point of $S$.
Then there exists an $\epsilon$-neighborhood $\map {N_\epsilon} z$ of $z$ such that either:
- $(1): \quad$ All elements of $\map {N_\epsilon} z$ are in $S$
or
- $(2): \quad$ All elements of $\map {N_\epsilon} z$ are not in $S$.
If $(1)$, then $z \in S$ which is a contradicts $z \notin S$.
If $(2)$, then $z$ is not a limit point of $S$, which also contradicts the conditions on $z$.
$\blacksquare$
Sources
- 1981: Murray R. Spiegel: Theory and Problems of Complex Variables (SI ed.) ... (previous) ... (next): $1$: Complex Numbers: Supplementary Problems: Point Sets: $126$