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 $z$.

$\blacksquare$


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