Equivalence of Definitions of Limit Point in Metric Space/Definition 2 implies Definition 3

Theorem
Let $M = \struct {S, d}$ be a metric space.

Let $\tau$ be the topology induced by the metric $d$.

Let $A \subseteq S$ be a subset of $S$.

Let $\alpha \in S$.

Let there exist a sequence $\sequence{\alpha_n}$ in $A \setminus \set \alpha$ such that $\alpha$ is a limit point of the sequence $\sequence{\alpha_n}$, considered as sequence in $S$.

Then:
 * $\alpha$ is a limit point in the topological space $\struct{S, \tau}$.

Proof
Let $U \in \tau$ be an arbitrary open set:
 * $\alpha \in U$

By definition of topology induced by metric:
 * $\exists \epsilon \in \R_{>0} : \map {B_\epsilon} \alpha \subseteq U$

By definition of limit point:
 * $\exists N \in \N : \forall n \ge N : d(\alpha, \alpha_n) < \epsilon$

By definition of open ball:
 * $\alpha_N \in \map {B_\epsilon} \alpha$

By hypothesis:
 * $\alpha_N \in A \setminus \set \alpha$

We have:

Hence:
 * $A \cap \paren{U \setminus \set \alpha} \ne \O$

Since $U$ was arbitrary, it follows that $\alpha$ is a limit point in the topological space $\struct{S, \tau}$ by definition.