Equivalence of Definitions of Limit Point in Metric Space/Definition 1 implies Definition 2

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:
 * $\forall \epsilon \in \R_{>0}: \paren {\map {B_\epsilon} \alpha \setminus \set \alpha} \cap A \ne \O$

Then:
 * there is a sequence $\sequence{\alpha_n}$ in $A$ such that $\alpha$ is a limit point of the sequence $\sequence{\alpha_n}$, considered as sequence in $S$.

Proof
A sequence $\sequence {\alpha_n}$ is constructed using finite recursion.

Let $\alpha_0$ be some arbitrary point in $S$ not equal to $\alpha$.

$\alpha_{n + 1}$ is some arbitrary point in $S \cap \map {B_{\frac {\map d {\alpha, \alpha_n} } 2} } p$ where $\map {B_{\frac {\map d {\alpha, \alpha_n} } 2} } \alpha$ denotes the open $\dfrac {\map d {\alpha, \alpha_n} } 2$-ball of $\alpha$.

We can be assured that such a point exists by.

Then:
 * $\map d {\alpha_n, \alpha} < 2^{-n} \map d {\alpha_0, \alpha}$

Since $\map d {\alpha_0, \alpha}$ is fixed:
 * $\map d {\alpha_n, \alpha} < 2^{-n} \map d {\alpha_0, \alpha} < r$

for some $n$ because the natural numbers are Archimedean.

Therefore $\alpha$ is the limit of the sequence $\sequence {\alpha_n}$ by definition.