Limit Point is Limit of Convergent Sequence

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

Let $E \subseteq X$ be a subset of $X$.

Let $p$ be a limit point of $E$.

Then there exists a sequence $\sequence {x_n} \subseteq E$ which converges to a limit:
 * $\ds \lim_{n \mathop \to \infty} x_n = p$

where $\ds \lim_{n \mathop \to \infty} x_n$ is the limit of the sequence $\sequence {x_n}$.

Proof
We may construct a sequence $\sequence {p_n}$ using finite recursion.

Let $p_0$ be some arbitrary point in $E$ not equal to $p$.

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

We can be assured that such a point exists by the definition of a limit point.

Then:
 * $\map d {p_n, p} < 2^{-n} \map d {p_0, p}$

Since $\map d {p_0, p}$ is fixed:
 * $\map d {p_n, p} < 2^{-n} \map d {p_0, p} < r$

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

Therefore $p$ is the limit of the sequence $\sequence {p_n}$ by definition.