Equivalence of Definitions of Convergent Sequence in Metric Space/Definition 4 implies Definition 2

Theorem
Let $M = \left({A, d}\right)$ be a metric space or a pseudometric space.

Let $l \in A$.

Let $\sequence {x_k}$ be a sequence in $A$. Let $\sequence {x_k}$ satisfy:
 * for every $\epsilon \in \R{>0}$, the open $\epsilon$-ball about $l$ contains all but finitely many of the $p_n$.

Then:
 * $\forall \epsilon > 0: \exists N \in \R_{>0}: \forall n \in \N: n > N \implies x_n \in \map {B_\epsilon} l$

where $\map {B_\epsilon} l$ is the open $\epsilon$-ball of $l$.

Proof
Let $\map {B_\epsilon} l$ be any open $\epsilon$-ball of $l$.

Let $A = \set {n : x_n \notin \map {B_\epsilon} l}$.

By assumption $A$ is finite.

From Finite Non-Empty Subset of Totally Ordered Set has Smallest and Greatest Elements, any finite subset of $\N$ has a maximum.

Let $N$ be the maximum of $A$.

Then for every $n > N$, $x_n$ must be in the open $\epsilon$-ball $\map {B_\epsilon} l$.