Limit of Sequence in Metric Space in Neighborhood

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

Let $\left\langle{a_n}\right\rangle$ be a sequence in $A$.

Then $\displaystyle \lim_{n \mathop \to \infty} a_n = a$ iff for each neighborhood $V$ of $a$:
 * $\exists N \in \N: n > N \implies a_n \in V$

Necessary Condition
Let $\displaystyle \lim_{n \mathop \to \infty} a_n = a$.

Let $V$ be a neighborhood of $a$.

By definition of neighborhood:
 * $\exists \epsilon \in \R_{>0}: B_\epsilon \left({a}\right) \subseteq V$

where $B_\epsilon \left({a}\right)$ denotes the open $\epsilon$-ball of $a$ in $M$.

By definition of limit:
 * $\exists N \in \N: n > N \implies d \left({a, a_n}\right) < \epsilon$

Hence $a_n \in V$.

Sufficient Condition
Let $\left\langle{a_n}\right\rangle$ be such that for each neighborhood $V$ of $a$:
 * $\exists N \in \N: n > N \implies a_n \in V$

Let $\epsilon \in \R_{>0}$.

Then by Open Ball is Neighborhood of all Points Inside, $B_\epsilon \left({a}\right)$ is a neighborhood of $a$.

Let $N \in \N$ be such that:
 * $\forall n > N: a_n \in B_\epsilon \left({a}\right)$

Then:
 * $d \left({a, a_n}\right) < \epsilon$

and so by definition of limit:
 * $\displaystyle \lim_{n \mathop \to \infty} a_n = a$