# Limit of Sequence in Metric Space in Neighborhood

Jump to navigation Jump to search

## Theorem

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

Let $\sequence {a_n}$ be a sequence in $A$.

Then $\ds \lim_{n \mathop \to \infty} a_n = a$ if and only if for each neighborhood $V$ of $a$:

$\exists N \in \N: n > N \implies a_n \in V$

## Proof

### Necessary Condition

Let $\ds \lim_{n \mathop \to \infty} a_n = a$.

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

By definition of neighborhood:

$\exists \epsilon \in \R_{>0}: \map {B_\epsilon} a \subseteq V$

where $\map {B_\epsilon} a$ denotes the open $\epsilon$-ball of $a$ in $M$.

By definition of limit:

$\exists N \in \N: n > N \implies \map d {a, a_n} < \epsilon$

Hence $a_n \in V$.

$\Box$

### Sufficient Condition

Let $\sequence {a_n}$ 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, $\map {B_\epsilon} a$ is a neighborhood of $a$.

Let $N \in \N$ be such that:

$\forall n > N: a_n \in \map {B_\epsilon} a$

Then:

$\map d {a, a_n} < \epsilon$

and so by definition of limit:

$\ds \lim_{n \mathop \to \infty} a_n = a$

$\blacksquare$