Limit of Sequence in Metric Space in Neighborhood
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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:
- $\ds \lim_{n \mathop \to \infty} a_n = a$
$\blacksquare$
Sources
- 1975: Bert Mendelson: Introduction to Topology (3rd ed.) ... (previous) ... (next): Chapter $2$: Metric Spaces: $\S 5$: Limits: Corollary $5.3$