Limit of Sequence in Metric Space in Neighborhood

From ProofWiki
Jump to navigation Jump to search

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 $a$:

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


Proof

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 $a_n \in V$.

$\Box$


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:

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

$\blacksquare$


Sources