Open Balls form Local Basis for Point of Metric Space

Theorem

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

Let $x \in A$.

Let $\BB_x$ be the set of all open balls of $M$ centered on $x$.

That is:

$\BB_x = \set {\map {B_\epsilon} x : \epsilon \in \R_{>0}}$

Then $\BB$ is a local basis of $x$.

Proof

Let $U$ be an open set of $M$ which has $x$ as an element.

Then by definition of an open set:

$\exists \epsilon \in \R_{>0}: \map {B_\epsilon} x \subseteq U$

From Open Ball of Metric Space is Open Set, $\BB_x$ is a set of open set which have $x$ as an element.

By definition of a local basis, $\BB_x$ is a local basis of $x$.

$\blacksquare$