Open Balls form Local Basis for Point of Metric Space

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

Let $x \in A$.

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

That is:
 * $\mathcal B_x = \set{\map {B_\epsilon} x : \epsilon \in \R_{>0}}$

Then $\mathcal B$ 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 is Open Set, $\mathcal B_x$ is a set of open set which have $x$ as an element.

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