# Open Ball is Open Set/Metric Space

## Theorem

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

Let $x \in A$.

Let $\epsilon \in \R_{>0}$.

Let $\map {B_\epsilon} x$ be an open $\epsilon$-ball of $x$ in $M$.

Then $\map {B_\epsilon} x$ is an open set of $M$.

## Proof

Let $y \in \map {B_\epsilon} x$.

From Open Ball of Point Inside Open Ball, there exists $\delta \in \R_{>0}$ such that:

$\map {B_\delta} y \subseteq \map {B_\epsilon} x$

The result follows from the definition of open set.

$\blacksquare$