Closed Ball contains Smaller Open Ball

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

Let $a \in A$.

Let $\epsilon, \delta \in \R_{> 0}$ such that $\epsilon \le \delta$.

Let $\map {B_\epsilon} a$ be the open $\epsilon$-ball on $a$.

Let $\map {B^-_\delta} a$ be the closed $\delta$-ball on $a$.

Then:
 * $\map {B_\epsilon} a \subseteq \map {B^-_\delta} a$

Proof
By definition of subset:
 * $\map {B_\epsilon} a \subseteq \map {B^-_\delta} a$