Open Ball contains Strictly Smaller Closed Ball

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

Let $a \in A$.

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

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

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

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

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