Open Ball is Subset of Open Ball

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

Let $x, y$ be points of $A$.

Then:
 * $\epsilon-\delta \ge \map d {x, y} \implies \map {B_\delta} y \subseteq \map {B_\epsilon} x$

where $\map {B_\epsilon} x$ denotes the open $\epsilon$-ball in $M = \struct {A, d}$.

Proof
Let $\epsilon - \delta \ge \map d {x, y}$.

Then $\epsilon \ge \map d {x, y} + \delta$.

If $z \in \map {B_\delta} y$, then $\map d {y, z} < \delta$.

So:

Thus $z \in \map {B_\epsilon} x$.

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