Open Ball of Point Inside Open Ball/Normed Vector Space

Theorem
Let $M = \struct {X, \norm {\, \cdot \,} }$ be a normed vector space.

Let $\map {B_\epsilon} x$ be an open $\epsilon$-ball in $M = \struct {X, \norm {\, \cdot \,} }$.

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

Then:
 * $\exists \delta \in \R: \map {B_\delta} y \subseteq \map {B_\epsilon} x$

That is, for every point in an open $\epsilon$-ball in a normed vector space, there exists an open $\delta$-ball of that point entirely contained within that open $\epsilon$-ball.

Proof
Let $\delta = \epsilon - \norm {x - y}$.

From the definition of open ball, this is strictly positive, since $y \in \map {B_\epsilon} x$.

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

So:
 * $\norm {x - z} \le \norm {x - y} + \norm {y - z} < \norm {x - y} + \delta = \epsilon$

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

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