Open Ball of Point Inside Open Ball/Normed Vector Space
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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 $\map {B_\delta} y \subseteq \map {B_\epsilon} x$.
$\blacksquare$
Sources
- 2017: Amol Sasane: A Friendly Approach to Functional Analysis: Chapter $1$: Normed and Banach spaces