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


Let $z \in \map {B_\delta} y$ be arbitrary.

We have:

\(\ds z\) \(\in\) \(\ds \map {B_\delta} y\) by hypothesis
\(\ds \leadsto \ \ \) \(\ds \norm {y - z}\) \(<\) \(\ds \delta\) Definition of Open Ball in Normed Vector Space

Then we have:

\(\ds \norm {x - z}\) \(\le\) \(\ds \norm {x - y} + \norm {y - z}\) Triangle Inequality
\(\ds \) \(<\) \(\ds \norm {x - y} + \delta\)
\(\ds \) \(=\) \(\ds \epsilon\)
\(\ds \leadsto \ \ \) \(\ds z\) \(\in\) \(\ds \map {B_\epsilon} x\) Definition of Open Ball in Normed Vector Space


That is:

$z \in \map {B_\delta} y \implies z \in \map {B_\epsilon} x$

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

$\blacksquare$


Sources

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