Open Ball is Open Set/Normed Vector Space
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Theorem
Let $M = \struct {X, \norm {\, \cdot \,}}$ be a normed vector space.
Let $x \in X$.
Let $\epsilon \in \R_{>0}$.
Let $\map {B_\epsilon} x$ be an open $\epsilon$-ball of $x$ in $M$.
Then $\map {B_\epsilon} x$ is an open set of $M$.
Proof
Let $y \in \map {B_\epsilon} x$.
From Open Ball of Point Inside Open Ball in Normed Vector Space, there exists $\delta \in \R_{>0}$ such that $\map {B_\delta} y \subseteq \map {B_\epsilon} x$
The result follows from the definition of open set.
$\blacksquare$
Sources
- 2017: Amol Sasane: A Friendly Approach to Functional Analysis ... (previous) ... (next): $\S 1.3$: Normed and Banach spaces. Topology of normed spaces