Open Ball is Open Set/Metric Space
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Theorem
Let $M = \struct {A, d}$ be a metric space.
Let $x \in A$.
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, 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
- 1967: George McCarty: Topology: An Introduction with Application to Topological Groups ... (previous) ... (next): Chapter $\text{III}$: Metric Spaces: Compactness
- 1975: W.A. Sutherland: Introduction to Metric and Topological Spaces ... (previous) ... (next): $2$: Continuity generalized: metric spaces: $2.3$: Open sets in metric spaces: Example $2.3.9$