Equality of Open Balls does not imply Equality of Centers
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Theorem
Let $M = \struct {A, d}$ be a metric space.
Let $x, y \in A$ and $r, s \in \R$ such that:
- $\map {B_r} x = \map {B_s} y$
Then it is not necessarily the case that their centers $x$ and $y$ are equal.
Proof
Let $A$ be arbitrary.
Let $d$ be the (standard) discrete metric on $A$.
Let $r \ge 1$ and $s \ge 1$.
Then from Open Ball in Standard Discrete Metric Space:
- $\forall x, y \in A: \map {B_r} x = \map {B_s} y = A$
whether $x = y$ or not.
$\blacksquare$
Sources
- 1975: W.A. Sutherland: Introduction to Metric and Topological Spaces ... (previous) ... (next): $2$: Continuity generalized: metric spaces: Exercise $2.6: 26$