Equality of Open Balls does not imply Equality of Radii

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 radii $r$ and $s$ are equal.

Proof

Let $A$ be arbitrary.

Let $d$ be the (standard) discrete metric on $A$.

Let $r \ge 1$ and $s \ge 1$ such that $r \ne s$.

Then from Open Ball in Standard Discrete Metric Space:

$\forall x, y \in A: \map {B_r} x = \map {B_s} y = A$

$\blacksquare$

Sources

• 1975: W.A. Sutherland: Introduction to Metric and Topological Spaces ... (previous) ... (next): $2$: Continuity generalized: metric spaces: Exercise $2.6: 26$