Open Balls whose Distance between Centers is Twice Radius are Disjoint/Proof 1
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Theorem
Let $M = \struct {A, d}$ be a metric space.
Let $x, y \in A$ such that $\map d {x, y} = 2 r > 0$.
Let $\map {B_r} x$ denote the open $r$-ball of $x$ in $M$.
Then $\map {B_r} x$ and $\map {B_r} y$ are disjoint.
Proof
Aiming for a contradiction, suppose $\map {B_r} x \cap \map {B_r} y \ne \O$.
Then:
\(\ds \exists z \in A: \, \) | \(\ds z \in \map {B_r} x\) | \(\text { and }\) | \(\ds z \in \map {B_r} y\) | Definition of Set Intersection | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds \map d {x, z} < r\) | \(\text { and }\) | \(\ds \map d {z, y} < r\) | Definition of Open Ball of Metric Space | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds \map d {x, z} + \map d {z, y}\) | \(<\) | \(\ds 2 r\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds \map d {x, y}\) | \(<\) | \(\ds 2 r\) | Metric Space Axiom $(\text M 2)$: Triangle Inequality: $\map d {x, y} \le \map d {x, z} + \map d {z, y}$ |
But this contradicts our initial assertion that $\map d {x, y} = 2 r$.
The result follows by Proof by Contradiction.
$\blacksquare$