Open Balls whose Distance between Centers is Twice Radius are Disjoint

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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 1

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.

$\blacksquare$


Proof 2

Let $z \in \map {B_r} x$.

Then:

\(\ds \map d {z, y}\) \(\ge\) \(\ds \size{ \map d {z, x} - \map d {y, x} }\) Reverse Triangle Inequality
\(\ds \) \(\ge\) \(\ds \map d {y, x} - \map d {z, x}\) Definition of Absolute Value
\(\ds \) \(=\) \(\ds \map d {x, y} - \map d {z, x}\) Metric Space Axiom $(\text M 2)$: Triangle Inequality
\(\ds \) \(=\) \(\ds 2 r - \map d {z, x}\) by hypothesis
\(\ds \) \(>\) \(\ds 2r -r\) as $z \in \map {B_r} x$
\(\ds \) \(=\) \(\ds r\)

That is:

$z \not \in \map {B_r} y$

$\blacksquare$


Sources