Open Balls whose Distance between Centers is Twice Radius are Disjoint/Proof 1

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.

$\blacksquare$