Open Balls whose Distance between Centers is Twice Radius are Disjoint/Proof 2
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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
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:
- $z \not \in \map {B_r} y$
$\blacksquare$