Three Points in Ultrametric Space have Two Equal Distances/Corollary 2

Theorem

Let $\struct {R, \norm {\,\cdot\,} }$ be a normed division ring with non-Archimedean norm $\norm{\,\cdot\,}$,

Let $x, y \in R$ and $\norm x \ne \norm y$.

Then:

$\norm {x + y} = \norm {x - y} = \norm {y - x} = \max \set {\norm x, \norm y}$

Proof

Let $d$ be the metric induced by the norm $\norm {\,\cdot\,}$.

By Non-Archimedean Norm iff Non-Archimedean Metric then $d$ is a non-Archimedean metric and $\struct {R, d}$ is an ultrametric space.

Let $x, y \in R$ and $\norm x \ne \norm y$.

By the definition of the non-Archimedean metric $d$ then:

$\norm x = \norm {x - 0} = \map d {x, 0}$

and similarly:

$\norm y = \map d {y, 0}$

By assumption then:

$\map d {x, 0} \ne \map d {y, 0}$

By Three Points in Ultrametric Space have Two Equal Distances then:

$\norm {x - y} = \map d {x, y} = \max \set {\map d {x, 0}, \map d {y, 0} } = \max \set {\norm x, \norm y}$

By Norm of Negative then:

$\norm {y - x} = \norm {x - y} = \max \set {\norm x, \norm y}$

By the definition of the non-Archimedean metric $d$ then:

$\norm y = \norm {0 - \paren {-y} } = \map d {0, -y} = \map d {-y, 0}$

By assumption then:

$\map d {x, 0} \ne \map d {-y, 0}$

By Three Points in Ultrametric Space have Two Equal Distances then:

$\map d {x, -y} = \max \set {\map d {x, 0}, \map d {-y, 0} } = \max \set {\norm x, \norm y}$

By the definition of the non-Archimedean metric $d$ then:

$\map d {x, -y} = \norm {x - \paren {-y} } = \norm {x + y}$

So $\norm {x + y} = \max \set {\norm x, \norm y}$.

The result follows.

$\blacksquare$

Sources

• 1997: Fernando Q. Gouvea: p-adic Numbers: An Introduction ... (previous) ... (next): $\S 2.3$: Topology: Proposition $2.3.3$