Closed Ball in Normed Division Ring is Closed Ball in Induced Metric

Theorem

Let $\struct{R, \norm {\,\cdot\,} }$ be a normed division ring.

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

Let $a \in R$.

Let $\epsilon \in \R_{>0}$ be a strictly positive real number.

Let $\map {{B_\epsilon}^-} {a; \norm {\,\cdot\,} }$ denote the closed ball in the normed division ring $\struct {R, \norm {\,\cdot\,} }$.

Let $\map {{B_\epsilon}^-} {a; d }$ denote the closed ball in the metric space $\struct {R, d}$.

Then:

$\map {{B_\epsilon}^-} {a; \norm {\,\cdot\,} }$ = $\map {{B_\epsilon}^-} {a; d }$

Proof

 $\displaystyle x \in \map { {B_\epsilon}^-} {a; \norm {\,\cdot\,} }$ $\leadstoandfrom$ $\displaystyle \norm {x - a} \le \epsilon$ Definition of closed ball in $\struct {R, \norm {\,\cdot\,} }$ $\displaystyle$ $\leadstoandfrom$ $\displaystyle \map d {x, a} \le \epsilon$ Definition of metric induced by $\norm {\,\cdot\,}$ $\displaystyle$ $\leadstoandfrom$ $\displaystyle x \in \map { {B_\epsilon}^-} {a; d }$ Definition of closed ball in $\struct {R, d}$

The result follows from Equality of Sets.

$\blacksquare$