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

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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.

$\blacksquare$


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