Sphere is Set Difference of Closed Ball with Open Ball/Normed Division Ring
Theorem
Let $\struct{R, \norm {\,\cdot\,} }$ be a normed division ring.
Let $a \in R$.
Let $\epsilon \in \R_{>0}$ be a strictly positive real number.
Let $\map {{B_\epsilon}^-} {a; \norm {\,\cdot\,} }$ denote the $\epsilon$-closed ball of $a$ in $\struct {R, \norm {\,\cdot\,} }$.
Let $\map {B_\epsilon} {a; \norm {\,\cdot\,} }$ denote the $\epsilon$-open ball of $a$ in $\struct {R, \norm {\,\cdot\,} }$.
Let $\map {S_\epsilon} {a; \norm {\,\cdot\,} }$ denote the $\epsilon$-sphere of $a$ in $\struct {R, \norm {\,\cdot\,} }$.
Then:
- $\map {S_\epsilon} {a; \norm {\,\cdot\,} } = \map { {B_\epsilon}^-} {a; \norm {\,\cdot\,} } \setminus \map {B_\epsilon} {a; \norm {\,\cdot\,} }$
Corollary
Let $p$ be a prime number.
Let $\struct{\Q_p,\norm{\,\cdot\,}_p}$ be the $p$-adic numbers.
Let $a \in \Q_p$.
Let $\epsilon \in \R_{>0}$ be a strictly positive real number.
Let $\map {{B_\epsilon}^-} a$ denote the $\epsilon$-closed ball of $a$ in $\Q_p$.
Let $\map {B_\epsilon} a$ denote the $\epsilon$-open ball of $a$ in $\Q_p$.
Let $\map {S_\epsilon} a$ denote the $\epsilon$-sphere of $a$ in $\Q_p$.
Then:
- $\map {S_\epsilon} a = \map { {B_\epsilon}^-} a \setminus \map {B_\epsilon} a$
Proof
The result follows directly from:
- Closed Ball in Normed Division Ring is Closed Ball in Induced Metric
- Open Ball in Normed Division Ring is Open Ball in Induced Metric
- Sphere in Normed Division Ring is Sphere in Induced Metric
- Sphere is Set Difference of Closed Ball with Open Ball
$\blacksquare$