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:

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

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:

$\map {S_\epsilon} a = \map { {B_\epsilon}^-} a \setminus \map {B_\epsilon} a$

The result follows directly from:

$\blacksquare$