Topological Properties of Non-Archimedean Division Rings/Open Balls are Clopen

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

Let $x \in R$.

Let $r \in \R_{>0}$.

Let $\map {B_r} x$ denote the open $r$-ball of $x$ in $\struct {R, \norm {\,\cdot\,} }$

Then:
 * The open $r$-ball of $x$, $\map {B_r} x$, is both open and closed in the metric induced by $\norm {\,\cdot\,}$.

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

By the definition of an open ball in $\norm {\,\cdot\,}$:
 * $\map {B_r} x$ is an open ball in the metric space $\struct {R, d}$.

By Open Ball of Metric Space is Open Set then $\map {B_r} x$ is open in $\struct {R, d}$.

So it remains to show that $\map {B_r} x$ is closed in $\struct {R, d}$.

Let $\map \cl {\map {B_r} x}$ denote the closure of $\map {B_r} x$.

Let $y \in \map \cl {\map {B_r} x}$.

By the definition of the closure of $\map {B_r} x$ then:
 * $\forall s > 0: \map {B_s} y \cap \map {B_r} x \ne \O$

In particular:
 * $\map {B_r} y \cap \map {B_r} x \ne \O$

Let $z \in \map {B_r} y \cap \map {B_r} x$.

By Centers of Open Balls:
 * $\map {B_r} y = B_r \paren{z} = \map {B_r} x$

By the definition of an open ball:
 * $y \in \map {B_r} y = \map {B_r} x$.

Hence:
 * $\map \cl {\map {B_r} x} \subseteq \map {B_r} x$

By Subset of Metric Space is Subset of its Closure then:
 * $\map {B_r} x \subseteq \map \cl {\map {B_r} x}$

So by definition of set equality:
 * $\map \cl {\map {B_r} x} = \map {B_r} x$

By Set is Closed iff Equals Topological Closure then $\map {B_r} x$ is closed.