Topological Properties of Non-Archimedean Division Rings/Centers of Closed Balls

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

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

Let $x, y \in R$.

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

Let ${B_r}^- \paren{x}$ be the closed $r$-ball of $x$ in $\struct {R,d}$

Let ${B_r}^- \paren{y}$ be the closed $r$-ball of $y$ in $\struct {R,d}$

Then:
 * If $y \in {B_r}^- \paren x$, then ${B_r}^- \paren x = {B_r}^- \paren y$

Proof
Let $y \in {B_r}^- \paren x$.

Let $a$ be any element of ${B_r}^- \paren x$.

By the definition of an open ball, then:
 * $\norm {y - x} \le r$
 * $\norm {a - x} \le r$

So:

By the definition of an open ball, then:
 * $a \in {B_r}^- \paren y$.

Since $a$ was arbitrary, then:
 * ${B_r}^- \paren x \subseteq {B_r}^- \paren y$

Now:

So $x \in {B_r}^- \paren y$ and the previous argument applies with the roles of $x$ and $y$ reversed.

Hence:
 * ${B_r}^- \paren y \subseteq {B_r}^- \paren x$

Finally, it has been shown:
 * ${B_r}^- \paren x \subseteq {B_r}^- \paren y$
 * ${B_r}^- \paren y \subseteq {B_r}^- \paren x$

So by definition of set equality:
 * ${B_r}^- \paren x = {B_r}^- \paren y$

which is what we needed to prove.