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

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Theorem

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

For $a \in R$ and $\epsilon \in \R_{>0}$ let:

$\map { {B_\epsilon}^-} a$ denote the closed $\epsilon$-ball of $a$ in $\struct {R, \norm {\,\cdot\,} }$


Let $x, y \in R$.

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


Then:

$y \in \map { {B_r}^-} x \implies \map { {B_r}^-} y = \map { {B_r}^-} x$


Proof

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


Let $a \in \map { {B_r}^-} y$.

By the definition of an closed ball, then:

$\norm {a - y} \le r$
$\norm {y - x} \le r$

Hence:

\(\ds \norm {a - x}\) \(=\) \(\ds \norm {a - y + y - x}\)
\(\ds \) \(\le\) \(\ds \max \set {\norm {a - y}, \norm {y - x} }\) Definition of Non-Archimedean Division Ring Norm
\(\ds \) \(\le\) \(\ds r\)


By the definition of a closed ball, then:

$a \in \map { {B_r}^-} x$.

Hence:

$\map { {B_r}^-} y \subseteq \map { {B_r}^-} x$


By Norm of Negative then:

$\norm {x - y} \le r$

By the definition of a closed ball, then:

$x \in \map { {B_r}^-} y$

Similarly it follows that:

$\map { {B_r}^-} x \subseteq \map { {B_r}^-} y$


By set equality:

$\map { {B_r}^-} x = \map { {B_r}^-} y$

$\blacksquare$


Sources