Closed Ball is Disjoint Union of Open Balls in P-adic Numbers

Theorem
Let $p$ be a prime number.

Let $\struct {\Q_p, \norm {\,\cdot\,}_p}$ be the $p$-adic numbers.

Let $\Z_p$ be the $p$-adic integers.

Let $a \in \Q_p$.

For all $\epsilon \in \R_{>0}$:
 * let $\map {{B_\epsilon}^-} a$ denote the closed ball of $a$ of radius $\epsilon$.


 * let $\map {B_\epsilon} a$ denote the open ball of $a$ of radius $\epsilon$.

Then:
 * $(1) \quad\forall n \in Z : \map {B^{\,-}_{p^{-n}}} a = \displaystyle \bigcup_{i = 0}^{p-1} \map {B_{p^{-n}}} {a + i p^n}$
 * $(2) \quad\forall n \in Z : \set{\map {B_{p^{-n}}} {a + i p^n} : i = 0, \dots, p-1}$ is a set of pairwise disjoint open balls

Proof
Let $n \in \Z$.

Condition $(1)$
Let $0 < i \le p-1$.

Let $x \in \map {B_{p^{-n}}} {a + i p^n}$.

$\leadsto \norm{ x - a - i p^n} < p^{-n}$

$\leadsto \norm{ x - a - i p^n} \le \max {\norm{x-a}, \norm{i p^n}}$

$\norm{x-a} = \norm{ x - a - i p^n} \lor \norm{x-a} = \norm{i p^n}$

$\norm{x-a} = \norm{ x - a - i p^n} \le p^{-n}$

$\norm{x-a} = \norm{i p^n} = \norm{i} \norm{p^n} = 1 \cdot p^{-n}$

$\norm{x-a} \le p^{-n}$

$x \in \map {B^{\,-}_{p^{-n}}} a$

$\displaystyle \bigcup_{i = 0}^{p-1} \map {B_{p^{-n}}} {a + i p^n} \subseteq \map {B^{\,-}_{p^{-n}}} a$

Also see

 * Leigh.Samphier/Sandbox/Sphere is Disjoint Union of Open Balls in P-adic Numbers