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 $0 \le i \le p-1$.

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

By definition of an open ball:
 * $\norm{ x - a - i p^n} < p^{-n}$

From Lemma:
 * $\norm{ x - a}_p \le p^{-n}$.

By definition of an open ball:
 * $x \in \map {B^{\,-}_{p^{-n}}} a$

Since $x$ was arbitrary:
 * $\map {B_{p^{-n}}} {a + i p^n} \subseteq \map {B^{\,-}_{p^{-n}}} a$

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

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

$\leadsto x \in a + p^n \Z_p$

$\leadsto p^{-n} \paren {x - a} \in \Z_p$

$Z$ is dense in $\Z_p$

$\leadsto \exists m \in \Z : p^{-n} \paren {x - a} - m \in p \Z_p$

$\leadsto \exists 0 \le i \le p-1 : p \divides m - i$

$\leadsto \exists 0 \le i \le p-1 : m - i \in p \Z_p$

$p \Z_p$ is a ring.

$\leadsto \exists 0 \le i \le p-1 : p^{-n} \paren {x - a} - m + m - i \in p \Z_p$

$\leadsto \exists 0 \le i \le p-1 : p^{-n} \paren {x - a} - i \in p \Z_p$

$\leadsto \exists 0 \le i \le p-1 : x \in a + ip^n + p^{n+1}\Z_p$

$\leadsto \exists 0 \le i \le p-1 : x \in \map {B_{p^{-n}}} {a + i p^n}$

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

$\leadsto \forall n \in Z : \map {B^{\,-}_{p^{-n}}} a = \displaystyle \bigcup_{i = 0}^{p-1} \map {B_{p^{-n}}} {a + i p^n}$

Condition $(2)$
Let $0 \le i,j \le p-1$.

$ x \in \map {B_{p^{-n}}} {a + i p^n} \cap \map {B_{p^{-n}}} {a + j p^n}$

$\leadsto p^{-n} \paren {x -a} - i, p^{-n} \paren {x -a} - j \in p \Z_p$

$\Z_p$ is a ring

$\leadsto p^{-n} \paren {x -a} - i, p^{-n} \paren {x -a} + j = j - i \in p \Z_p$

$\leadsto p \divides \paren{j - i}$

$\leadsto j \equiv i \mod p$

$\leadsto j = i$

$\map {B_{p^{-n}}} {a + i p^n} = \map {B_{p^{-n}}} {a + j p^n}$

Also see

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