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

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Theorem

Let $p$ be a prime number.

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

Let $a \in \Q_p$.

For all $\epsilon \in \R_{>0}$:

let $\map {{B_\epsilon}^-} a$ denote the closed $\epsilon$-ball of $a$.
let $\map {B_\epsilon} a$ denote the open $\epsilon$-ball of $a$.


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 $n \in \Z$.

From Closed Ball is Disjoint Union of Smaller Closed Balls in P-adic Numbers:

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

where $\set{\map {B^{\,-}_{p^{-\paren{n+1}}}} {a + i p^n} : i = 0, \dots, p-1}$ is a set of pairwise disjoint closed balls.

From Open Ball in P-adic Numbers is Closed Ball:

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

where $\set{\map {B_{p^{-n}}} {a + i p^n} : i = 0, \dots, p-1}$ is a set of pairwise disjoint Open balls.

The result follows.

$\blacksquare$

Also see