# Closed Ball is Disjoint Union of Smaller Closed Balls in P-adic Numbers/Union of Closed Balls

## 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 $n, m \in Z$, such that $n < m$.

Then:

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

## Proof

#### Lemma

$\forall y \in \Q_p: \norm{y}_p \le p^{-n}$ if and only if there exists $i \in \Z$:
$(1)\quad0 \le i \le p^\paren{m-n}-1$
$(2)\quad\norm{y - i p^n}_p \le p^{-m}$

Let $0 \le i \le p^\paren{m-n}-1$.

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

By definition of a closed ball:

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

From Lemma:

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

By definition of a closed ball:

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

Since $x$ was arbitrary:

$\map {B^{\,-}_{p^{-m}}} {a + i p^{-n}} \subseteq \map {B^{\,-}_{p^{-n}}} a$

Since $i$ was arbitrary:

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

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

By definition of a closed ball:

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

From Lemma:

$\exists i \in \N : 0 \le i \le p^\paren{m-n}-1 :\norm{ x - a - i p^{-n}} \le p^{-m}$

By definition of a closed ball:

$\exists i \in \N : 0 \le i \le p^\paren{m-n}-1 : x \in \map {B^{\,-}_{p^{-m}}} {a + i p^{-n}}$

Hence:

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

It follows that:

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

$\blacksquare$