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

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

Then:

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

## Proof

#### Lemma

$\forall y \in \Q_p: \norm y_p \le p^{-n}$ if and only if there exists $i \in \Z$ such that:
$(1)\quad 0 \le i \le p^\paren {m - n} - 1$
$(2)\quad \norm {y - i p^n}_p \le p^{-m}$

$\Box$

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:

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

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:

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

It follows that:

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

$\blacksquare$