# Closed Ball is Disjoint Union of Smaller Closed Balls in P-adic Numbers/Disjoint 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 open $\epsilon$-ball of $a$.

Then:

- $\forall n \in Z : \set{\map {B^{\,-}_{p^{-m}}} {a + i p^n} : i = 0, \dots, p^\paren{m-n}-1}$ is a set of pairwise disjoint open balls

## Proof

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

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

From Characterization of Open Ball in P-adic Numbers:

- $\norm{\paren {x -a} - ip^n}_p \le p^{-m}$

and

- $\norm{\paren {x -a} - jp^n}_p \le p^{-m}$

Then:

\(\displaystyle \norm{ip^n - jp^n}_p\) | \(\le\) | \(\displaystyle p^{-m}\) | Corollary to P-adic Metric on P-adic Numbers is Non-Archimedean Metric | ||||||||||

\(\, \displaystyle \leadsto \, \) | \(\displaystyle \norm{p^n}_p \norm{i - j}_p\) | \(\le\) | \(\displaystyle p^{-m}\) | Norm axiom (N2) : (Mulitplicativity) | |||||||||

\(\, \displaystyle \leadsto \, \) | \(\displaystyle p^{-n} \norm{i - j}_p\) | \(\le\) | \(\displaystyle p^{-m}\) | Definition of $p$-adic norm | |||||||||

\(\, \displaystyle \leadsto \, \) | \(\displaystyle \norm{i - j}_p\) | \(\le\) | \(\displaystyle p^{n-m}\) | Multiplying both sides by $p^n$. | |||||||||

\(\, \displaystyle \leadsto \, \) | \(\displaystyle p^\paren{m-n}\) | \(\divides\) | \(\displaystyle \paren{i - j}\) | Definition of $p$-adic norm | |||||||||

\(\, \displaystyle \leadsto \, \) | \(\displaystyle j\) | \(\equiv\) | \(\displaystyle i \mod p^\paren{m-n}\) | Definition of congruence modulo p | |||||||||

\(\, \displaystyle \leadsto \, \) | \(\displaystyle i\) | \(=\) | \(\displaystyle j\) | Integer is Congruent to Unique Integer less than Modulus | |||||||||

\(\, \displaystyle \leadsto \, \) | \(\displaystyle \map {B^{\,-}_{p^{-m} } } {a + i p^n}\) | \(=\) | \(\displaystyle \map {B^{\,-}_{p^{-m} } } {a + j p^n}\) |

The result follows.

$\blacksquare$