# Closed Balls of P-adic Number

## 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 $\epsilon$.

Then:

$\forall n \in Z : \map {B^{\,-}_{p^{-n} } } a = a + p^n \Z_p$

## Proof

Let $n \in \Z$.

Then:

 $\displaystyle x \in \map {B^{\,-}_{p^{-n} } } a$ $\leadstoandfrom$ $\displaystyle \norm {x - a}_p \le p^{-n}$ Definition of Closed Ball of Normed Division Ring $\displaystyle$ $\leadstoandfrom$ $\displaystyle p^n \norm {x - a}_p \le 1$ $\displaystyle$ $\leadstoandfrom$ $\displaystyle \norm {p^{-n} }_p \norm {x - a}_p \le 1$ Definition of $p$-adic norm $\displaystyle$ $\leadstoandfrom$ $\displaystyle \norm {p^{-n} \paren {x - a} }_p \le 1$ Norm axiom (N2) : (Mulitplicativity) $\displaystyle$ $\leadstoandfrom$ $\displaystyle p^{-n} \paren {x - a} \in \Z_p$ Definition of $p$-adic integers $\displaystyle$ $\leadstoandfrom$ $\displaystyle x - a \in p^n \Z_p$ $\displaystyle$ $\leadstoandfrom$ $\displaystyle x \in a + p^n \Z_p$

From set equality:

$\map {B^{\,-}_{p^{-n} } } a = a + p^n \Z_p$

The result follows.

$\blacksquare$