# Characterization of Open Ball in P-adic Numbers

## 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.

For any $\epsilon \in \R_{>0}$ and $a \in \Q_p$ let $\map {B_\epsilon} a$ denote the closed ball of center $a$ of radius $\epsilon$.

Let $n \in Z$.

Let $x, y \in \Q_p$.

The following are equivalent::

$(1)\quad x \in \map {B_{p^{-n}}} y$
$(2)\quad \norm{x -y}_p < p^{-n}$
$(3)\quad \map {B_{p^{-n}}} x = \map {B_{p^{-n}}} y$
$(4)\quad x - y \in p^{n+1} \Z_p$
$(5)\quad x + p^{n+1} \Z_p = y + p^{n+1} \Z_p$

## Proof

By definition of the $p$-adic numbers, $\norm {\,\cdot\,}_p$ is a non-Archimedean norm.

### Condition $(1)$ iff Condition $(2)$

This follows directly from the definition of a open ball in the $p$-adic numbers.

$\Box$

### Condition $(1)$ iff Condition $(3)$

By definition, $\map {B_{p^{-n}}} y$ is an open ball in a non-Archimedean norm $\norm {\,\cdot\,}_p$.

$x \in \map {B_{p^{-n}}} y \leadsto \map {B_{p^{-n}}} x = \map {B_{p^{-n}}} y$
$\map {B_{p^{-n}}} x = \map {B_{p^{-n}}} y \leadsto x \in \map {B_{p^{-n}}} x = \map {B_{p^{-n}}} y$

$\Box$

### Condition $(2)$ iff Condition $(4)$

 $\displaystyle \norm{x - y}_p < p^{-n}$ $\leadstoandfrom$ $\displaystyle \norm{x - y}_p \le p^{-\paren{n+1} }$ $p$-adic norm of $p$-adic number is power of $p$ $\displaystyle$ $\leadstoandfrom$ $\displaystyle \norm{x - y}_p \le \norm{p^{n+1} }_p$ Definition of $p$-adic norm on integers $\displaystyle$ $\leadstoandfrom$ $\displaystyle \dfrac {\norm{x - y}_p} {\norm{p^{-\paren{n+1} } }_p} \le 1$ Dividing both sides of equation by $\norm{p^{-\paren{n+1} } }$ $\displaystyle$ $\leadstoandfrom$ $\displaystyle \norm{p^{-\paren{n+1} } \paren{x-y} }_p \le 1$ Norm of Quotient in Division Ring $\displaystyle$ $\leadstoandfrom$ $\displaystyle p^{-\paren{n+1} } \paren{x-y} \in \Z_p$ Definition of $p$-adic integers $\displaystyle$ $\leadstoandfrom$ $\displaystyle x-y \in p^{n+1}\Z_p$

$\Box$

### Condition $(3)$ iff Condition $(5)$

$\map {B_{p^{-n}}} x = x + p^{n+1} \Z_p$

and

$\map {B_{p^{-n}}} y = y + p^{n+1} \Z_p$

Hence:

$\map {B_{p^{-n}}} x = \map {B_{p^{-n}}} y$ if and only if $x + p^{n+1} \Z_p = y + p^{n+1} \Z_p$

$\blacksquare$