# Open Ball in P-adic Numbers is Closed Ball

## 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 $a$
Let $\map {B^-_\epsilon} a$ denote the closed $\epsilon$-ball of $a$.

Then:

$\forall n \in \Z : \map {B_{p^{-n} } } a = \map {B^-_{p^{-\paren {n + 1} } } } a$

## Proof

Let $n \in \Z$.

Then:

 $\ds x \in \map { B_{p^{-n} } } a$ $\leadstoandfrom$ $\ds \norm {x - a}_p < p^{-n}$ Definition of Open Ball in $p$-adic Numbers $\ds$ $\leadstoandfrom$ $\ds \norm {x - a}_p \le p^{-\paren {n + 1} }$ $p$-adic Norm of $p$-adic Number is Power of $p$ $\ds$ $\leadstoandfrom$ $\ds x \in \map {B^-_{p^{-\paren {n + 1} } } } a$ Definition of Closed Ball in $p$-adic Numbers

By set equality:

$\map {B_{p^{-n} } } a = \map {B^-_{p^{-\paren {n + 1} } } } a$

$\blacksquare$