# Open Ball in P-adic Numbers is Closed Ball

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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 $a$
- 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:

\(\displaystyle x \in \map { B_{p^{-n} } } a\) | \(\leadstoandfrom\) | \(\displaystyle \norm {x - a}_p < p^{-n}\) | Definition of Open Ball of Normed Division Ring | ||||||||||

\(\displaystyle \) | \(\leadstoandfrom\) | \(\displaystyle \norm {x - a}_p \le p^{-\paren {n + 1} }\) | P-adic Norm of p-adic Number is Power of p | ||||||||||

\(\displaystyle \) | \(\leadstoandfrom\) | \(\displaystyle x \in \map { B^{\,-}_{p^{-\paren {n + 1} } } } a\) | Definition of Closed Ball of Normed Division Ring |

By set equality:

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

$\blacksquare$