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:

\(\displaystyle x \in \map { B_{p^{-n} } } a\)

\(\leadstoandfrom\)

\(\displaystyle \norm {x - a}_p < p^{-n}\)

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:

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

$\blacksquare$