Open and Closed Balls in P-adic Numbers are Totally Bounded

Theorem
Let $p$ be a prime number.

Let $\struct {\Q_p, \norm {\,\cdot\,}_p}$ be the $p$-adic numbers.

Let $a \in \Q_p$.

Let $n \in \Z$.

Then the open ball $\map {B_{p^{-n}}} a$ and closed ball $\map {B^-_{p^{-n}}} a$ are totally bounded subspace.

Proof
We begin by proving theorem for the closed ball $\map {B^-_{p^{-n}}} a$.

From Open Ball in P-adic Numbers is Closed Ball then the theorem will be proved.

Let $d$ denote the subspace metric induced by the norm $\norm {\,\cdot\,}_p$.

That is, $d: \map {B^-_{p^{-n}}} a \times \map {B^-_{p^{-n}}} a \to \R_{\gt 0}$ is the metric defined by:
 * $\forall x, y \in \map {B^-_{p^{-n}}} a: \map d {x,y} = \norm{x - y}_p$

By the definition of totally bounded it needs to be shown that:
 * for every $\epsilon \in \R_{>0}$ there exist finitely many points $x_0, \dots, x_n \in \map {B^-_{p^{-n}}} a$ such that:
 * $\displaystyle \inf_{0 \mathop \le i \mathop \le n} \norm {x_i - x}_p \le \epsilon$
 * for all $x \in \map {B^-_{p^{-n}}} a$.

From :
 * $\exists N \in \N: \forall k \gt N: p^{-k} \le \epsilon$.

Let $m = \max \set {n + 1, N}$.