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 subspaces.

Proof
We begin by proving the 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 $\norm {\,\cdot\,}_p$ on $\map {B^-_{p^{-n}}} a$.

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 $x_0, \dots, x_k \in \map {B^-_{p^{-n}}} a$ such that:
 * $\displaystyle \inf_{0 \mathop \le i \mathop \le k} \norm {x_i - x}_p \le \epsilon$
 * for all $x \in \map {B^-_{p^{-n}}} a$.

From Sequence of Powers of Reciprocals is Null Sequence:
 * $\exists N \in \N: \forall k \gt N: p^{-k} \le \epsilon$.

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

Then $m > n$.

From Closed Ball is Disjoint Union of Smaller Closed Balls in P-adic Numbers:
 * $\map {B^{\,-}_{p^{-n}}} a = \displaystyle \bigcup_{i = 0}^{p^\paren{m-n}-1} \map {B^{\,-}_{p^{-m}}} {a + i p^n}$

Let $x \in \map {B^{\,-}_{p^{-n}}} a$.

Let $k = p^\paren{m-n}-1$.

Then there exists $0 \le j \le k$ such that:
 * $x \in \map {B^{\,-}_{p^{-m}}} {a + j p^n}$

That is:
 * $\norm{x - a + j p^n}_p \le p^{-m} \le \epsilon$

Hence:
 * $\displaystyle \inf_{0 \mathop \le i \mathop \le k} \norm {a + i p^n - x}_p \le \norm{x - a + j p^n}_p \le \epsilon$

Since $x \in \map {B^{\,-}_{p^{-n}}} a$ was arbitrary, then:
 * $\forall x \in \map {B^{\,-}_{p^{-n}}} a : \displaystyle \inf_{0 \mathop \le i \mathop \le k} \norm {a + i p^n - x}_p \le \epsilon$

The result follows.