Closed Balls Centered on P-adic Number is Countable

Theorem
Let $p$ be a prime number.

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

Let $a \in \Q_p$.

Then the set of all closed balls centered on $a$ is the countable set:
 * $\displaystyle \mathcal B^{\, -} = \set{\map {B^{\, -}_{p^{-n}}} a : n \in \Z}$

Proof
Let $\epsilon \in \R_{\gt 0}$.

From Sequence of Powers of Reciprocals is Null Sequence:
 * $\exists n_1 \in \N: \forall k \ge n_1 : p^{-k} < \epsilon$

Similarly:
 * $\exists n_2 \in \N: \forall k \ge n_2 : p^{-k} < \dfrac 1 \epsilon$

Hence:
 * $p^{-n_1} < \epsilon$ and $ p^{-n_2} < \dfrac 1 \epsilon$

That is:
 * $p^{-n_1} < \epsilon < p^{n_2}$

From Power Function on Integer between Zero and One is Strictly Decreasing:
 * $-n_2 < n_1$.

Let:
 * $\displaystyle n = \min_{-n_2 \mathop \le k \mathop \le n_1} \set {k : p^{-k} \le \epsilon}$

Hence:
 * $\exists n \in Z: p^{-n} \le \epsilon < p^{-\paren{n+1}}$

By definition of closed balls:
 * $\map {B^{\, -}_{p^{-n}}} a \subseteq \map {B^-_\epsilon} a$