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

Then:

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

Let $x \in \map {B^-_\epsilon} a$.

By definition of a closed balls in $p$-adic numbers:
 * $\norm {x - a}_p \le \epsilon$

From P-adic Norm of p-adic Number is Power of p:
 * $\exists m \in \Z: \norm {x - a}_p = p^{-m}$

Hence $p^{-m} \le \epsilon$.

Aiming for a contradiction, suppose $m < n$.

Hence $m \le n - 1$.

From Power Function on Integer between Zero and One is Strictly Decreasing:
 * $p^{-n + 1} \le p^{-m}$.

By choice of $n$:
 * $\epsilon < p^{-n + 1} \le p^{-m} = \norm {x - a}_p$

This contradicts the assumption $\norm {x - a}_p \le \epsilon$.

Hence $m \ge n$.

From Power Function on Integer between Zero and One is Strictly Decreasing:
 * $p^{-m} \le p^{-n}$.

By definition of a closed balls in $p$-adic numbers:
 * $x \in \map {B^{\, -}_{p^{-n} } } a$

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

By definition of set equality:
 * $\map {B^-_\epsilon} a = $\map {B^{\, -}_{p^{-n} } } a$