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:
 * $\mathcal B^{\, -} = \set{\map {B^{\, -}_{p^{-n}}} a : n \in \Z}$

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

Lemma
From Leigh.Samphier/Sandbox/Closed Ball contains Smaller Closed Ball:
 * $\map {B^-_{p^{-n}}} a \subseteq \map {B^-_\epsilon} a$

From Leigh.Samphier/Sandbox/Open Ball contains Strictly Smaller Closed Ball:
 * $\map {B^-_\epsilon} a \subseteq \map {B_{p^{-\paren{n-1}}}} a$

From Open Ball in P-adic Numbers is Closed Ball
 * $\map {B_{p^{-\paren{n-1}}}} a = \map {B^-_{p^{-n}}} a $

Hence:
 * $\map {B^-_\epsilon} a \subseteq \map {B^-_{p^{-n}}} a $

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

Since $\epsilon \in \R_{\gt 0}$ was arbitrary then:
 * $\forall \epsilon \in \R_{\gt 0} : \exists n \in \Z : \map {B^-_\epsilon} a = \map {B^{\, -}_{p^{-n} } } a$

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

From Surjection from Countably Infinite Set iff Countable, it follows that $\mathcal B^{\, -}$ is a countable set.