Closed Balls Centered on P-adic Number is Countable/Lemma

Theorem
Let $p$ be a prime number.

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

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

Proof
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:
 * $n = \min \set {k : -n_2 \mathop \le k \mathop \le n_1 \text{ and } p^{-k} \le \epsilon}$

By choice of $n$:
 * $p^{-n} \le \epsilon < p^{n_2}$

From Power Function on Integer between Zero and One is Strictly Decreasing:
 * $n > -n_2$

Also by the choice of $n$:
 * $n \le n_1$

So:
 * $n_1 > n - 1 \ge -n_2$

Aiming for a contradiction, suppose:
 * $p^{-\paren{n-1}} \le \epsilon$

Then:
 * $n - 1 \in \set {k : -n_2 \mathop \le k \mathop \le n_1 \text{ and } p^{-k} \le \epsilon}$

So:
 * $n = \min \set {k : -n_2 \mathop \le k \mathop \le n_1 \text{ and } p^{-k} \le \epsilon} \le n - 1$

which is a contradiction.

Hence:
 * $\epsilon < p^{-\paren{n - 1}} = p^{-n + 1}$

The result follows.