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

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## Theorem

Let $p$ be a prime number.

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

Then:

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

## Proof

$\exists n_1 \in \N: \forall k \ge n_1 : p^{-k} < \epsilon$

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:

$p^{-n_1} < \epsilon < p^{n_2}$
$-n_2 < n_1$.

Let:

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

By choice of $n$:

$p^{-n} \le \epsilon < p^{n_2}$
$n > -n_2$

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:

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

So:

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

which is a contradiction.

Hence:

$\epsilon < p^{-\paren {n - 1} }$

The result follows.

$\blacksquare$