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

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## Theorem

Let $p$ be a prime number.

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

Then:

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

## Proof

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

Suppose $\epsilon \neq p^{-m}$.

Then:

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

and the theorem is proved with $n = m - 1$.

Now suppose $\epsilon = p^{-m}$.

$p^{-\paren{m + 1}} < p^{-m}$

So:

$p^{-\paren{m + 1}} < \epsilon \le p^{-m}$

and the theorem is proved with $n = m$.

$\blacksquare$