Is the set of integers with respect to the p-adic metric compact? Given the integers and a prime $p$.  I thought I had successfully shown that $\mathbb{Z}$ was compact with respect to the metric $|\cdot |_p$, by showing that the open ball centered at zero contained all integers with more than a certain number of factors of $p$, and then showing that the remaining integers took on a finite number of possible p-adic absolute values and thus fell into a finite number of balls.
Now if the integers are compact with respect to $|\cdot |_p$, then that means they are complete with respect to $|\cdot |_p$.  
But then I read that the p-adic integers $\mathbb{Z}_p$ are defined to be the completion of the integers with respect to $|\cdot |_p$, and include in their completion all the rational numbers with p-adic absolute value less than or equal to one.  So this means that the integers with respect to the p-adic metric are not complete, and thus not compact, and hence there must be something wrong with my proof, correct?
Edit: Ok upon typing this up I realized that my proof is most likely wrong as there's no reason to conclude that two elements with the same absolute value are necessarily in the same ball.

You don't prove compactness "by showing that the open ball centered at zero contained all integers with ..., and then showing that the remaining integers ... fell into a finite number of balls", i.e. by showing that there is a finite number of open balls covering the space.
Actually you can show more : $\Bbb Z$ with the $p$-adic metric is a bounded metric space : every integer is at a distance less than $1$ from $0$.
Instead, to prove compactness you have to show that for any covering of $\Bbb Z$ by open balls, you can select a finite number of those open balls and still cover $\Bbb Z$. For example, pick the covering of $\Bbb Z$ by placing an open ball on $n$ with radius $p^{-|n|}$. For most $p$ ($p \ge 5$) , you can't extract a finite cover of $\Bbb Z$ from this cover.
