Prove that $ \mathbb{Z}_p \subset \mathbb{Q}_p $ is a compact subgroup and show that this is the maximal compact subring.
Hint: use that every sequence of integers had a Cauchy sub-sequence with respect to the $p$-adic norm and that if $R \subset \mathbb{Q}_p$ is a compact subring then $ \left| x \right| \leq 1 $ for all $x \in R$.
Definitions: $\mathbb{Q}_p$ is the completion of $\mathbb{Q}$ with respect to the $p$-adic norm. Morover we let $\mathbb{Z}_p :=\{ x \in \mathbb{Q}_p : \left| x \right| \leq 1 \}$
My idea:: We clearly have by definition that $ \mathbb{Z}_p$ is a subring of $\mathbb{Q}_p$, since if $x,y \in \mathbb{Z}_p$ then we have $\left| x y \right|_p \leq 1 $ and $\left| x+y\right|_p \leq 1$ using the ultrametric property. In particular is a subgroup. Moreover by the hint we have that any subring is contained in $\mathbb{Z}_p$ hence is maximal. We are left to prove the compactness part. We can see $\mathbb{Z}_p$ as a metric space, with the metric induced by the $p$-adic norm, then I want to use the fact that in a metric space sequentially compact and compactness are equivalent notions. Then I want to prove that $\mathbb{Z}_p$ is complete (but don't know how), i.e. every Cauchy sequence converges, and I think that this is sufficient to prove that it is sequentially compact since every sequence of integers had a Cauchy subsequence which converge to some $\mathbb{Z}_p$, and I think that if it is true for integers then it is also true that any sequence of $p$-adic integers possesses a Cauchy subsequence as well. Anyway I found a little bit strange how it is stated the exercise.