Let $p$ be a prime and $(Z, +)$ the additive group of intergers. Consider the filter
$$V_p = \{U \subset \mathbb{Z} \mid \ \text{exists} \ n \in \mathbb{N} \ \text{such that}\ p^n\mathbb{Z} \subset U\}$$.
and the toplogy
$$\tau \ = \ \{V \subset \mathbb{Z} \mid \ \text{for all} \ v \in V, \ \text{exists} \ U \in V_p \ \text{such} \ v + U \subset V\}.$$
We say that $(\mathbb{Z}, +, \tau)$ is a topological group. This topology is called p-adic topology. By the Lemma 3.4.2 of D. Dikranjan (Introduction to Topological groups), we have that $\overline{\{0\}}$ is a closed normal subgroup of $\mathbb{Z}$. My question is: Who is $\overline{\{0\}}$? I'm trying to prove that: $\overline{\{0\}} = \cap \ p^n\mathbb{Z}$, unsuccessfully. Below is what I've been trying so far. If anyone knows an easier way to prove it, I would be grateful.
We say that every subgroup of $\mathbb{Z}$ is of the form $m\mathbb{Z}$, where $z \in \mathbb{Z}$. We want to prove that the only closed subgroups of $\mathbb{Z}$ are $p^n\mathbb{Z}$, with $n \in \mathbb{N}$. Let $m \in \mathbb{N}$.
If $m = p^n$ for some $n \in \mathbb{N}$, there's nothing to prove, because $p^n\mathbb{Z}$ is open by definition and every open subgroup is also closed.
Suppose $m \ne p^n$ for all $n \in \mathbb{N}$. Then we have that
$$\mathbb{Z} - m\mathbb{Z} = \{z' \in \mathbb{Z} \mid m \not\mid z'\}.$$
If it were true that $\mathbb{Z} - m\mathbb{Z}$ is open, then for every $z' \in \mathbb{Z} - m\mathbb{Z}$, there exists $U \in V_p$ such that
$$z' + U \subset \mathbb{Z} - m\mathbb{Z}.$$
Take $p^nz \in \mathbb{Z} - m\mathbb{Z}$. Exists $p^s\mathbb{Z} \subset U \in V_p$ such that $p^nz + U \subset \mathbb{Z} - m\mathbb{Z}$. If $n > z$ is absurdity, because
$$p^nz + p^s\mathbb{Z} \ = \ p^s\mathbb{Z} \subset p^nz + U \subset \mathbb{Z} - m\mathbb{Z},$$
but $0 \in p^s\mathbb{Z}$ and $0 \not\in \mathbb{Z} - m\mathbb{Z}$. The problem starts when n < s.