I'm studying $p$-adic integers and in the proof of the fact that closed subgroups of the additive group $\Bbb{Z}_p$ are ideals (see Robert's "A course in $p$-adic analysis", pp.23) I've found the following inclusion for $x\in \Bbb{Z}_p$: $$\Bbb{Z}_px\subseteq\overline{\Bbb{Z}x}$$ where the overline denotes the closure of $\Bbb{Z}x$ in the $p$-adic integers.
What I know is that $$\Bbb{Z}_p=\{\sum_{i\ge0}a_ip^i\mid a_i\in \Bbb{Z}\}$$ and the topology is the one generated by the $p$-adic metric $d(x,y)=\frac{1}{p^{v(x-y)}}$, where $v(x-y)=\inf\{i\ge0\mid x_i\ne y_i\}$ is the $p$-adic order.
Can you help to figure out why this holds? Maybe it depends on the fact that It's difficult for me to understand what the closure looks like.
Thank in advance for your help.
Here's the idea.
Let $a = \sum_{i\in\Bbb{N}} a_i p^i$ be a $p$-adic integer. Let $s_n = \sum_{i=0}^n a_ip^i$.
Observe that $s_n\in \Bbb{Z}$, and $(s_n)_{n\in\Bbb{N}}\to a$ in the $p$-adic topology, since $$a-s_n = \sum_{i=n+1}^\infty a_ip^i.$$
Then if $ax\in \Bbb{Z}_px$, $(s_nx)_{n\in\Bbb{N}}\to ax$, since multiplication is continuous. Thus $ax$ lies in the closure of $\Bbb{Z}x$. However $ax$ was an arbitrary element of $\Bbb{Z}_px$. Thus $\Bbb{Z}_px\subseteq \overline{\Bbb{Z}x}$.
Side note
If you don't yet know why multiplication is continuous, it's because if the first $n$ terms of $x_1$ and $x_2$ and $y_1$ and $y_2$ agree (thinking of them as power series), then the first $n$ terms of $x_1y_1$ and $x_2y_2$ agree. (Proof. Take both expressions mod $p^n$.)