Let $U=\mathbb{Z}_p^{\ast}$ be the group of p-adic units. We put $U_n=1+p^n\mathbb{Z}_p$. $U_n$ form a decreasing sequence of open subgroups of $U$, and $U=\displaystyle{\lim_{\leftarrow}}~ U/U_n$
That $U_n\supseteq U_m$ for $m>n$ is clear. Choose $x\in U_m$, then $x\equiv 1\mod p^m$. Since $p^m>p^n$ we get $p^{m}=p^{n+k}=p^{n}p^{k}$. Hence $x\equiv 1\mod p^n$ and $x\in U_n$.
How can I see, that the $U_n$ are open in $U$? What is the topology? The only metric (in the p-adic context) I know is for $\mathbb{Q}_p$ with $d(x,y)=e^{-v_p(x-y)}$. Is it the same metric?
How can we see, that $U=\displaystyle{\lim_{\leftarrow}}~ U/U_n$.
Thanks in advance for your comments.
For $a\in\Bbb Z_p$ and $r>0$ let $D(a,r)=\{x\in\Bbb Z_p:|x-a|_p<r\}$ be the open disc of center $a$ and radious $r$ in $\Bbb Z_p$. For each $n\in\mathbb N$ we have $U_n=1+p^n\Bbb Z_p=D(1,r)$ for $|p^n|_p<r\leq|p^{n-1}|_p$, thus it's an open disc. The projections $U\to U/U_n$ induces a group homomorphism $$\varphi:U\to\lim_\leftarrow U/U_n$$ by universal property of projective limit. The kernel of this homomorphism is $\bigcap_n U_n=\{1\}$, hence it's injective.
For surjectivity, first note that $U=\Bbb Z_p\setminus p\Bbb Z_p$, that's a $p$-adic integer is invertible if and only if it's not divisible by $p$, thus $U$ is a closed subset of $\Bbb Z_p$, because $p\Bbb Z_p$ is an open disc. As closed subset of the complete space $\Bbb Z_p$, $U$ is also complete. Let $v_n\in\lim U/U_n,n\in\Bbb N$ be a sequence. Write $v_n=u_nU_n$ with $u_n\in U$. Then for $m\geq n$ we have $$|u_m-u_n|_p=\left|\frac{u_m}{u_n}-1\right|_p\leq e^{-n}$$ because $u_m/u_n\in U_n$ being $u_m\equiv u_n\pmod{U_n}$. Thus $u_n$ is a Cauchy sequence which converges to some $u\in U$ and $\varphi(u)=v$.
Alternatively, if you defined $\Bbb Z_p$ as $\lim\Bbb Z/p^n\Bbb Z$, then you get the isomorphism $U\cong\lim(\Bbb Z/p^n\Bbb Z)^\times$ because the forgetful functor (which takes a ring into the group of invertible elements) preserve limits. But $U_n$ is the kernel of surjective group homomorphism $U\to(\Bbb Z/p^n\Bbb Z)^\times$ induced by $\Bbb Z_p\to\Bbb Z/p^n\Bbb Z$, hence $U/U_n\cong\Bbb Z/p^n\Bbb Z$, hence again $U\cong\lim U/U_n$.