Show that if $H$ is a closed subgroup of a profinte group $G$, then $H$ is the inverse limit of the open subgroups of $G$ containing $H$.
I was able to show that $H$ is profinite. I use a constructive proof, but in my approach I used subgroups $H_{i}$ of $G_{i}$, where $(G_{i},\varphi_{ij})$ is the inverse system with inverse limit $(G,\varphi_{i})$. Probabily this approach does not work.
So I would like some hints to solve the problem. My attempt was write $L = \varprojlim_{i \in I}U_{i}$ where $\{U_{i} \mid i \in I, U \supset H, U\text{ open in }G\}$ and show that $L \subset H$ and $H \subset L$. But I dont know how to proceed. Using this idea, I'm stuck in the first point: what are $\varphi_{ij}: U_{j} \to U_{i}$ that make $(U_{i},\varphi_{ij})$ an inverse system? I dont know if this is the best approach.
PS: I can to write a sketch of my proof for "$H$ is profinite" if necessary.
The set $\mathcal U = \{ U_i \}$ of all open subgroups containing $H$ is partially ordered by inclusion: $U_i \le U_j$ iff $U_j \subset U_i$. This makes $(\mathcal U, \le)$ a directed set. For $U_i \le U_j$ we define $\phi_{ij} : U_j \to U_i$ to be the inclusion map. The inverse limit of this system is nothing else than $\bigcap_i U_i$.