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I have currently read a proof (existence of sections for pro-finite groups (in the book profinite groups of Ribes)) and I did not understand the following two facts used (without mentioning any details):

1) We are given a pro-finite group $G$ (that is, a compact, Hausdorff and totally disconnected topological group)and two closed subroups $K\leq H$. We first assume that the quotient $H/K$ is finite. Then there exists an open subgroup $U$ of $G$ such that $U\cap H\subset K$. I do not understand how to find such an open subgroup $U$.

2) Next we construct a closed subgroup $T$ of $G$ such that $K\leq T\leq H$ and we assume that $T\neq K$. Then there exists an open subgroup $U$ of $G$ such that with $K\subseteq (U\cap T)$. Again I do not understand how to find such an open subgroup $U$.

Any help is much appreciated. Thanks!

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(1) follows from the fact that any profinite group has a fundamental system of neighbourhoods of the identity made of (normal) open subgroups.
Thus, since $H$ is closed (hence compact), it is a profinite group, too (check!).
$H\setminus K$ is closed because it is a union of finite cosets of $K$ (in $H$), hence $K$ is open in $H$. So we can write $K=H\cap V$ for some open set $V\subseteq G$.
Finally for the fact stated at the beginning there exists an open subgroup $U\subseteq V$, so that $H\cap U\subseteq K$, as wanted.
(2): as you state it, it's trivial: just take $U=G$. If you wanted the reverse inclusion, proceed exactly as in (1).

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