Question: Let $G$ be a totally disconnected compact Hausdorff topological group, and $V_\alpha$ be a neighborhood basis of the identity. Fixed a closed subgroup $H$ of $G$, and for any $\alpha$ we know if $V_\alpha$ is contained in $H$, can we determine $H$ uniquely?
Thanks for any answers !
No, certainly not. Indeed, if $H$ is not open, then it will not contain any $V_\alpha$ at all, and there are typically many different closed subgroups that are not open. For instance, if $F$ is a nontrivial finite group and $G=F^S$ for some set $S$, then $H=F^T$ is a closed subgroup of $G$ for any $T\subseteq S$, and is not open unless $T$ is cofinite in $S$.
Even if you restrict to open subgroups, it is not true. For instance, if $G$ is a finite group, then just $\{1\}$ alone is a neighborhood base, and every subgroup is open but contains the same sets from the base (namely, every subgroup contains $\{1\}$).