Let $G$ be a connected compact separable Hausdorff metric group, which is monothetic, i.e., has a dense subgroup generated by a single element. Such a group is necessarily Abelian.


**Question:**


Can the cardinality of the set of all closed subgroups of $G$ be uncountable?


Thank you.



The number of distinct closed subgroups of a compact monothetic group
Yes. Take $G=\prod\_{i<\omega}S^1$. It admits a metric which is compatible with the group action. An element $x=(e^{i\pi a\_0},e^{i\pi a\_1},\ldots,e^{i\pi a\_n},\ldots)$ generates a dense subgroup if the set $\{1,a\_0,a\_1,\ldots,a\_n,\ldots\}$ is $\mathbb{Q}$-linearly independent.


For every subset $I\subseteq\omega$ you have a closed subgroup $G=\prod\_{i\in I}S^1$.


