Zero Dim Topological group I have this assertion which looks rather easy (or as always I am missing something):
We have $G$ topological group which is zero dimensional, i.e it admits a basis for a topology which consists of clopen sets, then every open nbhd that contains the identity element of G also contains a clopen subgroup.
I naively thought that if I take $\{e\}$, i.e the trivial subgroup, it's obviously closed, so it's also open in this topology, i.e clopen, and it's contained in every nbhd that contains $e$, but isn't it then too trivial.
Missing something right?
:-)
 A: This is certainly false. Take $G=\mathbb{Q}$ with the standard topology and additive group structure. The topology is zero-dimensional since intersecting with $\mathbb{Q}$ the countably many open intervals whose endpoints are rational translates of $\sqrt 2$ gives a clopen basis for the topology on $G$. The trivial subgroup $\{0\}$ is certainly not open since the standard topology on $\mathbb{Q}$ is not discrete. So any clopen subgroup $H \subset G$ contains a nonzero rational number $q$ and so also $2q,3q,4q$ and so on. This shows $H$ is unbounded so, for example, the open neighbourhood $(-1,1) \cap \mathbb{Q}$ of $0 \in G$ contains no nontrival subgroup.
A: First of all, in a general topological space, a single point need not be closed.  But even if you are assuming that points are closed (e.g., this is a Hausdorff space), it does not necessarily follow from the zero-dimensional assumption that $\{e\}$ is open:
$\{e\}$ closed $\implies$ complement of $\{e\}$ open $\implies$ complement of $\{e\}$ is the union of clopen sets
But you cannot conclude from here (as you may want to) that the complement of $\{e\}$ is closed, because there may be infinitely many clopen sets involved in the union, and the infinite union of closed sets need not be closed.
