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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? :-)

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In every topological group, an open subgroup is also closed. But if in a Hausdorff topological group every closed subgroup is open, then the subgroup $\{e\}$ is open and therefore all singleton sets are open and $G$ is discrete. Just because you have a base consisting of clopen sets doesn't mean that every closed set is open! So that explains your mistake...are you also asking for a correct proof? –  Pete L. Clark Nov 25 '11 at 19:47
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Regarding Mike's answer, you probably want your group to be locally compact as well as zero-dimensional. –  Matt E Nov 25 '11 at 21:57

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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.

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Ah, I thought that it meant that the topology is composed of clopen sets. Anyway, it doesn't say that $G$ is locally compact, but if there's no other way to prove it, I guess I would like to see how to show this with the added assumption of local compactness. Thanks. –  MathematicalPhysicist Nov 26 '11 at 7:32

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.

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Probably the group is supposed to be locally compact as well. (This is a pretty standard assumption in topological group theory.) –  Matt E Nov 25 '11 at 21:58

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