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Hello,

Let $G$ denote a locally compact group and $S(G)$ the chabauty space of $G$, that is the set of closed subgroups of $G$ equiped with the chabauty topology, it is a compact space.

My question is : if $G$ is a locally profinite group, for example $G=GL(n,F)$ where $F$ is a non-archimidian field, and $\mathcal{OK}(G)$ the set of open compact subgroup of $G$, it is true that $\mathcal{OK}(G)$ is closed in $S(G)$ ?

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I just notive that you already created twice a tag that is a number. I have a vague idea why this happens. Thus the follwing explanation: the number displayed with the tag in the list is not part of the tag. It only states the current number of questions already tagged with this tag. You can essentially ignore it. In any case, please do not include it (or create any tag that is a number except if there should be a clear mathematical meaning to it in the given context). –  quid Jan 3 '13 at 16:29
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1 Answer 1

If $G$ is nondiscrete, the answer is clearly no, since the trivial group is approximable by compact open subgroups.

If $G$ is discrete (this is not the kind of groups you had in mind), the answer is yes unless $G$ has an infinite locally finite subgroup.

The terminology "locally profinite" (not used by group theorists), which stands for "locally compact totally disconnected", is lame because for discrete groups it does not mean locally finite (which means every finite subset generates a finite subgroup).

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Thank you, but it is possible that $\mathcal{OK}(G)$ is open in $S(G)$ ? –  Rajkarov Jan 3 '13 at 15:33
    
That $OK(G)$ be open is not always true (e.g., if $G$ is an infinite product of nontrivial finite groups) but sounds plausible when $G$ is the groups of points of a reductive group over a local field (of characteristic zero maybe). But I don't know. The first question is to consider an open profinite subgroup $H$ in such a group and determine whether $H$ is isolated in $S(H)$. –  YCor Jan 3 '13 at 22:12
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