It is well-known that a connected topological group can be generated by any neighborhood of the identity. There are non-connected topological groups for which this is still true, such as $\mathbb{Q}$. My question is: is there any characterization of those non-connected topological groups for which the result is true? (I am specifically thinking about p-adics $\mathbb{Q}_p$, where $\mathbb{Q}$ is dense. I think the result is not true in this case, but I do not know how to prove/disprove it).
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$\begingroup$ The subgroup ${\mathbb Z}_p$ is open in ${\mathbb Q}_p$. $\endgroup$– Andreas ThomCommented May 26, 2017 at 6:58
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$\begingroup$ OK, I think I should have been more specific. Actually, what I would like to see is if I could generate the whole $\mathbb{Q}_p$ from the unit ball $B_1(0)$. This led me to think about the more general question I asked (still I am interested in knowing if such characterization exists), but I guess I should drop the ``any'' in the case of $\mathbb{Q}_p$. It is still possible to generate $\mathbb{Q}_p$ from the unit ball $B_1(0)$?. $\endgroup$– LandlordCommented May 26, 2017 at 7:29
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1$\begingroup$ It depends on the way you define the metric but for the usual metric the unit ball in $\mathbb{Q}_p$ is $\mathbb{Z}_p$. Any larger ball (ie containing at least one more element) will generate $\mathbb{Q}_p$ since it will also contain $p^{-1}$. $\endgroup$– Simon WadsleyCommented May 26, 2017 at 7:56
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$\begingroup$ Your comments about $\mathbb{Q}$ are somewhat confusing. You first say that it satisfies the property, which certainly means you endow it with the real topology. Then you say that it's dense in $\mathbb{Q}_p$, but this embedding is not continuous for the real topology on $\mathbb{Q}$. Endowed with the $p$-adic topology, obviously $\mathbb{Q}$ does not satisfy the property. $\endgroup$– YCorCommented May 26, 2017 at 8:33
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$\begingroup$ Immediate restatements: let $G$ be a topological group. Then $G$ is generated by any of its neighborhoods of 1 $\Leftrightarrow$ the only open subgroup of $G$ is $G$ $\Leftrightarrow$ $G$ has no nontrivial continuous action on any discrete set. $\endgroup$– YCorCommented May 26, 2017 at 8:34
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Immediate restatements: let $G$ be a topological group. Then $G$ is generated by any of its neighborhoods of 1 $\Leftrightarrow$ the only open subgroup of $G$ is $G$ $\Leftrightarrow$ $G$ has no nontrivial continuous action on any discrete set. All these properties hold if $G$ is connected and you're asking about when the converse holds.
For a locally compact group $G$, $G$ has no proper open subgroup iff $G$ is connected.
On the other hand, there exists an abelian Polish group $G\neq\{1\}$ that is totally disconnected (hence not connected) but has no proper open subgroup. Reference: T. Christine Stevens, Connectedness of complete metric groups. Colloquium Mathematicae 50(2) (1986) 233-240 (eudml link).
I guess that even in the realm of Polish groups, you can't expect more than trivial restatements, except if looking at very restricted subclasses.
