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Let $G$ be a divisible, locally compact abelian group and $L_{1}\supseteq L_{2}\supseteq L_{3}\supseteq ...$ be a sequence of compactly generated, open subgroups of $G$. Can we deduce that there exist $n$ such that $L_{i}=L_{n}$ for $i\geq n$?

Aliakbar
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Not without additional assumptions.

Let $G=\mathbb{Q}$ with the discrete topology, or $(\mathbb{R}, +)$ with the usual topology. Let $L_n=2^n\mathbb{Z}$.

As far as what additional assumptions might make this true, I can't think of any off the top of my head. Perhaps if you gave some motivation I might be able to steer you in the right direction.

Owen Sizemore
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  • Additional assumptions? say what? – Aliakbar Aug 06 '13 at 07:29
  • Not sure, maybe if you explain why you need this that will give ideas about what kind of conditions might be helpful. One might be if $G$ is compact and (connected?). But I think then the $L_n$ will have to be compact and discrete, so finite. But that might be too restrictive. – Owen Sizemore Aug 06 '13 at 07:46