Let $K_{1}\supseteq K_{2}\supseteq K_{3}\supseteq \cdots$ be a descending sequence of compact subgroups of compact, torsion-free group $G$. Is $\bigcap_{r=1}^\infty n^r K\neq 0$? (for a positive integer $m$, $mK=\{mx;x\in K\}$)
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Where did the $K_i$ coem in? Did you mean $\displaystyle \bigcap_r n^r K_r$? – Alex Youcis Oct 07 '13 at 04:23
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$K$ is an arbitrary compact open subgroup. – Aliakbar Oct 07 '13 at 16:00
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It seems that the answer is negative. Let $\mathbb{Z}_p$ be the group of $p$-adic integers, which should be compact and torsion-free. Then $\mathbb{Z}_p\supset p\mathbb{Z}_p\supset p^2\mathbb{Z}_p\supset\dots$ and $\bigcap_{k=1}^\infty p^k\mathbb{Z}_p=\{0\}$, isn't it?
Alex Ravsky
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