Question about $\langle b\rangle$ in a Hausdorff Topological Group

Question

Let $G$ be a topological Hausdorff group, if $b\in G$, does there exist an open neighborhood of $b$ such that $U\cap \langle b\rangle$ is finite.

I know this is a really odd question, but I would really appreciate to know whether this is true or not.

Thanks

score 3 · Accepted Answer · answered Jun 24 '13 at 03:20

3

Not necessarily. For example, let $G$ be the unit circle in the complex plane, that is, the points $e^{it}$, where $t$ ranges over the reals, under the usual multiplication. Let $b=e^{it_0}$, where $t_0$ is not a rational multiple of $\pi$. It can be shown that the group generated by $b$ is dense in the unit circle.

You may be more familiar with the result in the following form. Suppose that $\tau$ is irrational. Then the set of fractional parts of $n\tau$, as $n$ ranges over the natural numbers, is dense in the unit interval.

answered Jun 24 '13 at 03:20

André Nicolas

507,029

could I ask something else?, does the property hold if $G$ is torsion-free? – Camilo Arosemena-Serrato Jun 24 '13 at 03:32
I certainly cannot think of an off-hand counterexample, and can sort of see how one might construct a proof. I think this would be an excellent separate question. – André Nicolas Jun 24 '13 at 03:43
A torsion-free example is $\mathbb{Z}_p$, the group of $p$-adic integers. – Julian Rosen Jun 24 '13 at 03:44
@PinkElephants, why? – Camilo Arosemena-Serrato Jun 24 '13 at 03:53
Suppose $b\in\mathbb{Z}_p$ is non-zero. The subgroup generated by $b$ contains the distinct elements $b+p^nb$ for $n=1,2,\ldots$, and every neighborhood of $b$ contains $b+p^nb$ for all sufficiently large $n$. Additionally, if $b\not\in p\mathbb{Z}_p$, then $\langle b\rangle$ is dense in $\mathbb{Z}_p$. – Julian Rosen Jun 24 '13 at 04:03

Question about $\langle b\rangle$ in a Hausdorff Topological Group

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