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
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.