For a fixed $n\in \mathbb{N}$ we consider the set of $n$-roots of unity $R(n)=\{z\in S^1; z^n=1\}$. It splits into mutually disjoint orbits. Let $R=\bigcup_{n=0}^{\infty} R(2^n-1)$. For each orbit in $R$ we choose an element closest to $1$ (i.e. the real part is maximal, if there are two such elements we choose one of them). The set of such selections is denoted by $S$. What are the condensation points of $S$? Is $1$ the only such point?
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4$\begingroup$ What do you mean by "orbit" in this context? $\endgroup$– Marco GollaAug 19, 2014 at 11:02
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$\begingroup$ Are you sure "condensation point" is the concept you want? If the set $S$ is countable, there are no condensation points, unless you're using "condensation point" in a way that differs from its standard meaning. $\endgroup$– Dave L RenfroAug 19, 2014 at 14:13
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2$\begingroup$ "Condensation point" is of course an accumulation point, while I take the orbits to be orbits of Galois (roots of a cyclotomic polynomial). But then $S$ consists of just $\exp( \pm 2\pi i/m)$ as $m$ ranges over the odd integers, and $1$ is clearly the only accumulation point. You can be more precise: as their sizes go to infinity, all orbits become equidistributed in the uniform measure $d\theta/2\pi$ of the circle. $\endgroup$– Vesselin DimitrovAug 19, 2014 at 15:19
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