(adsbygoogle = window.adsbygoogle || []).push({}); Conjecture Regarding Rotation of a Set by a Sequence of Angles.

Consider the following sequence, where the elements are rational numbers mulriplied by [itex]\pi[/itex]:

[itex] (\alpha_{i}) = \hspace{2 mm}\pi/4,\hspace{2 mm} 3\pi/8,\hspace{2 mm} \pi/4,\hspace{2 mm} 3\pi/16,\hspace{2 mm} \pi/4,\hspace{2 mm} 3\pi/8,\hspace{2 mm} \pi/4,\hspace{2 mm} 3\pi/32,\hspace{2 mm} \pi/4,\hspace{2 mm} 3\pi/8,\hspace{2 mm} \pi/4,\hspace{2 mm} 3\pi/16,\hspace{2 mm} \pi/4,\hspace{2 mm} 3\pi/8,\hspace{2 mm} \pi/4,\hspace{2 mm} 3\pi/64,\hspace{2 mm} \pi/4,\hspace{2 mm} \cdots[/itex]

Let [itex]K \subset ℝ^{2}[/itex] be a compact set. Also let [itex]R_{\alpha_{i}}[/itex] denote the rotation by [itex]\alpha_{i}[/itex].

Suppose [itex]R_{\alpha_{i}}K = \hspace{2 mm} K[/itex] for each [itex]\alpha_{i} \in (\alpha_{i})[/itex].

Question: Is it true that for all [itex]\theta \in [0, 2\pi)[/itex] [itex]R_{\theta}K = \hspace{2 mm} K[/itex].

Note:

If instead we had the sequence [itex](n\alpha)[/itex] where [itex]\alpha[/itex] is an irrational number, it is trivial that the conjecture holds. This is trivial due to the following fact from the study of continued fractions:

Given any real number on a circle, it can be approximated arbitrarily close by multiples of an irrational number.

But if [itex]\alpha[/itex] is a rational number this doesn't hold since after a finite number of rotations you will get back to where you started from. However in the question above we don't have rotations by a fixed rational number and the answer is not immediate!

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# Conjecture Regarding rotation of a set by a sequence of rational angles.

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