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

mehr1methanol

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!

haruspex

There are some things not making sense to me in this question.
Most obviously, the sequence clearly consists of irrationals, so I guess you mean rational multiples of pi. Secondly, I don't see any significance in showing it as a sequence, including repeats. It seems to be used only as a set - so why the repeats?
That said, if I've understood the question...
K is closed under rotations by 3nπ.2^-m, for all positive integers m, n.
Given θ, let x_i be the bits of the binary fraction expressing θ/3π. From this you can construct a sequence of rotations converging on θ.

mehr1methanol

That's exactly right. Thank you for pointing that out. I corrected the question.


The reason for the repeats is the following:
I'm preforming a symmetrization on the set K and the algorithm is such that it produces the above sequence.

I got confused myself because once I got the sequence, I thought the rule is that I must follow the sequence to get arbitrary close to a θ. But you're absolutely right. Once I show [itex] R_{\alpha_{i}}K = K[/itex] for each [itex] \alpha_{i}[/itex], I'm done. This is because of the role that "n" is playing in your solution.

When I get a little too excited I need someone to check I'm not doing something stupid. Thanks for the comment!