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I'm given following assignment. I previous assignment i just showed for rational r, that there is always a cluster point, since it's periodic.

Problem:

show that for $ r \in \mathbb{R} $\ $ \mathbb{Q} $ then $ [e^{i \pi 2rn}]_{n \in \mathbb{N}} $ still at least one cluster point.

$ x_n = ( cos(2 \pi r n ) , sin( 2 \pi r n )) = e^{i2 \pi rn} $

My thoughts: I know that the irrational numbers is epsilon close to a rational one, since they are show to have cluster points, they irrational must too. But I can't show any for $ r \cdot n $

I have plotted it, and it seems like, it's not periodic for irrational n.

Can anybody help?

O. T.
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Viktor Jeppesen
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    The sequence is in fact not periodic for irrational $r$.

    A way to proceed is to argue that the sequence is bounded; therefore it has a convergent subsequence. Yet for this you would need to know the result that yields the "therefore." Do you?

     – quid Jun 06 '16 at 20:02
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    Suppose the sequence takes on only finitely many values, Then by the Pigeonhole Principle some value is repeated. In that case, $r$ is rational. So if $r$ is irrational, the sequence takes on an infinite number of values, and then Bolzano-Weierstrass takes care of things. One can prove a far stronger result. – André Nicolas Jun 06 '16 at 20:03
  • @quid sounds like a easy way, since it's pretty obvious bounded. But the result you are referring to, is not ringing any bells, could you elaborate? – Viktor Jeppesen Jun 06 '16 at 20:11
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    It is the result that @AndréNicolas mentioned. Another relevant keyword would be "compact". The sequence has values in a compact set and therefore has a convergent subsequence. (Yet it is plausible you do not know this, and the solution in the answer is the intended one.) – quid Jun 06 '16 at 20:17
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    No you are on point, I just haven't heard of the pigeonhole principle (probably just language problems). It makes a lot of sense. Thank you for the help @quid – Viktor Jeppesen Jun 06 '16 at 20:22

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We can show in particular that $1$ is a cluster point.

Let $\epsilon > 0$ be arbitrary. Let $N$ be such that $\epsilon > 2 \pi /N$.

By the pigeonhole principle (hopefully you can figure out what I mean here), there must exist $k_1 < k_2 \leq N$ such that $|e^{2 \pi i r k_2} - e^{2 \pi i k_1}|<\epsilon$. So, we have $$ |e^{2 \pi ir (k_2 - k_1)} - 1| = |e^{2 \pi ir k_1}||\xi^{2 \pi i r(k_2 - k_1)} - 1| < \epsilon $$ Thus, setting $n = k_2 - k_1$, we may now state that for any $\epsilon > 0$, there exists an $n \in \Bbb N$ for which $|e^{2 \pi i r n} - 1| < \epsilon$. The conclusion follows.

Ben Grossmann
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