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We know that any sequence on $S^1$ must have a converging extracted subsequence, as $S^1$ is compact. Now, consider the sequence $a_n=(\cos(n),\sin(n))$. Could you find explicitly a subset of the natural numbers such that the corresponding subsequence converges? I don't even know whether it is possible to work it out, or whether there exists a nice representation of the solution. I thought about it and I found out a proof, but later I noticed that it actually contains an error. I'd prefer not to write here the proof, at least in this first time, since I could influence your reasoning. Thank you in advance,

Bye!

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Consider the continued fraction for $2\pi$. Let $\frac{h_n}{k_n}$ be the $n^{th}$ convergent. Then the numbers $h_n$ will give you a convergent subsequence which converges to $(1,0)$.

Since $|2\pi -\frac{h_n}{k_n}|\leq \frac{1}{k_n k_{n+1}},$ (see this theorem) it follows that $|2k_n \pi -h_n|\leq \frac{1}{k_n+1}.$ Since $k_{n+1}\rightarrow \infty$, and both $\sin$ and $\cos$ are continuous, it follows that $\lim_{n\rightarrow \infty} (\cos (h_n), \sin (h_n))=(1,0).$

Hope that helps,

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    @fatoddsun: I suggest taking at look at: http://en.wikipedia.org/wiki/Continued_fraction The short answer to why $k_n\rightarrow \infty$ is because k_n and they are integers. A more precise answer is that $k_n$ satisfies the recurrence relation $k_n =a_nk_{n-1}+k_{n-2}$ for all $n$, where $a_n\geq 1$. (When we write a continued fraction down, we write it in the short form $[a_0;a_1,a_2,a_3,\cdots $ which refers to $a_0 +\frac{1}{a_1+\frac{1}{a_2+\frac{1}{a_3+\cdots}}}.$ The coefficient $a_0$ is the only one that can be zero for an irrational)2011-06-02