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Actually I can prove the fact that $\sin(\mathbb{Z})$ is dense in $[-1,1]$ using the result that "any non trivial subgroup of the additive group of $\mathbb{R}$ is either cyclic or is dense in $\mathbb{R}$." But my problem is to prove that $\sin(\mathbb{N})$ is dense in $[-1,1]$. Here $\mathbb{N}$ is the set of Natural numbers and $\mathbb{Z}$ is the set of integers.

HSN
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Adimathematica
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  • http://math.stackexchange.com/questions/4764/sine-function-dense-in-1-1 – Jonas Meyer Jul 10 '14 at 09:15
  • Since $\sin(-x)=-\sin x$, you would use the fact that $\sin(\mathbb{Z})$ is the union of $\sin(+\mathbb{N})$, $\sin(-\mathbb{N})$ and ${ \sin(0) }$. The first two sets are each other's reflections, so if one is not dense, neither is the other. – Jeppe Stig Nielsen Jul 10 '14 at 09:28
  • BT I can't show {a+2bpi:a, b belong to N} is a subgroup of the additive group of R. BT if a, b belong to Z then it will be a subgroup. – Adimathematica Jul 10 '14 at 09:30
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    BT is it true that if the union of two set is dense then one of the set has to be dense??? – Adimathematica Jul 10 '14 at 09:35
  • @Adimathematica No, the union of two non-dense sets my be dense so there is something missing in my hint. – Jeppe Stig Nielsen Jul 10 '14 at 09:36
  • for any irrational number α, the set A={a+bα∣a∈N,b∈Z} is dense in R. I can't prove this result. My problem immediately follows from this result. – Adimathematica Jul 10 '14 at 09:44

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There exists a sequence $\alpha_n=a_n+2\pi b_n$ with $a_n,b_n\in \mathbb{Z}$, not $0$, such that $\alpha_n\to 0$ (take $\alpha_n\in ]0,1/n[$). Now by multiplying by $\pm 1$, one can suppose that $a_n\in \mathbb{N}$ and we have that $a_n\to +\infty$.

Now let $y\in \mathbb{R}$. There exists $\beta_k=u_k+2\pi v_k$, $u_k,v_k\in \mathbb{Z}$ such that $\beta_k\to y$. But as $a_m\to +\infty$ if $m\to \infty$, you can find a subsequence $a_{n_k}$ of $a_n$ such that $u_k+a_{n_k}\in \mathbb{N}$. As $\beta_k+\alpha_{n_k}\to y$, we have proven that $\mathbb{N}+2\pi \mathbb{Z}$ is dense in $\mathbb{R}$.

Kelenner
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