Closure of the set $\{\sin(n): n\in\mathbb{N}, n > 0\}$ is the interval of reals $[-1,1]$? Please prove or disprove.
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Thanks in advance
 A: In fact, it is sufficient to show that $\sin(\mathbb{Z})$ is dense in $[-1,1]$. But $\sin(\mathbb{Z})= \sin(\mathbb{Z}+ \pi \mathbb{Z})$; $\mathbb{Z}+ \pi \mathbb{Z}$ is a nondiscrete subgroup of $(\mathbb{R},+)$, because $\pi$ is not rational, so $\mathbb{Z}+ \pi \mathbb{Z}$ is dense in $\mathbb{R}$. Because $\sin$ is continuous, we deduce that $[-1,1]= \sin(\mathbb{R})= \sin \left( \overline{\mathbb{Z}+ \pi \mathbb{Z}} \right) \subset \overline{\sin (\mathbb{Z}+\pi \mathbb{Z})}= \overline{\sin(\mathbb{Z})}$, so $\sin(\mathbb{Z})$ is dense in $[-1,1]$.
A: Hint: Show that $A=\{n\beta+k:n,\,\,k\in\mathbb{Z}\}$ is dense in $\mathbb{R}$ for any irrational $\beta$, and conclude by using the periodic properties and continuity of $\sin$ function with $\beta=\pi$.
To do the density part, you may for example consider the group $(\mathbb{R},+)$ and show that any proper subgroup $G\subset \mathbb{R}$ is either dense or $G=a\mathbb{Z}$ for some $a\in\mathbb{R}$. Take $a=\inf \{x>0:x\in G\}$, and show that if $a>0$ then $G=a\mathbb{Z}$, and if $a=0$, then you find members of $G$ arbitrarily close to zero, and thus arbitrarily close to any real number with proper translations. I.e. that if $a=0$, then $G$ is dense. Now note that $(A,+)$ is a subgroup of $(\mathbb{R},+)$. If $A$ was not dense, then $A=a\mathbb{Z}$ for some $a\in\mathbb{R}$. Now $\beta\in A$ and $1\in A$ imply that $\beta=ak_{1}$ and $1=ak_{2}$ for some $k_{1},k_{2}\in\mathbb{Z}$. Hence $\beta=\frac{\beta}{1}=\frac{a k_{1}}{a k_{2}}=\frac{k_{1}}{k_{2}}\in\mathbb{Q}$, which is a contradiction. Hence $A$ is dense in $\mathbb{R}$.
