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Please prove or disprove.

Any help is appreciated Thanks in advance

T. Eskin
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Raffaele
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  • $|\sin x| \leq 1 \forall x \in \mathbb{R}$ Any number in $\mathbb{N}$ can never be differed by a integral multiple of $\pi$. – hjpotter92 Feb 25 '13 at 14:15

2 Answers2

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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 $[-1,1]$.

Seirios
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  • Thank you! I didn't know that non discrete subgroups of (R,+) are dense in R. Can you tell me a reference to this property? – Raffaele Feb 25 '13 at 14:42
  • @RaffaeleValente: I don't have a specific reference in mind, but the proof is as follow: Let $G$ be a subgroup of $(\mathbb{R},+)$ and $a = \inf {x \in G : x >0 }$. If $a \neq 0$ then $G=a \mathbb{Z}$; if $a=0$, $G$ is dense in $\mathbb{R}$. – Seirios Feb 25 '13 at 15:35
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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 $\mathbb{R}$.

T. Eskin
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  • @RaffaeleValente: If you wish, I may expand the density part into a precise solution. However, I believe you may also enjoy verifying the details and completing the proof yourself. – T. Eskin Feb 25 '13 at 14:50
  • Thank you! I'm working on it... – Raffaele Feb 25 '13 at 19:26
  • @Raffaele: The case $a=0$ is harder, but it goes e.g. like following. Assume $a=0$, whence there exists $(x_{n})\subseteq G$ s.t. $x_{n}\to 0$. Fix $x\in \mathbb{R}$ and $\varepsilon>0$. Choose $n\in\mathbb{N}$ with $x_{n}<\varepsilon$. There now exists $k\in\mathbb{Z}$ s.t. $\frac{x}{x_{n}}\in[k,k+1)$. Thus $kx_{n}\leq x\leq x_{n}(k+1)$, whence $|x-kx_{n}|<x_{n}<\varepsilon$. Since $G$ is a group, $kx_{n}\in G$, and thus $kx_{n}\in B(x,\varepsilon)\cap G$. Since $x$ and $\varepsilon$ were arbitrary, we have shown that $G$ is dense in $\mathbb{R}$. – T. Eskin Feb 25 '13 at 20:39