infimum, supremum of the sequence $\{\sin n\}$

Can someone please explain how should I proof that the supremum and infimum of this: $$\left\{\sin n\right\}_{n=1}^\infty$$ I just know that the supremum is $1$ and infimum is $-1$ because $$\mid \sin n\mid \leq 1$$but how should I proof that ? Any help or suggestion would be great.

Solution 1:

Hint: Note that the set $\Bbb Z+2\pi\Bbb Z$ is dense in $\Bbb R$, as it is a non-cyclic subgroup of $(\Bbb R,+)$. For a proof, see For an irrational number $\alpha$, prove that the set $\{a+b\alpha: a,b\in \mathbb{Z}\}$ is dense in $\mathbb R$

We show, the set of all limit points of $\{\sin n\}_{n\in \Bbb N}$ is exactly the set $[-1,1]$.

Now, for each $y\in[-1,1]=\text{image}(\sin)$, choose $x\in \Bbb R$ with $\sin x=y$.

Since the set of all limit points is a closed set it is enough to show each $y\in[-1,0)\cup (0,1]$ is a limit point of $\{\sin n\}_{n\in\Bbb N}$.

Next, find a sequence $\{x_n\}\subseteq\Bbb Z+2\pi\Bbb Z$ such that $x_n\to x$. write, $x_n=p_n+2\pi q_n$ for $p_n,q_n\in\Bbb Z$. In particular, $p_n\not=0$ for infinitely many $n$ as $y\not=0($otherwise $x_n$ would diverge$)$; so passing to the subsequence assume $p_n\not=0$ for all $n$. Now, $\sin x_n=\sin p_n\to \sin x=y$.

If all $p_n>0$ then we are done.

So, let $p_n>0$ for finitely many $n\geq 1$. Negelecting first few terms, assume $p_n<0$ for all $n\geq 1$. But, $\pi-x_n=-p_n-(2q_n-1)\pi\longrightarrow (\pi-x)$. So, $\sin(\pi-x_n)\to \sin(\pi-x)=\sin x=y$ and we are done.