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Is $1$ and $-1$ the Supremum and Infimum of the set $A=\left\{\sin(n), n \in \mathbb{N}\right\}$

I actually assumed, let $M=Sup(A)$ and $M <1$, now if we could find a $k \in \mathbb{N}$ such that $\sin k>M$, then we can say that $1$ is the supremum of the set.But how to go about finding $k$?

Ekaveera Gouribhatla
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    See https://math.stackexchange.com/q/4764, https://math.stackexchange.com/q/63526, https://math.stackexchange.com/q/3156727, https://math.stackexchange.com/q/484131, https://math.stackexchange.com/q/313943. – Martin R Jun 11 '21 at 12:30
  • As $n$ can get arbitary close to numbers of the form $(2k+\frac{1}{2})\pi$ and to numbers of the form $(2k+1+\frac{1}{2})\pi$, the infimum is $-1$ and the supremum $1$. – Peter Jun 11 '21 at 12:35
  • This post linked by @MartinR contains a complete proof, including a proof of the Kronecker density theorem (subject to knowing that $\pi$ is irrational). – Lee Mosher Jun 11 '21 at 12:39

1 Answers1

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By continuity of sine, it suffices to show that for any $\epsilon > 0,$ there's some integer $n$ so that $|n - (2k + 0.5)\pi| < \epsilon$ for some integer $k,$ since then $\sin n$ can be made arbitrarily close to $\sin (2k+0.5)\pi = \sin 0.5\pi = 1.$ A similar trick works for $-1.$

By dividing by $2\pi,$ it suffices to find some $n$ such that there exists a $k$ for which $$\left| \frac{n}{2\pi} - k -\frac{1}{4}\right| < \frac{\epsilon}{2\pi}.$$

Fortunately, $1/2\pi$ is irrational, and the fractional parts of integer multiples of an irrational number are dense mod $1$ (this is Dirichlet's theorem), so it is possible to find some $n$ and $k$ positive integers making this inequality hold.