Problem 1. (Do not need to complete) Dirichlet’s approximation theorem says that for any irrational number $\alpha$ there are an infinite number of integers $p$ and $q$ so that
$$\left|\alpha-\frac{p}{q}\right|<\frac{1}{q^{2}}$$
a) Show that when $\alpha$ is irrational, for each rational $r \in \mathbb{Q}$ the set $\left\{(p, q): r=p / q,\left|\alpha-\frac{p}{q}\right|<\frac{1}{q^{2}}\right\}$ is finite. What happens when $\alpha$ is rational?
b) Using Dirichlet’s approximation theorem, the fact that $|\sin x| \leq|x|$ and the fact that $\pi$ is irrational, show that there is a subsequence $\left\{\sin n_{k}\right\}_{k=1}^{\infty}$ of $\{\sin n\}_{n=1}^{\infty}$ so that $\lim _{k \rightarrow 0} \sin \left(n_{k}\right)=0 .$

(Hint: $\mid \sin (m \pi-x)|=| \sin (x) \mid \text { for } m \in \mathbb{N})$

Proof . a)如果$\alpha$是无理数，那么由于$|\alpha-r|>0$，因此这样的$\left\{(p, q): r=p / q,\left|\alpha-\frac{p}{q}\right|<\frac{1}{q^{2}}\right\}$ 是有限的。 而如果$\alpha$是有理数，那么当且仅当$|\alpha-r|=0$也就是$\alpha=r$的时候$\left\{(p, q): r=p / q,\left|\alpha-\frac{p}{q}\right|<\frac{1}{q^{2}}\right\}$ 是无限的

b)直接套用第一题结论加极限的定义

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