\(\{\lfloor n\alpha\rceil\mid n=1,2,3,\ldots\}\,\textbf{在} (0,1) \text{稠密}\)

elim

取定 \(\alpha\in (0,\infty)-\mathbb{Q}\),
【定义】\(f_\alpha(n)=\lfloor n\alpha\rceil:=n\alpha-\lfloor na\rfloor.\)
试证 \(\;\,(1)\quad f_\alpha:\mathbb{N}^+\to (0,1)\) 是单射;
\(\qquad\;(2)\quad\)对任意\(\varepsilon\in(0,1)\), 存在正整数\(n_\varepsilon\)使\(f_\alpha(n_\varepsilon)\in(0,\varepsilon)\);
\(\qquad\;(3)\quad\)对任意\((a,b)\subseteq(0,1),\;\{\lfloor n\alpha\rceil\mid n\in\mathbb{N}^+\}\cap(a,b)\ne\varnothing\)

APB先生

谢谢了 。

elim

【注记】显然\(\alpha\)是正无理数，\(\lfloor n\alpha\rceil\)是\(n\alpha\)的小数部分. 我们有
\((\dagger)\quad (k\in\mathbb{N})\wedge(k\lfloor x\rceil< 1)\implies (k\lfloor x\rceil = \lfloor kx\rceil)\)
\((^*)\quad (\lfloor x\rceil\in(0,1))\implies (\lfloor x\rceil+\lfloor -x\rceil=1) \)
\(\qquad\;\color{blue}{\small-\lfloor x\rfloor -1< -x < -\lfloor x\rfloor\implies \lfloor -x\rceil = -x+1+\lfloor x\rfloor=1-\lfloor x\rceil}\)
下面证明主贴的 \((1),(2),(3).\):
\((1)\quad\)设\(m,n\in\mathbb{N}^+,\;m\ne n\),
\(\qquad\;\)则\(\lfloor n\alpha\rceil-\lfloor m\alpha\rceil=(n-m)\alpha-\lfloor n\alpha\rfloor+\lfloor m\alpha\rfloor\)
\(\qquad\;\)是无理数因而\(\ne 0,\;\lfloor n\alpha\rceil\ne\lfloor m\alpha\rceil.\)
\((2)\quad\)据\((1),\;E_\alpha=\{\lfloor n\alpha\rceil\mid n\in\mathbb{N}^+\}\)是无穷集, 取\(k=\lceil 1/\varepsilon\rceil\)
\(\qquad\;\)则有正整数\(j\le k\)使\(E_\alpha\cap(\frac{j-1}{k},\frac{j}{k})\)为无穷集. 有\(m,n\in\mathbb{N}^+\)
\(\qquad\;\)使\(\;\frac{j-1}{k}< \lfloor m\alpha\rceil< \lfloor n\alpha\rceil < \frac{j}{k}.\) 令 \(n_\varepsilon=n-m,\)
\(\qquad\;\)则\(|\lfloor n_\varepsilon\alpha\rceil-(\lfloor n\alpha\rceil- \lfloor m\alpha\rceil)|< 1\)是整数. 故\(0< \lfloor n_\varepsilon\alpha\rceil<\varepsilon.\)
\(\qquad\;\)若\(n_\varepsilon < 0\) 则有\(\eta\in\mathbb{N}_+\)使\(\,1-\varepsilon< \lfloor\eta n_\varepsilon\alpha\rceil < 1\).
\(\qquad\;\)取\(m_\varepsilon =-\eta n_\varepsilon,\) 则 \(\lfloor m_\varepsilon\alpha\rceil\overset{(^*)}{=}1-\lfloor \eta n_\varepsilon\alpha\rceil< \varepsilon\)
\((3)\quad\)对 \(\phi\ne(a,b)\subset(0,1),\)取\(\varepsilon=b-a,\) 据(2), 有\(n_0\in\mathbb{N}_+\)使
\(\qquad\;0< \lfloor n_0\alpha\rceil< \varepsilon< b.\) 取 \(m=\max\{k\in\mathbb{N}_+: k \lfloor n_0\alpha\rceil < b\}\)
\(\qquad\;\)于是 \(b-m\lfloor n_0\alpha\rceil\le\lfloor n_0\alpha\rceil< b-a\) 故 \(a< \lfloor mn_0\alpha\rceil< b\)
\(\qquad\;\)故 \(\phi\ne(a,b)\subset(0,1)\implies \{\lfloor n\alpha\rceil\mid n\in\mathbb{N}^+\}\cap(a,b)\ne\varnothing\)

elim

易见 \(S_\alpha=\{n\alpha+m\mid n\in\mathbb{N}_+,\; m\in\mathbb{Z}\},\;S_{-\alpha}\) 均为\(\mathbb{R}\)的稠密子集.
但 \(S_\alpha\cap S_{-\alpha}=\varnothing\)