## Dense Circle

Proposition 1. Let $\alpha$ be an irrational number and denote $\{\mathbb{N}\alpha\}$ the set of fractional parts of $n\alpha$ for all $n\in\mathbb{N}$. Then $\{\mathbb{N}\alpha\}$ is dense in $[0,1]$.

Proof. We shall first prove that $\{\mathbb{N}\alpha\}$ is dense at $0$, then proceed to the whole range of $[0,1]$. Without loss of generality, $\alpha$ is assumed to be within $(0,1)$.

To show that $\{\mathbb{N}\alpha\}$ is dense at $0$, we claim that there exists a $m\in\mathbb{N}$ such that $\{m\alpha\} \leq \frac12\alpha$. Denote $n_0$ the largest integer satisfying $n_0\alpha < 1$, then $(n_0+1)\alpha \geq 1$.  If $\{(n_0+1)\alpha\}=(n_0+1)\alpha-1 \leq \frac12\alpha$, then $m=n_0+1$ is what we need, else let $\beta=1-\{n_0\alpha\}=1-n_0\alpha<\frac12\alpha$. Denote $n_1$ the largest integer satisfying $n_1\beta \leq 1$, then $(n_1+1)\beta > 1$ and $\{n_1n_0\alpha\}=\{n_1(1-\beta)\}=1-n_1\beta < \beta <\frac12\alpha$, that is, $m=n_1n_0$ is what we need. Then apply this procedure to $\{m\alpha\}$ and so on, which leads to a sequence approaching to $0$ at least as fast as $2^{-n}\alpha$.

Further, for any $\epsilon > 0$, we can find a $N_\epsilon\in\mathbb{N}$ such that $\{N_\epsilon\alpha\} < \epsilon$ by the above argument. Then for any $x\in[0,1]$, $|\{\lfloor\frac{x}{\epsilon}\rfloor N_\epsilon\alpha\} - x| < \epsilon$, thus $\{\mathbb{N}\alpha\}$ is dense at $x$. This completes the proof.                                                                                                                                           $\square$

Alternatively, we can prove Prop. 1 by the Dirichlet’s approximation theorem, which states that for any real number $\alpha$ and positive integer $N$, there exit integers $p$ and $q$ such that $1\leq q \leq N$ and $|q\alpha-p|<1/(N+1)$. Thus $\{\mathbb{N}\alpha\}$ is dense at $0$ and then dense everywhere on $[0,1]$.

Remark 1. Prop. 1 can be formulated on the unit circle $S^1$, that is, the set $e^{2\pi{i}\mathbb{N}\alpha}$ for any irrational number $\alpha$ is dense on $S^1$.

Remark 2. The interval $[0,1]$ can be replaced with $[0,\beta]$ where $\beta>0$ as long as $\alpha/\beta$ is irrational ($\alpha$ might be rational).

Remark 3. If we replace $\mathbb{N}$ with its subset $\mathbb{N}q+r$, where $q,r\in\mathbb{N}$, $\{(\mathbb{N}q+r)\alpha\}$ will still be dense in $[0,1]$. This is because $\{(\mathbb{N}q+r)\alpha\}=\{\mathbb{N}(q\alpha)+r\alpha\}$, in which $\{\mathbb{N}(q\alpha)\}$ is dense in $[0,1]$ and $r\alpha$ amounts to a constant (circular) shifting. On the other hand, for any infinite subset $H \subset \mathbb{N}$, $\{H\alpha\}$ will be dense somewhere in $[0,1]$, since $[0,1]$ is compact, but $\{H\alpha\}$ is not necessarily dense everywhere in $[0,1]$, for example, $H=\{n~|~\{n\alpha\}-\frac12|<\frac14\}$.

Problem 1. Let $d_\alpha(n)$ be the largest interval between two adjacent points from $\{\alpha\}, \{2\alpha\}, \ldots, \{n\alpha\}$ on $[0,1]$. By Prop. 1, $d_\alpha(n)$ monotonically goes to $0$. But what is the speed of decaying of $d_\alpha(n)$ as $n$ goes to infinity? For example, could it be $\mathcal{O}(n^{-1})$?

Problem 2. Let $\alpha$ be an irrational number. Then, for what kind of $f:\mathbb{N}\to\mathbb{N}$, $\{f(\mathbb{N})\alpha\}$ is dense in $[0,1]$? Equivalently, we could ask for what kind of subset $H \subset \mathbb{N}$ such that the set $\{H\alpha\}$ is dense in $[0,1]$. Specifically, is $\{\mathbb{N}^2\alpha\}$ dense on $[0,1]$? Let $P$ be the set of prime numbers, is $\{P\alpha\}$ dense on $[0,1]$?