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]?


About Shuhua Zhang

Born in Anhui, China. Awarded a doctor's degree in electronic engineering from Tsinghua University. Now a postdoc researcher working on audio and speech signal processing, especially source separation, in gipsa-lab, Grenoble-INP, France.
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