The density --- or otherwise --- of $\{\{2^N\,\alpha\}:N\in\mathbb{N}\}$ for ALL irrational $\alpha$.

Question

Problem

Is there an irrational $\alpha\in\mathbb{R}\backslash\mathbb{Q}$ such that the set $S= \{\,\{2^N\alpha\} :N\,\in\mathbb{N}\}$ is not dense in $[0,1]$.

Here $\{x\}=x-\lfloor x\rfloor$ is the fractional part of $x$.

Questions very similar to this have been asked in the past. For example

Multiples of an irrational number forming a dense subset

However right now I can't adjust them and put the two sticks together.

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Context

I was trying to prove the following proposition --- which isn't actually true when $n>2$.

Let $n>2$ and let $O_n$ be the (proper) subset of $[0,1]$ of irrational numbers containing only $0$s and $1$s in their base-$n$ expansion. If we take $\alpha\in O_n$ we then have that

$$\{n^N\alpha\}\in O_n\text{ for all }N\in\mathbb{N}$$

and so $S$ can not be dense in $[0,1]$ when $n>2$.

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Proposition

Suppose that $f:\mathbb{C}\rightarrow\mathbb{Z}$ is a power mapping $f(z)=z^n$ for some $n\geq 2$.

Then the dynamical system $(\mathbb{C},f)$ exhibits the following four behaviours:

1. If $z_0\in\mathbb{D}:=\{z\in\mathbb{C}:|z|<1\}$ then $z_n\rightarrow 0$.

2. If $z_0\in\mathbb{C}\backslash\bar{\mathbb{D}}:=\{z\in\mathbb{Z}:|z|>1\}$ then $z_n\rightarrow\infty$.

3. If $z_0\in\mathbb{T}=\{z\in\mathbb{C}:|z|=1\}$ then there are two behaviours:

a. If $z_0=e^{q\,2\pi i}$ with $q\in\mathbb{Q}$ then $z_0$ is eventually periodic.

b. If $z_0=e^{\alpha\,2\pi i}$ with $\alpha\in\mathbb{R}\backslash\mathbb{Q}$ then $z_0$ has a dense orbit.

In fact, $(\mathbb{T},f)$ is a chaotic mapping.

Proof: 1, 2 and the first part of 3 were illustrated in class for $n=4$. Here we prove for a general $n\geq 2$.

Assume that $z_0=e^{\alpha\cdot\,2\pi i}$ with $\alpha\in\mathbb{R}\backslash \mathbb{Q}$

Claim 1: $z_0$ is not eventually periodic.

Proof by Contradiction: Suppose that $z_0=e^{\alpha\cdot2\pi i}$ is eventually periodic. That is there exists an iterate of $z_0$, say $f^N(z_0)$, that is periodic. Note that

$$f^N(z_0)=z_0^{n^N}=\left(e^{\alpha\cdot 2\pi i}\right)^{n^N}=e^{n^N\alpha\cdot2\pi i}.$$

Now if this point is periodic then there exists an $M\geq 1$ such that

$$f^M(e^{n^N\alpha\cdot2\pi i})=e^{n^N\alpha\cdot2\pi i}$$

$$\Rightarrow e^{n^Nn^M\alpha\cdot2\pi i}=e^{n^N \alpha\cdot2\pi i}$$

If these two complex numbers are equal then their argument/angle must differ only by a multiple of $360^\circ$; i.e. $k\cdot 2\pi$ for $k\in\mathbb{Z}$:

$$n^{M}n^N\alpha\cdot 2\pi-n^N\alpha\cdot 2\pi=k\cdot 2\pi$$

$$\Rightarrow n^{N+M}\alpha-n^N\alpha=k$$

$$\displaystyle\alpha=\frac{k}{n^{N+M}-n^N},$$

but this is a contradiction as $\alpha$ is not a fraction. Hence $z_0$ is not eventually periodic $\bullet$

Claim 2: Two iterates of $z_0$ can be found that are arbitrarily close together.

Proof : Note that the length of the unit circle is $2\pi$. Suppose that we divide the unit circle into $N$ arcs $A_1,A_2,\dots,A_N$ each of equal length $\displaystyle \frac{2\pi}{N}$ (to be careful suppose that they contain their 'right' endpoints but not their 'left'). Now, we know that $z_0$ is not eventually periodic so the first $n$ iterates of $z_0$ are all distinct:

$$z_0,z_1,z_2,z_3,\dots,z_N.$$

Note that there are $N$ arcs but $N+1$ iterates so by the Pigeonhole Principle there is an arc $A_i$ containing two iterates of $z_0$ --- say $z_i$ and $z_j$ --- that are on the same arc. Take $N\rightarrow \infty$ so that the arcs becoming arbitrarily small and the points become arbitrarily close together $\bullet$

Claim 3: There is a number $M$ such that $z_0$ and $f^M(z_0)$ are arbitrarily close together.

Proof: We know that for any $N$ we can find two iterates of $z_0$ that are at most (radially) apart by $\displaystyle \frac{2\pi }{N}$. Denote these by

$z_i=f^i(z_0)$ and $z_j=f^j(z_0)$... what I was hoping to do here actually doesn't work. Hmmm..

Answer 1

The sequence $\{2^n \alpha \}$ is dense if and only if its binary development contains every possible finite sequence of $0$s and $1$s. (furthermore, it is uniformly distributed if and only if $\alpha$ is normal in base $2$. Almost all numbers are normal in base $2$)

If you have a finite sequence $d_1 \ldots d_k$ of length $d_k$, it is the binary expansion of an integer $a$, and irrational numbers whose binary development begin with $d_1 \ldots d_k$ are irrational numbers in the interval $[a/2^k ; (a+1)/2^k)$.

So if the sequence is dense, than $\alpha$ is irrational and for any finite sequence, there is a rank $n$ such that $\{2^n \alpha\} \in [a/2^k ; (a+1)/2^k)$. This means that the binary development of $\alpha$ contains the sequence $d_1 \ldots d_k$ starting at the $n$th digit.

Conversely, if the binary development contains all possible finite sequences, then the sequence $\{2^n \alpha\}$ visits every dyadic interval (intervals of the form $[a/2^k ; (a+1)/2^k)$), and because any open interval contains such a dyadic interval, it visits any open interval, so it is dense in $[0;1]$.

Then, it is not hard to see that the sequence generated by $\alpha = 0.1101001000100001\ldots$ is not dense since the binary development doesn't contain $111$, and $\alpha$ is obviously irrational.

Answer 2

Mercio's answer is correct and complete. From a dynamical point of view, what we can say is the following : every $z=e^{i \pi \theta}$, with $\theta$ rational, is eventually periodic.

But among the points that are not eventually periodic, not all have a dense orbit in all of $\mathbb{T}$. Indeed, there are invariant Cantor sets contained in $\mathbb{T}$ with orbits dense in those sets.