# 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.

# 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 $n>2$.

## 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..

• I do not know how to prove it, but the Wikipedia article "Equidistribution theorem" contains this sentence: "One noteworthy result is that the sequence $2^ka \mod 1$ is uniformly distributed for almost all, but not all, irrational $a$." (It might still be dense, though.) – Yoni Rozenshein Mar 27 '13 at 13:13
• @YoniRozenshein Thank you. – JP McCarthy Mar 28 '13 at 10:24

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.

• Thank you. I suppose something like $\displaystyle \alpha=\sum_{i=1}^\infty 2^{-i!}$ will do the job also. – JP McCarthy Mar 28 '13 at 11:07
• yup. with the "contains all finite sequence" criterion, it's easy to write examples not containing some finite sequence. In this chaotic system, you can pretty much choose the behaviour of the sequence as you go along writing the binary development of $\alpha$. – mercio Mar 28 '13 at 11:11
• Yes I am happy enough now --- these students understand the Dyadic Mapping quite well so the rest is fine. What threw me was that the previous lecture notes --- in as many words --- said that the behaviour of $(\mathbb{T},z^2)$ was similar to $(\mathbb{T},\alpha\cdot z)$ and I took this a bit too far. I ended up just giving partial results: results 1., 2., 3.(a), a seed with a dense orbit for the squaring function ($\alpha=0.0100011011000001010100110101011111\dots_2$) and used the points in $O_n$ to show that $f(z)=z^n$ for $n\geq 3$ has seeds with non-dense orbits. – JP McCarthy Mar 28 '13 at 11:17

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.

• Thank you GG we still have $z_0=e^{i\pi\theta}$ is eventually periodic if and only if $\theta\in\mathbb{Q}$ which is something. Would you have a nice word to describe the orbit of the non-periodic yet non-dense orbits? – JP McCarthy Mar 28 '13 at 11:19
• not to my knowledge... you might say that those orbits stay in a (closed) invariant set – Glougloubarbaki Mar 28 '13 at 15:09
• "All orbits stay in the orbit." -- hehe! Thanks again. – JP McCarthy Mar 28 '13 at 16:03