For $x\in\mathbb R\setminus\mathbb Q$, the set $\{nx-\lfloor nx\rfloor: n\in \mathbb{N}\}$ is dense on $[0,1)$

Question

Let $x\in \mathbb{R}$ an irrational number. Define $X=\{nx-\lfloor nx\rfloor: n\in \mathbb{N}\}$. Prove that $X$ is dense on $[0,1)$.

Can anyone give some hint to solve this problem? I tried contradiction but could not reach a proof.

I spend part of the day studying this question Positive integer multiples of an irrational mod 1 are dense and its answers. Only one answer is clear and give clues to solve the problem. This answer is the first one. However, this answer does not answer the question nor directly, nor the proof follows from this answer.

This answer has some mistakes, he use that $[(k_1-k_2)\alpha]=[k_1\alpha]-[k_2\alpha]$ which is not true. Consider $k_1=3, k_2=1, \alpha=\sqrt{2}$ we have $[(k_1-k_2)\alpha]=2\not= 3=[k_1\alpha]-[k_2\alpha] $. We only can assure that $[k_2\alpha]-[k_1\alpha]-1\leq [(k_2-k_1)\alpha]\leq[k_2\alpha]-[k_1\alpha]$.

Who answered said something interesting about additive subgroups of $\mathbb{R}$, but unfortunately the set $X=\{nx-[nx] : n\in \mathbb{N} \}$ is not a subgroup. Considering the additive subgroup $G=\langle X \rangle$, if we prove the part (a) of the link, we get that indeed $G$ is dense on $\mathbb{R}$ but we can not conclude that $X$ is dense on $[0,1)$.

I think this problem has not been solved.

Thanks!

Answer 1

Ok, since you've asked and it doesn't fit into a comment, there you go. I'll do it on a circle since it's slightly easier to explain and I'll leave it to you to complete it in the case of an interval. let's say you have a circle of length $1$. you take 'steps' along the circle of an irrational length, let's say counter-clockwise. you'll never hit the same spot twice so for any fixed $\epsilon > 0$ you'll eventually find two 'steps' $a_n$ and $a_m$ such that $0 < |a_n - a_m| < \epsilon$. the distance from $a_n$ to $a_m$ is the same as between $a_{n-m}$ and $a_0 = 0$ and so on. therefore if you let $k:= n-m$ and you only consider each $k$-th step you'll be going around the circle travelling a distance smaller than $\epsilon$ hence if you divide your circle into arcs of equal lengths greater than $\epsilon$ (but just slightly, say smaller than $2 \epsilon$) you'll have to land in each one of those in order to make your way all around the circle (because your steps are to small to jump over them). Every point of the circle is in at least one of those intervals which means that for each point of the circle you can find a number $a_j$ in your sequence that is closer than $2 \epsilon$ to it. Now conclude taking smaller and smaller $\epsilon$'s.

edit: oh, just note that I'm taking the distance along the circle, not the euclidean one

Answer 2

A pictoral representation of mm-aops's proof. I also avoid use of: "the distance from $a_n$ to $a_m$ is the same as between $a_{n-m}$ and $a_0 = 0$."

Let $\ \varepsilon = \frac{1}{200}.\ $ Then, we split the circle up into $201$ arcs of equal length. Think of these arcs as "boxes" in the sense of the pigeonhole principle: then there must be $\ n,m \leq 201\ $ such that $\ \mid a_{n} - a_m\mid \leq \frac{1}{201} <\varepsilon.$

Suppose the first two such $\ n,m\ $ are $\ a_{45}\ $ and $\ a_{162}\ .$

Any point on the circle must lie within $\ \varepsilon=\frac{1}{200}\ $ distance from $\ a_{45}\ $ or $\ a_{162}\ $ or $\ a_{279}\ $ or $\ a_{396}\ $ or ...

Answer 3

Here we use this notation: $\{z\}=frac(z)=z-[z]$.
It's easy to show that $\left\{z\pm y \right\}= \left\{ \left\{ z\right\}\pm \left\{ y\right\}\right\}$ and therefore, for every $n\in \mathbb{N}$, $\left\{ nz\right\}= \left\{ n\left\{ z\right\}\right\}$.
Let's split the interval $[ 0,1 [$ into $k>0$ subintervals $[\frac{i}{k},\frac{i+1}{k}[$ for $0\leqslant i< k$.
The function $f:\mathbb{N}\to [0,1[$ defined as $f(m)=\{mx \}$ is injective. In fact, if $m_1,m_2\in\mathbb{N}$ and $m_1\neq m_2$, then $(m_1-m_2)x\notin\mathbb{Z}$ and therefore $\left\{ m_1x\right\}-\left\{m_2x \right\}= \left ( m_1-m_2 \right )x-\left [m_1x \right ]+\left [m_2x \right ]\neq 0$. Hence, for the pigeonhole principle, there are $a,b\in\mathbb{N}$ distinct, such that $\{ax\}$ and $\{bx\}$ are in the same subinterval. We can always assume that $\{bx\}>\{ax\}$ and then $0<\{bx\}-\{ax\}<\frac{1}{k}$; since $\{bx\}-\{ax\}=\{\{bx\}-\{ax\}\}=\{(b-a)x\}$, we can conclude that $0<\{(b-a)x\}<\frac{1}{k}$.
Therefore, we have proved that for every $k\in\mathbb{N}\setminus \left\{0 \right\}$, there is $m\in\mathbb{N}\setminus \left\{0 \right\}$ such that $0<\{mx\}<\frac{1}{k}$.
This is enough to prove that $\{ nx-[nx]:n\in\mathbb{N}\}$ is dense in $[ 0,1 [$. In fact, if $z,y\in[0,1[$ with $z< y$, there is $k\in\mathbb{N}\setminus \left\{0 \right\}$ such that $0< \frac{1}{k} < y-z $, and so there is $m\in\mathbb{N}\setminus \left\{0 \right\}$ such that $0< \{mx\} < y-z $. Therefore there is $n\in\mathbb{N}$ such that $z< n\{mx\} < y$, so we have $z< \{nmx\} < y$.

Answer 4

I'll do it by contrapositive (i.e. not dense implies rational). Also, I want to mention that it's equivalent to show it for $\{nx-\lfloor nx \rfloor : n\in\mathbb{Z}\}$.

Suppose that there is a point in $[0,1)$ that isn't a limit point of $\{nx-\lfloor nx \rfloor : n\in\mathbb{Z}\}$. Thus, there are $0<a<b<1$ s.t. $(a,b) \cap \{nx-\lfloor nx \rfloor : n\in\mathbb{Z}\} = \emptyset$

This implies that $\bigcup_{n\in\mathbb{Z}}(a+n,b+n) \cap \{nx : n\in\mathbb{Z}\} = \emptyset$

Shifting by any multiple of $x$, we obtain the sets $\bigcup_{n\in\mathbb{Z}}(a+lx+n,b+lx+n)$ and $\{nx + lx : n\in\mathbb{Z}\} = \{nx : n \in\mathbb{Z}\}$

Thus, $\bigcup_{n\in\mathbb{Z}}(a+lx+n,b+lx+n) \cap \{nx : n \in\mathbb{Z}\} = \emptyset$

Let $B_{l,n} = (a + lx + n, b + lx + n)$. We obtain $\bigcup_{l,n\in\mathbb{Z}} B_{l,n} \cap \{nx : n\in\mathbb{Z}\} = \emptyset$

Now note that if any of the $B_{l,n}$ overlap, you can actually show that $\bigcup_{l,n\in\mathbb{Z}} B_{l,n}$ is the entirety of $\mathbb{R}$, which is a contradiction. Intuitively, you could think of it as given how cyclic these intervals are, if you have two intervals overlapping, you could find another to tack at the end of one of them, and so on and so forth.

Therefore, keep in mind that the only way for any two of these intervals to intersect is to be the same interval.

We will use measure theory to show that there are two balls with different $l$'s that are the same.

Consider $F_l = \text{frac}(B_{l,0})$. You may show these are measurable and that $m(F_l) = b-a$. If we choose $L$ s.t. $(b-a)L > 1$, you may show that $\{F_l : 0 \leq l < L\}$ is not a disjoint collection of sets because otherwise, $m(\bigcup_{l=0}^{L-1} F_l) = (b-a)L > 1$ which is a contradiction since $\bigcup_{l=0}^{L-1}F_l \subseteq [0,1)$

Thus, there are $l_1,l_2\in\mathbb{Z}$ s.t. $F_{l_1} \cap F_{l_2} \neq \emptyset$. By the definition of the balls, you may show there are also $n_1,n_2$ s.t. $B_{l_1,n_1} \cap B_{l_2,n_2}\neq\emptyset$.

Since $B_{l_1,n_1},B_{l_2,n_2}$ are not disjoint, are the same length, and can't overlap, we obtain that the start and end points of the two balls are the same, i.e, $a + l_1x + n_1 = a+ l_2x + n_2$ and thus $x = \frac{n_1-n_2}{l_2-l_1}$

Answer 5

$f \colon x \rightarrow x-\lfloor x \rfloor$

$F=\{f(nx),n \in \Bbb{Z}\}$

$F \neq \emptyset$ and $x \ge 0$ $\exists d=\inf(F)$

We can prove that $d=0$ (suppose a $\gt$ 0 and ...)

hence $\forall$ y $\gt$ 0 $\exists$ n $\in$ $\Bbb{N}$ , $y \gt f(nx) \gt 0$

$a,b \in [0;1[$ if $b-a \gt 0$ $\exists n \in \Bbb{N}$, $b-a \gt f(nx) \gt 0$

$\exists p \in \Bbb{N}$ , pf(nx) $\gt$ a (1) , if we suppose that p is the smallest number who verifies (1) we have that (p-1)fn(x) $\lt$ a therefore pf(nx) $\lt$ a+f(nx) $\lt$ b

therefore pf(nx) $\in$ [a;b] and pf(nx)=f(pnx)

we can conclude about the density

First time I use this way of writing I hope I have been clear