Density of positive multiples of an irrational number

Question

Let $x$ be irrational. Use $\{r\}$ to denote the fractional part of $r$: $\{r\} = r - \lfloor r \rfloor$. I know how to prove that the following set is dense in $[0,1]$: $$\{\{nx\} : n \in \mathbb{Z}\}.$$ But what about $$\{\{nx\} : n \in \mathbb{N}\}?$$ Any proof that I’ve seen of the first one fails for the second one.

Answer 1

A minor modification of the pigeonhole argument works. Let $m$ be any positive integer. By the pigeonhole principle there must be distinct $i,j\in\{1,\dots,m+1\}$ and $k\in\{0,\dots,m-1\}$ such that $\frac{k}m\le\{ix\},\{jx\}<\frac{k+1}m$; clearly $\{|(j-i)x|\}<\frac1m$. Let $\ell$ be the largest positive integer such that $\ell\{|(j-i)x|\}<1$, let $A_m=\{n|j-i|x:0\le n\le\ell\}$, and let $D_m=\big\{\{y\}:y\in A_m\big\}$.

If $x>0$, every point of $[0,1)$ is clearly within $\frac1m$ of $A_m=D_m$. If $x<0$, then

$$D_m=\{1-|y|:y\in A_m\}\;,$$

so every point of $[0,1)$ is again within $\frac1m$ of the set $D_m$. Since $D_m\subseteq\big\{\{nx\}:n\in\Bbb N\big\}$, we’re done.

Answer 2

Here is a proof where the second statement follows from the first with a minor step. I wrote this answer for another question, but then I saw it fit here better, other than being along very similar lines to Brian M. Scott's answer.

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Let $r$ be irrational. Choose an arbitrary $n\in\mathbb{N}$. Partition $[0,1)$ into $n$ subintervals $$ \left\{I_k=\left[\frac{k-1}n,\frac kn\right):1\le k\le n\right\}\tag1 $$ Consider the discrete set $\{\{kr\}:0\le k\le n\}\subset[0,1)$. There are $n+1$ elements in this set, so two of them must lie in the same subinterval. That means that we have $k_1\ne k_2$ so that $\{(k_1-k_2)r\}\in\left(0,\frac1n\right)$, we leave $0$ out of the interval because $r$ is irrational.

Let $m=\left\lfloor\frac1{\{(k_1-k_2)r\}}\right\rfloor$, then because $m$ is the greatest integer not greater than $\frac1{\{(k_1-k_2)r\}}$, $$ m\{(k_1-k_2)r\}\!\!\overset{\substack{r\not\in\mathbb{Q}\\\downarrow}}{\lt}\!\!1\lt(m+1)\{(k_1-k_2)r\}\tag2 $$ which implies that $\{m(k_1-k_2)r\}=m\{(k_1-k_2)r\}\in\left(\frac{n-1}n,1\right)$.

Since $\{(k_1-k_2)r\}\in\left(0,\frac1n\right)$, if $j\{(k_1-k_2)r\}\in I_k$, then $(j+1)\{(k_1-k_2)r\}\in I_k\cup I_{k+1}$; that is, $\{j(k_1-k_2)r\}=j\{(k_1-k_2)r\}$ cannot skip over any of the $I_k$. Therefore, $$ \left\{\{j(k_1-k_2)r\}:1\le j\le m\vphantom{\frac12}\right\}\tag3 $$ must have at least one element in each $I_k$ for $1\le k\le n$.

The only problem is that we don't know that $k_1-k_2\gt0$. However, this is not a problem since $\{j(k_1-k_2)r\}\in I_k\iff\{j(k_2-k_1)r\}\in I_{n+1-k}$. Therefore, $$ \left\{\{j(k_2-k_1)r\}:1\le j\le m\vphantom{\frac12}\right\}\tag4 $$ must also have at least one element in each $I_k$ for $1\le k\le n$.

Since $n$ was arbitrary, we have shown that $\left\{\mathbb{N}r\right\}$ is dense in $[0,1]$.

Answer 3

Really? I thought exactly the same proof worked for $\Bbb N$.

Let $k\in\Bbb Z$ with $k\ne 0$ and define $$f(t)=e^{2\pi ikt}.$$

Then $f$ has period $1$, and $$\frac1N\sum_{n=0}^{N-1}f(nx) =\frac1N\sum_{n=0}^{N-1}\left(e^{2\pi ikx}\right)^n=\frac1N \frac{e^{2\pi ikxN}-1}{e^{2\pi ikx}-1}\to0\quad(N\to\infty).$$ So the usual approximation shows that $$\frac1N\sum_{n=0}^{N-1}f(nx)\to\int_0^1 f(t)\,dt$$for $f\in C(\Bbb T)$ and you're done, as usual.

How is this any different from the case $n\in\Bbb Z$?