$\{m \alpha, m \in \mathbb Z\}$\; is dense in $[0,1]$ for irrational $\alpha$.

Question

Let $\alpha$ be irrational and $S=\{\{n\alpha\}:n\in \mathbb{Z}\}$.

I proved that for any positive integer $N, \exists m\in \mathbb{Z}$ such that $\{m\alpha\}<\frac{1}{N}.$

But how do I use the above fact to show that for $x\in [0,1]$ and $\forall \varepsilon>0,\;\big((x-\varepsilon, x+\varepsilon)\cap S\big)\neq \varnothing?$

Can anyone help me to answer my question.

See also Mariano's answer to an older question. The question there is more specific, but his chosen line of attack works for any irrational $\alpha$. Anyway, this kind of questions are more often encountered in number theory (the first lemmas in Diophantine approximation) rather than real analysis. After all, real analysis does not really concern about whether the numbers are rational or not. — Jyrki Lahtonen, Oct 04 '15 at 07:16

score 5 · Accepted Answer · 2015-10-04T07:24:45.660

5

Let $x\in [0,1]$, $\epsilon >0$. Then there is $N \in \mathbb N$ large so that

$$ \frac 1N <2\epsilon.$$

Now we know that there is $m \in \mathbb Z$ so that $0< \{ m\alpha\} <\frac 1N$. Let $K \in \mathbb N$ so that $K\{m\alpha\} <1$ and $(K+1)\{m\alpha\} \ge 1$. Then the set

$$C = \{ \{m\alpha\}, \{2m\alpha\}, \{3m \alpha\}, \cdots, \{ K m\alpha\}\} \subset [0,1]$$

has $K$ elements and $ \{im\alpha\} = i\{m\alpha\}$ for $i=1, 2, \cdots, K$. In particular, as $\{m\alpha\}$, which is the distance between adjacent elements in $C$, is less than $\frac 1N <2\epsilon$, there must be $J \in \{1, 2, \cdots, K\}$ so that $$\{Jm \alpha\} \in (x-\epsilon, x+\epsilon)$$ as $\{Jm \alpha\}\in S$, $(x-\epsilon, x+\epsilon) \cap S \neq \emptyset$.

edited Oct 04 '15 at 07:24

answered Oct 04 '15 at 07:12

1

What means index $n$ here? $C = { {m\alpha}, {2m\alpha}, {3m \alpha}, \cdots, { K m\alpha}}_{n\in \infty} \subset [0,1]$ – RFZ Oct 04 '15 at 07:16
@RFZ : There shouldn't be. Edited. – Oct 04 '15 at 07:17
3

Dear John Ma, very nice solution. Everywhere people write that it's sufficient to prove that for any $N$ exists ${m\alpha}<1/N$ but nobody explain how any number in $[0,1]$ can be approximated by elements of this set. – RFZ Oct 04 '15 at 07:23

Koro · Answer 2 · 2023-08-10T23:46:48.500

Define $P_n (a,b):=|\{1\le r\le n: \{r \alpha\}\in [a, b]\}|, |X|$ denotes cardinality of the set $X$.

By Weyl's equidistribution theorem, given any $a, b \in [0,1], a<b$ the following holds: $\lim_{n\to \infty} P_n/n =b-a$. $(1)$.

Now take any open set $(p, q)\subset [0,1], p<q$. If we show that $P_n\left(\frac{p+q}4,\frac{p+q}2\right)\ne0$ for some $n$, then we are done. But this is true by $(1)$. $\square$

$\{m \alpha, m \in \mathbb Z\}$\; is dense in $[0,1]$ for irrational $\alpha$.

2 Answers2

Linked