How to prove $(\{2^n3^m\alpha\})_{m,n\in\mathbb{N}}$ is dense in [0,1]?

Question

$\forall \alpha\in [0,1]\setminus\mathbb{Q}$, how to prove $(\{2^n3^m\alpha\})_{m,n\in\mathbb{N}}$ is dense in [0,1]? $\{x\}$ is the fractional part of x.

Any hint would be appreciated!

Answer 1

@zhoraster suggested me writing the proof as an answer. It is consisted of one definition, one lemma, then subsequently two pages of proof.

I will write the definition and the lemma, but I would not include full 2-page proof here.

First, we identify the fractional parts of real numbers as a one-dimensional torus $\mathbb{T}$.

Definition For a positive integer $n$, a set $X\subset \mathbb{T}$ is called $n$-invariant if $nx \ \mathrm{mod} \ 1$ is in $X$ whenever $x$ belongs to $X$.

Lemma Let $\mathcal{M}=(m_i)_{i\geq 1}$ be an infinite sequence of distinct positive integers, arranged in increasing order, such that $m_{i+1}/m_i\rightarrow 1$ as $i\rightarrow\infty$. Let $X$ be a closed infinite subset of $\mathbb{T}$ that is $m_i$-invariant for every $i\geq 1$. If $0$ is a limit point of $X$, then $X=\mathbb{T}$.

The first point in the proof is that for any positive integer $u$, $\{2^{um}3^{un}\}_{m, n\in \mathbb{N}}$ can be arranged in increasing order and the sequence satisfies the hypothesis for $\{m_i\}$ of the lemma. (This is where multiplicative independence of $2$ and $3$ is used.)

The second point in the proof is setting $X$ the closure of $\{2^m3^n\alpha \ \mathrm{mod} \ 1\}$, then proving that $X$ has a rational point. (Proof by contradiction, the lemma is applied.)

Lastly, if $X$ has a rational point, then $X$ satisfies the hypothesis for $X$ of the lemma. By the lemma, we have $X=\mathbb{T}$.

The second point is the most lengthy part of the proof. Rejecting the hypothesis '$X$ does not have any rational point' requires almost 1.5 pages of constructing a sequence of sets in $\mathbb{T}$.