What are the subsets of $\{\{n\alpha\} \:|\: n\in\mathbb{N}\}$ which are dense in $[0,1)$? If $\alpha\in\mathbb{R}\setminus\mathbb{Q}$, we know that
$$A=\{\{n\alpha\} \:|\: n\in\mathbb{N}\}$$
is dense in $[0,1)$. There are non-trivial subsets of $A$ which are still dense in $[0,1)$? By non-trivial, I mean a subset which is not
$$\{\{kn\alpha\} \:|\: n\in\mathbb{N}\}$$
for some $k\in\mathbb{N}\setminus\{0\}$. For example, is
$$\{\{2^n\alpha\} \:|\: n\in\mathbb{N}\}$$
dense in $[0,1)$?
 A: For $\left\{\{2^n \alpha\} \mid n\in\mathbb{N}\right\}$, it depends on $\alpha$. For simplicity, we can assume $0<\alpha<1$, because 
$$\alpha=\left \lfloor \alpha \right \rfloor + \{\alpha\} \Rightarrow
2^n \alpha=2^n \left \lfloor \alpha \right \rfloor + 2^n\{\alpha\} \Rightarrow
\left\{2^n \alpha\right\}=\left\{2^n \{\alpha\}\right\}$$
Also, we can assume $\alpha=\left(0.a_1a_2a_3...\right)_2, a_i\in\{0,1\}$ in base $2$. Then multiplying by $2^n$ simply shifts the bits to the left by $n$ or 
$$\left\{2^n \alpha\right\}=\left(0.a_{n+1}a_{n+2}...\right)_2$$
Now, let's look at the binary Liouville's constant, which is a transcendental number
$$\alpha=\sum\limits_{k=1}\frac{1}{2^{k!}}$$
and 
$$1=\frac{1}{2} \cdot 2=
\frac{1}{2}\cdot\frac{1}{1-\frac{1}{2}}=
\frac{1}{2}\cdot \left(1+\sum\limits_{k=1}\frac{1}{2^k}\right)=
\frac{1}{2} + \sum\limits_{k=2}\frac{1}{2^k}=
\sum\limits_{k=1}\frac{1}{2^k}$$
then
$$\Delta=\left|1-\left\{2^n \alpha\right\} \right|=
\left|\sum\limits_{k=1}\frac{1}{2^k} - \left\{2^n\sum\limits_{k=1}\frac{1}{2^{k!}}\right\}\right|=
\left|\sum\limits_{k=1}\frac{1}{2^k} - \left\{\sum\limits_{k=1}\frac{1}{2^{k!-n}}\right\}\right|=\\
\left|\sum\limits_{k=1}\frac{1}{2^k} - \sum\limits_{k!>n}\frac{1}{2^{k!-n}}\right|=...$$
or let's note by $k_0$ the very first of $k$'s such that $k!>n$, then (still base $2$)
$$\begin{array}{c|c|c|c} 
\text{pos} & 0 & . & 1 & 2 & ... & k_0!-n-1 & k_0!-n & k_0!-n+1 & ... \\ \hline
1 & 0 & \text{,} & 1 & 1 & 1 & 1 & 1 & 1 & ... \\ \hline
\{2^n\alpha\} & 0 & \text{,} & 0 & 0 & 0 & 0 & 1 & 0 & ... \\ \hline
\Delta & 0 & \text{,} & 1 & 1 & 1 & 1 & 0 & 1 & ... \\ \hline
\end{array}$$
and
$$...=\left|\sum\limits_{k=1}^{k_0!-n-1}\frac{1}{2^k} + \frac{1}{2^{k_0!-n+1}}+...\right|>
\sum\limits_{k=1}^{k_0!-n-1}\frac{1}{2^k} + \frac{1}{2^{k_0!-n+1}} \tag{1}$$
Now


*

*if $k_0!-n=1$, then from $(1) \Rightarrow \Delta > \frac{1}{2^2}$

*if $k_0!-n>1$, then from $(1) \Rightarrow \Delta > \frac{1}{2}$
As a result $\Delta > \frac{1}{2^3}$ (to cover the case when $n=0$ since $\alpha=\left(0.110...\right)_2$). So, $\left\{\{2^n\alpha\} \mid n\in\mathbb{N}\right\}$ will never be close enough to $1$ and thus, won't be dense on $[0,1)$ for $\alpha$ - binary Liouville's constant.
