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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 $[0,1)$?

Gabriel
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    Related:https://math.stackexchange.com/questions/1756649/powers-of-10-multiples-of-pi-or-any-irrational-are-dense see comments and answers – Tito Eliatron Nov 01 '18 at 18:58
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    There's this paper R. C. Buck "Limit points of subsequences" which establishes that for a given sequence almost all its subsequences share all its limit points. It doesn't give a way to validate such subsequence though. Anyway worth reading I guess. – freakish Nov 01 '18 at 22:15

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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.

rtybase
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