For a given positive real number $\alpha$, define the set $T_\alpha$ by $T_\alpha = \{ [n\alpha] \mid n = 1,2,\dots \}$. What is a necessary and sufficient condition (in terms of $\alpha$ and $\beta$) to have $T_\alpha \subseteq T_\beta$ ?
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1$\begingroup$ What is $\ [x]\ $ ? $\endgroup$– Wlod AACommented Jul 23, 2017 at 17:39
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2$\begingroup$ WHy do you think there is such a condition? Is this a problem you were given? $\endgroup$– Igor RivinCommented Jul 23, 2017 at 19:11
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2$\begingroup$ Not sure there is a "closed form" of such a condition. Say, one has $T_\alpha=T_\beta$ for any $\alpha,\beta\in(0,1)$, and $T_\alpha\subseteq T_\beta$ whenever $\beta\in(0,1)$ (and any $\alpha>0$). $\endgroup$– SevaCommented Jul 23, 2017 at 19:33
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2$\begingroup$ It is better to ask about $\alpha,\beta>1$. $\endgroup$– Alexey UstinovCommented Jul 24, 2017 at 3:25
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3$\begingroup$ I thought at first that $T_\alpha\subseteq T_\beta$ might be equivalent to “$\beta\le1$ or $\alpha\in\beta\mathbb N$”, but this is wrong, e.g., $T_2\subseteq T_{4/3}$. $\endgroup$– Emil JeřábekCommented Jul 24, 2017 at 15:41
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1 Answer
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A partial answer. It is clear that we should consider only $\alpha,\beta>1$.
Suppose we want to find $n$ such that $n\in T_\beta$, $n\notin T_\alpha$. It is sufficient to find integers $l$ and $k$ such that $$n\le l\beta<n+1,\qquad k\alpha<n,\quad n+1\le (k+1)\alpha,$$ or equivalently $$\left\{\frac n\beta\right\}\in\{0\}\cup\left(1-\frac 1\beta,1\right),\qquad \left\{\frac n\alpha\right\}\in\left(0,1-\frac 1\alpha\right].$$ If $\alpha$ and $\beta$ are linearly independent with $1$ over $\mathbb{Q}$ then points $\left(\big\{\frac n\beta\big\},\big\{\frac n\alpha\big\}\right)$ are uniformly distributed in $[0,1)^2$ and corresponding $n$ does exist. So the situation $T_\alpha \subseteq T_\beta$ is possible only if $c_1\alpha+c_2\beta+c_3=0$ for some integers $c_1$, $c_2$, $c_3$, $(c_1, c_2, c_3)\ne(0,0,0)$.
answered Jul 24, 2017 at 12:12