Japan 2007

First, we prove that any irrational number $\alpha >2$ satisfies the condition. Let $A(\alpha )\supset A(\beta )$, $[\beta ]=[m\alpha ]$ ($m$: positive integer). We can prove this by proving that $[k\beta ]=[km\alpha ]$ for any positive integer $k$. We are going to prove this by induction on $k$. Assume that $[k\beta ]=[km\alpha ]$, and $[(k+1)\beta ]=[l\alpha ]$ ($l$: positive integer). Then we only have to prove that $l=(k+1)m$. From $[m\alpha ]+[km\alpha ]\le \beta +k\beta <[m\alpha ]+[km\alpha ]+2$, $[l\alpha ]\le (k+1)\beta <[l\alpha ]+1$, we get $$[l\alpha ]<[m\alpha ]+[km\alpha ]+2<[l\alpha ]+3.$$ Note that all the terms are integers. We get $[l\alpha ]-1\le [m\alpha ]+[km\alpha ]\le [l\alpha ]$. Hence, $$\begin{gathered} (k+1)m\alpha -2<[m\alpha ]+[km\alpha ]\le [l\alpha ]\le l\alpha \\ <[l\alpha ]+1\le [m\alpha ]+[km\alpha ]+2\le (k+1)m\alpha +2 \end{gathered}$$ So, $-2<(l-(k+1)m)\alpha <2$. Since $\alpha >2$, we get $l=(k+1)m$. With that the induction is completed. Second, we prove that any irrational number $\alpha >2$ doesn't satisfy the condition. Let $\beta =\frac{\alpha }{2-\alpha }$. Then $\frac{\beta }{\alpha }=\frac{1}{2-\alpha }$ is not integer, because $\frac{1}{2-\alpha }$ is an irrational number. Let $m=[n\beta ]\in A(\beta )$ ($n$ is positive integer). Then $m\le n\beta <m+1\iff 2m-m\alpha \le n\alpha <(2m+2)-(m+1)\alpha$. We get the following inequality. $$m\le \frac{n+m}{2}\alpha <\frac{n+m+1}{2}\alpha <m+1$$ Since either $n+m$ or $n+m+1$ is even, we get $m\in A(\alpha )$, and $A(\alpha )\supset A(\beta )$. With that, the answer is $\alpha <2$.