, 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,
$$(k+1)m\alpha - 2 < [m\alpha] + [km\alpha] \le [l\alpha] \le l\alpha \\ < [l\alpha] + 1 \le [m\alpha] + [km\alpha] + 2 \le (k+1)m\alpha + 2$$
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 $m \le n\beta < m + 1 \Leftrightarrow 2m - m\alpha \le n\alpha < (2m + 2) - (m + 1)\alpha$. 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 $\alpha < 2$.