Suppose $\alpha > 2$. Let F be the set of real numbers $x \in [0,1]$ for which the inequality $||qx|| \le q^{1-\alpha}$ is satisified by infinitely many positive integers q. For each q, let $G_q$ denote the set of $x \in [0,1]$ satisifying $||qx|| \le q^{1-\alpha}$ for infinitely many positive $q \in N$.
My questions is: Why $G_q$ consists of q-1 intervals of length $2q^{-\alpha}$ and two 'end' intervals of length $q^{-\alpha}$?
Because $G_q$ can (almost) be written as the union of some closed intervals with centers $a/q$, where $a$ ranges from $0$ to $q$. After all, if $x = a/q + \epsilon$, with $\epsilon$ some small positive number the inequality reduces to
$$q^{1 - \alpha} \ge \|qx\| = |q\epsilon|$$ or $$\epsilon < q^{-\alpha}$$ This shows that $[a/q, a/q + q^{-\alpha}] \subseteq G_q$. The study of $\epsilon < 0$ and proving actual equality is similar.
Note that the two edge cases of $a = 0$ and $a = q$ have to be handled separately because $G_q \subseteq [0, 1]$.