It is possible, and even quite a bit more. Let $\ f(m)\ $ be the number of the $m$-th partition intervals which intersect $\ A.\ $ **Theorem** It's possible to have $\ A\ $ which is not dense in any non-trivial interval whenever $$ \lim_{m=\infty} \frac {f(m)}m\ =\ 0 $$ **Proof** The required $\ A\ $ can be given as follows: $$ A\ :=\\ \ \left\{ \frac {2\cdot k-1}{2\cdot m}: \ m\in\Bbb N\ \ \text{and} \ \ k\in\Bbb Z\ \ \text{and}\ \ 0<k\le f(m)\,\right\} $$ **End of Proof**