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**