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:
$$ A\ :=\ \left\{ \frac {2\cdot k-1}{2\cdot m}: \,\ k\in\Bbb Z\ \ \text{and}\ \ 0<k\le f(m)\,\right\} $$
End of Proof