Solution
= Solution
For one real coordinate, a union of $m$ intervals labels a fixed $n$-point sample by choosing at most $2m$ endpoints among the $n+1$ gaps, so the number of labelings is at most $(n+1)^{2m}$. Membership in a Cartesian product is the product of the $p$ coordinate-membership indicators. Applying part (c) repeatedly gives
$$
\boxed{s(\mathcal F_3,n)\leq\bigl((n+1)^{2m}\bigr)^p
=(n+1)^{2mp}}.
$$