Let $S\_1, \dots, S\_k \subset \mathbb{R}^n$ be a set of (non-regular) simplices. Let $m\_i$ indicate the number of vertices of simplex $S\_i$ (we do not assume it is equal to $n-1$).


Is there a simple upper bound on the maximum number of vertices of the intersection $\bigcap\_i S\_i$, stated in terms of the set $\{ m\_i\}\_{i=1..k}$? 


What about the maximum number of vertices of the convex hull of the union, $\mathrm{Conv}\big(\bigcup\_i S\_i\big)$?


(Cross-posting from [math.SE](https://math.stackexchange.com/questions/3210365/upper-bound-on-number-of-vertices-in-intersection-and-union-of-complex-polytopes))



Upper bound on number of vertices in intersection (and union) of simplices
In the literature, the dimension is usually $d$ (rather than your $n$),
and the number objects is $n$ (rather than your $k$).


The intersection of $n$ halfspaces in dimension $d$ can have 
$n^{\lfloor d/2 \rfloor}$ vertices.
This is achieved by the dual of cyclic polytopes.
See the MO question 
[How many vertices can a convex polytope have?](https://mathoverflow.net/q/127423/6094).


For the union, although the union itself can be complicated, the
convex hull of the union is not: All the vertices of the simplices
could fall on the hull. So the union can have at
most $O(n d)$ vertices, because each of the $n$ simplices
can have at most $d+1$ vertices.


