What is the maximum size of a set system where the intersection of any two incomparable members is not in the set? Let the set $\mathcal{F}$ consist of subsets of $[n]$.  Suppose that for any incomparable $A$ and $B$ in $\mathcal{F}$, we have $A \cap B \notin \mathcal{F}$.  What is the largest possible size of $\mathcal{F}$?
 A: It was proved by Kleitman that $|\mathcal F|\le {n\choose n/2}+2^n/n$, see http://www.sciencedirect.com/science/article/pii/0097316576900376. 
A: If you restrict yourself to incomparable subsets, you can take, e.g., all $t+1$-subsets of for $n=2t+1$. All intersections are now of size at most $t$ and don't belong to the family. For even $n=2t$ you'd need to take all $t+1$-subsets as well, but this wouldn't be as large as a fraction of the total family, but asymptotically essentially the same in relative size, compared to $2^n.$ 
Thus, asymptotically, if 
$${\cal F}=\{A \subset [n]: |A|=\lfloor{n/2}\rfloor+1\},$$ this will give $$|{\cal F}|=\binom{n}{\lfloor n/2 \rfloor+1}=  O(2^{n}/\sqrt{n}).$$ If this family is enlarged by adding any other set which is smaller, this would destroy the intersection avoidance property since all smaller sets already occur as intersections of incomparable sets.
Edit: As pointed out in the comments by Seva, it is in fact fine a set of size larger than $t+1$ and preserve the required property, e.g., add $[n]$. Can we add more than one such set? If we add two distinct $t+1+a$ and $t+1+b$ sets, where $a,b\geq 1,$ as long as the intersection of these two sets is of size $\leq t$ we're fine. So the only problem would occur if these two sets intersected on a $t+1$ set. So we're not completely free to add all larger sets to the ensemble which would result in a total of $O(2^n)$ sets.
Thus it seems we can greedily pack larger sets, ensuring any two pairs intersect in at most $t$ points. For example we can use a sunflower set whereby a $t-$set intersection is extended to distinct $t+2$ sets by adjoining 2 points to obtain each new set, so this would add $\lfloor (t+1)/2 \rfloor$ sets to the family.
In any case, clearly the largest family size obtainable this way would be $O(2^n).$
