The Bollobas'1965 theorem is the following:
If $A_1,...,A_n$ and $B_1,...,B_n$ are two sequences of subsets of $X=\{1,...,r\}$ such that $A_i\cap B_j = \emptyset$ if and only if $i=j$, then $$\sum_{i=1}^n\binom{|A_i|+|B_i|}{|A_i|}^{-1}\leq 1.$$
First of all, I am interested in the case where $A_i$ and $B_i$ are complementary.
It means we obtain $$\sum_{i=1}^n\binom{r}{|A_i|}^{-1}\leq 1.$$ from which we have the Sperner's theorem since $\binom{r}{\lfloor \frac{r}{2} \rfloor}^{-1}\leq \binom{r}{k}^{-1}$ for all $k$: $$n\leq \binom{r}{\lfloor \frac{r}{2} \rfloor}.$$ For a given $n$, it is therefore easy to find the smallest value $r$ for which the previous inequality is verified.
However, I am interested in a special case where there are additional conditions on the $A_i$'s and $B_i$'s:
$$|(A_i\cap B_j) \cup (A_{i+1}\cap B_{j+1})|\geq 2, \text{ for all } i\neq j.$$
How could I adapt the previous results with this additional condition? That is, for a given $n$, how could I find the smallest integer $r$ such that all the conditions are verified?
I suppose that for a given $n$, the smallest integer $r$ for which there are such sets $A_i$ and $B_i$ will be bigger than without the additional conditions.
A way to prove the original problem is by counting the number of permutations separating a pair $(A,B)$ (see this link for example). However, I do not see how to adapt this approach.
Thanks for your help!
