This question is more of a graph problem. Suppose $A=\{1,2,..,n\}$, and we want to construct the sets $B(i)$,$1\leq i\leq n$, such that $i \not\in B(i)$ and the following constraints on the set $B(i)$ are satisfied:

1. If $j \in B(i)$ then $ i \notin B(j)$,

2. For all $i \in B(j)$, $B(j) \setminus \{i\} \not \subset B(i)$,

3. For all $i \notin B(k)$, $B(i) \not \subset {\{k\} \cup B(k)}$

Can we construct such subsets for all values of $n$? I can prove that you can construct such subsets for $n=4,5,6,..10$, but I can not generalize the proof for all values of $n$. I believe that such subsets cannot be constructed in general.