A combinatorial question. Is this a known result, false, or open? Let $X$ be a set of $n-1$ elements. Does there exist a family $S_1,S_2...S_n\in 2^X$ such that $$|S_i\cap S_j|\le 1$$ and $$|\overline S_i\cap \overline S_j|\le 1$$?
That is, neither the sets themselves, nor their complements, have any pairwise intersections bigger than a singleton.
EDIT: i,j, distinct, of course.
EDIT2: Now revolved as false, see below. A more top-down proof that this cannot hold for $n\ge8$ was given on Reddit http://www.reddit.com/r/math/comments/231i2w/a_combinatorial_problem_intersecting_families/cgsih68
 A: I think this suffices for $n=6$.  Certainly not the most elegant solution, though.
Proof of nonexistence for $n=6$.  We shall use, as Will Jagy, the alphabet $X=\{a,b,c,d,e\}$.  We shall call the desired family $S$ and the set of all the complements as $S'$.
Clearly we cannot have more than one singleton, as if we did, say $\{a\}, \{b\}$, then $\overline{\{a\}}=\{b,c,d,e\}$ and $\overline{\{b\}}=\{a,c,d,e\}$ and clearly these have intersection with three elements.
Similarly, we cannot have more than one four-element set, as these would clearly have intersection containing three elements.
It is not hard to see further that the empty set and/or the entire set cannot lie in this family, as from the above if the entire set were an element, then we know that there is at least one two or three-element set, and their intersection will be two or three-elements, respectively.  If the empty set were an element, then by symmetry $S'$ will have $X$ and we know that this leads to a contradiction.
Next, we can have at most two three-element sets, as since there are only five elements, any two three-element sets must have at least one element in common, so the first two possible ones will be of the form $\{a,b,c\}$ and $\{a,d,e\}$.  But then any third will have intersection with one of these that has more than a single element:  Since the remaining three element sets are $$\{a,b,d\},\{a,b,e\}, \{a,c,d\},\{a,c,e\},\{b,c,d\},\{b,c,e\},\{b,d,e\},\{c,d,e\}$$ all of which can be checked to intersect one of $\{a,b,c\}$ or $\{a,d,e\}$ at two elements.
This shows us that $S$ contains at most one singleton and the rest two and three-element sets, where there are at most two three-element sets.  Thus, there are at least three two-element sets, meaning that $S'$ has at least three three-element sets, giving us a contradiction since we know that $S'$ cannot have the desired property.
A: $n=5,$ alphabet $abcd$
$$ ab,bc,cd,da,ac  $$
$$ cd,da,ab,bc,bd   $$
