Give an example of four different subsets A, B, C and D of {1, 2, 3, 4} such that all intersections of two subsets are different.

Question

My work,

Suppose E={1,2,3,4} then power set of E is

P(E)={ {}, {1}, {2}, {3}, {4} {1,2}, {2,3}, {3,4}, {1,3}, {1,4}, {2,4}, {1,2,3},{2,3,4}, {1,2,4}, {1,3,4}, {1,2,3,4} } Shows the possible subsets of E.

But i cant figure out what A,B,C,D should be. It seems like no matter what I put there are bound to be two intersections where they both produce an empty set.

Please help.

Answer 1

$\{1,2\},\{2,3\},\{3,4\},\{1,2,3,4\}$

---

The most subsets you can take is $5$, achievable by taking all the subsets of size $4$ and the complete subset.

Proof: Let $A$ be the size of the smallest subset you take. Then you can take at most $\binom{|A|}{2}$ other subsets. When $|A|<4$ this is less than $4$. If the smallest size is $4$ the maximum is clearly taking all the subsets of size $4$ or more.

Answer 2

The collection $A=\{1,2\},B=\{2,3\},C=\{1,2,3\},D=\{1,3,4\}$ seems to work.

Answer 3

If they asked the dual problem, find four subsets of $\{1,2,3,4\}$ whose unions are all different, the obvious answer is $\{1\},\{2\},\{3\},\{4\}$. So take the complements: $\{2,3,4\},\{1,3,4\},\{1,2,4\},\{1,2,3\}$. The intersection of the complements is the complement of the union.