Let be a positive integer. Suppose we are given distinct sets, each containing finitely many objects. Place each set into one of two categories, the red sets and the blue sets, so that there is at least one set in each category. We define the symmetric difference of two sets as the set of objects belonging to exactly one of the two sets. Prove that there are at least different sets which can be obtained as the symmetric difference of a red set and a blue set.
(This problem was suggested by Max Alekseyev.)
Solution
Call an element a diverse if there is a red set containing , a red set not containing , a blue set containing , and a blue set not containing . We consider two cases based on whether a diverse element exists. We use to denote the symmetric difference of sets and .
Case 1: Suppose no element is diverse. We now show that if are two distinct red sets and are two distinct blue sets, then . Since are distinct, there is an element in one and not the other. Since is not diverse, is in both or in neither of . Either way will lie in exactly one of and , so the two cannot be equal. Thus, all of the symmetric differences of red sets and blue sets are distinct. If there are red sets and blue sets, then , so there are at least symmetric differences.
Case 2: Suppose there is an element that is diverse. Let be the set of red sets that do not contain , be the set of red sets that do contain , the set of blue sets that do not contain , and the set of blue sets that do contain . By the diversity of all four of these sets are non-empty.
By the pigeonhole principle, one of or has at least sets. Without loss of generality, let it be . Apply the inductive hypothesis to obtain a set of at least distinct sets that are the result of symmetric differences between and . Likewise, by the pigeonhole principle one of or has at least sets. Without loss of generality, let it be . Apply the inductive hypothesis to obtain a set of distinct sets that are the result of symmetric differences between and .
Notice that all of the sets in do not contain while all of the sets in do contain . Therefore, and together contain distinct sets obtained from the symmetric difference of red sets and blue sets, completing the induction.