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Combinatorics Difficulty 6.9 National Olympiad Prove it United States

Let nn be a positive integer. Suppose we are given 2n+12^n + 1 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 2n2^n 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 aa, a red set not containing aa, a blue set containing aa, and a blue set not containing aa. We consider two cases based on whether a diverse element exists. We use AΔBA\Delta B to denote the symmetric difference of sets AA and BB.

Case 1: Suppose no element aa is diverse. We now show that if R,RR, R' are two distinct red sets and B,BB, B' are two distinct blue sets, then RΔBRΔBR\Delta B \neq R'\Delta B'. Since R,RR, R' are distinct, there is an element ss in one and not the other. Since ss is not diverse, ss is in both B,BB, B' or in neither of B,BB, B'. Either way ss will lie in exactly one of RΔBR\Delta B and RΔBR'\Delta B', so the two cannot be equal. Thus, all of the symmetric differences of red sets and blue sets are distinct. If there are rr red sets and bb blue sets, then r+b=2n+1r+b = 2^n+1, so there are at least rb2n1=2nrb \geq 2^n \cdot 1 = 2^n symmetric differences.

Case 2: Suppose there is an element aa that is diverse. Let R0R_0 be the set of red sets that do not contain aa, R1R_1 be the set of red sets that do contain aa, B0B_0 the set of blue sets that do not contain aa, and B1B_1 the set of blue sets that do contain aa. By the diversity of aa all four of these sets are non-empty.

By the pigeonhole principle, one of R0B0R_0 \cup B_0 or R1B1R_1 \cup B_1 has at least 2n1+12^{n-1} + 1 sets. Without loss of generality, let it be R0B0R_0 \cup B_0. Apply the inductive hypothesis to obtain a set A0A_0 of at least 2n12^{n-1} distinct sets that are the result of symmetric differences between R0R_0 and B0B_0. Likewise, by the pigeonhole principle one of R0B1R_0 \cup B_1 or R1B0R_1 \cup B_0 has at least 2n1+12^{n-1} + 1 sets. Without loss of generality, let it be R0B1R_0 \cup B_1. Apply the inductive hypothesis to obtain a set A1A_1 of 2n12^{n-1} distinct sets that are the result of symmetric differences between R0R_0 and B1B_1.

Notice that all of the sets in A0A_0 do not contain aa while all of the sets in A1A_1 do contain aa. Therefore, A0A_0 and A1A_1 together contain 2n2^n distinct sets obtained from the symmetric difference of red sets and blue sets, completing the induction.

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