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Iranian Mathematical Olympiad

Combinatorics Difficulty 5.7 AIME, harder Prove it Iran

Let SS be a set with nn elements and P(S)P(S) be the set of all subsets of SS. We want to partition P(S)P(S) to mm parts such that if AA, BB and A∪BA \cup B are in the same part then A=BA = B. Find the minimum value of mm so that such a partition exists.

Solution

The answer is m=n+1m = n + 1. To give an example, for all integers 0≤i≤n0 \le i \le n define
Ai={E∣E⊂S,∣E∣=i}. A_i = \{E \mid E \subset S, |E| = i\}.
This partition satisfies the condition of problem, because if AA, BB and A∪BA \cup B are in the same partition it means that they have the same number of elements but A,B⊂A∪BA, B \subset A \cup B, so we have A=A∪B=BA = A \cup B = B.

To prove that it is impossible for m≤nm \le n, we give n+1n + 1 sets such that no two of them can be in the same part. Assume that S={1,2,3,…,n}S = \{1, 2, 3, \dots, n\}. Consider A0=∅A_0 = \emptyset and for all integers 1≤i≤n1 \le i \le n,
Ai={1,2,…,i}. A_i = \{1, 2, \dots, i\}.
For i<ji < j we have Ai∪Aj=AjA_i \cup A_j = A_j so they can not be in the same partition.

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