MathLabs

Problem 2

Let n≥2n \ge 2 be an integer. Let A1,A2,…,A2nA_1, A_2, \dots, A_{2^n} be the 2n2^n subsets of an nn-element set, listed in some order. Prove that ∣A1∖A2∣+∣A2∖A3∣+⋯+∣A2n−1∖A2n∣+∣A2n∖A1∣≥2n−2.|A_1\setminus A_2| + |A_2\setminus A_3| + \cdots + |A_{2^n-1}\setminus A_{2^n}| + |A_{2^n}\setminus A_1| \ge 2^{n-2}.
Step 5 of 5: Conclude the bound
#{i:ai+bi=0}≤2n−1⟹∑x(ax+bx)≥2n−1⟹∑i∣Ai∖Ai+1∣≥2n−2\#\{i:a_i+b_i=0\}\le2^{n-1}\Longrightarrow\sum_x(a_x+b_x)\ge2^{n-1}\Longrightarrow\sum_i|A_i\setminus A_{i+1}|\ge2^{n-2}
Detailed analysis

The zero terms form an independent set in the cycle of length 2n2^n, so at most 2n−12^{n-1} terms vanish. Every other term is at least 11, hence ∑x(ax+bx)≥2n−1\sum_x(a_x+b_x)\ge2^{n-1}. Step 3 gives the required lower bound 2n−22^{n-2}.