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 1 of 5: Set up transition counts
ax:=#{i:x∈Ai∖Ai+1},bx:=#{i:x∈Ai+1∖Ai},i=1,…,2n (indices mod 2n)a_x:=\#\{i:x\in A_i\setminus A_{i+1}\},\quad b_x:=\#\{i:x\in A_{i+1}\setminus A_i\},\quad i=1,\dots,2^n\ (\text{indices mod }2^n)
Detailed analysis

Index the sets cyclically modulo 2n2^n. For each element xx, let axa_x count transitions in which x belongs to AiA_i but not Ai+1A_{i+1}, and let bxb_x count transitions in the opposite direction. Then the target sum is ∑i∣Ai∖Ai+1∣=∑xax\sum_i|A_i\setminus A_{i+1}|=\sum_x a_x.