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 4 of 5: No two consecutive terms vanish
ai+bi=0  ⟺  Ai+1=Aica_i+b_i=0\iff A_{i+1}=A_i^c
Detailed analysis

For a transition Ai→Ai+1A_i\to A_{i+1}, ai+bia_i+b_i counts the elements whose membership changes. It is zero exactly when Ai+1=AicA_{i+1}=A_i^c. Two consecutive zero terms would give Ai+2=AiA_{i+2}=A_i, impossible because all listed subsets are distinct.