Progression-Free Subset Sums Force Distinct Subset Sums

Let $A \\subseteq \\{1, \\dots, N\\}$ have $n$ elements and suppose the set of its subset sums contains no three-term arithmetic progression. We show that the $2^n$ subset sums are then pairwise distinct, and hence that $2^n \\le nN + 1$. The mechanism is that a repeated sum manufactures a progression out of an intersection and a union. If $\\sum B = \\sum C$ for distinct subsets $B$ and $C$, then the three values $\\sum(B \\cap C)$, $\\sum B$ and $\\sum(B \\cup C)$ are in arithmetic progression, because the lattice identity $\\sum(B \\cap C) + \\sum(B \\cup C) = \\sum B + \\sum C$ makes the middle term the average of the outer two. A base-three construction gives the opposite bound $g_3(n) \\le (3^n-1)/2$, and the exact values for $n \\le 5$ are $1, 3, 8, 22, 60$.