No, I mean that you can just choose the origin for your first point and go straight to \gamma_1 \cup \zeta_1 when constructing N, and run the same logic.
Yes, precisely – it is another way of saying the same thing, I think it’s just easier to reason about (horizontal line segments that decrease when you’ve found the optimal solution).
Translating the N is better than L proof becomes: look at the feerate at \gamma_i; now consider the feerate for \gamma_i + \zeta_i – it’s higher, because \zeta_i \ge f and the size of that set is greater, and an optimal chunking will extend that feerate leftwards, so N chunks better up to i, for each i, done. You don’t have to deal with line intersections.