Problem 6
Following the arrows from any starting point traces out an increasing chain, and every positive integer lies on exactly one such chain.
Draw an arrow from to for each . Since is injective, every positive integer has at most one arrow pointing into it, so following the arrows forward and backward from any point traces out an ascending chain, terminating backward only at the finitely many start-points with no arrow pointing in. Together these chains partition the positive integers into disjoint ascending chains , each skipping forward by at most per step. There are at most chains, since among any consecutive positive integers every chain must contain at least one element (as gaps between consecutive elements of one chain are at most ), so at most chains can fit.