A note on maximal triangle-free graphs

We show that a maximal triangle-free graph on n vertices with minimum degree δ contains an independent set of 3δ − n vertices which have identical neighborhoods. This yields a simple proof that if the binding number of a graph is at least 3/2 then it has a triangle. This was conjectured originally by Woodall. We consider finite undirected graphs on n vertices with minimum degree δ. A maximal triangle-free graph is one which does not contain the triangle K3 but the addition of any edge would create a triangle. Equivalently, it is a triangle-free graph of diameter two. We say that two (nonadjacent) vertices of a graph are similar if they have the same neighborhoods. Similarity is obviously an equivalence relation. In this paper we show that in a maximal triangle-free graph there is a similarity class of size at least 3δ − n. As a consequence we obtain a short proof that if the binding number of a graph is at least 3/2 then the graph contains a triangle. We denote the set of neighbors of a vertex x by N(x) and the degree of x by deg (x). For a set S of vertices, the neighborhood of S, denoted N(S), is given by the set of all vertices which are adjacent to a vertex in S (i.e. ⋃ v∈S N(v)). Then the binding number of the graph is the minimum of |N(S)|/|S| taken over all nonempty sets S of vertices such that N(S) is not the whole graph. Further, we denote the number of vertices in the similarity class of vertex x by s(x). Theorem 1 Let G be a maximal triangle-free graph on n vertices with minimum degree δ. Then there is a vertex v such that s(v) ≥ δ + 2deg (v)− n. In particular, if G has no pair of similar vertices then δ ≤ (n+ 1)/3. Research supported by NSF grant DMS 9108403 and NSA grant MDA 904-92-H-3029.