By B. Bollobás (Eds.)

ISBN-10: 0720408431

ISBN-13: 9780720408430

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I) Extremal graphs without large forbidden subgraphs 37 Clearly, and 0 s e(G")- e(G')c 2/3"2n2 If p is sufficiently small. Therefore instead of proving Theorem 7 for G" it is sufficient to prove it for G'. Hence we may and shall assume that G' = G". (B) Let R k be the graph whose vertices are the classes V, ( i = 1 , .. We prove that R k does not contain a triangle ( K " ) . Let us assume that V,, V, and V, from a triangle in R k . Put U' U" v,: d(x, V , ) c p>, E v,: d(x, V,)G p}. = {x E '{X For any x E U = V,- U'- Utt there exist a U,,x and a U2,xin V, and V, respectively, joined to x completely, where I UL,x 1 z=prn.

D,J>k - s - ' r. Proof. We may assume that r + 2s S 2k - 1 since otherwise there is nothing to prove. Suppose s = 0. Then A n B = A so A U C = V(L) and the vertices of A and C alternate around L. However, this is impossible since then r = 2 m - 2 ( m - k ) = 2 k . Thus sal. For each vertex b , E B - A there is an interval on L consisting of vertices bl, cl, f , , c2, f 2 , . . ,fi ,, c,, a,, where c, E C, f, E A n B and Q, E A -B. There are s such intervals and so the vertices of D also form S intervals, some of which may be empty.

If now we omit all the edges (x, y ) for which x E B,, y E Bt+2p for some z and p then we change the graph into a bipartite one. Hence there exists a pair ( I , p ) for which at least (c'n2- pn2)/l: edges were omitted between B, and Hence there exists a K,(T, T ) joining B, to in the sense that the first (second) class of it is contained in B, (B,+2p). Let these classes be denoted by 0, and ECtzp, respectively. If 0, is already defined, D,-l can also be defined as follows: ID,/ = T, we find t vertices in 0, and 2T vertices in B,-, joined to each other completely.