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Source. Theorem 1, p. 4, of Stanisław P. Radziszowski and Xu Xiaodong, On the most wanted Folkman graph, Geombinatorics 16 (2007), no. 4, 367--381, read in the authors' manuscript named on the source card; pages here are the manuscript's printed pages 1--15, and the journal pagination was not compared. The theorem is J. Folkman's, Graphs with monochromatic complete subgraphs in every edge coloring, SIAM J. Appl. Math. 18 (1970), 19--24, the paper's reference [5].

Statement

Setting (pp. 3--4, Definitions 1 and 2). Fe(s,t;k)\mathcal{F}_e(s,t;k) is the set of graphs GG with no KkK_k such that every red/blue coloring of the edges of GG has a red KsK_s or a blue KtK_t; Fe(s,t;k)F_e(s,t;k) is the least order of a graph in Fe(s,t;k)\mathcal{F}_e(s,t;k). The vertex versions Fv(s,t;k)\mathcal{F}_v(s,t;k) and Fv(s,t;k)F_v(s,t;k) are defined in the same way with the vertices 2-colored instead of the edges.

Theorem 1 (Folkman 1970, p. 4, quoted). "For all k>max(s,t)k>max(s,t), edge- and vertex- Folkman numbers Fe(s,t;k)F_e(s,t;k), Fv(s,t;k)F_v(s,t;k) exist."

That is, for k>max⁡(s,t)k>\max(s,t) the sets Fe(s,t;k)\mathcal{F}_e(s,t;k) and Fv(s,t;k)\mathcal{F}_v(s,t;k) are nonempty. The paper gives no proof. It adds that k>R(s,t)k>R(s,t) gives Fe(s,t;k)=R(s,t)F_e(s,t;k)=R(s,t) (p. 4), and that Folkman's theorem, instantiated to two colors, settles the existence question of Erdős and Hajnal (1967) with a very large bound for Fe(3,3;4)F_e(3,3;4) (p. 8).

Read depth

Claims checked: Definitions 1 and 2 and the statement of Theorem 1 were read on the page images of the manuscript. The paper cites the theorem; Folkman's paper was not read for this page. Nothing here is independently reviewed.

Dependencies

Folkman's paper, cited above; nothing in the corpus.

Bears on

  • Problem 582: the problem asks whether some K4K_4-free graph has a monochromatic triangle in every 2-coloring of its edges. The case s=t=3s=t=3, k=4k=4 of Theorem 1 says Fe(3,3;4)F_e(3,3;4) exists, which is a yes; the paper says so on p. 8, citing Folkman rather than proving it.