Research notes on every problem, and a library of the papers behind them. Built from the open erdos repository.
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Statement
Notation as on the definitions page.
Theorem 4 (p. 367, quoted). "Let be regular. Then there exists a graph with , such that, fo. [sic] every and for every vertex-decomposition , of it, for some ."
No GCH is assumed. says that contains a complete -graph for every finite and no infinite complete graph. The paper offers it as the case of a possible strengthening of Theorem 3 in which some member keeps (p. 367), and says it does not know whether the result extends to limit cardinals .
Problem 1 (p. 367). Assume GCH. Is there a graph with and such that every vertex-decomposition of type has a member with ? The paper calls this the simplest unsolved case.
Source. P. Erdős and A. Hajnal, On decomposition of graphs, Acta Math. Acad. Sci. Hungar. 18 (1967), 359--377, doi:10.1007/BF02280296; the edition read is named on the source card.
Read depth. Claims checked: Theorem 4, Problem 1 and the proof on pp. 367--368 were read clause by clause on the page images. Nothing here is independently reviewed.
Proof pointer
Pp. 367--368. Take as vertices the pairs of ordinals below and join when and , the Sierpiński-type graph of two orderings. An infinite complete subgraph would give an infinite decreasing sequence of ordinals, so . For a decomposition into classes, Lemma 2/A (p. 363) gives a class of full lexicographic type , and Lemma 3 (p. 363) finds in it, for every , pairs with , a complete -graph.
Dependencies
Lemmas 2 and 3 of the same paper (p. 363).
Bears on
None directly among the problems this corpus records.