Research notes on every problem, and a library of the papers behind them. Built from the open erdos repository.
Updated
Claim. The answer to Problem 118 is no, and the failure at occurs already at . Larson's abstract states that whenever with there is a graph on the ordinal with no independent set of order type and no pentagram, a set of five points pairwise joined by edges; in arrow notation, . At this gives , which with the positive relation of Schipperus and Darby makes a partition ordinal whose two-colorings need not contain a red copy of itself or a blue . Schipperus (Countable partition ordinals, p. 1197) reports that Larson also proved , so that is the exact boundary for this ; that positive relation is not stated in the abstract and is recorded here as Schipperus's report.
Depends on. Schipperus 1999 for the positive relation , which the counterexample needs and which Larson's abstract does not claim.
Sources. The abstract, as the publisher's record gives it, states the negative relation only, and the statement above is the abstract's. Nothing on this page is independently reviewed by this project.
Source. Jean A. Larson, An ordinal partition avoiding pentagrams, J. Symbolic Logic 65 (2000), no. 3, 969--978, doi:10.2307/2586684. The issue is dated September 2000 and carries no day, so this page's date is the first of that month.
Acceptance. Reviewed: the curator of erdosproblems.com, T. F. Bloom,
marks Problem 118 disproved and records that Larson showed the statement
false at and (problem page last edited 17
January 2026). The paper appeared in a refereed journal,
but refereed is not listed because the abstract does not show that the
paper proves the positive relation the counterexample needs.