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Research notes on every problem, and a library of the papers behind them. Built from the open erdos repository.

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Claim. Assume GCH, let κ=ℵωω+1\kappa=\aleph_{\omega_{\omega+1}} and λ=κ+=ℵωω+1+1\lambda=\kappa^+=\aleph_{\omega_{\omega+1}+1}, so that 2κ=λ2^\kappa=\lambda. Then λ↛(λ,4,4)3\lambda\not\to(\lambda,4,4)^3, and so, with more colors available, λ↛(λ,(4)ℵ0)3\lambda\not\to(\lambda,(4)_{\aleph_0})^3. The claimant's summary gives the construction: identify λ\lambda with 2κ2^\kappa, carry both a well-order of type λ\lambda and the lexicographic order of 2κ2^\kappa, and color a triple by where its middle point in the well-order sits lexicographically relative to the other two. The lemma that every subset of 2κ2^\kappa which the lexicographic order well-orders in either direction has size at most κ\kappa means that the first color carries no homogeneous set of size λ\lambda, and a four-point argument rules out homogeneous sets of size 44 in the other two. This page records the construction from the claimant's posted summary.

Submission note. Posted to erdosproblems.com as a proof claim by Yanyang Li (account yanyang_li) on 2 September 2026, giving "GPT 5.6 Sol" as the AI used:

Assuming GCH, let

>λ=ℵωω+1+1andκ=ℵωω+1,>> \lambda=\aleph_{\omega_{\omega+1}+1} \quad\text{and}\quad \kappa=\aleph_{\omega_{\omega+1}}, >

so that 2κ=λ2^\kappa=\lambda. The proof identifies λ\lambda with 2κ2^\kappa and combines a well-order of type λ\lambda with the lexicographic order on 2κ2^\kappa. A colouring of triples is defined according to the lexicographic position of the middle point in the well-order. The key lemma is that every well-ordered or reverse well-ordered subset of (2κ,<lex)(2^\kappa,<_{\rm lex}) has cardinality at most κ\kappa. This rules out a homogeneous set of size λ\lambda in the first colour. A direct four-point argument shows that the other two colours contain no homogeneous set of size 44. Hence

>λ↛(λ,4,4)3,>> \lambda\not\to(\lambda,4,4)^3, >

and therefore, a fortiori,

>λ↛(λ,4,4,…)3.>> \lambda\not\to(\lambda,4,4,\ldots)^3. >

Why it is rejected. The claim was posted as a full proof of Problem 1218 in the wording the site carried until 7 September 2026, with λ=ℵωω+1+1\lambda=\aleph_{\omega_{\omega+1}+1} as both the resource and the first target. On that day the site corrected the first target to ℵωω+1\aleph_{\omega_{\omega+1}}, citing Erdős, Hajnal and Rado (1965, p. 131), after a thread comment of 3 September 2026 reported the transcription error. The site's curator, Thomas Bloom, wrote under the claim on 8 September 2026 that it proves the earlier version of the problem, which contained a typo and already follows from the 1965 work of Erdős, Hajnal and Rado, and that the statement had been updated to the correct, harder version that Erdős and Hajnal asked. The same-cardinal relation is Corollary 13 of P. Erdős, A. Hajnal and R. Rado, Partition relations for cardinal numbers, Acta Math. Acad. Sci. Hungar. 16 (1965), 93--196, p. 138 (on its source card): under GCH, ℵδ↛(ℵδ,4)3\aleph_\delta\not\to(\aleph_\delta,4)^3 for every non-inaccessible ℵδ\aleph_\delta, and the countably many further colors may be left unused. The claim settles no instance of the corrected relation ℵβ+1↛(ℵβ,(4)ℵ0)3\aleph_{\beta+1}\not\to(\aleph_\beta,(4)_{\aleph_0})^3 with β=ωω+1\beta=\omega_{\omega+1}, which has the smaller first target and is the stronger statement; the thread comment of 3 September 2026 notes that a coloring of this type has homogeneous sets of size κ\kappa in its first color. As a claim about Problem 1218 it is therefore rejected, and the problem's standing is untouched by it. The record is kept so that the site's listing of a full proof claim is not misread.

Standing. The claimant is Yanyang Li, who posted the claim on the site's proof-claims tab on 2026-09-02 under the forum name yanyang_li as a full proof of the wording then on the site, naming GPT 5.6 Sol as the system used. The one comment under the claim is the curator's of 8 September 2026, summarized above. The site's label is OPEN (page last edited 7 September 2026, the day of the correction); no one has reviewed or refereed the result, and this corpus has not built it.

Depends on. No page of this wiki.