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OP-9383OpenLogic and foundations

Erdős #1167 · set theory / ramsey theory

Let r≥2r\geq 2 be finite and λ\lambda be an infinite cardinal. Let κα\kappa_\alpha be cardinals for all α<γ\alpha<\gamma.

Is it true that

2λ→(κα+1)α<γr+12^\lambda \to (\kappa_\alpha+1)_{\alpha<\gamma}^{r+1}

implies

λ…\lambda…
MSC 03 · Mathematical logic and foundations

Inferred from the community field; finer classification has not been reviewed.

Source reports OPENAs of 2026-09-08Source & evidence ↓
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Problem statement

Let r≥2r\geq 2 be finite and λ\lambda be an infinite cardinal. Let κα\kappa_\alpha be cardinals for all α<γ\alpha<\gamma.

Is it true that

2λ→(κα+1)α<γr+12^\lambda \to (\kappa_\alpha+1)_{\alpha<\gamma}^{r+1}

implies

λ→(κα)α<γr?\lambda \to (\kappa_\alpha)_{\alpha<\gamma}^{r}?

Here ++ means cardinal addition, so that κα+1=κα\kappa_\alpha+1=\kappa_\alpha if κα\kappa_\alpha is infinite.

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Background

A problem of Erdős, Hajnal, and Rado, sometimes known as the 'negative stepping-up lemma'.

In [ErHa71] they note that the most difficult case is when r=2r=2, one κα\kappa_\alpha is singular, and the others are finite; in this case they could not prove the statement even assuming GCH.

In [EHMR84] they prove this when

  • All κα\kappa_\alpha are finite,
  • κ0\kappa_0 and κ1\kappa_1 are infinite and κ0\kappa_0 is regular,
  • r≥3r\geq 3, κ0\kappa_0 is infinite and regular,
  • r≥3r\geq 3 and κ0\kappa_0 and κ1\kappa_1 are infinite,
  • r≥4r\geq 4 and κ0\kappa_0 is infinite.

Progress & source assessment

The source labels this entry OPEN in the 2026-09-08 snapshot. Editorial progress notes are reproduced as source commentary, not independently verified results. Prize amounts and Lean badges at the source are not local awards or verification.

References
  1. [1]Thomas Bloom (ed.), Erdős Problems #1167 link
  2. [2]Upstream Lean 4 statement for Erdős #1167 linkExternal statement reference. Not compiled or verified by this platform; upstream files may contain open proof obligations.
  3. [3]Formal Conjectures — Erdős Problem 1167 link

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