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Level A · Machine-checkable Hard Combinatorics P-erdos-1167

Erdős Problem #1167

Erdős Problem 1167. Let r ≥ 2 be finite, γ ≥ 2, and λ be an infinite cardinal. Let κ_α > r be cardinals for all α < γ. Is it true that 2^λ → (κ_α + 1)_α < γ^r+1 implies λ → (κ_α)_α < γ^r? Here + means cardinal addition, so that κ_α + 1 = κ_α if κ_α is infinite. A problem of Erdős, Hajnal, and Rado.

From the catalogue. Imported from The Formal Conjectures Authors (Google DeepMind and contributors) (Apache-2.0) — original. Nobody has started on it here yet: tasks are created as soon as someone asks for one or submits a claim.

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@misc{cairn-erdos-1167,
  title        = {Erdős Problem #1167},
  author       = {{Cairn Commons contributors}},
  howpublished = {\url{https://cairn-commons.com/problems/erdos-1167}},
  year         = {2026},
  note         = {Open problem on Cairn Commons, CC BY 4.0. Accessed 2026-09-29}
}

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Current state

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The problem

The question

erdos_1167. Erdős Problem 1167. Let be finite, , and be an infinite cardinal. Let be cardinals for all . Is it true that implies Here means cardinal addition, so that if is infinite.

A problem of Erdős, Hajnal, and Rado.

finite_targets. Finite-target case. When all are finite, is the ordinary natural-number successor. Special case of erdos_1167.

binary_colors. Binary-color case. The specialization (two color classes).

infinite_targets. Infinite-target case. When all are infinite and bounded by , , so the hypothesis simplifies to a "pure" stepping-down lemma: The condition is needed to avoid a size obstruction: without it, the conclusion would require a subset of of size , which is impossible (see infinite_targets_needs_bound).

r_eq_two. case. The stepping-down from 3-uniform to 2-uniform partition relations: implies . Generalises the classical Erdős–Rado stepping-up/down theorem for pairs.

Formal statement (Lean 4)

From Formal Conjectures, module FormalConjectures.ErdosProblems.«1167» (5 statements). answer(sorry) marks a yes/no question: a proof of the statement on the right of ↔, or of its negation, answers it.

theorem erdos_1167 : answer(sorry) ↔
    ∀ (r : ℕ), 2 ≤ r →
    ∀ (lam : Cardinal.{u}), ℵ₀ ≤ lam →
    ∀ (γ : Ordinal.{u}), 2 ≤ γ →
    ∀ (κ : γ.ToType → Cardinal.{u}), (∀ α, (r : Cardinal.{u}) < κ α) →
      cardinalPartitionRel ((2 : Cardinal.{u}) ^ lam) (r + 1) γ (fun α => κ α + 1) →
      cardinalPartitionRel lam r γ κ
theorem finite_targets (r : ℕ) (hr : 2 ≤ r) (lam : Cardinal.{u}) (hlam : ℵ₀ ≤ lam)
    (γ : Ordinal.{u}) (hγ : 2 ≤ γ) (n : γ.ToType → ℕ) :
    cardinalPartitionRel ((2 : Cardinal.{u}) ^ lam) (r + 1) γ
        (fun α => (n α : Cardinal.{u}) + 1) →
    cardinalPartitionRel lam r γ (fun α => (n α : Cardinal.{u}))
theorem binary_colors (r : ℕ) (hr : 2 ≤ r) (lam : Cardinal.{u}) (hlam : ℵ₀ ≤ lam)
    (κ : (2 : Ordinal.{u}).ToType → Cardinal.{u}) (hκ : ∀ α, (r : Cardinal.{u}) < κ α) :
    cardinalPartitionRel ((2 : Cardinal.{u}) ^ lam) (r + 1) 2 (fun α => κ α + 1) →
    cardinalPartitionRel lam r 2 κ
theorem infinite_targets (r : ℕ) (hr : 2 ≤ r) (lam : Cardinal.{u}) (hlam : ℵ₀ ≤ lam)
    (γ : Ordinal.{u}) (hγ : 2 ≤ γ) (κ : γ.ToType → Cardinal.{u}) (hκ : ∀ i, ℵ₀ ≤ κ i)
    (hκ_le : ∀ i, κ i ≤ lam) :
    cardinalPartitionRel ((2 : Cardinal.{u}) ^ lam) (r + 1) γ κ →
    cardinalPartitionRel lam r γ κ
theorem r_eq_two (lam : Cardinal.{u}) (hlam : ℵ₀ ≤ lam)
    (γ : Ordinal.{u}) (hγ : 2 ≤ γ) (κ : γ.ToType → Cardinal.{u})
    (hκ : ∀ α, (2 : Cardinal.{u}) < κ α) :
    cardinalPartitionRel ((2 : Cardinal.{u}) ^ lam) 3 γ (fun α => κ α + 1) →
    cardinalPartitionRel lam 2 γ κ

What counts as progress

  • A Lean proof of one of the statements above, pinned as the claim's formal statement.
  • Partial results: special cases, weaker bounds, reductions — as verified claims.
  • Computations and numerical evidence with published code (reproducible).
  • Literature: the problem may have been solved or partly solved already — check erdosproblems.com/1167. Report it as a literature claim.
  • A precise flaw in the formal statement (a misformalisation) — report it upstream too.

References

erdosproblems.com/1167

The original Erdős–Hajnal problem list gives the additional conditions , , and . Without , the statement is false: taking and with gives a counterexample, since the partition relation with one color degenerates to a cardinality comparison (see erdos_1167.unrestricted_is_false). Without it is also false: with , and targets , the premise holds, but a constant colouring of the pairs of has neither a red pair nor a blue set of size .

Source and licence

Imported from Formal Conjectures (Erdős problems), commit e6d1743831c2. Statements and descriptions © The Formal Conjectures Authors, Apache License 2.0; reformatted for this page.