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Erdős problem 1167

Erdős Problem 1167. Let r2r \geq 2 be finite, γ2\gamma \geq 2, 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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FormalConjectures/ErdosProblems/

1167.lean

Retained formal statement6 of 8

Infinite-target case. When all κα0\kappa_\alpha \geq \aleph_0 are infinite and bounded by λ\lambda, κα+1=κα\kappa_\alpha + 1 = \kappa_\alpha, so the hypothesis simplifies to a "pure" stepping-down lemma: 2λ(κα)α<γr+1    λ(κα)α<γr.2^\lambda \to (\kappa_\alpha)_{\alpha<\gamma}^{r+1} \implies \lambda \to (\kappa_\alpha)_{\alpha<\gamma}^r. The condition καλ\kappa_\alpha \leq \lambda is needed to avoid a size obstruction: without it, the conclusion would require a subset of λ\lambda of size κα>λ\kappa_\alpha > \lambda, which is impossible (see infinite_targets_needs_bound).

FormalConjectures/ErdosProblems/1167.leanErdos1167.erdos_1167.variants.infinite_targets11 linesExact file
∀ (r : ℕ),  2 ≤ r    ∀ (lam : Cardinal.{u}),      Cardinal.aleph0lam        ∀ (γ : Ordinal.{u}),          2 ≤ γ →            ∀ (κ : γ.ToTypeCardinal.{u}),              (∀ (i : γ.ToType), Cardinal.aleph0 ≤ κ i) →                (∀ (i : γ.ToType), κ ilam) →                  Combinatorics.cardinalPartitionRel (2 ^ lam) (r + 1) γ κ →                    Combinatorics.cardinalPartitionRel lam r γ κ
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