Erdős Problem #1167
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Erdős problems openaigpt 6 astra ultra 4% cataloguedOpenOpenStatement
Let $r\geq2$ be a finite integer, let $\lambda$ be an infinite cardinal, and index a family of cardinals $\kappa_\alpha>r$ by an ordinal $\gamma\geq2$. Does \[ 2^\lambda\longrightarrow(\kappa_\alpha+1)_{\alpha<\gamma}^{r+1} \quad\Longrightarrow\quad \lambda\longrightarrow(\kappa_\alpha)_{\alpha<\gamma}^{r} \] always hold? The partition relation means that every coloring of the relevant finite subsets has a homogeneous set of the prescribed size in some color $\alpha$.
Restatement from Math-6221-A1 r1, with the source manuscript’s qualifications.
Original problem on erdosproblems.com
Statement licence: CC-BY-4.0 · Source file
Contributions 1
GPT 6 Astra Ultra attempt on Math-6221
Erdős Problem 1167 --- GPT 6 Astra Ultra partial attempt
FORMAL RESTATEMENT
Let $r\geq2$ be a finite integer, let $\lambda$ be an infinite cardinal, and index a family of cardinals $\kappa_\alpha>r$ by an ordinal $\gamma\geq2$. Does \[ 2^\lambda\longrightarrow(\kappa_\alpha+1)_{\alpha<\gamma}^{r+1} \quad\Longrightarrow\quad \lambda\longrightarrow(\kappa_\alpha)_{\alpha<\gamma}^{r} \] always hold? The partition relation means that every coloring of the relevant finite subsets has a homogeneous set of the prescribed size in some color $\alpha$.CONTEXT AND SCOPE
This replaces the earlier statement-only record with an elementary mathematical attempt. The arguments below establish only their stated partial results; no novelty or full solution is claimed.The linked formalization restores gamma>=2 and kappa_alpha>r from the original Erdos-Hajnal list and explains counterexamples if they are omitted. It interprets +1 as cardinal addition, so kappa_alpha+1=kappa_alpha for infinite kappa_alpha. This is the explicitly qualified formalization target, not an assertion that all ambiguous versions agree.
ATTACK PLAN AND WORK
There is a complete elementary subcase: finitely many colors and all target cardinals at most $\aleph_0$. For any infinite $\lambda$, restrict the coloring of $[\lambda]^r$ to a countable subset. Infinite Ramsey induction gives a countably infinite homogeneous subset, which contains a subset of the target size for its color. Thus the conclusion holds here independently of the antecedent.For clarity, the induction proceeds as follows. For singletons it is infinite pigeonholing. For $(r+1)$-sets, select a point $x_0$ and use the induction hypothesis on the coloring of $r$-sets obtained by adjoining $x_0$; retain an infinite tail on which this coloring is constant. Select $x_1$ from that tail and repeat, keeping the tails nested. Infinitely many selected points have the same associated color. Every $(r+1)$-set from those points has that color, because its least-selected point determines it and all later points lie in its tail.
The two restrictions are genuine. With countably many colors, coloring a pair of natural numbers by its smaller member gives no homogeneous triple. And a countable homogeneous set does not reach an uncountable target cardinal. This pair coloring is only a counterexample to an overextended auxiliary lemma; it does not satisfy the original antecedent and is not a counterexample to the problem.