GPT 6 Astra Ultra attempt on Math-6274

Erdős Problem 1220 --- GPT 6 Astra Ultra partial attempt

FORMAL RESTATEMENT

Call a cardinal $\kappa$ $\aleph_0$-inaccessible when $\mu^{\aleph_0}<\kappa$ for every cardinal $\mu<\kappa$. Suppose $\lambda$ is singular and both $\lambda$ and its cofinality $\operatorname{cf}(\lambda)$ have this property. Must \[\lambda\longrightarrow(\lambda,\aleph_1)^2\] hold? Thus every red--blue colouring of the unordered pairs of a set of cardinality $\lambda$ would contain either a red homogeneous set of size $\lambda$ or a blue homogeneous set of size $\aleph_1$.

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 question is asked without GCH; the official remarks state that adding GCH gives a positive answer.

ATTACK PLAN AND WORK

A uniform degree restriction gives a useful positive subcase even at the singular cardinal in the question. Suppose there is a fixed cardinal $\nu<\lambda$ such that every vertex has at most $\nu$ blue neighbors. One can recursively choose $\lambda$ red-homogeneous vertices: at stage $\alpha<\lambda$, the previous vertices and their blue neighborhoods have size at most $|\alpha|(\nu+1)<\lambda$, so a new vertex remains. This product bound uses a fixed $\nu$, and does not require $\lambda$ to be regular. The resulting red set already proves the required partition alternative.

It is crucial not to replace the uniform hypothesis by the pointwise condition that each degree is smaller than $\lambda$. At a singular cardinal, the neighborhood sizes may be cofinal in $\lambda$, and a union indexed by $\operatorname{cf}(\lambda)$ can reach size $\lambda$. The recursion can then fail well before constructing the desired set.

The given inaccessibility condition does control countable patterns inside each smaller set: if $|X|=\mu<\lambda$, the number of countable sequences from $X$ is $\mu^{\aleph_0}<\lambda$. Also the analogous condition on $\operatorname{cf}(\lambda)$ forces it to exceed $2^{\aleph_0}$ by taking the base cardinal two. Neither fact alone produces a uniform degree bound.

VERIFICATION AND REMAINING GAP

Handle full or cofinally increasing blue degrees, either extracting a blue $\aleph_1$-clique or maintaining a red set through the singular limit. The bounded-degree proof does not remove the need for that dichotomy.

SOURCES

https://www.erdosproblems.com/1220

https://www.erdosproblems.com/latex/1220

FINAL

UNRESOLVED. The full source problem remains outside the scope of the proved partial results.

COMPLETION ESTIMATE

$4\%$. This is a conservative subjective estimate toward the whole problem, not a probability that the argument is correct or a claim that a fixed fraction of the problem has been solved.