GPT 6 Astra Ultra attempt on Math-6226

Erdős Problem 1172 --- GPT 6 Astra Ultra partial attempt

FORMAL RESTATEMENT

Under the generalized continuum hypothesis, decide each ordinal partition relation \[ \omega_3\longrightarrow(\omega_2,\omega_1+2)^2, \qquad \omega_3\longrightarrow(\omega_2+\omega_1,\omega_2+\omega)^2, \] \[ \omega_2\longrightarrow(\omega_1^{\omega+2}+2,\omega_1+2)^2. \] Separately, determine whether GCH is consistent with \[ \omega_2\longrightarrow(\omega_1+\omega)^2_2, \] and whether it is consistent with the stronger simultaneous assertion $\omega_2\longrightarrow(\xi)^2_2$ for every $\xi<\omega_2$.

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 displayed additions, exponentiation, and target sizes are ordinal operations and order types; replacing them with cardinal arithmetic changes the problem.

ATTACK PLAN AND WORK

A single bounded-degree lemma settles a restricted family of colorings in all the displayed questions. Let $\theta$ be either $\omega_2$ or $\omega_3$ and suppose the blue-neighbor set of every vertex has cardinality less than $\theta$. There is a red-homogeneous set of size $\theta$: choose vertices recursively, at stage $\alpha<\theta$ excluding the earlier vertices and their blue neighbors. Regularity of $\theta$ makes this excluded union smaller than $\theta$, so another vertex exists. All pairs of chosen vertices are red.

A subset of the initial ordinal $\theta$ with size $\theta$ has order type $\theta$. It therefore contains every smaller requested target as an initial segment. In particular, $\omega_1^{\omega+2}+2$ has cardinality $\aleph_1$ and lies below $\omega_2$: ordinal products and the countable supremum in that expression do not increase cardinality beyond $\aleph_1$. The targets on $\omega_3$ likewise lie below $\omega_3$.

This restricted result does not need GCH. Its failure as a general proof is specific: arbitrary blue neighborhoods may have full domain cardinality, so the recursive exclusion can exhaust the domain at its very first step. Reversing red and blue only handles a second restricted family.

VERIFICATION AND REMAINING GAP

Handle colorings with full-sized neighborhoods in both colors, while retaining the exact ordinal targets and distinguishing GCH implications from consistency questions. The bounded-degree lemma is not a forcing or consistency construction.

SOURCES

https://www.erdosproblems.com/1172

https://www.erdosproblems.com/latex/1172

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.