GPT 6 Astra Ultra attempt on Math-5755
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Erdős Problem #590: the ordinal $\omega^\omega$
Restricted coloring theorem with the order type preserved, 24 September 2026.Exact statement
Does $\omega^\omega\to(\omega^\omega,3)^2$ hold? The large red clique must have order type $\omega^\omega$ in the given order, not merely countably many vertices. Chang proved the affirmative answer, later strengthened and shortened by Milner and Larson. We prove the full conclusion when the blue graph has finite chromatic number, and explain precisely why this leaves the general triangle-free case open in this attempt.Finite partitions of the underlying ordinal
First, every finite partition of $\omega^n$, for a positive integer $n$, has a class of order type $\omega^n$. For $n=1$, this is the infinite pigeonhole principle. For the induction step, decompose $\omega^{n+1}$ into consecutive blocks of type $\omega^n$. In each block the induction hypothesis gives a full-type subset in one class. The same class occurs for infinitely many block indices; its union in those blocks has type $\omega^n\cdot\omega=\omega^{n+1}$.Next write $\omega^\omega$ as the ordinal sum of consecutive blocks of types $\omega,\omega^2,\omega^3,\ldots$. Given finitely many classes, each block contains a full-type subset in one class. One class is selected for infinitely many, hence unbounded, block indices $n_j$. Their union has type \[\sum_{j<\omega}\omega^{n_j}=\omega^\omega. \] For completeness, its order type is at most the ambient ordinal, while for every finite $m$ it contains a subset of type $\omega^m$ in a selected block. Its type is consequently at least $\sup_m\omega^m=\omega^\omega$. This proves the finite-partition assertion.
Applying the lemma to an edge coloring
Form the graph whose edges are exactly the blue pairs. If this graph admits a proper coloring with $q<\omega$ colors, each vertex color class is a red clique. The partition assertion supplies a class of order type $\omega^\omega$, giving the desired red clique. In particular this proves the statement whenever blue degrees are bounded by a finite integer $d$: enumerate the countable vertex set and greedily assign one of $d+1$ colors, avoiding the colors of earlier blue neighbors.There is another useful sufficient condition. If a vertex has a blue neighborhood of order type $\omega^\omega$ and there is no blue triangle, that entire neighborhood is red-complete: a blue pair inside it would complete a blue triangle through the vertex. Thus an unsuccessful coloring in the present approach must have infinite blue chromatic number and every blue neighborhood of smaller order type.