gpt pro 5.2 attempt on Math-5756

Let $\alpha$ be the infinite ordinal $\omega^{\omega^2}$. Is it true that in any red/blue colouring of the edges of $K_\alpha$ there is either a red $K_\alpha$ or a blue $K_3$? For comparison, Specker [Sp57] proved this property holds when $\alpha=\omega^2$ and false when $\alpha=\omega^n$ for $3\leq n<\omega$. Chang proved this property holds when $\alpha=\omega^\omega$ (see [590] ). See [592] for the general case. References [Sp57] Specker, Ernst, Teilmengen von Mengen mit Relationen . Comment. Math. Helv. $1957$, 302-314.

FORMAL RESTATEMENT

Fix the ordinal $\alpha := \omega^{\omega^2}$. Let $[\alpha]^2 := \{\{x,y\}: x,y<\alpha,\ x\neq y\}$. A  red/blue edge-colouring is a map $c:[\alpha]^2\to\{\text{red},\text{blue}\}$. Write $K_\alpha$ for the complete graph on vertex set $\alpha$ with edge-set $[\alpha]^2$.

A  red $K_\alpha$ means: there exists a subset $X\subseteq \alpha$ such that $1$ $(X,<)$ is order-isomorphic to $(\alpha,<)$ (so $X$ has order-type $\alpha$), and $2$ for all distinct $x,y\in X$, $c(\{x,y\})=\text{red}$.

A  blue $K_3$ means: there exist $x<y<z<\alpha$ with $c(\{x,y\})=c(\{x,z\})=c(\{y,z\})=\text{blue}$.

Question: Is it true that every $c:[\alpha]^2\to\{\text{red},\text{blue}\}$ produces either a red $K_\alpha$ or a blue $K_3$? (Equivalently: does $\alpha\to(\alpha,3)^2$ hold for $\alpha=\omega^{\omega^2}$?)

QUICK LITERATURE/CONTEXT CHECK

I will not import any external results beyond what is explicitly stated in the problem text. The text states: the property holds for $\alpha=\omega^2$ (Specker), fails for $\alpha=\omega^n$ for $3\le n<\omega$ (Specker), and holds for $\alpha=\omega^\omega$ (Chang). The present case $\alpha=\omega^{\omega^2}$ is presented as open.

ATTACK PLAN

Proof-track ideas.

• (Inductive block construction) Use the Cantor normal form structure of $\omega^{\omega^2}=\sup_{m<\omega}\omega^{\omega\cdot m}$ and try to build a red copy by iterating a "build a red $\omega^{\omega\cdot(m+1)}$ from many red $\omega^{\omega\cdot m}$" step under the hypothesis "no blue triangle".
• (Canonical/closure argument) Attempt to define, for each vertex, a large red-closed set using the triangle-free-blue constraint and then perform transfinite recursion to obtain a red set of the desired order-type.

Disproof-track ideas.

• (Specker-style colouring) Try to generalize the known counterexamples for $\omega^n$ ($n\ge 3$) by defining a blue triangle-free graph on $\omega^{\omega^2}$ whose red edges avoid a red copy of $\omega^{\omega^2}$.
• (Block-diagonal construction) Decompose $\omega^{\omega^2}$ into $\omega$ many intervals and define colors by relative positions so that every large-order-type subset picks up enough blue edges to prevent being all red, yet without creating a blue triangle.

I did not succeed in either direction. Below are rigorous partial lemmas and sanity checks.

WORK

FAST REALITY CHECK (finite and simplest infinite cases)

• Finite analogue fails quickly: a brute-force check over all $2^6$ red/blue colourings of the edges of $K_4$ finds counterexamples. For instance, colour a single edge blue and all other edges red. Then there is no blue triangle (only one blue edge) and there is no red $K_4$ (since not all edges are red). So "$4\to(4,3)^2$" is false.
• For $\alpha=\omega$ the analogue does hold: any red/blue coloring of $K_\omega$ contains either an infinite red clique (hence a red $K_\omega$) or a blue triangle. A full proof is given as Lemma lem:omega_case.

Lemma 1 (blue neighborhood is a red clique). Let $\alpha$ be any ordinal and let $c:[\alpha]^2\to\{\text{red},\text{blue}\}$ be a colouring with no blue triangle. For any vertex $v<\alpha$, define the blue neighborhood \[ N_{\text{blue}}(v):=\{u<\alpha: u\neq v\text{ and }c(\{u,v\})=\text{blue}\}. \] Then every edge inside $N_{\text{blue}}(v)$ is red; i.e. for all distinct $u,w\in N_{\text{blue}}(v)$ we have $c(\{u,w\})=\text{red}$.

Proof. Fix $v$ and take distinct $u,w\in N_{\text{blue}}(v)$. By definition $c(\{u,v\})=c(\{w,v\})=\text{blue}$. If $c(\{u,w\})=\text{blue}$ also held, then $\{u,v,w\}$ would be a blue $K_3$, contradicting the hypothesis. Therefore $c(\{u,w\})=\text{red}$. Since $u,w$ were arbitrary, all edges inside $N_{\text{blue}}(v)$ are red.$\square$

Lemma 2 (the $\omega$ case, constructive). Let $c:[\omega]^2\to\{\text{red},\text{blue}\}$ be a colouring with no blue triangle. Then there exists an infinite increasing sequence $(x_n)_{n<\omega}$ of naturals such that every edge $\{x_i,x_j\}$ ($i<j$) is red. Equivalently, $K_\omega$ contains a red $K_\omega$.

Proof. We build recursively an increasing sequence $x_0<x_1<x_2<\cdots$ together with infinite sets $A_n\subseteq\omega$ satisfying:

1. $x_n=\min(A_n)$,
2. $A_{n+1}\subseteq A_n\setminus\{x_n\}$,
3. for every $m\le n$ and every $y\in A_{n+1}$, the edge $\{x_m,y\}$ is red.

Start with $A_0:=\omega$ and $x_0:=\min(A_0)=0$. Suppose $A_n$ and $x_n$ are defined and $A_n$ is infinite. Partition $A_n\setminus\{x_n\}$ into \[ B_n:=\{y\in A_n: y\neq x_n,\ c(\{x_n,y\})=\text{blue}\},\qquad R_n:=\{y\in A_n: y\neq x_n,\ c(\{x_n,y\})=\text{red}\}. \] If $B_n$ is infinite, then by Lemma lem:blue_nhood_red the set $B_n$ is a red clique. Also, by the inductive property $3$ with $A_n$ in place of $A_{n+1}$, every earlier $x_m$ ($m<n$) is red-adjacent to every vertex of $A_n$, hence to every vertex of $B_n$. Therefore $\{x_0,\dots,x_{n-1}\}\cup B_n$ is an infinite red clique. In particular, we can extract from it an increasing red clique of order type $\omega$, and we are done.

If instead $B_n$ is finite, then $R_n$ is infinite (since $A_n\setminus\{x_n\}$ is infinite). Define $A_{n+1}:=R_n$ and $x_{n+1}:=\min(A_{n+1})$. Condition $1$ and $2$ hold by construction. Condition $3$ holds because for $m<n$, every $x_m$ was red-adjacent to every element of $A_n$ by the inductive hypothesis, and by definition every element of $A_{n+1}=R_n\subseteq A_n$ is red-adjacent to $x_n$.

Thus either the process terminates at some stage (giving an infinite red clique immediately), or it continues for all $n<\omega$. In the latter case, for any $i<j$, we have $x_j\in A_i$ by repeated application of $2$, hence $c(\{x_i,x_j\})=\text{red}$ by $3$. Therefore $(x_n)_{n<\omega}$ is an infinite increasing red clique as required.$\square$

Lemma 3 (restriction gives a red $\omega^\omega$ inside $\omega^{\omega^2}$ under "no blue triangle"). Assume the theorem stated in the problem text for $\omega^\omega$: every red/blue colouring of $[\omega^\omega]^2$ contains either a red $K_{\omega^\omega}$ or a blue $K_3$. Then for $\alpha=\omega^{\omega^2}$, any colouring $c:[\alpha]^2\to\{\text{red},\text{blue}\}$ with no blue triangle contains a red $K_{\omega^\omega}$.

Proof. Since $\omega^\omega<\omega^{\omega^2}$, the initial segment $S:=\{\xi<\alpha: \xi<\omega^\omega\}$ has order type $\omega^\omega$. Restrict $c$ to $[S]^2$. This restricted colouring has no blue triangle (any blue triangle inside $S$ would also be a blue triangle in $\alpha$). By the assumed theorem for $\omega^\omega$, the restriction must therefore contain a red $K_{\omega^\omega}$. That is, there exists $T\subseteq S$ of order-type $\omega^\omega$ such that all edges in $[T]^2$ are red. Since $T\subseteq\alpha$, this is also a red $K_{\omega^\omega}$ in $K_\alpha$.$\square$

VERIFICATION

• Lemma lem:blue_nhood_red: checked that the only needed hypothesis is "no blue triangle"; no ordinal-specific assumption is used.
• Lemma lem:omega_case: the recursion maintains that $A_n$ is infinite unless we already have an infinite red clique from an infinite $B_n$. The conclusion produces an increasing red clique, hence order type $\omega$.
• Lemma lem:restrict_omegaomega: the only nontrivial input is the $\omega^\omega$ statement, which is explicitly mentioned in the problem text (attributed to Chang). The rest is a restriction argument.

FINAL

UNRESOLVED

(i) Strongest proved partial result. If a colouring $c:[\omega^{\omega^2}]^2\to\{\text{red},\text{blue}\}$ has no blue triangle, then (by restricting to an initial segment of type $\omega^\omega$ and using the $\omega^\omega$ result stated in the problem text) it contains a red $K_{\omega^\omega}$ (Lemma lem:restrict_omegaomega). In particular, any counterexample to $\omega^{\omega^2}\to(\omega^{\omega^2},3)^2$ must still contain very large red cliques (of type $\omega^\omega$).

(ii) First gap (crisp). Given a blue-triangle-free colouring $c:[\omega^{\omega^2}]^2\to\{\text{red},\text{blue}\}$, prove or refute that there exists $X\subseteq\omega^{\omega^2}$ of order type $\omega^{\omega^2}$ with all edges in $[X]^2$ red.

(iii) Top 3 next moves.

1. Prove an "upgrade lemma": assuming no blue triangle, show that from cofinally many red copies of $\omega^{\omega\cdot m}$ one can build a red copy of $\omega^{\omega\cdot(m+1)}$.
2. Develop a canonical form for triangle-free-blue colourings on $\omega^{\omega^2}$ (e.g. by analyzing colors between blocks in the natural decomposition $\omega^{\omega^2}=\sup_{m<\omega}\omega^{\omega\cdot m}$).
3. Search for an explicit Specker-type triangle-free-blue colouring on $\omega^{\omega^2}$ that destroys red copies of order type $\omega^{\omega^2}$; even a concrete candidate on the subordinal $\omega^{\omega\cdot 2}$ would be informative.

(iv) Minimal counterexample structure. A minimal counterexample would be a colouring on $\omega^{\omega^2}$ with no blue triangle and no red $K_{\omega^{\omega^2}}$; by Lemma lem:restrict_omegaomega it would still contain a red $K_{\omega^\omega}$. Thus any counterexample must allow very large red cliques while preventing their extension to order type $\omega^{\omega^2}$, likely by forcing many blue edges between different "$\omega^\omega$-blocks" without ever creating a blue triangle.