GPT 6 Astra Ultra attempt on Math-5407

Erdős Problem 70: countable order types in triple colourings

1. Statement and scope

Let $\mathfrak c$ be the initial ordinal of cardinality $2^{\aleph_0}$. The relation $\mathfrak c\to(\beta,n)^3_2$ means that every colouring of three-element subsets by $0,1$ has either a colour-zero homogeneous subset of order type $\beta$, or a colour-one homogeneous $n$-element subset. Is this true for every countable $\beta$ and every finite $n\ge2$? We prove the cases $\beta\le\omega$ for every $n$, and all countable $\beta$ for $n\le3$. The general question remains unresolved.

2. Infinite Ramsey's argument, with order type accounted for

We prove by induction on $r\ge1$ that every two-colouring of the $r$-subsets of an infinite subset of $\mathbb N$ has an infinite homogeneous subset. For $r=1$ this is the infinite pigeonhole principle. Given the result for $r-1$, choose an element $x_1$, and apply the induction hypothesis to the colouring $A\mapsto c(\{x_1\}\cup A)$ on $(r-1)$-subsets of the remaining infinite tail. Obtain an infinite tail $X_1$ on which this colouring is constant, say $e_1$. Choose $x_2\in X_1$, remove its finite initial segment, and repeat inside that tail, producing a nested infinite set $X_2$ and a colour $e_2$. Continuing gives $x_1<x_2<\cdots$, with all later selected points in $X_i$. Infinitely many $e_i$ have one colour $e$. Restrict to their corresponding $x_i$. In any $r$-subset of this subsequence, the other $r-1$ elements lie in the stabilized tail of its least element, so the subset has colour $e$.

Apply the result for $r=3$ to the initial copy of $\omega$ inside $\mathfrak c$. If the resulting infinite set is colour one, choose any $n$ of its elements. Otherwise it has order type exactly $\omega$ and contains an initial segment of every order type $\beta\le\omega$. This proves the claimed range without confusing countability with arbitrary countable order type.

3. Small $n$ and monotonicity

For $n=2$, any two-element set is vacuously colour-one homogeneous for triples. For $n=3$, either a colour-one triple exists, or all triples have colour zero, in which case the initial segment of type $\beta$ works. Every countable $\beta$ is below $\mathfrak c$, so there is no size obstruction. If a partition relation holds, decreasing either target order type or finite target size preserves it, by taking a subset of the corresponding homogeneous set. Increasing the ambient ordinal also preserves it, by restriction to the original initial segment. The problem page records the external Erdős--Rado result $\mathfrak c\to(\omega+j,4)^3_2$ for every finite $j\ge2$. Monotonicity extends that recorded result to every $\beta<\omega\cdot2$ when $n=4$; we do not reproduce its proof here.

4. The exact obstruction to a naive limit step

Finding colour-zero sets of type $\omega+j$ for each finite $j$ does not produce one of type $\omega\cdot2$: the sets supplied for different $j$ need not be nested or compatible, and their mixed triples need not have colour zero. Even an increasing union of chosen homogeneous pieces must control triples meeting several pieces. The elementary Ramsey construction above produces only type $\omega$, whatever the ambient cardinality. Thus the first order type not covered by the recorded $n=4$ range is $\omega\cdot2$, not merely $\omega^2$.

5. Assessment

The elementary ranges and the precise monotonicity deductions are proved. There is no construction of the compatible transfinite homogeneous blocks required for the general case, and no countercolouring. This is a checked extension of earlier GPT-5.2 partial write-ups, with the infinite Ramsey argument supplied directly and the external range distinguished from it. Source: https://www.erdosproblems.com/70.

Completion rate estimate: 15%.