GPT 6 Astra Ultra attempt on Math-5474

Erdős Problem 171: density Hales--Jewett

1) Statement and scope

For each $t\geq1$ and $\epsilon>0$, must every $A\subseteq\{1,\ldots,t\}^N$ of size at least $\epsilon t^N$ contain a combinatorial line when $N$ is sufficiently large? A line has a nonempty set $S\subseteq\{1,\ldots,N\}$ of variable coordinates: its $i$th word has symbol $i$ in every coordinate of $S$ and a fixed common symbol at each remaining coordinate.

The general assertion is the established density Hales--Jewett theorem of Furstenberg and Katznelson, with a later combinatorial proof by the Polymath project. This attempt gives a complete elementary proof for $t\leq2$. It does not reconstruct the theorem for all alphabets.

2) Two symbols reduce to an antichain problem

Identify a word in $\{1,2\}^N$ with the set of positions containing $2$. Two different words form a combinatorial line precisely when their associated sets $B,C$ satisfy $B\subsetneq C$: the variable coordinates are $C\setminus B$, with the other coordinates fixed. Consequently a family without a line is an antichain of subsets of $[N]$.

Choose a permutation of $[N]$ uniformly at random, and consider the chain consisting of its initial sets of lengths $0,1,\ldots,N$. For a fixed $k$-set $B$, the probability that this chain contains $B$ is $1/\binom Nk$, since exactly $k!(N-k)!$ of the $N!$ permutations put its members first. A chain meets an antichain $\mathcal A$ at most once, so taking expectations gives \[ \sum_{B\in\mathcal A}\frac1{\binom N{|B|}}\leq1. \] Every binomial coefficient is at most $\binom N{\lfloor N/2\rfloor}$. Therefore \[ |\mathcal A|\leq\binom N{\lfloor N/2\rfloor}. \] The middle layer attains this bound. Stirling's formula now gives \[ 2^{-N}\binom N{\lfloor N/2\rfloor} \sim\sqrt{\frac2{\pi N}}\longrightarrow0. \] For fixed $\epsilon>0$ this is smaller than $\epsilon$ for all large $N$. Thus any family of density at least $\epsilon$ has a line when $t=2$. This proof also gives a threshold of order $\epsilon^{-2}$, up to an absolute constant. For $t=1$ and $N\geq1$, the unique word itself is a line under the stated definition, so the assertion is immediate.

3) The missing general step

For $t\geq3$, an inclusion chain of sets no longer encodes all $t$ words of a combinatorial line. The preceding antichain argument does not yield a density bound tending to zero for these alphabets. One needs the general density-increment or ergodic argument; none is provided here. The published theorem resolves the original problem, but citing its conclusion would not extend the proof just given.

4) Outcome and sources

Outcome of this attempt: UNRESOLVED in full; the cases $t=1,2$ are proved. The estimate describes this attempt's mathematical coverage, not the current status of the already solved problem. No new result or local Lean verification is claimed.

• D. H. J. Polymath, A new proof of the density Hales--Jewett theorem, Annals of Mathematics 175 $2012$, 1283--1327, https://annals.math.princeton.edu/2012/175-3/p06, https://arxiv.org/abs/0910.3926.
• T. F. Bloom, Problem 171, https://www.erdosproblems.com/171, checked 23 September 2026.

5) COMPLETION ESTIMATE

COMPLETION: 20%