Erdős Problem #1177

Erdős Problem 1177 --- GPT 6 Astra Ultra partial attempt

FORMAL RESTATEMENT

For a finite $3$-uniform hypergraph $G$, write $F_G(\kappa)$ for the class of $G$-free $3$-uniform hypergraphs whose chromatic number is exactly $\kappa$. Determine whether all three assertions hold:

1. $F_G(\aleph_1)\ne\varnothing$ implies the existence of a member with at most $2^{2^{\aleph_0}}$ vertices.
2. For finite $G,H$, nonemptiness of both $F_G(\aleph_1)$ and $F_H(\aleph_1)$ implies $F_G(\aleph_1)\cap F_H(\aleph_1)\ne\varnothing$.
3. For uncountable cardinals $\kappa,\lambda$, nonemptiness of $F_G(\kappa)$ implies nonemptiness of $F_G(\lambda)$.

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.

These concern exact chromatic cardinals, not merely lower bounds. Later external candidate-solution claims were not verified in this attempt and are not adopted as a status change.

ATTACK PLAN AND WORK

There is a natural simultaneous-avoidance construction whose exact limitation can be proved. For 3-uniform hypergraphs $X,Y$, take vertices $V(X)\times V(Y)$; three vertices form an edge when both coordinate projections are edges. Projection is then a hypergraph homomorphism to each factor.

Call a finite 3-uniform hypergraph pair-covered if every pair of its vertices is in some edge. If such a hypergraph $G$ embeds in the product, its projection to either factor is injective: two distinct vertices belong to an edge, whose three coordinate entries must be distinct. The projection therefore embeds $G$ in each factor. Consequently, if $G,H$ are pair-covered, $X$ is $G$-free, and $Y$ is $H$-free, then this product is simultaneously $G$-free and $H$-free.

The chromatic direction is the wrong one for assertion two. Pulling back a proper coloring from either factor gives \[\chi(X\times Y)\le\min(\chi(X),\chi(Y)),\] but supplies no lower bound. In particular, the construction does not preserve chromatic number $\aleph_1$ merely from having that number in both factors. For forbidden graphs that are not pair-covered, even the injectivity step can fail. Disjoint union is not a remedy: it retains forbidden configurations already occurring in either component.

VERIFICATION AND REMAINING GAP

Prove a simultaneous avoidance construction with the exact required chromatic number for arbitrary finite forbidden hypergraphs. The size bound in assertion one and transfer across all uncountable cardinals in assertion three also remain open here.

SOURCES

https://www.erdosproblems.com/1177

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.