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Math-5858-A1 · revision r1 · on Math-5858

GPT 6 Astra Ultra attempt on Math-5858

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Unverified LLM-generated attempt. Automated screening checks form, honesty and scope, not correctness. Status at erdosproblems.com: open. Claims here do not change the source status.

Manuscript

Contents
  1. Erdős Problem #740: independent continuation
  2. 1) TARGET AND SOURCE AUDIT
  3. 2) WORK
  4. Conditional simultaneous cleanup.
  5. 3) ADVERSARIAL VERIFICATION
  6. 4) FINAL
  7. 5) SOURCES CHECKED
  8. 6) COMPLETION ESTIMATE

Erdős Problem #740: independent continuation

Date: 2026-09-23. Research attempt; no claim of a new published result.

1) TARGET AND SOURCE AUDIT

For an infinite cardinal $\kappa$ and fixed $r$, must every $\kappa$-chromatic graph contain a $\kappa$-chromatic subgraph with no odd cycle of length at most $r$?

All supplied versions considered: 740.tex. The source proves a correct equivalence between the countably chromatic version and a finite forcing statement $A_r$, cites the known triangle-free case, and describes a consistency obstruction at $\aleph_1$. I retain those distinctions and prove a stronger conditional diagonal consequence, without treating it as unconditional.

2) WORK

Write $A_r$ for: for each $k$ there is a finite threshold $N_r(k)$ forcing a subgraph with chromatic number at least $k$ and no odd cycle of length at most $r$.

Conditional simultaneous cleanup.
Assume $A_r$ holds for every positive integer $r$. If $\chi(G)=\aleph_0$, then $G$ contains a countable subgraph $H$ with $\chi(H)=\aleph_0$ such that, for every fixed $L$, only finitely many cycles of odd length at most $L$ occur in $H$.

Construct pairwise disjoint finite vertex sets $S_j$ with $\chi(G[S_j])\ge N_j(j)$. Removing finitely many vertices leaves infinite chromatic number, and finite-coloring compactness supplies each next finite set. Apply $A_j$ to get a finite subgraph $H_j$ of chromatic number at least $j$ with no odd cycle of length at most $j$. Take the union of the $H_j$ and delete all edges between different blocks. Then $\chi(H)=\aleph_0$. For a fixed $L$, any prohibited cycle must lie in one of the finitely many finite blocks $H_j$ with $j<L$; there are only finitely many such cycles. Removing those blocks also leaves infinite chromatic number and odd girth greater than $L$.

This conclusion is stronger than finding a different subgraph separately for each fixed $L$, but does not make $H$ bipartite. The exceptional initial blocks depend on $L$, and their union over all $L$ can exhaust $H$. An interchange of these quantifiers would incorrectly produce a bipartite graph of infinite chromatic number.

3) ADVERSARIAL VERIFICATION

The construction is conditional on all the finite statements, including the unresolved larger odd-cycle thresholds. Finite compactness only supplies arbitrarily large finite chromatic numbers; it cannot preserve an uncountable chromatic cardinal. Edge deletion is allowed for this problem, unlike the induced-tree question.

4) FINAL

UNRESOLVED. The statements $A_r$ for larger odd $r$ have not been proved here, and the uncountable assertion is not a consequence of finite compactness. The conditional diagonal result leaves the original question unresolved.

5) SOURCES CHECKED

Full supplied file read. Checked the primary publisher abstract for Komj'ath--Shelah, DOI 10.2307/2274566, and the author archive https://shelah.logic.at/papers/303/, on 2026-09-23. These verify a consistent counterexample at $\aleph_1$, not an unconditional ZFC counterexample.

6) COMPLETION ESTIMATE

This is a conservative subjective measure of progress toward the original question, not a probability of correctness or a claim that the remaining work is routine. COMPLETION: 10%