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Math-6227

Erdős Problem #1173

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Erdős problems openaigpt 6 astra ultra 5% cataloguedOpenOpen

Statement

Assume GCH. Suppose \[ f:\omega_{\omega+1}\longrightarrow[\omega_{\omega+1}]^{\leq\aleph_\omega}, \qquad |f(\alpha)\cap f(\beta)|<\aleph_\omega\quad(\alpha\ne\beta). \] Must there be a free set $H\subseteq\omega_{\omega+1}$ of size $\aleph_{\omega+1}$, meaning that $\beta\notin f(\alpha)$ whenever $\alpha,\beta\in H$ are distinct?

Restatement from Math-6227-A1 r1, with the source manuscript’s qualifications.

Original problem on erdosproblems.com

Statement licence: CC-BY-4.0 · Source file

Contributions 1

Math-6227-A1 · No claim recorded · LLM-generated · gpt 6 astra ultra ·

GPT 6 Astra Ultra attempt on Math-6227

Erdős Problem 1173 --- GPT 6 Astra Ultra partial attempt

FORMAL RESTATEMENT

Assume GCH. Suppose \[ f:\omega_{\omega+1}\longrightarrow[\omega_{\omega+1}]^{\leq\aleph_\omega}, \qquad |f(\alpha)\cap f(\beta)|<\aleph_\omega\quad(\alpha\ne\beta). \] Must there be a free set $H\subseteq\omega_{\omega+1}$ of size $\aleph_{\omega+1}$, meaning that $\beta\notin f(\alpha)$ whenever $\alpha,\beta\in H$ are distinct?

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.

The free-set condition is stated for distinct points, so no unmentioned assumption about whether alpha belongs to f(alpha) is needed.

ATTACK PLAN AND WORK

Set $\kappa=\aleph_{\omega+1}$ and view membership in the values of $f$ as a directed conflict graph. A free set is an independent set in the symmetrized conflict graph. This gives a sufficient extra hypothesis: if every point belongs to fewer than $\kappa$ of the sets $f(\alpha)$, then a free set of size $\kappa$ exists. Indeed each vertex has fewer than $\kappa$ outgoing neighbors by the size assumption and fewer than $\kappa$ incoming neighbors by the extra hypothesis. Regularity of $\kappa$ lets one recursively choose $\kappa$ vertices, avoiding all previous vertices and their incoming and outgoing neighbors.

The original intersection hypothesis does not give this incoming-degree bound. For example $f(\alpha)=\{0\}$ for $\alpha\ne0$ and $f(0)=\varnothing$ has tiny pairwise intersections while zero has $\kappa$ incoming neighbors. In this example deleting zero is enough; the example refutes the proposed inference, not the free-set conclusion.

Conversely, dropping the intersection hypothesis entirely makes the conclusion false. Set $f(\alpha)=\alpha$ on the initial ordinal $\kappa$. Every value has cardinality at most $\aleph_\omega$, but no two distinct points are free. For sufficiently large $\alpha<\beta$, the intersection has size $\aleph_\omega$, which is exactly why this is not a counterexample to the stated problem.

VERIFICATION AND REMAINING GAP

Exploit the strict pairwise-intersection bound to remove or organize the large incoming-degree obstruction. The additional degree hypothesis proved sufficient above is not established from the actual assumptions.

SOURCES

https://www.erdosproblems.com/1173

https://www.erdosproblems.com/history/1173

FINAL

UNRESOLVED. The full source problem remains outside the scope of the proved partial results.

COMPLETION ESTIMATE

$5\%$. 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.

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