Reused writeup on Math-5758 (gpt pro 5.2)

Erdős Problem 598 --- collection record

PROVENANCE AND STATUS

Official problem: https://www.erdosproblems.com/598

Saved collaborative-database status: open. Snapshot checked: 2026-09-23T17:22:24Z. Latest status and formalization links: https://teorth.github.io/erdosproblems/

This is an attributed collection of existing writeups. The model-folder names below identify their original attribution; placement in GPT_6_Astra_Ultra does not attribute their mathematical work to that model. All locally available source versions selected for this problem are reproduced below, without editing their text.

Proof claims, computations, literature statements, and completion estimates in the reused text have not been independently verified in this collection pass. A source claiming a solution does not change the saved upstream status.

Primary source (latest numbered version in the preferred model folder): attacks/erdos/gpt_pro_5.2/598.tex.

REUSED SOURCE 1

Original attribution: gpt_pro_5.2. Original file: attacks/erdos/gpt_pro_5.2/598.tex.

Let $m$ be an infinite cardinal and $\kappa$ be the successor cardinal of $2^{\aleph_0}$. Can one colour the countable subsets of $m$ using $\kappa$ many colours so that every $X\subseteq m$ with $\lvert X\rvert=\kappa$ contains subsets of all possible colours?

1) FORMAL RESTATEMENT

Let $m$ be an infinite cardinal, viewed as an initial ordinal, and let \[ \kappa := (2^{\aleph_0})^+, \] the successor cardinal of the continuum. Let $[m]^{\aleph_0}$ denote the set of all countable subsets of $m$.

The question asks whether there exists a function (a colouring) \[ c : [m]^{\aleph_0} \to \kappa \] such that for every subset $X\subseteq m$ with $|X|=\kappa$ we have \[ c\big([X]^{\aleph_0}\big) = \kappa, \] i.e., for every colour $\xi<\kappa$ there exists a countable $A\subseteq X$ with $c(A)=\xi$.

Edge case: if $m<\kappa$ then there is no $X\subseteq m$ of size $\kappa$, so the requirement is vacuous.

2) QUICK LITERATURE/CONTEXT CHECK

No additional results are stated in the problem text. To respect the integrity constraint, I do not import external set-theoretic results here.

3) ATTACK PLAN

• Proof-track idea: attempt an explicit construction of $c$ from a well-ordering of $[m]^{\aleph_0}$, ensuring each colour class "hits" every $X$ of size $\kappa$.
• Disproof-track idea: given an arbitrary colouring $c$, try to diagonalize and build an $X$ of size $\kappa$ that omits some colour, by ensuring $[X]^{\aleph_0}$ avoids a specified colour class.

I did not reach a construction or a diagonal counterexample; the WORK gives necessary conditions and sanity checks.

4) WORK

Fast reality check (vacuity). If $m<\kappa$, then the requirement "for every $X\subseteq m$ with $|X|=\kappa$" has no instances, so any function $c:[m]^{\aleph_0}\to\kappa$ works.

Lemma 598.1 (vacuous case).
If $m<\kappa$, then the answer to the literal question is "yes" (trivially).

Proof. If $m<\kappa$ then there is no subset $X\subseteq m$ with $|X|=\kappa$, so the stated requirement on $X$ is vacuously satisfied by any colouring. $\square$

Lemma 598.2 (each colour class must be $\kappa$-hitting, and large when $m=\kappa$).
Assume $m=\kappa$ and let $c:[\kappa]^{\aleph_0}\to\kappa$ satisfy the requirement. For each colour $\xi<\kappa$ let \[ F_{\xi}:=\{A\in[\kappa]^{\aleph_0}: c(A)=\xi\}. \] Then:

1. For every $X\subseteq\kappa$ with $|X|=\kappa$, there exists $A\in F_{\xi}$ with $A\subseteq X$ (i.e., $F_{\xi}$ hits every $\kappa$-subset).
2. In particular, $|F_{\xi}|\ge\kappa$ for every $\xi<\kappa$.

Proof. $1$ is just a restatement of the defining requirement applied to a fixed colour $\xi$.

$2$ Suppose for contradiction that $|F_{\xi}|<\kappa$ for some $\xi$. Since each $A\in F_{\xi}$ is countable, the union \[ U := \bigcup F_{\xi} \] has cardinality $|U|<\kappa$ because $\kappa$ is a successor cardinal and hence regular: a union of $<\kappa$ many countable sets has size $<\kappa$. Therefore $X:=\kappa\setminus U$ still has $|X|=\kappa$. But then no $A\in F_{\xi}$ can satisfy $A\subseteq X$ (every such $A$ is contained in $U$), contradicting $1$. Hence $|F_{\xi}|\ge\kappa$. $\square$

5) VERIFICATION

• Lemma 598.2 uses regularity of $\kappa=(2^{\aleph_0})^+$, which is standard for successor cardinals: if $\lambda<\kappa$, then $\lambda\cdot\aleph_0<\kappa$.
• The vacuous case $m<\kappa$ is a genuine ambiguity/edge case in the literal wording; in most nontrivial readings one assumes $m\ge\kappa$.

6) FINAL

UNRESOLVED

(i) Strongest proved partial result. If $m<\kappa$ the requirement is vacuous (Lemma 598.1). If $m=\kappa$ and such a colouring exists, then each colour class must be $\kappa$-hitting and has size at least $\kappa$ (Lemma 598.2).

(ii) First gap (crisp). For the nonvacuous case $m\ge\kappa$, either construct a colouring $c:[m]^{\aleph_0}\to\kappa$ with $c([X]^{\aleph_0})=\kappa$ for every $X\subseteq m$ of size $\kappa$, or construct (from an arbitrary $c$) an $X$ of size $\kappa$ and a colour $\xi$ such that no countable subset of $X$ has colour $\xi$.

(iii) Top 3 next moves.

1. Diagonal attempt: given $c$, try to build $X={\langle x_{\xi}:\xi<\kappa\rangle}$ recursively so that for a fixed colour $\xi_*$ every candidate $A\in F_{\xi_*}$ is blocked by omitting at least one element of $A$ from $X$.
2. Constructive attempt: explicitly partition $[m]^{\aleph_0}$ into $\kappa$ many families each of which hits every $X$ of size $\kappa$; Lemma 598.2 shows each family must have size at least $\kappa$.
3. Specialize to $m=\kappa$ first and attempt to exploit the well-ordering of $[\kappa]^{\aleph_0}$ to define $c$ by a transfinite recursion guaranteeing the hitting property for each colour.

(iv) What a minimal counterexample would look like. A counterexample (to existence) would be a proof that for every colouring $c:[m]^{\aleph_0}\to\kappa$ there exists some $X\subseteq m$ with $|X|=\kappa$ and some colour $\xi<\kappa$ such that $F_{\xi}$ fails to hit $X$ (i.e., no $A\subseteq X$ has colour $\xi$).

COMPLETION ESTIMATE

The primary source supplies no extractable numerical completion estimate. Its mathematical progress is not numerically assessed here.

Newly verified progress in this collection pass: $0\%$. This measures new verification work, not the amount of mathematics achieved in the reused text.