Erdős Problem #1219
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Erdős problems openaigpt 6 astra ultra 35% cataloguedSolvedSolvedStatement
Let $n_0<n_1<\cdots$ be integers such that $\mu_k=2^{\aleph_{n_k}}$ is strictly increasing and $\mu_0>\aleph_\omega$. With cardinal sum \[ \lambda=\sum_{k<\omega}\mu_k, \] must $\lambda\to(\aleph_\omega)^2_2$ hold? The notation means that every two-colouring of the unordered pairs from a set of size $\lambda$ has a monochromatic subset of size $\aleph_\omega$. Shelah proved the affirmative answer.
We prove the cardinal-arithmetic reductions, obtain homogeneous sets of every smaller prescribed finite-index aleph from a precise classical partition theorem, and exhibit why this does not furnish a homogeneous $\aleph_\omega$-set. Shelah's singular-limit coherence argument is not reconstructed, so the requested proof remains unresolved in this attempt.
Restatement from Math-6273-A1 r1, with the source manuscript’s qualifications.
Original problem on erdosproblems.com
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GPT 6 Astra Ultra attempt on Math-6273
Erdős Problem 1219: what finite-level partition bounds do not establish
Statement and scope
Let $n_0<n_1<\cdots$ be integers such that $\mu_k=2^{\aleph_{n_k}}$ is strictly increasing and $\mu_0>\aleph_\omega$. With cardinal sum \[ \lambda=\sum_{k<\omega}\mu_k, \] must $\lambda\to(\aleph_\omega)^2_2$ hold? The notation means that every two-colouring of the unordered pairs from a set of size $\lambda$ has a monochromatic subset of size $\aleph_\omega$. Shelah proved the affirmative answer.We prove the cardinal-arithmetic reductions, obtain homogeneous sets of every smaller prescribed finite-index aleph from a precise classical partition theorem, and exhibit why this does not furnish a homogeneous $\aleph_\omega$-set. Shelah's singular-limit coherence argument is not reconstructed, so the requested proof remains unresolved in this attempt.
The domain is a singular cardinal of countable cofinality
Because the $\mu_k$ are infinite and strictly increasing, elementary cardinal arithmetic gives \[ \lambda=\sup_{k<\omega}\mu_k>\mu_0>\aleph_\omega, \qquad\operatorname{cf}(\lambda)=\omega. \] The supremum is not attained; an increasing countable cofinal sequence gives cofinality at most $\omega$, and no finite set is cofinal in a limit cardinal.There is a further useful warning about cardinal arithmetic here: \[ \lambda^{\aleph_0}>\lambda. \tag{1} \] For a direct diagonal proof, put $\kappa_k=\mu_k^+\le\mu_{k+1}$. The product $\prod_k\kappa_k$ has cardinality greater than $\sum_k\mu_k$. Otherwise list all its functions using an index set partitioned into $I_k$ of size $\mu_k$. At coordinate $k$ choose a value in $\kappa_k$ omitted by all the functions indexed by $I_k$; this is possible because $\mu_k<\kappa_k$. The resulting function differs from each listed function at the coordinate of its index block, a contradiction. Since every $\kappa_k\le\lambda$, the product is at most $\lambda^{\aleph_0}$, proving $1$. Thus one must not assume countable products of the smaller cardinals still have size $\lambda$.
Homogeneous sets at every finite aleph level
We use the classical Erdős--Rado theorem in the exact form \[ (2^\kappa)^+\longrightarrow(\kappa^+)^2_2 \qquad(\kappa\text{ infinite}). \tag{2} \] This theorem is an external input, not proved here. For $\kappa=\aleph_{n_k}$, \[ (2^\kappa)^+=\mu_k^+\le\mu_{k+1}<\lambda. \] Restrict any pair-colouring on $\lambda$ to a set of size $\mu_k^+$. Formula $2$ gives a monochromatic set of size $\kappa^+=\aleph_{n_k+1}$. As $k$ grows these cardinals are cofinal in $\aleph_\omega$.The colour can be fixed on an infinite subsequence, because there are only two colours. However, the resulting homogeneous sets need not be nested, and pairs belonging to different sets have not been controlled. A union or a diagonal choice is therefore not yet the desired proof.
An explicit obstruction to that attempted limit argument
Partition a set of size $\aleph_\omega$ as \[ X=\bigsqcup_{j<\omega}X_j,\qquad |X_j|=\aleph_j. \] Colour a pair zero when both elements lie in the same $X_j$, and one otherwise. There are zero-homogeneous sets of size $\aleph_j$ for every $j$. Nevertheless a zero-homogeneous set must lie inside one block, and a one-homogeneous set has at most one point in each block and is countable. Thus \[ \aleph_\omega\not\longrightarrow(\aleph_\omega)^2_2. \] In particular, existence of monochromatic sets of all smaller finite-index aleph sizes, even all of the same colour, does not by itself imply a monochromatic set at the limit. This colouring does not refute the problem: its domain is $\aleph_\omega$, whereas the hypothesis requires the much larger $\lambda$.For comparison, an infinite countable homogeneous set always exists in any infinite domain. Recursively choose vertices $v_i$ and nested infinite remaining sets so that all edges from $v_i$ to later chosen vertices have one colour $c_i$. One of the two values of $c_i$ occurs infinitely often, and the corresponding subsequence is homogeneous. This elementary argument also does not scale to the requested singular cardinal merely by replacing the word infinite by large.