Problem Statement
Let $G$ be a graph on at most $\aleph_1$ vertices which contains no $K_4$ and no $K_{\aleph_0,\aleph_0}$ (the complete bipartite graph with $\aleph_0$ vertices in each class). Is it true that\[\omega_1^2 \to (\omega_1\omega, G)^2?\]What about finite $G$?
Categories:
Graph Theory Ramsey Theory Set Theory
Progress
Erdős and Hajnal proved that $\omega_1^2 \to (\omega_1\omega,3)^2$. Erdős originally asked this with just the assumption that $G$ is $K_4$-free, but Baumgartner proved that $\omega_1^2 \not\to (\omega_1\omega, K_{\aleph_0,\aleph_0})^2$.Source: erdosproblems.com/597 | Last verified: January 15, 2026