It's well-know that:

If $\kappa$ is a cardinal and $\mu$ is a infinite cardinal, and if we partition $\mu$ into $\kappa$ sets ($\kappa < cof(\mu)$), then one set contains $\mu$ members. In notacion of Ramsey, $\mu \rightarrow (\mu)_{\kappa}^{1}$ is true for $\kappa < cof(\mu)$.

I know that $\mu \rightarrow (\mu)_{\kappa}^{2}$ it's not true in general. Are there condition for $\mu \rightarrow (\mu)_{\kappa}^{2}$ be true?

Actually, I want to prove that $\mu \rightarrow (\mu)_{2^{|I|}}^{2}$ for $|I|$ small for $\mu$.

Note: $\lambda$ is small for $\kappa$ wherever $\{\lambda_j : j \in J\}$ a family of cardinal such that $<\kappa$ and$|J|\le \lambda$, then \begin{equation*} \prod_{j \in J}\lambda_j < \kappa. \end{equation*}


The condition $\kappa\to(\kappa)^2_2$ is quite significant: Any such $\kappa$ larger than $\aleph_0$ is a weakly compact cardinal. These are strongly inaccessible cardinals that are a limit of strongly inaccessible cardinals. In fact, they are a Mahlo limit of Mahlo cardinals, and much more. Weak compactness is a natural rung in the large cardinal ladder. As such, the existence of weakly compact cardinals cannot be established in ZFC (unless ZFC is inconsistent).

The condition $\kappa\to(\kappa)^2_2$ implies the apparently stronger $\kappa\to(\kappa)^n_\lambda$ for any $n<\omega$ and $\lambda<\kappa$. In turn, the condition is implied by natural stronger large cardinal assumptions, such as measurability. Jech's set theory book and Kanamori's monograph on large cardinals are good references to learn the basic properties of weakly compact cardinals, including the equivalence at the beginning of this paragraph.

  • $\begingroup$ And it's obviosly that $\kappa \rightarrow (\kappa)_{\lanbda}^2$ implies $\kappa \rightarrow (\kappa)_2^2$, right? Thank you so much, Andres! $\endgroup$ – Tom Ryddle Mar 10 at 15:52
  • $\begingroup$ Yes, reducing the number of colors preserves positive partition relations. Likewise, in this case, reducing the superscript also preserves them (this is not true in general, but here it is, since $\kappa$ is a limit ordinal). $\endgroup$ – Andrés E. Caicedo Mar 10 at 16:39

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