# Erdős–Rado theorem generalisation for infinitely many colours

I'll ask my question and then explain the background. Thanks in advance.

Question: Is it the case that $$\beth_{\omega}\rightarrow (\beth_{\omega})_{{\aleph_{0}}}^{n}$$ for finite $$n$$? If not, what more generally can be said about cardinals $$\kappa$$ such that $$\kappa\rightarrow (\kappa)_{{\aleph_{0}}}^{n}$$ holds, for finite $$n$$?

Background:

Wikipedia succinctly states the Erdős–Rado theorem as follows: $$\beth_{n}^{+}\rightarrow (\aleph_{1})_{{\aleph_{0}}}^{{n+1}}$$ In words and paraphrasing slightly this says: if we countably colour the $$n{+}1$$-size subsets of a set $$X$$ of cardinality greater than $$\beth_{n}$$, then there exists a homogeneous subset of uncountable cardinality.

Then it re-states a more general form: $$\exp_{n}(\kappa )^{+}\longrightarrow (\kappa^{+})_{\kappa }^{{n+1}}$$ An $$\aleph_0$$-colouring is also a $$\kappa$$-colouring for $$\kappa\geq\aleph_0$$, and $$\beth_\omega > \exp_n(\beth_i)$$ for finite $$i$$, so from this it seems to me that we can derive for every finite $$i$$ that $$\beth_{\omega}\longrightarrow (\beth_i^{+})_{\aleph_0}^{{n+1}} .$$ This suggests that we might try to take a limit as $$i$$ rises towards $$\omega$$ --- and we arrive at my question.

• The answer to your first question is no, even for $n=1$, as $\beth_\omega\not\to(\beth_\omega)^1_{\aleph_0}$.
• To start with, if $\kappa$ is an infinite cardinal such that $\kappa\to(\kappa)^2_2$, then $\kappa$ must be strongly inaccessible; and increasing the number of colours from $2$ to $\aleph_0$ makes no difference except in the case $\kappa=\aleph_0$.