# Partition calculus question

For $m,n,k < \omega$, consider the equation

$X \to (\omega \times k)^{m}_{n}$

What is the smallest $X$ known to satisfy it?

Baumgartner-Hajnal theorem gives a satisfactory answer for $m=2$, but it fails badly for $m>2$, and actually I hardly managed to find anything in the literature concerning $m>2$. The only bound I know is the one given by Erdos-Rado - $\beth_{\omega}$. So $\aleph_{\omega}$ works assuming $GCH$. Can it be proved to work in $ZFC$? Any improvements and/or references are very welcome.

Also I would be happy to hear any results for the same equation with $\omega$ replaced by an arbitrary ordinal $\lambda$.

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It is known that $\mathbb Q\not\rightarrow(\omega+1,4)^3$. I.e., there is a coloring of the unordered triples of rationals with two colors $0$ and $1$ such that no set of rationals of ordertype $\omega+1$ is homogeneous of color $0$ and no set of size 4 is homogeneous of color 1.

On the other hand, Milner and Prikry showed $\omega_1\rightarrow(\omega\cdot 2,4)^3$ and Jones proved $\omega_1\to (\omega+m,n)^3$ for all $n,m\in\omega$. Milner and Prikry conjectured that $\omega_1\rightarrow(\alpha,n)^3$ holds for all $n\in\omega$ and all countable $\alpha$.

Note that all the positive results are unsymmetric, i.e., they only give finite homogeneous sets in one of the colors.

Also, there is a definable (continuous, actually) coloring of the unordered triples of $2^\omega$ (ordered lexicographically) with two colors such that every infinite homogeneous set has ordertype either $\omega$ or $\omega^*$.

Finally, let us consider this coloring: for each $\alpha\lneq\omega_1$ fix a wellordering $\leq_\alpha$ of $\alpha$ of ordertype $\leq\omega$.
Given three ordinals $\alpha\lneq\beta\lneq\gamma\lneq\omega_1$, let $c(\alpha,\beta,\gamma)=0$ if $\leq_\alpha$ agrees with the usual ordering of ordinals on $\{\alpha,\beta\}$ and otherwise let $c(\alpha,\beta,\gamma)=1$.

I think there is no homogeneous set of color $0$ of ordertype $\omega+2$ and no homogeneous set of color $1$ of ordertype $\omega+1$. This would show $\omega_1\not\rightarrow(\omega+2)^3_2$.

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Thanks for your answer, Stefan. However, it does not quite address my concern. I mean, these are explications of saying that for $m>2, k>1$ we can not take $\omega_1$ as $X$. But then what can we take? How about $\omega_2$? –  Artem Chernikov Jul 29 '11 at 13:00
I thought for a while that it might be possible to show $\omega_n\not\rightarrow(\omega+n+1)_2^{n+2}$, but the proof I had in mind doesn't seem to work. Hence I removed my conjecture about this from my answer. –  Stefan Geschke Jul 30 '11 at 21:22
I see. But do you still expect it to be true? Also, now I know that what I actually want to solve is $X \to (\omega+k)^{m}_{n}$. Aren't there any positive results at all with bounds better than the one given by Erdos-Rado, that is $\beth_{\omega}$? –  Artem Chernikov Jul 30 '11 at 21:42

I think that the general conjecture $\omega_1\to(\alpha,n)^3$ goes back to Erdos and Rado.

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Thanks, Peter, I was not aware of that, but I have not personally looked at the Milner-Prikry paper either, I have to confess. –  Stefan Geschke Jul 29 '11 at 13:48