8 Removed higher dimensional conjecture

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 $\omega_1\not\rightarrow(\omega+2)^3_2$.

Edited after Artem's comment:: I think my example can actually be iterated, and I guess that $\omega_n\not\rightarrow(\omega+n+1)^{n+2}_2$, but I have to think about this some more.

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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 $\omega_1\not\rightarrow(\omega+2)^3_2$.

Edit

Edited after Artem's comment:: I think my example can actually be iterated, as follows:

For every countable ordinal $\alpha$ there is a coloring $c:[\alpha]^2\to 2$ such and I guess that there are no $c$-homogeneous sets of ordertype $\omega+1$. Why? Fix a wellordering $\leq_\alpha$ of $\alpha$ of type $\leq\omega$. For $\beta\lneq\gamma\lneq\alpha$ let $c(\beta,\gamma)=0$ if the usual order agrees with $\leq_\alpha$ on $\{\beta,\gamma\}$.

Now we iterate \omega_n\not\rightarrow(\omega+n+1)^{n+2}_2$, but I have to think about this by showing that if for each ordinal$\alpha$of size$\kappa$there is a coloring$c_\alpha^n$on$[\alpha]^n$with no homogeneous set of type$\omega+m$, then on each$\beta$of size$\kappa^+$there is a coloring$c_\beta^{n+1}:[\beta]^{n+1}\to 2$without homogeneous sets of type$\omega+m+1$. This gives$\omega_n\not\rightarrow(\omega+n+1)^{n+2}_2$.some more. 6 edited body Here is what I know about this (and mostly learned from my student Thilo Weinert): It is known that$\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$\omega_1\not\rightarrow(\omega+2)^3_2$. Edit: I think my example can actually be iterated, as follows: For every countable ordinal$\alpha$there is a coloring$c:[\alpha]^2\to 2$such that there are no$c$-homogeneous sets of ordertype$\omega+1$. Why? Fix a wellordering$\leq_\alpha$of$\alpha$of type$\leq\omega$. For$\beta\lneq\gamma\lneq\alpha$let$c(\beta,\gamma)=0$if the usual order agrees with$\leq_\alpha$on$\{\beta,\gamma\}$. Now we iterate this by showing that if for each ordinal$\alpha$of size$\kappa$there is a coloring$c_\alpha^n$on$[\alpha]^n$with no homogeneous set of type$\omega+m$, then on each$\beta$of size$\kappa^+$there is a coloring$c_\beta^{n+1}:[\beta]^{n+1}\to 2$without homogeneous sets of type$\omega+m+1$. This gives$\omega_n\not\rightarrow(\omega+n+1)^{n+2}_3$.\omega_n\not\rightarrow(\omega+n+1)^{n+2}_2$.

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