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edited Jul 30 2011 at 21:22 |
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 $\omega_1\not\rightarrow(\omega+2)^3_2$.
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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7
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edited Jul 29 2011 at 13:56 |
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
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.
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edited Jul 29 2011 at 13:47 |
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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5
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edited Jul 29 2011 at 13:15 |
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 $\omega_n\not\rightarrow(\omega+n+1)^{n+2}_3$.
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4
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edited Jul 29 2011 at 12:38 |
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
now 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 ${\alpha,\beta}$ \{\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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edited Jul 29 2011 at 12:30 |
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
now 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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edited Jul 29 2011 at 12:18 |
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
now 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 coloring of the unordered triples of $2^\omega$ (ordered lexicographically) with two colors such that
every infinite homogeneous set has ordertype either $\omega$ or $\omega^*$.
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1
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answered Jul 29 2011 at 12:12 |
Here is what I know about this (and 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
now 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 $\alpha$.
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