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Is it consistent with ZFC to have a cardinal $\kappa$ which is not Ramsey and $\kappa \rightarrow [\kappa]^{<\omega}_{\omega,n}$ holds for some $n\in \omega$?

The partition relation $\kappa \rightarrow [\kappa]^{<\omega}_{\omega,n}$ holds iff for every function $f:[\kappa]^{<\omega}\rightarrow \omega$, there is an $A\in [\kappa]^\kappa$ such that for each $l\in \omega$, $|f''[A]^l|\leq n$.

Ramsey cardinals have $\kappa \rightarrow (\kappa)^{<\omega}_\omega$, which is equivalent to $\kappa \rightarrow [\kappa]^{<\omega}_{\omega,1}$.

It seems to be similar to a "$n$-Rowbottom cardinal", but any definition I've seen of $\gamma$-Rowbottom cardinals are for $\gamma>\omega$, which makes me wonder if "$n$-Rowbottom" is just equivalent to Ramsey.

For example the definition from Kanamori's book:

"If $\omega < \gamma < \kappa$, then $\kappa $ is $\gamma$-Rowbottom iff $\kappa \rightarrow [\kappa]^{<\omega}_{\lambda,<\gamma}$ for any $\lambda<\kappa$"

I guess I'm asking if $n>1$ and $\kappa \rightarrow [\kappa]^{<\omega}_{\omega,n}$ implies $\kappa \rightarrow (\kappa)^{<\omega}_\omega$. I don't think it does, but it would be nice to see an example where cardinals with partition relations of this form are used.

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Suppose (for example), that for every coloring $c:[\kappa]^{< \omega} \rightarrow \omega$, there is a set $H \in [\kappa]^\kappa$ such that $(\forall n \in \omega)|c([H]^n)| \leq 16$. Note that this implies that cofinality of $\kappa$ is at least $\omega_1$. Let us try to bring $16$ down to $1$. Assume that this is impossible for some coloring $c$. Then for every $\kappa$ sized subset $A \subseteq \kappa$, there is some $m$ and there are arbitrarily long sequences $s_1, s_2, \dots s_k \in [A]^{m}$ such that max$(s_i) <$ min$(s_{i+1})$ and $c(s_i) \neq c(s_{i+1})$. –  Ashutosh May 2 '13 at 21:57
    
Define a new coloring $d:[\kappa]^{< \omega} \rightarrow \omega^{< \omega}$ such that for $s \in [\kappa]^{< \omega}$, $d(s)$ encodes the $c$-colors of all finite increasing sequences in $s$. Get an $H \in [\kappa]^{\kappa}$ such that, $(\forall n \in \omega)|d([H]^n)| \leq 16$. Let $s_1^{0}, s_1^{1}, s_2^{0}, s_2^1, \dots , s_5^0, s_5^1 \in [H]^m$ be a sequence of sets such that max of each set is less than the min of next one and $c(s_i^0) \neq c(s_i^1)$, for $1 \leq i \leq 5$. But this meanss $|d([H]^{5m}| \geq 32$. –  Ashutosh May 2 '13 at 21:57
    
Thank you, this helped me. –  Greg May 28 '13 at 18:10
    
I'm not sure whether these are related to your question, but you might look at the Ramsey-like cardinals Victoria Gitman studies in her JSL papers: boolesrings.org/victoriagitman/2008/08/31/ramsey-like-cardinals. –  Zach N May 31 '13 at 6:27

2 Answers 2

Suppose (for example), that for every coloring $c:[\kappa]^{< \omega} \rightarrow \omega$, there is a set $H \in [\kappa]^\kappa$ such that $(\forall n \in \omega)|c([H]^n)| \leq 16$. Note that this implies that cofinality of $\kappa$ is at least $\omega_1$. Let us try to bring $16$ down to $1$. Assume that this is impossible for some coloring $c$. Then for every $\kappa$ sized subset $A \subseteq \kappa$, there is some $m$ and there are arbitrarily long sequences $s_1, s_2, \dots s_k \in [A]^{m}$ such that max$(s_i) <$ min$(s_{i+1})$ and $c(s_i) \neq c(s_{i+1})$. Define a new coloring $d:[\kappa]^{< \omega} \rightarrow \omega^{< \omega}$ such that for $s \in [\kappa]^{< \omega}$, $d(s)$ encodes the $c$-colors of all finite increasing sequences in $s$. Get an $H \in [\kappa]^{\kappa}$ such that, $(\forall n \in \omega)|d([H]^n)| \leq 16$. Let $s_1^{0}, s_1^{1}, s_2^{0}, s_2^1, \dots , s_5^0, s_5^1 \in [H]^m$ be a sequence of sets such that max of each set is less than the min of next one and $c(s_i^0) \neq c(s_i^1)$, for $1 \leq i \leq 5$. But this meanss $|d([H]^{5m}| \geq 32$.

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In what follows there's no math so don't read unless you are jobless. I had posted this answer a long time ago but for some reason only half of the text would show up so I deleted it out of frustration and copy pasted it in two comments. For some strange reason the question has showed up again on my list of "active" questions and amazingly, the deleted answer is showing the whole text hence this undeletion. –  Ashutosh Jul 14 '13 at 0:28

This is not a complete answer, but clears up some of the confusion.

After a bit more research, the definition of $\kappa \rightarrow [\kappa]^{<\omega}_{\alpha,\beta}$ is not the one I stated. This is where the confusion over Rowbottom cardinals stemmed from.

The real definition is for every $f:[\kappa]^{<\omega}\rightarrow \alpha$ there is an $A\in [\kappa]^\kappa$ such that $|f"[A]^{<\omega}|\leq \beta$.

With this new definition, $f:[\kappa]^{<\omega}\rightarrow \omega$, $f(\eta)=\mbox{ length of } \eta$,

contradicts $\kappa \rightarrow [\kappa]^{<\omega}_{\omega,n}$, (as for any $A\in [\kappa]^\kappa$, $f"[A]^{<\omega}=\omega$). So that this never holds.

However, the question still stands, with the old definition in the last post, (which is a bit weaker) does this imply $\kappa$ is Ramsey? (this seems to be a slight generalization of $\kappa\rightarrow (\kappa)^{<\omega}_\omega$)

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You shuold edit your question rather than post the edit as an answer! –  Benoît Kloeckner May 2 '13 at 17:41

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