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If $\kappa \rightarrow (\alpha)^r_2$ holds for every $r\in \omega$, then is $\kappa$ an $\alpha$-Erdős cardinal? (or rather, does $\kappa \rightarrow (\alpha)^{<\omega}_2$ hold?)

$\kappa \rightarrow (\alpha)^r_2$ means that $\forall f: [\kappa]^r \rightarrow 2$, $\exists A\in[\kappa]^\alpha$ such that $|f"[A]^r|=1$.

$\kappa \rightarrow (\alpha)^{<\omega}_2$ is $\forall f: [\kappa]^{<\omega} \rightarrow 2$, $\exists A\in[\kappa]^\alpha$ such that $\forall l \in \omega$, $|f"[A]^l|=1$.

Clearly if $\kappa \rightarrow (\alpha)^r_2$ holds for every $r\in \omega$, then splitting $f:[\kappa]^{<\omega}\rightarrow 2$ into functions $f_r:[\kappa]^r\rightarrow 2$ and applying the partition relations, we get that for each $l\in \omega$ we can find an $A_l\in[\kappa]^\alpha$ where $|f"[A_l]^l|=1$. But we can't simply take the intersection of these $A_l$ as this may not have size $\alpha$. (It could easily be empty!)

We can also find the $A_l$ such that $A_i\supseteq A_{i+1}$ for all $i$, by restricting the $f_r:[A_{r-1}]^r\rightarrow 2$ and finding homogenous sets for these. But again, the intersection of all of these may not have size $\alpha$.

In particular, does this hold when $\alpha=\kappa$?

What about if $\kappa \rightarrow (\alpha)^r_\gamma$ holds for every $r\in \omega$? Then does $\kappa \rightarrow (\alpha)^{<\omega}_\gamma$ hold?

If this is not the case, then is there a name for these $\kappa$ where $\kappa \rightarrow (\alpha)^r_2$ holds for every $r\in \omega$?

EDIT: In view of the answers; $\kappa$ is weakly compact implies $\kappa\rightarrow (\kappa)^r_\gamma$ for any $r<\omega$, and $\gamma<\kappa$. But does $\forall \alpha<\kappa$, $\kappa \rightarrow (\alpha)^2_2$ imply $\forall \alpha<\kappa,r<\omega, \gamma<\kappa, \kappa\rightarrow (\alpha)^r_\gamma$?

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2 Answers

up vote 8 down vote accepted

The two properties are not the same.

If $\kappa$ is weakly compact, then $\kappa\rightarrow(\kappa)^r_\lambda$ for any $r<\omega$ and $\lambda<\kappa$. These cardinals are compatible with $V=L$.

A cardinal $\kappa$ for which $\kappa\rightarrow(\kappa)^{<\omega}_2$ is called a Ramsey cardinal, and these cardinals imply that $0^\sharp$ exists.

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Thanks. But why does $\kappa$ weakly compact imply $\kappa\rightarrow (\kappa)^r_\lambda$ for all $r$? How do you get this from $\kappa\rightarrow (\kappa)^2_2$? –  user33625 May 31 '13 at 13:06
    
Oh, that's an old induction argument using the "inaccessible + tree property" characterization of weakly compact cardinals. Kanamori's "Higher Infinite" and Jech's "Set Theory" have it as an exercise, but Hajnal and Hamburger's "Set Theory" has it written out on page 220 if you've got access to that. –  Todd Eisworth May 31 '13 at 13:33
    
Thanks again, I must have just missed this page in Kanamori! –  user33625 May 31 '13 at 13:38
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Unfortunately $\kappa\rightarrow(\alpha)^r_2$ for every $r$ is much weaker than $\kappa\rightarrow(\alpha)^{<\omega}_2$. In fact, the least $\kappa$ for which $\kappa\rightarrow(\alpha)^r_2$ holds is $\exp_{r-1}(|\alpha|)^+$ (the general Erdos-Rado theorem), and so the least $\kappa$ for which this holds for all $r$, is $\exp_\omega(|\alpha|)$, a singular cardinal. Here $\exp_r(\tau)$ is the $r$-fold iterated power of $\tau$.

If $\alpha$ is limit, then the least $\kappa$ for which $\kappa\rightarrow(\alpha)^{<\omega}_2$ holds, is strongly inaccessible.

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What about when $\kappa\rightarrow(\alpha)^r_2$ for all $\alpha<\kappa$ and $r<\omega$? I guess it won't make $\kappa \rightarrow (\alpha)^{<\omega}_2$. Though by your answer such a $\kappa$ would be strongly inaccessible since it'd be larger than $\exp_\omega(|\alpha|)$ for every $\alpha<\kappa$. Would $\kappa\rightarrow(\alpha)^2_2$ for all $\alpha<\kappa$ imply $\kappa\rightarrow(\alpha)^r_2$ for all $\alpha<\kappa$ and $r<\omega$? –  user33625 May 31 '13 at 14:25
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