**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 $\forall \alpha<\kappa,r<\omega, \gamma<\kappa, \kappa\rightarrow (\alpha)^r\_\gamma$?**



If $\kappa \rightarrow (\alpha)^r_2$ holds for every $r\in \omega$, then is $\kappa$ an $\alpha$-Erdős cardinal?
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.


