Erdös cardinals in $L$ I've readed in Jech's book that the existence of the $\omega-$Erdös cardinal $\kappa(\omega)$ (that is the minimumm cardinal $\kappa$ for which $\kappa\rightarrow (\omega)^{<\omega}$) it is compatible with the axiom of constructibility $V=L$. More precisely, it can be shown that if $\kappa=\kappa(\omega)$ exists then $L\vDash\kappa\rightarrow (\omega)^{<\omega}$.
In this regard, I would like to ask some things about some of the details of that proof. First of all assuming the existence of a monochromatic set $H$ (in $V$) for a colouring $c:[\kappa(\omega)]^{<\omega}\longrightarrow 2$ in $L$, we want to see that there exists a monochromatic set $H'\in L$ for $c$. For this purpouse Jech defines a tree $T$ formed by finite increasing sequences $\langle \alpha_1,\ldots,\alpha_{n-1}\rangle$ such that for every $k\leq n$ then $[\{\alpha_1,\ldots,\alpha_{n-1}\}]^k$ is monochromatic, and endows it with the reverse order $\supset$.
I feel that I understand the reasons why $T\in L$ and $c$ has a monochromatic set iff $T$ is illfounded. Despite of this, I'm not sure of which is the role played by the term ''increasing sequences'' and $\supset$. 
More precisely, my questions are the following:


*

*Why couldn't we take as the elements of the tree any finite sequence like before paying no attention to their order and $\subset$ instead of $\supset$?

*If $\alpha<\omega_1^L$ then is it true that the existence of $\kappa(\alpha)$ implies $L\models\kappa(\alpha)\to(\alpha)^{<\omega}$? If this would be true, how could I prove it?


Best, 
Cesare.
 A: Whether to use $\subset$ or $\supset$ is just a matter of convention: Do your trees grow upwards or downwards? All you want is to have your tree consist of finite approximations to the homogeneous set you are after. 
Similarly, to have the sequences increase does not matter much. The intention is simply that if at the end you have a branch through the tree, that gives you an increasing sequence $\alpha_0<\alpha_1<\dots$, and the set $H=\{\alpha_n\mid n<\omega\}$ is homogeneous and of order type $\omega$. Had we not paid attention to the sequences being increasing, your set $H$ would at the end still be infinite, so its initial segment of type $\omega$ would have been the homogeneous set you were after, but $H$ could in principle be better and have a larger order type.
Actually, that should suggest you how to modify the proof to answer your second question: Fix a bijection $\pi$ in $L$ between $\omega$ and $\alpha$. Now you could take as elements of your tree finite sequences $(\alpha_0,\dots,\alpha_k)$ such that for all $i,j$, we have $\alpha_i<\alpha_j$ iff $\pi(i)<\pi(j)$ (plus, the appropriate homogeneity requirement, of course). At the end, an infinite sequence gives you a (homogeneous) set of order type $\alpha$.
