# 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:

1. 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 $\supset$?
2. 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.

• What you say at the end of the first paragraph is not true: $\kappa(\omega)$ is in general much larger than $\kappa(\omega)^L$. In fact, for all we know the latter is countable in $V$. Presumably what was meant is that if $\kappa\to(\omega)^{<\omega}$ then $L\models\kappa\to(\omega)^{<\omega}$. – Andrés E. Caicedo Mar 19 '16 at 16:07
• Thank you @AndrésCaicedo. I've just edited my question :) – Cesare Mar 19 '16 at 16:37
• Your second question should be fixed similarly. – Andrés E. Caicedo Mar 19 '16 at 16:38

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 $\alpha$.
• Thank you very much @Andrés Caicedo! Which is the reason for what that proof fails when we consider $\kappa(\omega^L_1)$? Is it because "being a branch of size $\omega_1$" is not absolute? – Cesare Mar 19 '16 at 17:55
• Yes, exactly. The existence of a branch of length $\omega$ through a tree is equivalent to the tree being ill-founded, which is an absolute condition. On the other hand, even if $\omega_1=\omega_1^L$, a tree may have a branch of length $\omega_1$ in $V$ but none in $L$. Such an example can be obtained using a Suslin tree. – Andrés E. Caicedo Mar 19 '16 at 21:29