Todorcevic's proof on existence of Aronszajn Trees I'm currently using Kanamori's book, The Higher Infinite, and am trying to understand the proof of Theorem 7.10 on page 79. However, I am unable to follow the construction of the $\beta^{\alpha}_i$.
Is this conceptually right: You start with $\beta$. Then you take all chains of length $\beta$, remove $\alpha$ many nodes from each and take the intersection of all the chains remaining. The length of this gives you $\beta^{\alpha}_1$?
If this is correct, what happens if $\beta$ is uncountable and $\alpha = 1$? It tells you to continue the construction until you end up with $\beta^{1}_n = 1$ and I don't understand how this would be possible. (For $\kappa_1$ infinite, $\kappa_1 - \kappa_2 = \kappa_3 \implies \kappa_1 = max(\kappa_2, \kappa_3)$?)
Thank you in advance for any help!
Edit: I can post the construction from the book here, but I don't know if I'm allowed.
 A: No, $\beta_0^\alpha=\beta$, so $\beta_1^\alpha$ is the ordinal $\min(C_\beta\setminus\alpha)$: it’s the smallest member of $C_\beta$ that is greater than or equal to $\alpha$. It exists because $C_\beta$ is cofinal in $\beta$, and it’s necessarily less than $\beta_0^\alpha=\beta$, since $C_\beta\subseteq\beta$. 
If this is greater than $\alpha$, repeat the process with $\beta_1^\alpha$ replacing $\beta_0^\alpha$: $\beta_2^\alpha=\min(C_{\beta_1^\alpha}\setminus\alpha)$ is the smallest member of $C_{\beta_1^\alpha}$ greater than or equal to $\alpha$. Since $C_{\beta_1^\alpha}\subseteq\beta_1^\alpha$, we have $\beta_2^\alpha<\beta_1^\alpha$. Continue in this fashion as long as $\beta_k^\alpha>\alpha$. Since the $\beta_k^\alpha$ are strictly decreasing with $k$, there must be an $n\in\omega$ such that $\beta_n^\alpha=\alpha$; at that point stop (since $C_{\beta_nT\alpha}\setminus\alpha=\varnothing$) and set $\rho_\beta(\alpha)$ equal to the $n$-sequence of the traces on $\alpha$ of the sets $C_{\beta_k^\alpha}$ for $k<n$:
$$\rho_\beta(\alpha)=\langle C_{\beta_k^\alpha}\cap\alpha:k<n\rangle\;.$$
