TITLE: Does being special on a club imply being special?
QUESTION [8 upvotes]: Let $T$ be an Aronszajn-tree, $C\subset \omega_1$ a club set and $f:\bigcup\limits_{\alpha\in C}T_\alpha\longrightarrow \mathbb Q$ a strictly increasing function (where $T_\alpha$ is the $\alpha$-level of $T$). Is $T$ special (i.e. there exists such an $f$ defined on the whole $T$)?
I suspect that this is true and that it is a known fact, but I have not found any reference.

REPLY [6 votes]: This is a classic result that I learned by reading The Souslin Problem by Devlin and Johnsbråten (Lecture Notes in Mathematics 405).
First, recall that another way to think of special trees is that they are the $\omega_1$-trees that admit a cover by countably many antichains. Suppose $C = \{\gamma_\alpha:\alpha\lt\omega_1\}$ is closed unbounded with $\gamma_0 = 0$ and that $\bigcup_{\alpha\lt\omega_1} T(\gamma_\alpha) = \bigcup_{n\lt\omega} A_n$ where each $A_n$ is an antichain. For $t \in T(\gamma_\alpha)$ let $\{t_m : m \lt \omega\}$ be the set of extensions of $t$  with height less than $\gamma_{\alpha+1}$. Observe that each $B_{m,n} = \{t_m : t \in A_n\}$ is an antichain in $T$ and $T = \bigcup_{m,n\lt\omega} B_{m,n}$.