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.



Does being special on a club imply being special?
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 $T = \bigcup\_{m,n\lt\omega} B\_{m,n}$.


