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.

  • $\begingroup$ Nice question! It is clear at least that there can be $f$ defined on $T|C$ that do not extend to $T$. For example, perhaps $f$ is $0$ on the root, and $f$ is defined on $T_\omega$ in such a way that there are extensions of the node $0$ and of $1$ with arbitrarily small values by $f$. In this case, there is no way to extend $f$ to define $f(0)$ and $f(1)$. $\endgroup$ Mar 6, 2014 at 0:30
  • $\begingroup$ @JoelDavidHamkins Certainly, it does not seem trivial to me. However, in his book "Proper and improper forcing", VII 3.20, Shelah says: "The small gain is that we directly find a function specializing $T$ rather than finding one specializing a closed unbounded set of levels, and then using a theorem saying this is equivalent". But I cannot find a proof or a reference of that theorem in the book or elsewhere. $\endgroup$
    – Carlos
    Mar 6, 2014 at 1:24

1 Answer 1


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 $T = \bigcup_{m,n\lt\omega} B_{m,n}$.


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