Construction of special $\omega_1$-Aronszajn tree Problem from Kunen II.40: The definition is the following: An $\omega_1$-Aronszajn tree $T$ called _special_ iff $T$ is the union of $\omega$ antichains. Need to prove that $T$ is special iff there is a map $f: T \rightarrow \mathbb{Q}$ such that for $x,y \in T, x < y \rightarrow f(x) < f(y)$, and show that a special Aronszajn tree exist. The hint is to construct $T$ and $f$ simultaneously by induction. It seems like I should somehow "pack" the antichains (equivalence classes?), to achieve a map from a "large" tree to a relatively small set. Couldn't advance any furher than that, though. Any help? Thanks in advance.

Suppose $T$ is a countable union of antichains. We are going to construct a map $g: T\to 2^\omega$ so that range of $g$ is countable and it is strictly increasing (with respect to the lexicographical ordering on $2^\omega$). Pick a function $f: T\to \omega$ so that $f^{-1}(n)$ is an antichain for all $n$. For $t\in T$, define $g(t)=x$ by: $x(n)=1$ if and only if $n\leq f(t)$ and $\\{ s\in T: s\leq t\\}\cap f^{-1}(n)\
e\varnothing$. It is easy to verify that $g$ is as required.

To construct a special A-tree. Consider the subtree of the A-tree constructed in Kunen's book which consists of nodes of successor height.