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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.

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1 Answer 1

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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 $\{ s\in T: s\leq t\}\cap f^{-1}(n)\ne\varnothing$. 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.

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Thanks for your answer and sorry for the late response (hadn't time to go over the proof this week). I must admit, I'm having trouble understanding it. First, I don't really understand why $g$ is strictly increasing: by choosing $s,t \in T$ all we seem to know, that they are mapped to different $n$ by $f^{-1}$. Next, am I missing something, or the construction of the special tree $T$ by existing $f: T \rightarrow \mathbb{Q}$ is trivial? Could you, please, collaborate more? –  Pavel Jul 13 '12 at 12:05
    
@Pavel If $f: T\to\mathbb Q$ is strictly increasing then $f^{-1}(q)$ is an antichain for all $q\in\mathbb Q$. The rest follows since $\mathbb Q$ is countable. –  azarel Jul 13 '12 at 17:36

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