# Aronszajn tree which cannot be a Suslin tree

I am doing some exercises from Kunen's book "Set Theory", I'm having problems with exercise 39 from page 90, the exercise goes like this: Show that any Aronszajn tree which is a subtree of $\{ s \in \omega^{< \omega_1}: s$ is $1-1\}$ cannot be Suslin tree. Hint: For each $n \in \omega$, $\{s \in T: \exists \alpha (dom(s)= \alpha + 1 \wedge s(\alpha)=n)\}$ is an antichain.

Can anyone help me with this exercise? PLEASE!

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Let $T_n=\{s\in T: \exists \alpha\ (dom(s)=\alpha+1 \wedge s(\alpha)=n)\}$. Let us show that $T_n$ is an antichain. Suppose $s,t \in T_n$ and $s\leq t$ then there are $\alpha< \beta$ so that $dom(s)=\alpha+1$ and $dom(t)=\beta+1$ but this implies that $t(\alpha)=t(\beta)=n$ which is a contradiction since $t$ is one-to-one. Now let us prove that there are no Suslin subtrees. If $S\subseteq \{s\in\omega^{<\omega_1}: s \ \text{is} 1\text{-}1\}$ then $S^+=\bigcup_{\alpha<\omega_1} Lev_{\alpha+1}( S)$ is a countable union of anti chains, since $S^+\subseteq \bigcup_{n\in\omega} T_n$ since $S$ is Aronszajn one of the chains must be uncountable. Thus, $S$ contains an uncountable anti chain so it is not Suslin.
Thank you very much, this helped me a lot, I have a question, I know that $S^+$ is a union of antichains but, why is it countable? and why $S^+ \subseteq \bigcup_{n\in \omega} T_n$ ? –  luciana Apr 12 '12 at 20:09
@analucia If $s:\alpha+1\to \omega$ is 1-1 and $s(\alpha)=n$ then $s\in T_n$. –  azarel Apr 12 '12 at 20:11