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Let $T$ be a normal Suslin tree. We define a set $S$ to be the collection of branches in $T$. Given an element $x \in T$, it has $\omega$ successors in the level above - order these in the same way as $\mathbb{Q}$. Now we construct a dense linear ordering for $S$: given two branches $a,b$ in $T$, let $\alpha$ be the smallest level where the branches differ. The two differing points $a_{\alpha}$ and $b_{\alpha}$ must lie on a successor level and are immediate succesors of a point in the level below, so we can order them as rational numbers - we say $a < b$ if $a_{\alpha} < b_{\alpha}$ and $a > b$ if $a_{\alpha} > b_{\alpha}$.

Now given an open interval $(a,b) \subseteq S$, I want to show there is some $x \in T$ such that $I_x \subseteq (a,b)$, where $I_x := \{ c \in S \mid x \in c\}$. I'm a little bit stumped here, can anybody help me out? My text (Jech) claims that $I_x$ is an interval, i'm not sure why this is true either. Any help would be appreciated.

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up vote 3 down vote accepted

Take $\alpha$ to be the smallest level where $a$ and $b$ differ. As you noticed, $\alpha = \beta + 1$ for some $\beta$, and $a_\beta = b_\beta$; call this node $z$. Since $a < b$, it follows that in the ordering of the successors of $z$ we have $a_\alpha < b_\alpha$. Since the successors of $z$ are densely ordered, there is a successor $x$ of $z$ such that $a_\alpha < x < b_\alpha$.

Consider any $c \in S$ containing $x$. Note that $c$, $a$ and $b$ all agree up to the $\beta$th spot, but since $c_\alpha = x$ we have that $a_\alpha < c_\alpha < b_\alpha$ and so by definition $a < c < b$. Therefore $I_x \subseteq (a,b)$.

I think that the $I_x$ are "intervals" in the following more abstract way: $(a,b) \subseteq I_x$ for all $a , b \in I_x$ with $a < b$. But you don't even need this.

Suppose you have an uncountable family $\{ ( a_\alpha , b_\alpha ) \}_{\alpha < \omega_1}$ of pairwise disjoint open intervals. To each you can associate some $x_\alpha \in T$ such that $I_{x_{\alpha}} \subseteq ( a_\alpha , b_\alpha )$, but only countably many of these $I_{x_{\alpha}}$ can be pairwise disjoint!

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Cool, I understand what you're saying :) Thanks! – Paul Slevin Mar 29 '12 at 13:38

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