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Let $T$ be a normal Suslin tree, so among other things for each $x \in T$, there is some $y > x$ at each higher level less than $\omega_1$, and every branch in $T$ is at most countable. My question is, how can both of these be true? Can we not construct a branch of length $\omega_1$ by induction starting at a root and choosing an element in the level above?

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    A simple example that may help you clarify things: Consider the collection of strictly increasing sequences of rationals, ordered by \sigma<\tau iff the sequence $\sigma$ is a proper initial segment of the sequence $\tau$. This is a tree. It has height $\omega_1$. It has no $\omega_1$ branches. (This is not quite an $\omega_1$-tree in that levels can be uncountable. But examples of Aronszajn trees can be obtained by thinning out this tree.)2012-05-23

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