Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
1  
What do you do when there isn't a 'level above'? IIRC, your idea works fine at successor stages, but it's not immediately clear that a path can be picked out at limit stages. (Also, I believe you have a typo and mean 'every branch in T is countable' in your characterization?) –  Steven Stadnicki May 23 '12 at 20:05
    
Fixed, sorry about the typo. I see what you mean now, obviously if I have chosen a path going to level $\beta$ for $\beta < \alpha$ a limit, I can't just make a simple choice at level $\alpha$. –  Paul Slevin May 23 '12 at 20:08
1  
The point is that all branches "decay" into limit levels. So at every limit level only a few of the branches reaching it will pass through. –  Asaf Karagila May 23 '12 at 20:09
2  
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.) –  Andres Caicedo May 23 '12 at 20:19

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.