Let $\lambda$ be a limit ordinal and let $T$ be a tree such that for every element $t \in T$ and every $\beta < \lambda$, there is a branch of length at least $\beta$ that contains $t$. Does it follow that $T$ has a branch of length $\lambda$?
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asked Sep 21, 2022 at 9:48
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No. Aronszajn trees are the classical example here. The formal definition of an Aronszajn tree is simply a tree of height $\omega_1$ where every level is countable. This tells you nothing about your requirement. However, the standard construction, indeed the one due to Aronszajn, is the "model tree", in a sense.
This is a tree of height $\omega_1$ (with countable levels), such that any node $t$ we can find a branch, in the form of a maximal chain, of height $\beta$, for any limit ordinal $\operatorname{ht}(t)<\beta<\omega_1$ which contains $t$. However, there are no branches of length $\omega_1$ in an Aronszajn tree.
In most, if not all, investigations of these sort of trees we tend to begin by making these assumptions: every point is a splitting point; every point has extensions arbitrarily high; any point in a limit level is the limit of a unique branch.
Aronszajn trees have the stricter condition that each level is countable, as well. We can generalise these to $\kappa>\omega_1$ which involves interesting combinatorics and large cardinal axioms.
answered Sep 21, 2022 at 9:59
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$\begingroup$ The definition of Aronszajn trees as I know it only requires that it has height $\omega_1$, rather than that the subtree above each node $t$ has height $\omega_1$. I don’t doubt that trees satisfying the stronger condition exist, but it’s not part of the definition. $\endgroup$ Commented Sep 21, 2022 at 10:05
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1$\begingroup$ Emil, while that is true, literally all work on trees start by making the additional harmless assumptions that eliminate oddities and trivialities: the tree is normal (i.e. any point is a splitting point, every point has an extension to any level above it) and Hausdorff (points at limit levels are always the limit of a unique branch). So while it's not "formally" part of the definition, it is, in some sense, part of the working definition. $\endgroup$– Asaf Karagila ♦Commented Sep 21, 2022 at 10:12
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1$\begingroup$ That’s all fine, but you should still clarify it in the answer. The OP is likely not aware of working conventions in this subfield, lest he would not have asked the question in the first place. $\endgroup$ Commented Sep 21, 2022 at 10:27