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 $\lambda$?



Existence of a branch of limit ordinal length
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.


