Trunk in a tree

Question

Just to make things clear: the author defines a trunk of a tree by formula in [...] in $4.3.1(a)$ in the snippet below. Is it true that since $\eta_0\in T$ is maximal then for all $\nu\in T$ it holds vacuously that $lg(\nu)\leq lg(\eta_0)$ ?

Answer 1

No, the definition allows $\nu\in T$ with $\operatorname{\ell g}(\nu)>\operatorname{\ell g}(\eta_0)$. It imposes just two requirements on them:

• $\nu\upharpoonright\operatorname{\ell g}(\eta_0)=\eta_0$, since $\operatorname{\ell g}\big(\nu\upharpoonright\operatorname{\ell g}(\eta_0)\big)\le\operatorname{\ell g}(\eta_0)$; and
• there is a $\mu\in T$ such that $\operatorname{\ell g}(\mu)\le\operatorname{\ell g}(\nu)$, and $\mu\ne\nu\upharpoonright\operatorname{\ell g}(\mu)$.

The first follows from the fact that $\eta_0$ is the trunk of $T$, and the second follows from the fact that $\nu$ is longer than the trunk.

The terminology really is descriptive: every sequence in $T$ that is no longer than $\eta_0$ is an initial segment of $\eta_0$, and every longer sequence in $T$ has $\eta_0$ as an initial segment. All of the branching occurs above $\eta_0$.