Countable tree ranks and monotonicity under extension maps

A branch provides the set of all its prefixes, every member of which has a child in that set; hence the child relation is not well-founded. Conversely, if well-foundedness fails, take a nonempty set D of nodes with no child-minimal member. Starting with one node in D, recursively choose its least-coded child in D. Such a child exists by the defining failure of minimality. Their union, together with the initial node's prefixes in the tree, is an infinite branch. This proves both implications, also for an empty tree, whose relation is vacuously well-founded and whose body is empty.

Induct over the well-founded child relation by F3. If all child ranks are countable, their successors are countable ordinals. There are at most countably many children; pad the sequence by zero at unused alphabet codes. F2, licensed by A1, bounds the supremum of these successor ranks below ${\omega }_1$. By F1 this supremum is the parent rank. Leaf ranks are the empty supremum zero, so the induction proves all node ranks countable, including the root; the empty-tree rank is zero by F1.

For an extension-preserving f, induct on the source child relation by F3. For each child u of s, induction gives $r_S(u)\le r_T(f(u))$. Since f(u) properly extends f(s), a finite chain of target child steps and F1 gives $r_T(f(u))<r_T(f(s))$. Thus $r_S(u)+1\le r_T(f(s))$. Taking the supremum over all children proves $r_S(s)\le r_T(f(s))$, including a leaf whose rank is zero.

Fix $\alpha <{\omega }_1$ and an injection $c:\alpha \to \mathbb{N}$. Let T consist of the root and the coordinatewise c-codes of finite strictly decreasing sequences of ordinals below $\alpha$. It is a tree. It has no infinite branch, since an infinite descending ordinal sequence would have a least value followed by a smaller value; hence it is well-founded by step 1.1. By F4, every node ending at $\beta$ has rank $\beta$: its children end at exactly $\gamma <\beta$, whose ranks by induction are $\gamma$, and ${\sup }_{\gamma <\beta }(\gamma +1)=\beta$. The same formula at the root gives rank $\alpha$. For $\alpha =0$ the tree is root-only and the supremum is zero. QED.