Loosely speaking, my question is "how high can a tree be?".
A tree is a partially ordered set $(N,<)$ such that
- for every element of $N$, the set of predecessors of this element is linearly ordered,
- there is an element of $N$ called the root that is a predecessor of every other element of $N$,
- every linearly ordered subset of $N$ is finite,
- every $n\in N$ has at most countably many immediate successors (as defined in the obvious way).
The height of each element $n\in N$ is defined as the following ordinal:
- If $n$ has no successors, then $n$ has height $1$
- If $n$ has successors, then the height of $n$ is the supremum of the heights of the successors of $n$ plus $1$
What is the smallest ordinal that is not the height of an element of a tree?
The smallest such ordinal is $0$ because you defined your rank (height) inappropriately (only successor ordinals are possible). You want to define the rank of a node without successors as $0$, and of a node $a$ with successors as the supremum of the set $\{\alpha+1\mid\alpha$ is the rank of an immediate successor of $a\}$. With this modification, the smallest ordinal not a rank is now $\omega_1$.
It cannot be smaller, since any countable ordinal $\alpha$ defines a tree whose root has rank $\alpha$ in the obvious way: $0=\emptyset$ defines such a tree with a unique node (the root). Having defined trees for all $\beta<\alpha$, consider the tree with $\alpha$ at the root, the elements of $\alpha$ as its immediate successors (indexed by these elements), and the corresponding tree structure inductively defined on each $\beta<\alpha$ attached to the node indexed $\beta$. It is easy to see this is a tree according to your definition (well-foundedness ensures the finiteness condition).
We cannot have trees with elements of uncountable height because of your fourth requirement. The point is that if $a$ is a node of uncountable height, then one of its successors also has uncountable height (although strictly smaller), because there are only countably many successors, and the supremum of a countable collection of countable ordinals is countable. This gives us a contradiction by considering a decreasing sequence of such nodes.