I came across the following definitions of tree and transitive tree, respectively, in the book Modal Logic by P. Blackburn:
Definition 1. A tree is a relational structure $(T, R)$, where:
- $T$, the set of nodes, contains a unique $r \in T$ (the root) st $$\forall t \in T\, (rR^*t), $$ where $R^*$ is the reflexive transitive closure of $R$.
- Every element of $T$ distinct from $r$ has a unique $R$-predecessor. (This is literally what the definition says. I assume it actually means there is a unique maximal element that precedes $t$)
- $R$ is acyclic.
Definition 2. A transitive tree is a strict partial order $(T,<)$ st:
There is a root r satisfying that if $t\neq r $, then $r<t$.
The set $\{s\in T: s<t\}$ of predecessors of $t$ is linearly ordered and finite.
Immediately after the second definition, there is the following
Problem. If $(T,S)$ is a tree, then $(T,S^+)$ is a transitive tree, where $S^+$ is the transitive closure of $S$.
My question is about the cardinality of the set of predecessors with the given definition. In particular, let us take $T=\omega+1$, and $S$ the inverted order given to the ordinal numbers, i.e. $$\alpha S \beta \iff \beta < \alpha$$
where $<$ is the usual order for ordinal numbers: $\beta <\alpha$ iff $\beta\in \alpha$.
My assertion is that this is a counterexample of the problem, since
- $S=S^+$, for $S$ is already transitive.
- $\omega$ is the root of $(T,S)$.
- Every element $\alpha \in T$ different from its root has a unique maximal predecessor, namely $\alpha +1$.
- $S$ is acyclic.
- For any $\alpha$, the set $\{\beta\in T:\beta S\alpha\}=\{\beta\in T:\alpha<\beta\}$ is infinite.
Thus, $(T,S)$ is a "tree" (according to this definition, because of 2, 3 and 4), but not a "transitive tree", because of 1 and 5.
Did I misunderstand something? Is that actually a counterexample? Or am I missing something?