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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:

  1. $T$, 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 $R$.
  2. 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 $t$)
  3. $R$ is acyclic.

Definition 2. A transitive tree is a strict partial order $(T,<)$ st:

  1. There is a root r satisfying that if $t\neq r $, then $r<t$.

  2. 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

  1. $S=S^+$, for $S$ is already transitive.
  2. $\omega$ is the root of $(T,S)$.
  3. Every element $\alpha \in T$ different from its root has a unique maximal predecessor, namely $\alpha +1$.
  4. $S$ is acyclic.
  5. 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?

Ricardo Lopez
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    I think you are correct. That is, if I'm not missing anything. Given the question in the title and the definitions, I mentally came up with the counter-example: ${0}\cup{1/n:n\geq1}$ with the order inherited from $\mathbb Q$; notice this is order-isomorphic to your's. – amrsa Aug 11 '23 at 08:21

1 Answers1

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You're right on target.

In your example, $\langle T,<\rangle$ satisfies 1.1, 1.2, and 1.3, so is a tree in this sense.

It is also a strict total order (and so a strict partial order) and satisfies 2.1, but as you've shown, it does not satisfy 2.2, so is not a transitive tree according to the definition given here.

In fact, it's not even a tree by most of the definitions I've seen before! (Many others can be found here.)

Cameron Buie
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