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Define that a tree in $X$ is a set of ordinal-indexed sequences with codomain $X$ that is closed under the operations of restricting to an ordinal. (I do not know if this definition is standard.)

Under this definition, every tree induces a poset in a natural way. Is there a nice characterization of posets induced by trees? I feel like there probably should be, but I can't think of one.

Furthermore, every such poset induces a comparability graph. Is there a nice characterization of the comparability graphs of posets induced by trees?

goblin GONE
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    It seems like a definition of trees that I've seen before. In particular, it looks like it's readily the same as definition (iii) in this question, if I understand you correctly. Let me know if you need help seeing the equivalence. The definition given there may be what you're looking for in terms of characterization of posets induced by trees. – Cameron Buie Jul 07 '13 at 15:52
  • @CameronBuie, yes that looks about right. I'll have a go at proving it tomorrow. – goblin GONE Jul 07 '13 at 16:02

1 Answers1

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As mentioned in the comments, a poset $P$ comes from a tree in this sense iff for each $p\in P$, the set $\{q\in P:q<p\}$ is well-ordered. The reverse direction is obvious: given a tree in your sense and an element $p:\alpha\to X$ of it, $\{q\in P:q\leq p\}$ is just the set of restrictions of $p$ to ordinals $\beta<\alpha$, and this set is isomorphic to $\alpha$ and hence well-ordered.

For the forward direction, suppose $P$ is a poset such that $\{q\in P:q<p\}$ is well-ordered for all $p\in P$. Let $\alpha_p\in Ord$ be the order-type of $\{q\in P:q<p\}$ and let $f_p:\alpha_p\to P$ be the unique order-preserving injection with image $\{q\in P:q<p\}$. The set $\{f_p:p\in P\}$ is then a tree in $P$ in your sense, which is isomorphic to $P$ via $p\mapsto f_p$.

Eric Wofsey
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