A Proof of the Alternate Thomass\'e Conjecture for Countable $NE$-Free Posets

TL;DR

This paper proves that countable $N$-free posets have either one or infinitely many siblings up to isomorphism, using well-quasi-order properties and labelled ordered trees, thus partially confirming Thomassé's conjecture.

Contribution

It provides a partial proof of Thomassé's conjecture for countable $N$-free posets by establishing the sibling count dichotomy.

Findings

Countable $N$-free posets have either one or infinitely many siblings.

The proof uses well-quasi-order properties and labelled ordered trees.

The result partially confirms Thomassé's conjecture for this class.

Abstract

An $N$-free poset is a poset whose comparability graph does not embed an induced path with four vertices. We use the well-quasi-order property of the class of countable $N$-free posets and some labelled ordered trees to show that a countable $N$-free poset has one or infinitely many siblings, up to isomorphism. This, partially proves a conjecture stated by Thomass\'e for this class.