Two linear orderings of a same set are perpendicular if every self-mapping of this set that preserves them both is constant or the identity. Two isomorphy types of linear orderings… Click to show full abstract
Two linear orderings of a same set are perpendicular if every self-mapping of this set that preserves them both is constant or the identity. Two isomorphy types of linear orderings are orthogonal if there exist two perpendicular orderings of these types. Our main result is a characterisation of orthogonality to ω : a countably infinite type is orthogonal toωif and only if it is scattered and does not admit any embedding into the chain of infinite classes of its Hausdorff congruence. Besides we prove that a countable type is orthogonal toω + n (2 ≤ n < ω) if and only if it has infinitely many vertices that are isolated for the order topology. We also prove that a typeτis orthogonal toω + 1 if and only if it has a decomposition of the formτ = τ1 + 1 + τ2withτ1orτ2orthogonal toω, or one of them finite nonempty and the other one orthogonal toω + 2. Since it was previously known that two countable types are orthogonal whenever each one has two disjoint infinite intervals, this completes a characterisation of orthogonality of pairs of types of countable linear orderings. It follows that the equivalence relation of indistinguishability for the orthogonality relation on the class of countably infinite linear orders has exactly seven classes : the classes respectively of ω, ω + 1, ω + 2, ω + ω, ωω, 3 ⋅ η and η, where η is the type of the ordering of rational numbers and 3 ⋅ η is the lexicographical sum along η of three element linear orders.
               
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