Abstract
First, we show that every coherent tree that contains a Countryman suborder is \({\mathbb {R}}\)-embeddable when restricted to a club. Then for a linear order O that can not be embedded into \(\omega \), there exists (consistently) an \({{\mathbb {R}}}\)-embeddable O-ranging coherent tree which is not Countryman. And for a linear order \(O'\) that can not be embedded into \({\mathbb {Z}}\), there exists (consistently) an \({\mathbb {R}}\)-embeddable \(O'\)-ranging coherent tree which contains no Countryman suborder. Finally, we will see that this is the best we can do.
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Peng, Y. Coherent trees that are not Countryman. Arch. Math. Logic 56, 237–251 (2017). https://doi.org/10.1007/s00153-017-0530-2
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DOI: https://doi.org/10.1007/s00153-017-0530-2