Every cutset meets every fibre in certain poset products
In the case of (finite usually) posets there are many subclasses of constructively interesting posets as well as many classes of axiomatically interesting posets. An example of the former is the class of posets of the form \(C^n_2\) (or here \(K^n_{1,1})\), i.e., the Boolean lattices. An example of the latter is the class of chain-meets-antichain posets. This class is known to be the same as the class of \(N\)-free (4-fence-free) posets according to Rival-Zaguia. It is almost always an interesting as well as a tricky problem to match a class of the first type precisely (i.e., by an if and only if result) with a class of the second type, usually via some \(P\)-freeness theorem as in the result quoted above. In this case the axiomatically interesting class is the class of skeletal, i.e., cutset-meets-fiber posets, where a cutset meets all maximal chains and a fiber meets all antichains and is thus seen to be somewhat analogous to the class of chain-meets-antichain posets. The constructively interesting class is the class of posets \(K_{1,n}\times---\times K_{1,n_k}\) where \(K_{1,n}=\underline 1\oplus\underline n\) is the linear (ordinal) sum of a singleton with an \(n\)-point antichain, of which it is conjectured that all elements of the class are skeletal posets. Like the series-parallel posets in the \(N\)-free case, the conjecture is only of the ``inclusion'' type, but it would produce a quite strong result if proven in its entirety. Important cases of the conjecture providing significant strengthening are provided in Theorem 5: \((K_{1,1})^r\times K_{1,m}\times K_{1,n}\) is skeletal; and Theorem 6: \(K_{1,l}\times K_{1,m}\times K_{1,n}\) is skeletal. The proofs are obtained after some delicate and rather complicated manoeuvers which indicate that a proof of the full conjecture could be of significant difficulty.
scientific article; zbMATH DE number 1321846