Question

Let $I$ be a poset and for any $i$ let $P_i$ be a poset. Let $P$ be the sum over $I$ of the sets $P_i$, and let $<_P$ be the relation defined on $P$ by $q<_Pr$ iff $q$ and $r$ are members of the same $P_i$ and $q<_ir$ or $q$ is member of $P_j$ $r$ is member of $P_k$ and $j<_ik$. Maybe that formally, I should have taken $Q_i$ as the cartesian product of the singleton of $i$ with $P_i$, and $Q$ as the union of the $Q_i$'s, and compared couples as $(i,q)$ and $(j,r)$. Anyway, it is clear that $P$, equipped with our relation is a poset, that is the poset sum over $I$ of the posets $P_i$. Particularly, in the case that $I$ and $P_i$ all are totally ordered sets , $P$ is a totally ordered set. And, in this particular case, a theorem of Schoenflies asserts that "every (totally) ordered set is the union of scattered sets over a densely ordered indexing set";

Question: does there exist a similar decomposition theorem in the general case of posets ?

Gérard Lang

Answer 1

I suggest first restricting to, say, suborders $P$ of the plane $\mathbb{R}^2$(with the product ordering where $(x_1,y_1)\leq(x_2,y_2)$ iff $x_1\leq x_2$ and $y_1\leq y_2$). See if you can find a satisfactory decomposition there, focusing on simple counterexamples to a straight generalization of Schoenflies' Theorem.

For example, let $$P=(\{0\}\times[0,1])\cup\bigcup_{n=0}^\infty(\{2^{-n} \}\times\{m\cdot 2^{-n}:m=1,\ldots,2^n\}).$$ Suppose you had a decomposition of $P$ into an ordered sum where the index has some property D reminiscent of denseness and each summand has property S reminiscent of scatteredness. If the linear order $[0,1]$ lacks property S, then you must decompose $P$ into singletons. Hence, either $[0,1]$ has property S or $P$ has property D. (Neither of these options looks appealing to me.)