# Ordered sum of posets

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

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@Gérard LANG, I've taken the liberty to edit your text into LaTeX and fixed a few spelling mistakes. Please re-edit if I've made the wrong choices in some places. – j.c. Jun 29 '11 at 16:14
Sorry, I did something similar as jc, in parallel. I then rolled back to jc's version as it was better than mine. – user9072 Jun 29 '11 at 16:17
That theorem of Schoenflies is quite interesting. Where can one find it? Could you perhaps spell out when subsets are scattered? – Chris Heunen Jun 29 '11 at 16:19
I rewrote my question because I thought it would not be readable. The theorem is given inside "Set Theory by Kuratowski and Mostowski(with an introduction to Descriptive set theory)" theorem 5 ,page 209. A linearly ordered set which contains no infinite densely ordered subset is said to be scattered; a set is densely ordered if for any two elements there exists another element strictly between them. Gérard lang – Gérard Lang Jun 29 '11 at 16:27
@Chris: see en.wikipedia.org/wiki/Scattered_order for two equivalent definitions. The theorem is then actually easy to prove: for $x\le y$, let $x\sim y$ iff the interval $[x,y]$ is scattered. This gives you an equivalence relation, and it is not hard to see that each equivalence class is a convex scattered subset, and $P/{\sim}$ is densely ordered. – Emil Jeřábek Jun 29 '11 at 16:31

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.)