Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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

share|improve this question
    
@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. –  quid 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
1  
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
2  
@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

1 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.)

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.