# Completion of a single totally ordered down-set

This is a follow-up question to Complete sets of incompatible totally ordered down-set in a partially ordered set.

Let $(P,\leq)$ be a partially ordered set such that for every $p\in P$ the set $\{q\in P: q\leq p\}$ is finite. A down-set is a set $d\subseteq P$ such that $x\in d$ and $x'\in P, x'\leq x$ imply $x'\in d$. If the down-set is totally ordered, we say it is a totally ordered down-set (tods).

Let $d_1, d_2$ be tods. We say that they are incompatible if neither $d_1\subseteq d_2$ nor $d_2\subseteq d_1$ holds. A set of pairwise incompatible tods is called a bunch. A bunch $B$ said to be complete if for every maximal chain $m\subseteq P$ there is $b\in B$ such that $b\subseteq m$.

Given a finite tods $t$, is there a complete bunch $B$ also consisting of finite sets only, and $t\in B$?

(Motivation: the answer to the linked post shows that bunches consisting of finite tods cannot always be completed, but it is still possible that if you start out with a singleton bunch consisting of 1 finite tods, the singleton bunch can be completed.)

• There is something wrong with the sentence where you define complete. Maybe $C = B$? – jmc Mar 2 '15 at 8:23
• That's right, thanks for noticing - I just corrected it. – Dominic van der Zypen Mar 2 '15 at 8:29
• Hmm, I think you need more assumptions. The integers do not contain a finite tods, right? This easily gives counter examples. So I think every maximal chain should have a minimal element. – jmc Mar 2 '15 at 8:37
• Right - I need that below every element there are only finitely many. Will add this. Sorry for the inconvenience and thanks for your precise reading! – Dominic van der Zypen Mar 2 '15 at 8:51

I think that in this case there will always be a complete bunch $B$.

• For every maximal chain $m$ with $m \cap t = \varnothing$, take the minimal element of $m$, and add it as tods to $B$.
• If $m \cap t \ne \varnothing$, and $t \not\subset m$, then take the minimal element $x$ of $m \setminus t$, and add $\{x\} \cup (m \cap t)$ to $B$.
• Finally, add $t$ to $B$, and now $B$ is complete.

$B$ is also a bunch of finite tods, because by construction all tods in $B$ are finite and pairwise incompatible.

• Hang on - $t$ and $\{x\}\cup t$ are comparable, so $B$ is no longer a bunch, isn't it...? (Thanks to Joel Adler for noticing this.) – Dominic van der Zypen Mar 3 '15 at 12:56
• @DominicvanderZypen — You are right, I made a typo. I'll fix it. Bjørn's answer is correct (and more concise.) – jmc Mar 3 '15 at 13:29

Yes. We can take the collection of all tods $s$ such that $s \backslash t$ is a singleton and $s$ is incompatible with $t$. Any maximal chain extends exactly one of these, or $t$.

• Ah, that's a very concise way of putting it. Nice. – jmc Mar 2 '15 at 9:05
• @jmc thanks, you were 5 minutes faster though – Bjørn Kjos-Hanssen Mar 2 '15 at 15:01