This is a follow-up question to [Complete sets of incompatible totally ordered down-set in a partially ordered set](https://mathoverflow.net/questions/198612/complete-sets-of-incompatible-totally-ordered-down-set-in-a-partially-ordered-se).


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.)



Completion of a single totally ordered down-set
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.


