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.)

complete. Maybe $C = B$? $\endgroup$ – jmc Mar 2 '15 at 8:23