Let $(P,\leq)$ be a partially ordered set. 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 club. A club $C$ said to be complete if for every maximal chain $m\subseteq P$ there is $c\in C$ such that $c\subseteq m$.

Given a club $D$ consisting of finite members only, is there a complete club $C$ also consisting of finite sets only, and $C \supseteq D$?

Let $(P,\leq)$ be a partially ordered set. 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 club. A club $C$ said to be complete if for every maximal chain $m\subseteq P$ there is $c\in C$ such that $c\subseteq m$.

Given a club $D$ consisting of finite members only, is there a complete club $C$ also consisting of finite sets only, and $C \supseteq D$?