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 $C \supseteq D$?



Complete sets of incompatible totally ordered down-set in a partially ordered set
The answer is No.


We define two sets of elements of $\{0,1,2\}^\omega$ in the following way:


1. For $n\in\omega$ let $u\_n$ be defined by $u\_n(k)=1$ for $k\leq n$ 
and $u\_n(k)=0$ for $k>n$;
2. For $n\in\omega$ let $t\_n$ be defined by $t\_n(k)=1$ for $k\leq n$ 
and $t\_n(n+1) = 2$ and $t\_n(k)=0$ for $k>n+1$;


Note that informally speaking, the $u\_n$ have the form $(1,1,1,\ldots, 1,0,0,\ldots)$
and the $t\_n$ have the form $(1,1,1,\ldots, 1,2,0,\ldots)$.


Then we set $$E:=\{u\_n:n\in \omega\}\cup\{t\_n: n\in \omega\}.$$ The ordering
on $E$ is componentwise (equivalently, the ordering inherited from the product
ordering on $\{0,1,2\}^\omega$).


**Step 1**. Let $c\_n = \downarrow t\_n$. Then $P = \{c\_n: n\in \omega\}$ is a club.


*Proof.* We have to show that for $m<n $ the tods $c\_m, c\_n$ are
incompatible. This is the case if we find incomparable elements
in $c\_m, c\_n$. This is easy: $t\_m\in c\_m$ and $t\_n\in c\_n$ are incomparable
in the ordering we chose for $E$.


**Step 2**. The set $m = \{u\_n: n\in \omega\}$ is a maximal tods.


*Proof.* It's easy to see that $m$ is a chain and a down-set. Next, maximality: if we consider $m\cup\{t\_n\}$ for some $n\in \omega$, the elements
$u\_{n+2}$ and $t\_n$ are not comparable, so $m\cup\{t\_n\}$
is not totally ordered.


**Step 3**. If $x$ is a finite tods with $x\subseteq m$, then 
$x$ is compatible with a member of $P$. 


*Proof.* Any finite tods $x\subseteq m$ has the form
$x = \downarrow u\_n$ for some $n\in \omega$. So $c\_n = \downarrow
t\_n \supseteq x$ therefore $c\_n \cup x = c\_n$ is a tods,
so by definition $c\_n$ and $x$ are compatible.


**Conclusion**. There is no complete club consisting of finite sets $P'\supseteq P$
such that $P$ contains a finite subset of $m$. Therefore the answer to the question is No.


