2
$\begingroup$

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 $C \supseteq D$?

$\endgroup$
  • 2
    $\begingroup$ I think that there is a typo; it should be $\ x' \le x\ $ rather than $\ x\le x$. $\endgroup$ – Włodzimierz Holsztyński Feb 27 '15 at 10:43
  • 3
    $\begingroup$ In case this is new terminology, I might suggest that you reconsider, since club is also widely understood to mean "closed and unbounded". $\endgroup$ – Joel David Hamkins Feb 27 '15 at 13:14
  • 1
    $\begingroup$ I suggest "bunch" and have used this in a follow-up question, see mathoverflow.net/questions/198820/… $\endgroup$ – Dominic van der Zypen Mar 2 '15 at 8:14
2
$\begingroup$

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.

$\endgroup$
5
$\begingroup$

Here's a simplified version of Dominic van der Zypen's counterexample: order finite binary strings by extension, with the empty string at the bottom. Consider the club $ D$ consisting of the tods generated by strings of the form $0^n 1$.

$\endgroup$
  • $\begingroup$ That's right - this is much more concise...! Thanks for this example. $\endgroup$ – Dominic van der Zypen Feb 27 '15 at 9:03
  • $\begingroup$ @Bjørn Kjos-Hanssen -- thank you for telling me about my ERROR in my answer which I already deleted after your comment (sorry, your comment goes down with my deletion, sorry). My brain malfunctions, too bad. $\endgroup$ – Włodzimierz Holsztyński Feb 27 '15 at 11:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy