4
$\begingroup$

Let $(P,\leq)$ be a directed set with uncountable cofinality. For every element $p\in P$, we are given a finite set $c_p\subset P\smallsetminus \{p\}$ of "incompatible elements". We say that a subset $Q\subseteq P$ is compatible if for every $q\in Q$ we have $c_q\cap Q = \emptyset$.

Question: is it always possible to find a cofinal compatible subset?

Remarks:

  • I can show that the answer is yes if there is a finite uniform upper bound on the cardinalities of the sets $c_p$.

  • It was pointed out to me that the assumption that the cofinality be uncountable is certainly necessary, as seen by considering $(\mathbb{N},\leq)$ with $c_n = \{1,2,...,n-1\}$.

  • I suspect that a (positive) solution would require some argument in infinite combinatorics.

  • I am very happy to assume the axiom of choice (or any other useful set of axioms).

$\endgroup$
2
  • $\begingroup$ Hi Federico, I've written an answer to your question, but it doesn't seem to be displaying properly on the MO homepage. I'm not sure what's wrong, or whether it's an issue with my computer or with MO. But I hope the answer is visible to you? $\endgroup$
    – Will Brian
    Commented Sep 27, 2022 at 19:58
  • $\begingroup$ Hi Will. Yes, I do see your answer correctly. Thanks a lot! $\endgroup$ Commented Sep 28, 2022 at 7:44

1 Answer 1

4
$\begingroup$

No, it is not always possible. For a counterexample, let $P$ be the set of all finite subsets of some uncountable set $X$, ordered by inclusion (that is, $a \leq b \Leftrightarrow a \subseteq b$). In this poset, a subset $D$ of $P$ is cofinal if and only if for every finite subset $a$ of $X$, there is some $b \in D$ with $b \supseteq a$.

For each $a \in P$, let $c_a$ be the (finite) set of all proper subsets of $a$. Suppose $D$ is cofinal in $P$, and fix some $a \in D$. There is some $x \in X \setminus a$, and (because $D$ is cofinal) there is some $b \in D$ with $b \supseteq a \cup \{x\}$. But now we have $b \in D$ and $a \in D$ and $a \in c_b$.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .