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Call an antichain (set of pairwise incomparable elements) $A$ of a poset $P$ strong if for every $p,q \in P$ with $p \leq q$ there exists an $a\in A$ which is comparable with both $p$ and $q$.

Consider the subsets of $\omega$ partially ordered by $\subseteq$, factorize it mod finite and then delete the smallest and largest element. A strong antichain of the resulting poset $P$ can be constructed by a straightforward transfinite recursion under the assumption that the reaping number is continuum. My question is if the existence of a strong antichain of $P$ is provable in ZFC alone?

The question has a great importance with respect to the existence of certain kind of infinite matroids. As far as I know it came up first in Bowler & Geschke's "Self-dual uniform matroids on infinite sets" paper.

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  • $\begingroup$ I don't quite get your definition of strong antichain: The way I understand your definition, it cannot exist in $\mathcal{P}(\omega) \, /\, \text{fin}$: Let $A \subset P$ be a strong antichain and pick an $a \in A$. Define $q$ to be a stronger condition such that $a \setminus q$ is infinite. Set $p:=\omega \setminus (a \setminus q)$. Then (if your notation is to force upwards) $p \leq q$, but there cannot exist $b \in A$ which is both comparable with $p$ and $q$. $\endgroup$
    – Johannes Schürz
    Commented Oct 19, 2020 at 14:00
  • $\begingroup$ If the poset has a largest (or smallest) element $a$, then $A:=\{a \}$ is a strong antichain since $a$ is comparable with everybody. (The deletion of the largest and smallest element is to exclude these trivial strong antichains.) $\endgroup$
    – Attila Joó
    Commented Oct 20, 2020 at 19:29
  • $\begingroup$ Sure, $a$ must be co-infinite in my comment. But I don't see why there should be a non-trivial strong antichain. My comment above seems to prove the opposite, right? $\endgroup$
    – Johannes Schürz
    Commented Oct 20, 2020 at 20:17
  • $\begingroup$ If I understood correctly in your first comment you consider a poset which has a largest and smallest element and claim that it has no strong antichain. In my reaction pointed out that it is wrong since in this case we always have. The existence of a strong non-trivial antichain in your poset is consistent with ZFC, my question is if it is actually provable in ZFC or independent of it. $\endgroup$
    – Attila Joó
    Commented Oct 21, 2020 at 21:38
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    $\begingroup$ I think I figured out what is the reason of the misunderstanding. An antichain of a poset is a set of pairwise incomparable elements in my question. In the context of forcing this term is used differently: set of pairwise incompatible elements. Since you wrote 'stronger condition' I guess you used the forcing related interpretation. $\endgroup$
    – Attila Joó
    Commented Oct 22, 2020 at 13:54

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