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Given a poset $L$, call it trivial if $\left|L\right| < 2$ and let $\mathcal I\left(L\right)$ be its poset of ideals, $\mathcal C\left(L\right)$ be its set of chains, and $\mathcal M\left(L\right)$ be its poset of (finitely) meet-irreducible elements. Frankl's union-closed sets conjecture can be stated equivalently as follows.

  1. In any nontrivial finite lattice $L$, there exists a nonempty meet-irreducible ideal $I \in \mathcal M\left(\mathcal I\left(L\right)\right)$ such that $\left|I\right| \le \left|L\setminus I\right|$.

This conjecture is notorious for thwarting attempts at generalization, including the obvious generalization to infinite lattices. However, all counterexamples I have seen to generalizations to infinite lattices involve an infinite strictly descending chain (see here and here). Are there any known/obvious counterexamples to either of the following generalizations?

  1. In any nontrivial lattice $L$ satisfying the descending chain condition, there exists a nonempty meet-irreducible ideal $I \in \mathcal M\left(\mathcal I\left(L\right)\right)$ such that $\left|I\right| \le \left|L\setminus I\right|$.
  2. In any nontrivial lattice $L$, there exists a chain of nonempty meet-irreducible ideals $C \in \mathcal C\left(\mathcal M\left(\mathcal I\left(L\right)\right)\right)$ such that $\left|\bigwedge C\right| \le \left|L\setminus \bigwedge C\right|$.

Edit: To clarify, here I am including the empty set in the lattice of ideals so $\mathcal I\left(L\right)$ is complete and $\bigwedge C$ in (3.) exists. However, the ideal $I$ in (1.) and (2.) and the ideals in $C$ in (3.) must be nonempty.

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  • $\begingroup$ Are you sure that 1 is stated correctly? I'd say you should be looking for $a\in\mathcal M(L)$ (so an element in $L$ rather than an ideal) such that $|I(a)|\le|L\backslash I(a)|$ where $I(a)$ denotes the principal ideal generated by $a$. $\endgroup$
    – Igor Makhlin
    Commented Jun 28 at 17:51
  • $\begingroup$ @IgorMakhlin Please correct me if any of this is wrong, but your function I is a lattice embedding and all nonempty ideals of a lattice satisfying ACC are principal, so if you exclude the empty set (which doesn't change which other ideals are meet-irreducible), I is a lattice isomorphism. Therefore, the principal ideals generated by meet-irreducibles and the nonempty meet-irreducibles in the lattice of ideals are the same for a finite lattice. $\endgroup$
    – Evan Bailey
    Commented Jun 28 at 18:54
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    $\begingroup$ Oh, okay, got it. Sorry, I'm too used to "ideal" meaning "lower set". If you require ideals to be directed, then yes, the two are equivalent. $\endgroup$
    – Igor Makhlin
    Commented Jun 28 at 19:33

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