(edit: I decided to simplify the question and only pose it for bounded posets first)

The Union-closed sets conjecture is equivalent for lattices P to:

There exists a join-irreducible element $a$ with $|[a,M]| \leq |P|/2$, when $M$ is the maximum of $P$.

Recall that an element a of a poset is join-irreducible if there is no subset $X \subseteq P$ with $a\not\in X$ and $a=\bigvee X$.

Call a (finite) bounded poset $P$ lattice-like in case an element $x \in P$ is join-irreducible iff $x$ is covers a unique element.

Every lattice is lattice-like but not every bounded poset is lattice-like.

Question 1: Is the above conjecture also true for lattice-like posets?

This is true for all such posets with at most 8 points. I would think there is a counterexample but I have not found one yet.

Question 2: Are there attempts in the literature already to generalise the Union-closed sets conjecture from lattices to a larger class of posets?


1 Answer 1


Here is a counterexample of size 23.

Let $m=6$ and let $$P=\{0,a_1,\dots,a_m,1\}\cup\{b_{ij}: 1\le i<j\le m\}$$ where $0<a_i<b_{jk}<1$ whenever $i$ is distinct from $j$ and $k$.

The cardinality of $P$ is $|P|=m+2+\binom{m}{2}=6+2+15=23$.

The join-irreducible elements are only the $a_i$ and $0$, since each $b_{ij}=\bigvee\{a_k:k\ne i, k\ne j\}$.

Each $|[a_i,1]|=2+\binom{m}{2}-(m-1)>|P|/2$ as long as $$2+\binom{m}2 - (m-1) > \frac12\left(m+2+\binom{m}2\right)$$ $$4+\binom{m}2>3m$$ which is true for $m=6$ but not for $m=5$.

  • 2
    $\begingroup$ I wonder if 23 is actually the minimum cardinality of an example... $\endgroup$ Oct 29, 2021 at 6:54

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