Union-closed sets conjecture

Consider four distinct bags of labels: an empty bag, {apple}, {banana}, and {apple, banana}. Combining any two bags, keeping one copy of each label, gives another bag on the list. Apple appears in two of the four bags, and banana does too. Must every finite collection closed under this combining rule, with at least one nonempty bag, have a label appearing in half or more of its bags?[1]

Combining two permitted sets must produce another permitted set. Does that local closure rule always force an element occurring in at least half the family? The four sets $\varnothing,\{a\},\{b\},\{a,b\}$ illustrate both closure and exact half frequency.

For a finite family $\mathcal F$ containing a nonempty set, the condition and target are

$$A,B\in\mathcal F\Longrightarrow A\cup B\in\mathcal F,\qquad \exists x:\ |\{A\in\mathcal F:x\in A\}|\geq |\mathcal F|/2.$$

Member sets are distinct; the empty set may be included. A family containing only the empty set has no candidate element and is excluded.[1] [2]

The four-bag example meets the half threshold exactly. The conjecture asks for one frequent label in every allowed collection, not for every label to be frequent. Smaller universal fractions and restricted cases are known; the general half-or-more guarantee remains unproved.[1] [3]

Gilmer's 2022 result guarantees an element in at least one percent of the sets; later refinements improve that fraction, without reaching the full half target in the reviewed sources. One percent is cited as the original contribution, not a current record.

The August 2026 preprint restricts inclusion-chain height, the largest number of members in a strictly nested chain. It reports the strictly-more-than-half form for empty-set-free families of height at most four; adding the empty set gives the usual at-least-half form at height at most five. The unrestricted conjecture has no such height bound.[1] [2] [3]

The combining rule links choices that might otherwise be unrelated. Three separate one-label bags would each make its label rare, but that collection fails the rule because its two-label unions are missing. The question measures how strongly the permission to combine choices constrains the whole collection.[1]

This is a bridge between closure, logic and information. Pick a set uniformly from the family; the chance it contains a given element is that element's frequency. Entropy measures uncertainty in this random choice. A uniform choice among $m$ sets has the maximum possible entropy $\log_2 m$.

The union of two independent choices need not be uniform, but closure keeps it inside the same family. Gilmer shows that if every element is sufficiently rare, taking a union increases entropy beyond that maximum, a contradiction. This explains why some element must recur. Extending such control to half frequency remains the substantive gap.[1] [3]