Frankl's union-closed sets conjecture has been proved for finite families whose inclusion structure has height at most four. Here, height means how many containment levels can be stacked under set inclusion. In the empty-set-free version, the conclusion is that some element appears in strictly more than half of the family's members.
This is a proof study, not an experiment. It examines finite, nonempty families of distinct nonempty sets that are closed under taking unions, so combining members by union stays within the family. Instead of working from an empirical sample, the analysis selects a smallest hypothetical counterexample by minimizing the number of members first, then the size of the underlying ground set.
What the proof establishes
The result has an equivalent standard form. Once a bottom element, the empty set, is adjoined to the inclusion structure, the usual at-least-half statement holds for finite nontrivial union-closed families with height at most five. This gives the bounded-height result at height five in the convention that includes the empty set.
The paper is more cautious about height five. It does not claim that the unrestricted empty-set-free case is solved. Instead, it describes what a smallest counterexample would have to look like if one exists. Writing its number of members as 2t and the number of coordinates as n, the analysis requires an even-sized family, at least three coordinates marked as critical, and no coordinate appearing in more than t members. It also requires t to be at least 2n - 1, or twice the number of coordinates minus one.
Why the next level is harder
That numerical rigidity is what closes the height-four proof. In the decisive equality case, a critical-pair argument identifies the unique height-two extremal double-avoidance family, meaning the sets that avoid two selected critical coordinates. A third critical coordinate then forces t = 2n - 2, in conflict with t at least 2n - 1. The proposed minimal counterexample therefore cannot exist at height four.
For a possible height-five counterexample, each critical coordinate is tied to a largest avoiding member. That member must be the coatom, the full ground set with that coordinate removed. For any two distinct critical coordinates, the proof gives an either-or result: either the union of all members avoiding both is the ground set with both coordinates removed, or at least the rounded-up value of (t + 1) divided by 2 members avoid both. The analysis says that this lower bound is at least n.
The second, large-fiber alternative brings another restriction. If a pair does not have the full double-avoidance union, each set's trace, or pattern, on the relevant coordinate set can only be empty, omit one coordinate, or include them all. The paper also derives exact counts for the layers created by those patterns, including a relation between the layer count h_p and the quantities t, g and δ_p. In a minimal counterexample, these patterns are forced normal forms, not optional examples.
Projection supplies a separate obstruction. Projection here means deleting one coordinate from every set. For each critical coordinate p, the projected family has exactly 2t - c_p members, and a surviving coordinate z has frequency t - δ_z - j_p(z). The proof also requires some surviving z to satisfy a defect inequality, while every matching group of original members that collapses to the same projected set has at least three members. This is a necessary condition, not a contradiction that finishes height five.
A narrower but unfinished problem
The last reduction counts the building blocks needed to cover the critical coordinates. The paper calls the least number of join-irreducible members whose union covers them the critical join-cover number. In the height-five minimal-counterexample framework, that number lies between three and five, while a minimal cover whose union is not the full ground set has at most four members. The five-member case is then excluded, leaving three-cover and four-cover possibilities.
In the nonfull four-cover branch, the structure tightens again. A minimal four-member cover leaves exactly one coordinate outside its union, and that coordinate is noncritical. The family splits into two fibers, or groups distinguished by that coordinate, with the lower fiber larger than the other. These reductions leave explicit three- and four-cover configurations, but they do not complete the unrestricted height-five proof. Further exclusion or transfer inequalities are still needed.
The document is an arXiv preprint, version 1, dated 25 Aug 2026. Its four-level result is complete within the stated finite-family class, while its height-five findings remain conditional on a counterexample existing and narrow the possibilities rather than resolve the conjecture.
Paper data and sources
Original title: Frankl's Conjecture at Height Four and the Structure of Height-Five Counterexamples
Authors: Chenxiao Tian
Journal/Repository: arXiv
Status: Preprint, not yet peer-reviewed
First online: 2026-08-25
DOI: Not available
Original paper · Full text