Suppose Frankl's union-closed sets conjecture is false. Choose a counterexample \(\mathcal F\) with as few member sets as possible, then with as few ground elements as possible.
If exactly three ground elements occur in
$$ \frac{|\mathcal F|-1}{2} $$
members, the projection of \(\mathcal F\) onto those three elements contains the entire Boolean cube:
$$ \boxed{\mathcal F\!\restriction_{\{a,b,c\}}=2^{\{a,b,c\}}.} $$
All eight patterns. Not merely the three singletons, not merely a generating subfamily: the whole cube.
This is not a proof of Frankl's conjecture. The fibers over those eight patterns can still be complicated. But a hypothetical smallest counterexample is forced into a much more rigid shape than "some separating union-closed family." It has an exact finite-lattice core, an odd number of members, at least three almost-majority coordinates, and a deletion inequality that makes the three-coordinate frontier snap into a cube.
That snap is the bit I like.
First remove the fake coordinates#
There are several familiar normalizations around union-closed families, and some of them are treacherous if used too casually.
Deleting unused ground points is harmless. Identifying two points with identical incidence columns is also harmless. But being separating is not the final canonical form.
Assume \(\varnothing\in\mathcal F\) and take complements:
$$ \mathcal I=\{U\setminus A:A\in\mathcal F\}. $$
Then \(\mathcal I\) is a finite lattice under inclusion. For each ground element \(x\), intersect all members of \(\mathcal I\) containing \(x\). This gives the singleton closure \(c_x\).
Every join-irreducible of \(\mathcal I\) is some \(c_x\), but the converse can fail: a coordinate may have a unique incidence column while its singleton closure is still join-reducible. So choose one coordinate for each actual join-irreducible and project onto those coordinates.
The projection is injective on members because every finite-lattice element is determined by the join-irreducibles below it. It preserves unions, family size, and the frequencies of the retained coordinates. Therefore a minimum-ground counterexample already equals this join-irreducible coordinate core.
The distinction is tiny and annoyingly real. The separating family
$$ \{\varnothing,\{1,2\},\{1,3\},\{1,2,3\}\} $$
has three distinct incidence columns, but its lattice core is only a two-dimensional cube.
Replacing one minimal member by the empty set#
What if \(\varnothing\notin\mathcal F\)?
Choose an inclusion-minimal member \(A\). No two other members can have union \(A\), so removing \(A\) preserves union closure. Replace it by \(\varnothing\):
$$ \mathcal F_0=(\mathcal F\setminus\{A\})\cup\{\varnothing\}. $$
The family has the same number of distinct members, remains union-closed, and no element frequency increases. Thus a counterexample stays a counterexample.
This same-cardinality replacement is the safe normalization. Simply adding or deleting \(\varnothing\) changes the family size and can move the exact half-frequency threshold.
The removable-set inequality#
Call \(R\in\mathcal F\) removable when \(\mathcal F\setminus\{R\}\) is still union-closed. Every inclusion-minimal member is removable, and every member of \(\mathcal F\) is a union of removable members.
More strongly, any subfamily \(D\) of removable members can be deleted simultaneously. If \(m=|\mathcal F|\), write
$$ f_x=|\{A\in\mathcal F:x\in A\}|, \qquad s_x=m-2f_x, $$
and for \(D\neq\varnothing\),
$$ d_x=|\{A\in D:x\in A\}|. $$
After deleting \(D\), the slack at \(x\) is exactly
$$ s_x-|D|+2d_x. $$
The smaller family cannot still be a counterexample, because \(\mathcal F\) was chosen with minimum cardinality. Hence some element witnesses Frankl's conclusion in the smaller family, which rearranges to
$$ \boxed{s_x+2d_x\le |D|.} $$
This one inequality contains the parity and the cube.
For a one-member deletion, some omitted element must have \(s_x=1\). Thus \(m\) is odd, and every removable member omits an element of frequency \((m-1)/2\).
For a two-member deletion, every pair of removable members jointly omits such an element.
Call the elements with \(s_x=1\) tight. Since removable members generate the top set, their traces cover all tight elements. There cannot be only one tight element, because every removable member omits it. There cannot be only two, because one removable member covering each would jointly cover both, contradicting the two-deletion condition.
So there are at least three.
Exactly three leaves no room between the singletons#
Now suppose the tight set is exactly
$$ T=\{a,b,c\}. $$
Every removable member has a proper trace on \(T\). No removable trace can contain two tight elements: if one contained \(a,b\), pairing it with any removable member covering \(c\) would make the pair cover all of \(T\).
But removable members generate the top, so their traces must cover \(a,b,c\). Therefore removable members with traces
$$ \{a\},\qquad\{b\},\qquad\{c\} $$
all exist.
Take their unions. Union closure produces the three two-point patterns and the full three-point pattern; \(\varnothing\) supplies the empty pattern. Hence every subset of \(T\) appears.
There is a little extra structure hiding here. The singleton-pattern members are removable, while each two-point-pattern member obtained as their pairwise union is not removable: it is explicitly the union of two other members. The cube arrives with prescribed behavior in its fibers.
Where the result stops#
The exact-three branch is now a fiber problem over \(2^3\), not an arbitrary union-closed family. One can ask how many members lie over each Boolean pattern, how union closure couples those fibers, and whether the tight frequency equations admit a Poonen-style certificate.
None of that is automatic. Four or more tight elements remain possible, and even the forced three-cube may have sufficiently elaborate fibers to survive all current inequalities.
The theorem does something narrower and cleaner: every hypothetical smallest counterexample can be put in one exact lattice-normalized form, and its smallest possible tight frontier is not fuzzy. It is a cube.
Notebook references: C-0230