Let \(\mathcal F\) be a finite union-closed family whose inclusion-minimal nonempty members are exactly three pairwise-disjoint triples
$$ M_1,M_2,M_3. $$
Put
$$ K=M_1\mathbin{\dot\cup}M_2\mathbin{\dot\cup}M_3. $$
Then one of the nine coordinates in \(K\) belongs to at least half the members of \(\mathcal F\):
$$ \boxed{ \max_{x\in K} f_x(\mathcal F)\ge \frac{|\mathcal F|}{2}. } $$
This remains true if the empty set is present.
The proof is a tiny local accounting argument, but it only becomes local after the right quotient. Outside coordinates may be arbitrarily complicated; the three minima still force every fiber over them to obey the same nine-point completion rules.
Freeze everything outside the nine points#
For each set \(G\) in the image of \(A\mapsto A\cup K\), collect the traces
$$ \mathcal H_G = \{A\cap K:A\in\mathcal F,\ A\cup K=G\}. $$
These carrier fibers partition \(\mathcal F\). Every fiber is nonempty, contains \(K\), and is closed under adjoining any complete minimum:
$$ S\in\mathcal H_G \quad\Longrightarrow\quad S\cup M_i\in\mathcal H_G. $$
Every nonempty trace contains at least one whole \(M_i\). Indeed, its source member contains some inclusion-minimal nonempty member of \(\mathcal F\). The empty trace can occur only when \(\varnothing\in\mathcal F\), and then only in the fiber \(G=K\).
So the whole theorem reduces to a statement about subsets of nine points with three completion operations.
The charge table is almost empty#
Give a trace \(S\subseteq K\) charge
$$ q(S)=2|S|-9. $$
Among permitted nonempty traces, the only negative shapes are
$$ q(M_i)=-3 $$
and
$$ q(M_i\cup\{x\})=-1, \qquad x\in M_j,\quad i\ne j. $$
Everything else has nonnegative charge.
If a bare block \(M_i\) appears, completion forces the two pair traces
$$ M_i\cup M_j,\qquad M_i\cup M_k, $$
each of charge \(3\). Split each pair trace's charge equally between its two possible bare-block demanders. Those payments cover every \(-3\) deficit.
Now fix \(x\in M_j\). There are at most two negative one-point extensions, one from each other block:
$$ M_i\cup\{x\},\qquad M_k\cup\{x\}. $$
Completing either by the remaining block forces the same seven-point trace
$$ (K\setminus M_j)\cup\{x\}, $$
whose charge is \(5\). One such recipient pays both possible unit deficits.
The pair traces and seven-point recipients are disjoint payment classes. After all deficits are paid, the top \(K\) still contributes \(9\). Therefore every nonempty carrier fiber satisfies
$$ \boxed{ \sum_{S\in\mathcal H_G}(2|S|-9)\ge 9. } $$
If the exceptional fiber also contains \(\varnothing\), its charge is \(-9\), so that fiber still has total charge at least zero.
Sum the fibers#
Adding the local inequalities gives
$$ 0\le \sum_G\sum_{S\in\mathcal H_G}(2|S|-9) = 2\sum_{x\in K}f_x(\mathcal F)-9|\mathcal F|. $$
Hence the average frequency of the nine coordinates in \(K\) is at least \(|\mathcal F|/2\). One of them is abundant.
I like how little global structure survives the quotient. Arbitrarily many outside coordinates collapse into the fiber label \(G\), and the proof only needs the fact that adjoining each real minimum is still a legal union.
The boundary is exact. Unequal or larger disjoint minima admit ordinary families with negative versions of this scalar charge. The next step cannot be "use the same average more carefully." It needs a block-adaptive vector charge or a frequency-preserving reduction of the minimum sizes.
Notebook references: R-0581