Union of shadows

NEW

Problem: Reconstructed statement: for $\mathcal A\subset [n]^{(3)}$, let $\partial\mathcal A\subset[n]^{(2)}$ be its lower shadow, and let

Equivalently, $V\mathcal A$ is the set of 4-sets containing two disjoint edges of the shadow graph $\partial\mathcal A$. The conjecture asks whether, if $\mathcal C$ is the first $|\mathcal A|$ triples in colex order, then always

Result: The statement is false. Take $n=7$. Let $S=\{1,2,3,4\}$, $T=\{5,6,7\}$, and

Then

Its shadow graph has edge set

with no edges inside $T$. A 4-set $Q\subset[7]$ fails to lie in $V\mathcal A$ exactly when $|Q\cap S|=1$, since then the induced shadow graph is a star. There are

such 4-sets. Hence

Now let $\mathcal C$ be the first 22 triples of $[7]^{(3)}$ in colex order. Since $\binom63=20$,

Thus $\partial\mathcal C$ is the complete graph on $[6]$, together with edges $71,72,73$. Therefore every 4-set inside $[6]$ is in $V\mathcal C$, giving $\binom64=15$ sets. A 4-set containing $7$ is in $V\mathcal C$ unless its other three vertices are $\{4,5,6\}$. Thus there are $\binom63-1=19$ further sets. Hence

So

contradicting the proposed “colex is best” inequality.

Verification audit: the example has the same ground set, the same family size $22$, and uses the standard union-of-shadows interpretation of $V\mathcal A$. The colex initial segment is computed explicitly, and both $V$-sizes are counted directly.

Citation: Problem source: Béla Bollobás and Imre Leader, “Union of shadows,” Theoretical Computer Science, 2003. The counterexample above is given directly here; no external disproof is used.