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Source. Ruiliang Li, On an Erdős--Lovász problem: 3-critical 3-graphs of minimum degree 7, arXiv:2512.24850v1 (31 December 2025), Proposition 3.5 and proof, printed pp. 6--7 (PDF pp. 6--7).

Dependencies. None.

Used in. Theorem 1.1.

Bears on. #834: shows that the degree bound six under the transversal reading of "$3$-critical" cannot be lowered.

Statement

The complete 3-uniform hypergraph $K_5^{(3)}$ is transversal-critical of order three and satisfies $\delta(K_5^{(3)})=6$.

Rewritten proof

No two vertices form a transversal: the other three vertices themselves form an edge. Every three vertices do form a transversal, since two 3-subsets of a five-element set must intersect. Hence $\tau(K_5^{(3)})=3$.

For an edge $e$, write $\{u,v\}$ for the two vertices outside $e$. Every other 3-subset contains $u$ or $v$, so $\{u,v\}$ is a transversal of $K_5^{(3)}-e$. No single vertex meets all edges of this deletion: among the four triples avoiding a given vertex, at most one is $e$. Therefore $\tau(K_5^{(3)}-e)=2$ for every $e$.

Finally, each vertex belongs to ${4\choose2}=6$ triples, so the hypergraph is 6-regular.