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A tripartite graph has a set partition of its vertices into three classes, with every edge joining different classes. Each pair of classes induces a bipartite graph, and every triangle in a graph uses one vertex from each class.

Let $S$ be a nonempty set and $\mathcal{P}\subseteq \mathcal{P}(S)$ be upward closed. An ultrafilter all of whose members lie in $\mathcal{P}$ exists exactly when every finite set partition of $S$ has a cell in $\mathcal{P}$. For sufficiency, the complements of sets outside $\mathcal{P}$ have the finite intersection property: a finite cover by such sets could be disjointified into a partition with no cell in $\mathcal{P}$. The ultrafilter lemma extends the generated proper filter. Necessity follows because an ultrafilter selects exactly one cell of every finite partition. This criterion requires partition richness of the whole ground set; it does not assert that the complement family is closed under unions.

For every $\epsilon >0$, all sufficiently large finite graphs admit a set partition into at most $M(\epsilon )$ nonempty cells, each of size at most $\epsilon n$, with the displayed ordered irregular-pair weight bound. Start with small cells. Use regularity energy including diagonal rectangles, refine simultaneously by the witness subsets of all irregular pairs, and apply the refinement variance identity for regularity energy. Each failing round raises energy by more than ${\epsilon }^5$, while energy lies in $[0,1]$. The recurrence $k\mapsto k{2}^k$ bounds the final cell count. Equitability is unnecessary for this weighted version.