Chapter 45: Modern Extremal Combinatorics

How can a few containers control an enormous family of independent sets?

A hypergraph $\mathcal H$ consists of a vertex set and a collection of subsets called edges. It is $r$-uniform if every edge has exactly $r$ vertices. An independent set contains no entire hyperedge. This language can encode a familiar constraint in an unfamiliar ambient space: take the graph edges of $K_n$ as the vertices of a three-uniform hypergraph, and make each triangle's three graph edges a hyperedge. Independent sets are then exactly the triangle-free graphs on the labeled vertex set $[n]=\{1,\ldots,n\}$.

A container family $\mathcal C$ is a collection of vertex subsets such that every independent set $I$ is contained in some $C\in\mathcal C$. A container need not itself be independent. Its usefulness depends on two budgets: how many containers there are and how much freedom remains inside each. If every container has at most $m$ vertices, then

$$\lvert\{I:I\text{ independent}\}\rvert\le\sum_{C\in\mathcal C}2^{\lvert C\rvert}\le\lvert\mathcal C\rvert2^m.$$

This is an upper bound, not an equality: sets may belong to several containers, and a container with hyperedges has subsets that are not independent. Coverage must be proved before the bound can be used.

For the triangle hypergraph on $K_4$, there are six hypergraph vertices and four hyperedges. Its independent sets can be covered by seven explicit triangle-free containers: the four stars $K_{1,3}$ and the three distinct spanning $K_{2,2}$ graphs. To justify coverage, a triangle-free graph on four vertices has no odd cycle, since the only possible odd cycle is a triangle. It is therefore bipartite. Extend a bipartition to a complete bipartite graph; the nonempty spanning possibilities are the listed stars and balanced bipartitions. The empty graph is contained in any of them.

These seven containers give the upper bound $4\cdot2^3+3\cdot2^4=80$. The exact number of independent sets is $41$, out of $2^6=64$ subsets. The bound is loose at this small scale, but the example exposes the mechanism and its overlap cost. In large instances, a theorem produces containers with a subquadratic logarithmic count while supersaturation bounds their sizes. That combination gives a sharp leading enumeration exponent.

The encoding, coverage argument, and union bound are the chapter foundation. They suffice for the finite part of the final investigation; its theorem-audit part restates the additional hypotheses it uses. The numbered sections are optional research extensions. Prior connections are extremal sets, extremal graphs, hypergraphs, the probabilistic method, random structures, and entropy methods.

45.1 Extremal stability and supersaturation

Optional extension; not required to complete the chapter foundation.

For a forbidden graph $F$, $\operatorname{ex}(n,F)$ is the maximum edge count of an $F$-free graph on $n$ vertices. Stability describes configurations close to this maximum; supersaturation says that exceeding the extremal scale forces many forbidden copies. They answer different questions: near-optimal structure versus the cost of crossing a threshold.

Here is an exact stability calculation for triangle-free graphs. Let $G$ have $n$ vertices and $m$ edges. Choose a vertex of maximum degree $d$, let $A$ be its neighborhood, and let $B$ be the remaining vertices. Since $G$ is triangle-free, $A$ is independent. Let $b=e(G[B])$. Summing degrees over $B$ gives

$$m+b\le d(n-d)\le n^2/4.$$

This proves Mantel's bound. If $m\ge n^2/4-\delta n^2$, then $b\le\delta n^2$, $\lvert d-n/2\rvert\le\sqrt\delta\,n$, and the number of missing $A$–$B$ edges is at most $2\delta n^2$. Deleting the $b$ internal edges and adding the missing cross edges takes at most $3\delta n^2$ edits to reach a complete bipartite graph, with the displayed imbalance bound. This is a complete small stability theorem, not an appeal to an unspecified stability principle.

For supersaturation, let $t(G)$ be the number of unordered triangles. Every graph edge $uv$ has at least $\deg u+\deg v-n$ common neighbors. Summing and applying Cauchy–Schwarz gives

$$3t(G)\ge\sum_v\deg(v)^2-nm\ge\frac{4m^2}{n}-nm.$$

Consequently, if $m\ge(1/4+\eta)n^2$ with $\eta\gt0$, then

$$t(G)\ge\frac{\eta+4\eta^2}{3}n^3.$$

The right-hand side may be negative below the threshold, where the inequality is simply uninformative. General fixed-forbidden-configuration supersaturation is established; a primary formulation is Proposition 9.4 in Saxton and Thomason, citing Erdős and Simonovits. The triangle inequalities above have been proved directly and do not require that general result.

Solved example. $K_{3,3}$ with one edge deleted is triangle-free with $n=6,m=8$, one below Mantel's maximum nine. A degree-three vertex gives a balanced cut with no internal edges and one missing cross edge. For contrast, adding an internal edge to $K_{3,3}$ gives exactly three triangles, one with each vertex in the opposite part.

Exercise. How many triangles are created by adding one internal edge to $K_{a,b}$? Hint: choose the third vertex. Solution: an edge added in the part of size $a$ creates exactly $b$ triangles. This is a concrete supersaturation trace, not the general minimum for every graph with the same edge count.

45.2 Graph and hypergraph removal lemmas

Optional extension; not required to complete the chapter foundation.

The established graph removal lemma has the following quantifier order. For every fixed graph $F$ and every $\varepsilon\gt0$, there exists $\delta\gt0$ such that every $n$-vertex graph with at most $\delta n^{v(F)}$ copies of $F$ can be made $F$-free by deleting at most $\varepsilon n^2$ edges. Copies here are injective copies; choosing labeled versus unlabeled counts merely rescales $\delta$ for fixed $F$. This is ordinary subgraph removal, not induced-subgraph removal.

For a fixed $r$-uniform hypergraph $F$, the corresponding hypergraph lemma replaces the deletion budget by $\varepsilon n^r$. It does not follow from applying graph regularity to the pairs of a hypergraph. Primary sources include Fox's graph-removal paper and Gowers's hypergraph regularity and counting theorem.

Proof sketch and omitted prerequisites. A regularity-based graph proof removes edges in exceptional, irregular, and low-density pairs. Any surviving reduced copy of $F$ then produces many actual copies by a counting lemma; choose $\delta$ small enough to rule that out. Hypergraph proofs require regularity and counting on a compatible hierarchy of lower-dimensional faces. The partition and embedding theorems are omitted, and no useful explicit numerical value of $\delta$ is claimed here. Fox also gives a graph proof avoiding the usual regularity route.

Exercise. For $k$ vertex-disjoint triangles and no other edges, find the triangle count and minimum deletions. Hint: an edit cannot affect two components. Solution: both are $k$. With $n=3k$, the copy density is $1/(27k^2)$ and the edit density is $1/(9k)$; both tend to zero. “Few triangles” is a scale-dependent statement, not synonymous with zero triangles.

45.3 Regularity and counting lemmas

Optional extension; not required to complete the chapter foundation.

For disjoint nonempty vertex sets $A,B$, let $d(A,B)=e(A,B)/(\lvert A\rvert\lvert B\rvert)$. A pair is $\varepsilon$-regular if all $X\subseteq A,Y\subseteq B$ of sizes at least $\varepsilon\lvert A\rvert,\varepsilon\lvert B\rvert$ satisfy $\lvert d(X,Y)-d(A,B)\rvert\le\varepsilon$. Thus regularity controls large subrectangles, not every individual degree.

The established dense regularity lemma says that, given $\varepsilon\gt0$ and a lower bound $m_0$ on the number of parts, sufficiently large graphs admit an equitable partition into $k$ nonexceptional parts with $m_0\le k\le M(\varepsilon,m_0)$, an exceptional set of size at most $\varepsilon n$, and at most $\varepsilon k^2$ irregular unordered pairs. Proof sketch: refine an irregular pair along witnesses that increase an energy index; the bounded index prevents unlimited refinement. The full quantitative partition construction is omitted.

A concrete counting statement is enough here. Fix $0\lt\varepsilon\lt1/2$. Suppose $A,B,C$ are disjoint, all three pairs are $\varepsilon$-regular, all their densities are at least $d$, and $d\ge2\varepsilon$. Then the number of triangles with one vertex in each part is at least

$$(1-2\varepsilon)(d-\varepsilon)^3\lvert A\rvert\lvert B\rvert\lvert C\rvert.$$

For all but at most $2\varepsilon\lvert A\rvert$ vertices of $A$, the neighborhoods in $B,C$ each have size at least $(d-\varepsilon)$ times the part size. Regularity of $(B,C)$ applies to these neighborhoods, because their relative sizes are at least $\varepsilon$, and supplies density at least $d-\varepsilon$. Summing proves the bound. The standard typical-degree assertion follows by taking a set of low-degree vertices as a regularity witness. This proves the displayed triangle form; it does not state a complete arbitrary-graph embedding lemma. See the primary regularity/counting discussion in Gowers.

Solved example. Let each part have two vertices and include every cross-part edge, with no internal edges. Each pair has density one and is regular for every positive $\varepsilon$: every nonempty subrectangle also has density one. There are exactly $2\cdot2\cdot2=8$ crossing triangles. The diagram displays all twelve graph edges and all eight triples, so the model can be audited without an asymptotic approximation.

Exercise. If $(A,B)$ and $(A,C)$ are complete but $(B,C)$ has no edges, what happens? Hint: identify the failed hypothesis. Solution: there are no crossing triangles. The third density is zero, so the positive lower-density condition fails. Regularity alone, without positive densities on every required pair, does not force the configuration.

45.4 Hypergraph container theorems

Optional extension; not required to complete the chapter foundation.

Let $\mathcal H$ be an $r$-uniform hypergraph on $N$ vertices with average degree $d=r e(\mathcal H)/N\gt0$. For $2\le j\le r$, define the maximum codegree $\Delta_j$ to be the largest number of hyperedges containing a specified $j$-vertex set. A large average degree does not prevent edges from concentrating around one pair, so these higher overlaps must be audited.

The following convenient sufficient-condition form is an established consequence of the container theorems of Saxton–Thomason and Balogh–Morris–Samotij. Fix $r\ge2$ and $0\lt\varepsilon\lt1/2$. There exist constants $c=c(r,\varepsilon)\in(0,1/2)$ and $K=K(r,\varepsilon)$ such that, if $0\lt\tau\lt c$ and

$$\Delta_j\le c\,d\,\tau^{j-1}\qquad(2\le j\le r),$$

there is a family $\mathcal C$ covering every independent set, with

$$e(\mathcal H[C])\le\varepsilon e(\mathcal H),\qquad \log\lvert\mathcal C\rvert\le K N\tau\log(1/\tau).$$

Logarithms in this display are natural. Containers can be specified by a bounded-length tuple of fingerprints, each a subset of the covered independent set and each of size $O_{r,\varepsilon}(N\tau)$. This form uses maximum codegrees as a stronger, easy-to-check sufficient condition; the original co-degree-function theorem is more flexible.

Proof mechanism, explicitly a sketch. A deterministic scanning algorithm records a small fingerprint when the independent set contains a vertex with enough surviving constraints. Recorded vertices create lower-uniformity constraints, and the codegree bounds keep these constraints from accumulating too unevenly. The reconstruction rule yields a cover while reducing the induced hyperedge count. Iteration gives the $\varepsilon$ budget, and counting possible fingerprints gives the entropy term. Omitted prerequisites: the algorithm's degree-measure invariants and the co-degree-function estimates; no universal numerical value of $c$ is assumed.

Solved example. For the triangle hypergraph $\mathcal T_n$ on graph edges of $K_n$,

$$N=\binom n2,\qquad e(\mathcal T_n)=\binom n3,\qquad d=n-2,\qquad\Delta_2=\Delta_3=1\quad(n\ge3).$$

Adjacent graph edges extend to exactly one triangle; disjoint ones extend to none. A triple is contained in one hyperedge exactly when it is a triangle. Choose $\tau=A n^{-1/2}$ with $A^2\ge2/c$. For $n\ge4$, the $j=3$ inequality follows from $c(n-2)A^2/n\ge1$. The $j=2$ inequality holds once $cA\sqrt n/2\ge1$, and $\tau\lt c$ holds once $n\gt(A/c)^2$. Thus, for sufficiently large $n$, all assumptions hold and $\log\lvert\mathcal C\rvert=O_\varepsilon(n^{3/2}\log n)$. The constant $A$ is chosen before $n$ tends to infinity.

Exercise. Why does $\tau=n^{-1}$ fail this sufficient criterion? Hint: use the full-edge codegree. Solution: $d\tau^2=(n-2)/n^2\to0$, whereas $\Delta_3=1$. The third-order overlap condition fails despite the small pair codegrees.

45.5 Random analogues of extremal theorems

Optional extension; not required to complete the chapter foundation.

In $G(n,p)$ each possible graph edge is included independently with probability $p=p(n)$. The random Mantel theorem is an established result: for every fixed $\eta\gt0$, there is $C=C(\eta)$ such that, if $p\ge C n^{-1/2}$, then with probability tending to one every triangle-free subgraph $F\subseteq G(n,p)$ satisfies

$$e(F)\le(1/2+\eta)e(G(n,p)).$$

This is a relative statement about retained host edges, not the dense inequality $e(F)\le n^2/4$. Primary transference sources include Conlon–Gowers; container methods give another route. Proof sketch and omitted prerequisites: combine sparse fingerprints, concentration for host intersections with containers, and a supersaturation/stability input. A naive union bound over a coarse fixed container family can introduce an extra logarithmic loss; it is not a proof of the sharp threshold displayed above.

The exponent comes from the two-density parameter $m_2(H)=\max_{J\subseteq H,\ v(J)\ge3}(e(J)-1)/(v(J)-2)$. For a triangle it equals two, giving scale $n^{-1/2}$. More general random Turán theorems have their own hypotheses on the fixed forbidden graph; this section states only the triangle case.

Solved example. Below this scale, when $n^{-2}\ll p\ll n^{-1/2}$, the expected edge count is asymptotic to $n^2p/2$ and the expected triangle count to $n^3p^3/6$. Their ratio is asymptotic to $np^2/3\to0$. Edge concentration and Markov's inequality imply that the triangle count is $o(e(G))$ with high probability. Delete at most one edge per original triangle. The resulting triangle-free subgraph retains $(1-o(1))e(G)$ edges, contradicting a one-half conclusion in this lower-density regime. This is a proof of the scale obstruction, not merely an expectation comparison.

Exercise. For $p=n^{-3/4}$, compare expected triangles and edges. Hint: substitute the powers. Solution: their orders are $n^{3/4}$ and $n^{5/4}$; the ratio is of order $n^{-1/2}$. Almost all host edges can be retained after removing triangles, so the dense extremal fraction does not transfer at this probability.

45.6 Sparse regularity and transference principles

Optional extension; not required to complete the chapter foundation.

For a reference density $p\gt0$, normalized density is $d_p(A,B)=e(A,B)/(p\lvert A\rvert\lvert B\rvert)$. A pair is $(\varepsilon,p)$-regular when all sufficiently large subpairs have normalized density within $\varepsilon$ of the whole pair's normalized density. This allows an absolute error of $\varepsilon p$, unlike dense regularity's fixed absolute error.

A graph is $(\gamma,p,D)$-upper-uniform if disjoint vertex sets of size at least $\gamma n$ have at most $Dp\lvert X\rvert\lvert Y\rvert$ edges between them. A standard sparse regularity lemma says: given $\varepsilon\gt0$, $D\ge1$, and $m_0$, there are $\gamma\gt0$ and $M$ such that upper-uniform graphs admit the analogous equitable partition with at most $\varepsilon k^2$ pairs failing $(\varepsilon,p)$-regularity, for sufficiently large $n$. Proof sketch: normalized energy increments use upper-uniformity to retain a bounded energy budget. The refinement proof and parameter hierarchy are omitted; Scott's primary sparse-regularity paper states the upper-uniform framework and develops a further generalization.

Sparse regularity alone does not supply a counting lemma. Neighborhoods may have relative size only about $p$, below the $\varepsilon$ scale, so dense typical-neighborhood reasoning may stop exactly where it is needed. A transfer theorem requires further pseudorandomness or a random-host embedding theorem. Primary treatments include Conlon–Gowers and the sparse counting applications in Saxton–Thomason. All are established results with explicit host hypotheses; the word “sparse” is not itself an assumption sufficient for transference.

Solved example. Suppose a pair has edge density $p/2$. Its normalized density is $1/2$. Dense $\varepsilon$-regularity permits a subpair density error $\varepsilon$, whose relative error is $\varepsilon/p$ and can diverge as $p\to0$. An $(\varepsilon,p)$-regularity guarantee permits error only $\varepsilon p$, giving normalized error at most $\varepsilon$. For $p=1/100$ and $\varepsilon=1/10$, the respective absolute tolerances are $1/10$ and $1/1000$.

Exercise. Does a matching on $n$ vertices look dense-regular on large vertex parts at fixed error tolerance? Hint: its total edge count is only $n/2$. Solution: every large subpair has density $O(1/n)$, so dense regularity becomes almost vacuous for fixed $\varepsilon$. Yet the matching has no triangles. Small absolute densities do not establish a relative triangle-counting theorem.

45.7 Extremal configurations in pseudorandom structures

Optional extension; not required to complete the chapter foundation.

A deterministic pseudorandom host needs a numerical discrepancy condition. For a $d$-regular graph on $n$ vertices with adjacency matrix $A$, let $\lambda$ bound the absolute values of all eigenvalues on the subspace orthogonal to the all-ones vector. For subsets $S,T$, let $e^{\mathrm{ord}}(S,T)$ count ordered adjacent pairs, allowing overlap of the subsets. The expander mixing inequality is

$$\left\lvert e^{\mathrm{ord}}(S,T)-\frac dn\lvert S\rvert\lvert T\rvert\right\rvert\le\lambda\sqrt{\lvert S\rvert\lvert T\rvert}.$$

Complete proof mechanism: center the two indicator vectors, use $A\mathbf1=d\mathbf1$, bound the operator on the centered subspace by $\lambda$, and apply Euclidean Cauchy–Schwarz. The sharper centered-norm form is also valid. For an independent set $S$, this gives $\lvert S\rvert\le\lambda n/d$ when $d\gt0$.

Triangle counts have a different sensitivity. Since $6t(G)=\operatorname{tr}(A^3)$,

$$\left\lvert6t(G)-d^3\right\rvert\le n\lambda^3.$$

Thus $n\lambda^3\ll d^3$ is a sufficient condition for $t(G)=(1+o(1))d^3/6$. It is not asserted to be a necessary or optimal condition. Spectral mixing and small-subgraph counting have different error budgets; being an expander does not automatically provide every sparse counting theorem. A primary pseudorandomness framework is Chung, Graham, and Wilson's quasi-random graph paper; the two displayed bounds are derived here directly.

Solved example. For $K_4$, $d=3,n=4$, and the remaining eigenvalues are all $-1$, so $\lambda=1$. The spectral independent-set bound is $4/3$, forcing integer size at most one, which is exact. The triangle trace is $(27-3)/6=4$. For a four-cycle, $d=2$ but an orthogonal eigenvalue is $-2$, so $\lambda=d$: bipartiteness prevents the small-relative-error condition, consistent with the absence of triangles.

Exercise. Why must absolute eigenvalues be used rather than only the second-largest eigenvalue? Hint: inspect a regular bipartite graph. Solution: it has eigenvalue $-d$, which detects the bipartition. Ignoring that negative eigenvalue would falsely suggest strong mixing across all large subsets.

45.8 Combinatorial discrepancy and partial coloring

Optional extension; not required to complete the chapter foundation.

For a set system $\mathcal S$ on $[n]$ and a coloring $x\in\{-1,1\}^n$, its discrepancy is $\max_{S\in\mathcal S}\lvert\sum_{i\in S}x_i\rvert$. The aim is to split every set almost evenly at once. A partial coloring lies in $\{-1,0,1\}^n$: zero means unassigned, not a third final color.

The brief bridge to later optional sections 52.1 and 52.4 is the translation to a zero-one incidence matrix and a sequence of partial assignments. No later material is an entry requirement. A full primary constructive treatment is Lovett and Meka; the classical result is Spencer's Six Standard Deviations Suffice.

An established partial-coloring principle gives, for an $m\times n$ real matrix with entries of absolute value at most one and $m\ge n$, a partial coloring fixing at least half the coordinates, with row discrepancy at most $C\sqrt{n\log(2m/n)}$ for an absolute constant $C$. Reapply to the remaining columns and sum the budgets. When $m=n$ initially, this yields a full coloring with discrepancy $O(\sqrt n)$ because the shrinking column counts make the budgets summable. Proof sketch and omitted prerequisites: the existential argument quantizes row sums and uses an entropy pigeonhole argument to obtain two well-separated sign vectors with similar quantized data; half their difference is a partial coloring. The quantitative entropy lemma is omitted. The cited constructive approach uses a constrained random walk rather than this abbreviated sketch.

Solved example. On four elements take $S_1=\{1,2\}$, $S_2=\{2,3\}$, $S_3=\{3,4\}$, $S_4=[4]$. The partial coloring $(1,0,0,-1)$ assigns half the coordinates and has row sums $1,0,-1,0$. Completing it as $(1,-1,1,-1)$ makes all four row sums zero. This verifies one completion, not a promise that any arbitrary half-coloring admits a zero-discrepancy completion.

Exercise. Can the system consisting of all singleton sets have discrepancy zero? Hint: each final color has absolute value one. Solution: no; its discrepancy is exactly one under every coloring. General discrepancy theorems provide bounded error, not exact simultaneous balance.

45.9 Container methods for independent-set enumeration

Optional extension; not required to complete the chapter foundation.

Let $T(n)$ be the number of labeled triangle-free graphs. Containers and supersaturation yield the established leading asymptotic

$$T(n)=2^{n^2/4+o(n^2)}.$$

Here is the deduction using the sufficient-condition theorem already stated, following the primary enumeration framework of Saxton–Thomason. Fix $0\lt\eta\lt1/4$ and choose container error $\varepsilon=\eta$. A container is a graph-edge set with at most $\eta\binom n3$ triangles. If it had at least $(1/4+\eta)n^2$ graph edges, section 45.1 would force at least $(\eta+4\eta^2)n^3/3$ triangles, greater than $\eta n^3/6$ and hence greater than the allowed count. Every container therefore has fewer than $(1/4+\eta)n^2$ edges.

The number of containers has logarithm $O_\eta(n^{3/2}\log n)=o(n^2)$. Applying the union bound gives $\log_2T(n)\le(1/4+\eta)n^2+o(n^2)$. For the lower bound, every subgraph of one fixed balanced complete bipartite graph is triangle-free, giving $T(n)\ge2^{\lfloor n^2/4\rfloor}$. Letting fixed $\eta$ become arbitrarily small proves the leading asymptotic. This argument is complete conditional on the established container theorem; it does not require claiming that every triangle-free graph is bipartite.

Solved example. For $n=4$, the fixed balanced cut gives sixteen triangle-free graphs. Exact enumeration gives $41$, and the seven-container cover gives the upper bound eighty. An asymptotic exponent need not be numerically sharp at four vertices, and an overlapping cover can even exceed the total number of all graphs.

An elementary fingerprint-count budget helps explain the theorem. If each fingerprint has at most $s$ of $N$ vertices, then there are at most $\sum_{i=0}^s\binom Ni$ possibilities. For $s\le N/2$, this is at most $\exp(Nh(s/N))$, where $h(u)=-u\log u-(1-u)\log(1-u)$ is binary entropy in natural units. A bounded number of fingerprints changes only the constant in the exponent. Reconstruction must still be specified: counting short subsets alone does not prove a cover.

Exercise. Why is the count of all bipartitions insufficient as the entire upper-bound proof? Hint: use a five-cycle. Solution: some triangle-free graphs are not contained in any bipartite graph. General containers cover them while allowing a small triangle count; supersaturation, not exact bipartiteness, supplies the size bound.

45.10 Open problems in Turán-type theory

Optional extension; not required to complete the chapter foundation.

For a fixed $r$-uniform hypergraph $F$, define its Turán density by

$$\pi(F)=\lim_{n\to\infty}\frac{\operatorname{ex}(n,F)}{\binom nr}.$$

The limit exists because averaging induced subhypergraphs makes the normalized extremal densities nonincreasing for $n\ge v(F)$. Graph chromatic-number theorems do not determine arbitrary hypergraph Turán densities.

The tetrahedron $K_4^{(3)}$ consists of all four triples on four vertices. The classical ordinary Turán conjecture predicts $\pi(K_4^{(3)})=5/9$. Status checked on 13 October 2026: the general ordinary-density problem remains unresolved in the primary literature located for this chapter. A 23 September 2026 preprint by Jeong and collaborators reports the upper bound

$$\pi(K_4^{(3)})\le\frac{312372062889819}{560000000000000}\lt0.557808.$$

This is a reported preprint result with an exact flag-algebra certificate and a Lean formal proof described by its authors; it is dated here rather than silently labeled a settled best-known bound for all future readers. Its value remains above $5/9$. The separate September 2026 preprint on uniform Turán density proves a value of $1/2$ under an additional uniform-distribution condition. Its title must not be interpreted as resolving the ordinary $5/9$ question. The uniform density is a different invariant.

Solved example: the lower-bound construction. Partition $3m$ vertices into equally sized classes $A,B,C$. Include all triples of type $ABC$, $AAB$, $BBC$, and $CCA$. There are

$$m^3+3m\binom m2$$

edges. Divide by $\binom{3m}3$ and let $m\to\infty$ to obtain density $5/9$. To check tetrahedron-freeness, classify four vertices by their part sizes. A $4$ or $3+1$ pattern misses a triple wholly inside one part. A $2+2$ pattern misses the reverse-oriented double-single triple. A $2+1+1$ pattern has a doubled class, but only one of its two cyclic double-single types is allowed, so it also misses a triple. These exhaust the patterns and prove the lower bound. This construction is established; the unproved step is a matching upper bound.

Proof scope for the reported upper bound: its semidefinite/flag-algebra certificate is not reproduced. Small enumeration, supersaturation, and containers do not by themselves supply it. Containers can enumerate constrained families relative to a density once an extremal/supersaturation input is known; they do not automatically determine an unknown density.

Exercise. Evaluate the construction at $m=2$. Hint: retain the exact finite denominator. Solution: it has $8+3\cdot2\cdot1=14$ edges on six vertices, out of $\binom63=20$ possible triples, for density $7/10$. This exceeds $5/9$ without contradiction: $5/9$ is a limiting density, not the construction's exact finite density for every $n$.

Apply a container theorem and audit every sparsity assumption

The investigation connects exact finite coverage with the asymptotic codegree audit. Its supersaturation step turns few forbidden configurations into a container-size bound, while the tetrahedron comparison identifies the additional extremal information that remains needed for an open density problem. A practical analogue is a constraint hypergraph for resource selections: hyperedges are prohibited combinations, independent sets are feasible selections, and containers compress the space of candidate solutions. Mathematical existence of containers is not an automatic efficient optimization algorithm.