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Partition both sides of a half graph into classes of size at least $100$. A class of size $s$ has at most $2\lceil s/10\rceil$ elements with fewer than $s/10$ class elements on one side in the underlying order. If there are $k_X$ and $k_Y$ classes, the union of these boundary elements has size at most $2n/5+2(k_X+k_Y)\le 11n/25$. Some index $m$ is therefore away from both boundaries. The class containing $m$ on each side has at least one tenth of its points strictly above $m$ and strictly below $m$. Choosing lower $X$ points with upper $Y$ points, or upper $X$ points with lower $Y$ points, yields edge density of a bipartite graph values one and zero. Both cannot be within $1/10$ of the same class-pair edge density of a bipartite graph, so that pair is not a $1/10$-regular pair of vertex sets.

A half graph has two vertex classes $X=\{x_1,\dots ,x_n\}$ and $Y=\{y_1,\dots ,y_n\}$, with $x_i$ adjacent to $y_j$ exactly when $i\le j$. Its nested vertex neighbourhoods give a useful obstruction to partitions in which every pair is a regular pair of vertex sets.

For disjoint nonempty vertex sets $U,V$ of a graph, an $\epsilon$-irregular pair is one that is not an $\epsilon$-regular pair of vertex sets. Equivalently there are witnesses $U^{\prime }\subseteq U$, $V^{\prime }\subseteq V$ with $|U^{\prime }|\ge \epsilon |U|$, $|V^{\prime }|\ge \epsilon |V|$ and $|d(U^{\prime },V^{\prime })-d(U,V)|>\epsilon$. Splitting both cells by these witnesses increases their contribution to the regularity energy by at least ${\epsilon }^4|U||V|/n^2$: the witness rectangle has relative weight at least ${\epsilon }^2$, its mean discrepancy has magnitude greater than $\epsilon$, and the Cauchy-Schwarz inequality bounds the weighted sum of squared discrepancies below by the rectangle weight times its squared mean. This is the local increment in the energy-increment proof of Szemerédi regularity.

Use the equitable form of the Szemerédi regularity lemma. For $0<\epsilon <1/2$ and a prescribed lower bound $m_0$, there are $M,n_0$ such that every sufficiently large graph admits a set partitionwith at most $\epsilon k^2$ unordered irregular pairs among the nonexceptional cells. A uniform pair here means a regular pair of vertex sets. Larger values of $\epsilon$ follow from a smaller parameter, so the stated restriction loses nothing.

For the proof, retain the exceptional vertices as singleton cells in an auxiliary set partition $\mathcal{P}$. Define its regularity energy by the ordered sumFor diagonal rectangles the adjacency matrix of a graph defines $d(A,A)$; diagonal terms are included. Thus $0\le q\le 1$. If $A\times B$ is split into rectangles of densities $d_{uv}$ and sizes $w_{uv}$, their weighted mean is $d=d(A,B)$. Expanding squares proves the refinement variance identity for regularity energy:Every refinement therefore increases or preserves energy.

Suppose equal cells $V_i,V_j$ of size $m$ are $\epsilon$-irregular. Choose witnesses $X\subseteq V_i$, $Y\subseteq V_j$ of sizes at least $\epsilon m$, whose density differs from $d(V_i,V_j)$ by more than $\epsilon$. Refine the cells so that these witnesses are unions of refined atoms. Applying the Cauchy-Schwarz inequality inside $X\times Y$ gives an increase, for this oriented rectangle, greater thanChoose witnesses simultaneously for every irregular pair. Each original cell is cut by at most $k-1$ witness sets and has at most $2^k$ atoms. If more than $\epsilon k^2$ unordered pairs are irregular, both orientations contribute, givingas long as the exceptional set has at most $n/2$ vertices.

To restore equal sizes without sacrificing this increase, use equalization with a controlled exceptional set. Cut every witness atom into blocks of size $\ell =\lfloor m/4^k\rfloor$, leaving fewer than $\ell$ unused vertices in each atom. Make each unused vertex a singleton exceptional cell. This is a further refinement, so there is no energy loss. The number newly exceptional is at mostWhen $m\ge 2\cdot 4^k$, the new nonexceptional cell count $k^{\prime }$ is at most $4k{4}^k$: indeed $\ell \ge m/(2\cdot 4^k)$ and $n/m\le 2k$. It is at least $k$, since each old cell supplies at least $4^k-2^k\ge 1$ blocks.

Let $L=\lceil 2{\epsilon }^{-5}\rceil +1$. Choose the initial count $k_0\ge m_0$ so large that $L{2}^{-k_0}\le \epsilon /2$, and begin with $k_0$ equal cells and fewer than $k_0$ leftover vertices. Take $n_0$ large enough that these initial leftovers are at most $\epsilon n/2$. Iterate the bound $k\mapsto 4k{4}^k$ for $L$ rounds to obtain a finite number $M$. Increase $n_0$ further to ensure $n\ge 4M{4}^M$. Then every cell size encountered satisfies $m\ge n/(2M)\ge 2\cdot 4^k$, so every equalization is legitimate. The total exception budget is at most $\epsilon n$, because the initial losses use half the budget and all later losses sum to at most $Ln{2}^{-k_0}$. Energy cannot increase by more than one, so $L$ failing rounds are impossible. The process must stop with at most $\epsilon k^2$ irregular pairs. This proves the Szemerédi regularity lemma, including the bounded cell count, equitability and exceptional-set control.

For the dense-pair application, apply this proof with $\eta =1/100$ and $m_0=100$. Suppose every $\eta$-regular pair of nonexceptional cells has density at most $1/4$. Edges touching $V_0$ number at most $\eta n^2$. Edges inside nonexceptional cells number at most $k{m}^2/2\le n^2/(2m_0)$. The at most $\eta k^2$ irregular cell pairs contribute at most $\eta n^2$. Finally the regular cross-cell pairs contribute at most $k(k-1)m^2/8\le n^2/8$. Thereforea contradiction. Some regular pair $(U,W)$ consequently has density greater than $1/4$. It is also a $1/8$-uniform pair, since subsets of relative size at least $1/8$ qualify in the stronger $\eta$-regularity condition. The common cell size satisfies $m\ge (1-\eta )n/M\ge n/(2M)$. Thus the dense regular pair from edge surplus givesThis first proves the assertion for $n\ge n_0$. For the finitely many smaller possible orders, choose the endpoints of any edge as singleton sets: the pair has density one and is regular for every positive parameter. Taking $c=\min \{1/(2M),1/n_0\}>0$ includes these cases as well.

For disjoint nonempty vertex sets $U,V$, write $d(U,V)=e(U,V)/(|U||V|)$. A regular pair of vertex sets is $\epsilon$-regular if every $U^{\prime }\subseteq U,V^{\prime }\subseteq V$ with $|U^{\prime }|\ge \epsilon |U|$, $|V^{\prime }|\ge \epsilon |V|$ satisfies $|d(U^{\prime },V^{\prime })-d(U,V)|\le \epsilon$.

The Szemerédi regularity lemma states that, for every $\epsilon >0$ and integer $m_0$, there are $M,n_0$ such that every finite simple graph on $n\ge n_0$ vertices has a set partitionwith $m_0\le m\le M$, $|V_0|\le \epsilon n$, $|V_1|=\cdots =|V_m|$, and at most $\epsilon m^2$ irregular unordered pairs among the nonexceptional classes. Prove it for $0<\epsilon \le 1/2$; larger parameters follow by using a smaller one.

For the energy-increment proof of Szemerédi regularity, keep every exceptional vertex as a singleton cell in the set partition used to measure regularity energy. Define the equitable regularity energyThe sum is ordered, and on diagonal cells subset density means the average of the adjacency indicator, with zero on loops. Thus $0\le q\le 1$. If $U,V$ are refined into $U_a,V_b$, expanding squares givesThe cross term vanishes because the weighted arithmetic mean of the refined subset densities is the original subset density. Consequently every refinement of a set partition increases or preserves regularity energy.

Suppose the current nonexceptional classes have common size $t$ and there are $k$ of them. For each irregular pair $V_i,V_j$, choose witnesses $A_{ij},B_{ij}$ of sizes at least $\epsilon t$ and discrepancy greater than $\epsilon$. Refine each class by all its incident witness subsets. There are at most $k-1$ cuts in each class and at most $2^k$ resulting atoms. The witness rectangle is a union of refined rectangles. By the Cauchy-Schwarz inequality, its contribution to the variance identity is at leastIndeed, the sum of squared deviations on the witness rectangle is at least its total weight times the square of their weighted arithmetic mean. If more than $\epsilon k^2$ unordered pairs are irregular, summing even just one orientation of each givesprovided the exceptional set has at most $n/2$ vertices. This is the strictly positive regularity energy increment.

Restore equal class sizes without losing that increment. Divide every witness atom into blocks of size $\ell =\lfloor t/4^k\rfloor$ and make every leftover vertex an exceptional singleton. This is a further refinement of a set partition, so it cannot decrease $q$. There are at most $k{2}^k$ atoms, hence fewer than $k{2}^k\ell \le n{2}^{-k}$ newly exceptional vertices. If $t\ge 2\cdot 4^k$, then $\ell \ge t/(2\cdot 4^k)$, so the number of full blocks is at most $2k{4}^k$. Each original class produces at least one full block: its leftovers occupy less than $2^k\ell$, whereas $t\ge 4^k\ell >2^k\ell$. Thus the new class count is at least $k$.

Choose $R_{\ast }=\lceil 4{\epsilon }^{-5}\rceil +1$, and choose an initial integer $k_0\ge m_0$ large enough that $R_{\ast }{2}^{-k_0}\le \epsilon /2$. Initially partition into $k_0$ equal classes and fewer than $k_0$ exceptional vertices. Take $n_0$ large enough that this initial exception is at most $\epsilon n/2$. During at most $R_{\ast }$ rounds, the additional exception is at most $R_{\ast }n{2}^{-k_0}\le \epsilon n/2$, so it never exceeds $\epsilon n\le n/2$.

A bound on class counts is obtained by iterating $k\mapsto 2k{4}^k$ at most $R_{\ast }$ times starting at $k_0$; call the resulting bound $M$. Increasing $n_0$ further to satisfy $n_0\ge 4M{4}^M$ guarantees $t\ge n/(2k)\ge 2\cdot 4^k$ at every round, so all full-block constructions are valid. If the desired regularity had not been reached after $R_{\ast }$ refinements, the regularity energy increments would total more than one, impossible. Therefore the process terminates with the required partition and bounds. The singleton treatment of discarded vertices is what preserves regularity energy monotonicity during equalization.

For disjoint nonempty vertex sets $U,W$, define their edge density of a bipartite graph by $d(U,W)=e(U,W)/(|U||W|)$. They form an $\epsilon$-regular pair of vertex sets if every $X\subseteq U$, $Y\subseteq W$ with $|X|\ge \epsilon |U|$, $|Y|\ge \epsilon |W|$ satisfies $|d(X,Y)-d(U,W)|\le \epsilon$.

The Szemerédi regularity lemma asserts that, given $\epsilon >0$ and a positive integer $m_0$, there are $M,n_0$ such that every graph on $n\ge n_0$ vertices has a set partition $V_0,V_1,\dots ,V_k$ withand at most $\epsilon k^2$ irregular ordered pairs $(V_i,V_j)$ with $1\le i\ne j\le k$.

Assume $0<\epsilon \le 1/2$, which suffices since a smaller regularity parameter gives the claim for a larger one. Treat the exceptional vertices as singleton cells in the full partition $\mathcal{P}$, and use equitable regularity energyHere the adjacency indicator function defines density even for diagonal cells. Clearly $0\le q\le 1$. If ${\mathcal{P}}^{\prime }$ refines $\mathcal{P}$, weighted averaging of densities gives the refinement variance identity for regularity energyTo verify it, expand each square and use ${\sum }_{A^{\prime },B^{\prime }}|A^{\prime }||B^{\prime }|d(A^{\prime },B^{\prime })=|A||B|d(A,B)$.

For every irregular ordered pair of nonexceptional cells, choose witnesses $X_{ij},Y_{ij}$ of relative sizes at least $\epsilon$ and density discrepancy greater than $\epsilon$. Split each $V_i$ along all witnesses involving it; symmetry allows at most one witness for each other cell, hence at most $2^k$ atoms in each $V_i$. In the witness rectangle, the weighted Cauchy-Schwarz inequality in the variance identity contributes more than ${\epsilon }^4|V_i||V_j|/n^2$.

If more than $\epsilon k^2$ pairs are irregular and the equal cell size is $L$, the energy increase is greater thanRestore equal cell sizes by cutting each witness atom into pieces of size $\ell =\lfloor L/4^k\rfloor$, with leftover vertices added to $V_0$ as singleton cells. This operation is a further refinement, so the energy does not decrease. There are at most $k{2}^k$ atoms, and fewer than $\ell$ leftover vertices in each, giving at most $k{2}^k\ell \le n{2}^{-k}$ new exceptional vertices.

Here is a uniform bound ensuring that the iteration is legitimate. Put $R=\lceil 4{\epsilon }^{-5}\rceil$ and chooseStart with $k_0$ equal classes and fewer than $k_0$ exceptional vertices. Taking $n\ge 2k_0/\epsilon$ makes the initial exception at most $\epsilon n/2$. At every refinement the number of nonexceptional cells does not decrease: each old cell loses at most $L{2}^{-k}\le L/2$ vertices and is cut into pieces of size at most $L/4^k$. Thus each of the at most $R$ rounds adds at most $\epsilon n/(2R)$ exceptional vertices.

When $L\ge 2\cdot 4^k$, the new cell count is at mostusing $|V_0|\le \epsilon n\le n/2$. Define $K_0=k_0$ and $K_{j+1}=4K_j{4}^{K_j}$, and take $M=K_R$. Choosing $n_0\ge \max \{2k_0/\epsilon ,4M{4}^M\}$ ensures $L\ge n/(2M)\ge 2\cdot 4^M$ at every stage. All rounds are therefore defined, with cell count between $m_0$ and $M$ and exceptional size at most $\epsilon n$.

If the lemma still failed after $R$ refinements, the energy would have increased by more than $R{\epsilon }^5/4\ge 1$, impossible because it lies in $[0,1]$. Hence some stage has the required regular pairs. This completes the energy-increment proof of Szemerédi regularity, including the equal-size and exceptional-set conditions.

We derive the triangle removal lemma from the permitted Szemerédi regularity lemma, then turn a corner in an integer grid into a triangle in a graph.

The triangle removal lemma says that for every $\eta >0$ there are $\rho >0$ and $N_0$ such that a graph on $N\ge N_0$ vertices with fewer than $\rho N^3$ triangles in a graph can be made triangle-free by deleting at most $\eta N^2$ edges. Here triangles in a graph are unordered triples of distinct vertices. We prove this consequence with the necessary quantitative dependence.

We may assume $0<\eta <1$. Put $d=\eta /4$, choose an integer $m_0\ge 4/\eta$, and chooseApply the Szemerédi regularity lemma to obtain an exceptional class $V_0$ of size at most $\epsilon N$ and equal-sized classes $V_1,\dots ,V_k$ of size $L$, where $m_0\le k\le M$, and at most $\epsilon k^2$ pairs are not regular pairs of vertex sets. The constant $M$ depends only on these chosen parameters. Delete all edges incident with $V_0$, all edges within a class, all edges between irregular pairs, and all edges between pairs whose edge density of a bipartite graph is less than $d$. The respective costs are at mostTheir sum is at most $\eta N^2/2$, hence certainly at most $\eta N^2$.

If a triangle in a graph survives, it lies in three distinct classes whose three pairs are $\epsilon$-regular pairs of vertex sets of edge density of a bipartite graph at least $d$ in the original graph. We now check the needed regular triangle counting lemma. In the first class all but at most $2\epsilon L$ vertices have at least $(d-\epsilon )L$ neighbours in each of the other two classes; otherwise the definition of a regular pair of vertex sets would be violated. For each such vertex $v$, its two neighbour sets each have size at least $\epsilon L$, so their mutual edge density of a bipartite graph is at least $d-\epsilon$. It follows that the original three classes contain at leasttriangles in a graph. The elementary lower bound holds because $\epsilon \le 1/4$ and $\epsilon \le d/4$. Also $L=(N-|V_0|)/k\ge N/(2M)$. Thus a surviving triangle in a graph implies at least $d^3{N}^3/(64M^3)$ original triangles in a graph. Set $\rho =d^3/(128M^3)$ and take $N_0$ large enough for the Szemerédi regularity lemma. Fewer than $\rho N^3$ original triangles in a graph force the cleaned graph to be triangle-free, proving the triangle removal lemma.

Now suppose $0<\delta \le 1$ and $A\subseteq [n{]}^2$. Use the tripartite graph encoding of a grid to construct a tripartite graph with disjoint labelled classesFor each $(a,b)\in A$, put in the three edges joining $a\in X$, $b\in Y$, and $a+b\in Z$. These yield $|A|$ canonical triangles in a graph. They are pairwise edge-disjoint triangles: an $XY$ edge determines $(a,b)$ directly, an $XZ$ edge determines $b=z-a$, and a $YZ$ edge determines $a=z-b$. Any deletion making the graph triangle-free must therefore remove at least $|A|$ edges.

On the other hand, an arbitrary triangle in a graph with labels $(x,y,z)$ gives three points of $A$:Writing $d=z-x-y$, these are $(x,y),(x,y+d),(x+d,y)$. If $d\ne 0$, they form the required corner in an integer grid. If there is no such corner in an integer grid, every triangle in a graph is canonical, and the total number is precisely $|A|\le n^2$.

Apply the triangle removal lemma with $\eta =\delta /32$. The constructed graph has $N=4n$ vertices. For sufficiently large $n$, $N\ge N_0$ andIn the absence of a corner in an integer grid, the triangle removal lemma would destroy all triangles in a graph by deleting at mostedges, contradicting the pairwise edge-disjoint triangles. Every sufficiently large grid therefore has the asserted corner in an integer grid in every subset of positive fixed density of a finite subset. For $\delta >1$ there is no eligible subset, so the assertion is vacuous. This proves the corners theorem, including the stipulated nonzero displacement; no positivity of $d$ was required.

For disjoint nonempty vertex sets $A,B$, write $d(A,B)=e(A,B)/(|A||B|)$. The regular pair of vertex sets $(A,B)$ is $ϵ$-uniform when every $X\subseteq A$, $Y\subseteq B$ with $|X|\ge ϵ|A|$ and $|Y|\ge ϵ|B|$ satisfiesThis is also called an $ϵ$-regular pair.

Count transversal cliques by maintaining common neighbourhoods. For $r=1$ there are exactly $n$ copies, so the claim is immediate. For $r\ge 2$ the hypotheses are vacuous if $\lambda >1$; assume $0<\lambda \le 1$. SetAfter choosing vertices in the first $i-1$ classes, let $C_j\subseteq V_j$ be their common neighbours in each later class. We maintain $|C_j|\ge (\lambda -\eta {)}^{i-1}n\ge an$. In a regular pair, whenever $|C_j|\ge \eta n$, fewer than $\eta n$ vertices of $V_i$ have fewer than $(d(V_i,V_j)-\eta )|C_j|$ neighbours in $C_j$: otherwise those bad vertices, together with $C_j$, contradict regularity. There are at most $r-1$ future classes, so their union excludes at most $(r-1)\eta n$ candidates from $C_i$. At least $an/2$ choices remain, and each maintains the stated bound for all new common neighbourhoods.

Making these choices through all $r$ classes gives at least $(an/2{)}^r$ transversal cliques. Different selections give different vertex sets, since the classes are disjoint. ConsequentlyAll constants depend only on $r,\lambda$, not on $n$.

The Szemerédi regularity lemma states that for every $\eta >0$ and integer $m_0$ there exist $M,n_0$ such that every graph of order $N\ge n_0$ has a partition $V_0,V_1,\dots ,V_k$ with $m_0\le k\le M$, $|V_0|\le \eta N$, equal-sized nonexceptional classes, and at most $\eta k^2$ irregular unordered pairs of classes.

Apply the partition to prove clique removal. For $r\ge 2$, choose $\lambda =\min \{ϵ/4,1/2\}$, $m_0\ge \max \{r,4/ϵ\}$, and a regularity parameter $\eta \le \min \{ϵ/8,1/2\}$ small enough for the preceding counting argument at density $\lambda$. The Szemerédi regularity lemma supplies $M$. Delete all edges incident with $V_0$, inside individual classes, between irregular pairs, and between regular pairs with density below $\lambda$. If the common class size is $L$, the number deleted is at mostA surviving $K_r$ would have its vertices in distinct classes; every pair of these classes is regular and has density at least $\lambda$. The counting result would then give at least ${\delta }_0{L}^r$ copies in the original graph. Since $L=(N-|V_0|)/k\ge N/(2M)$, choose, for example,This would exceed $\delta N^r$, contradicting the assumed copy count. Thus fewer than $\delta N^r$ cliques can be destroyed by removing at most $ϵN^2$ edges, leaving no $K_r$. This is the clique removal lemma.

For $r=1$, take $\delta <1$: a graph of order $N$ always has $N$ copies of $K_1$, so the hypothesis is impossible. The assertion in that degenerate case is vacuous, rather than an edge-removal procedure capable of deleting vertices.

Write $d(U,V)=e(U,V)/(|U||V|)$ for the edge density of a bipartite graph between nonempty disjoint vertex sets. A regular pair of vertex sets $(U,V)$ is $\epsilon$-regular if $|d(U^{\prime },V^{\prime })-d(U,V)|\le \epsilon$ whenever $U^{\prime }\subseteq U$, $V^{\prime }\subseteq V$, $|U^{\prime }|\ge \epsilon |U|$, and $|V^{\prime }|\ge \epsilon |V|$.

The Szemerédi regularity lemma says that for every $\epsilon >0$ and positive integer $m_0$ there is $M=M(\epsilon ,m_0)$ such that every finite simple graph on $N\ge m_0$ vertices has a set partitionwhere $V_1,\dots ,V_k$ have equal positive size and at most $\epsilon k^2$ unordered pairs $(V_i,V_j)$, $i<j$, fail to be regular pairs of vertex sets. The version requiring only sufficiently large $N$ is equivalent, since the finitely many smaller admissible sizes can use singleton sets as classes. It suffices to prove this for $0<\epsilon \le 1/2$: reducing a larger parameter to $1/2$ gives stronger conclusions.

Here is a complete energy proof, including restoration of equal class sizes. For a set partition with exceptional set $V_0$, use the following version of equitable regularity energy, omitting pairs incident with $V_0$:For $i=j$, define $d(V_i,V_i)$ using the ordered adjacency indicator function, with diagonal entries zero. Thus the same formula works on every rectangle. The refinement variance identity for regularity energy iswhere $A,B$ run over the refined cells. Expanding the square proves the identity because their weighted mean edge density of a bipartite graph is $d(U,V)$. In particular, refinement never decreases the energy. If a union of refined cells $U^{\prime }\times V^{\prime }$ has discrepancy greater than $\epsilon$, the Cauchy-Schwarz inequality bounds its contribution below by $|U^{\prime }||V^{\prime }|{\epsilon }^2/N^2$. For an irregular equal-class pair of size $L$, witnesses have sizes at least $\epsilon L$, so the increase exceeds ${\epsilon }^4{L}^2/N^2$ in each orientation.

Choose $k_0\ge \max \{m_0,2\}$ so large that $2^{-k_0}\le {\epsilon }^6/100$. Initially partition into $k_0$ equal classes and fewer than $k_0$ exceptional vertices. For sufficiently large $N$ this exceptional set has size at most $\epsilon N/2$. Suppose there are more than $\epsilon k^2$ irregular unordered pairs at a subsequent step, with equal class size $L$. Choose one witnessing pair of subsets for every such pair. Split each class according to membership in all its witness subsets; each class produces at most $2^k$ atoms. The preceding refinement variance identity for regularity energy shows that the energy increase exceedsas long as the exceptional set has size at most $\epsilon N\le N/2$.

Apply equalization with a controlled exceptional set. Set $\ell =\lfloor L/4^k\rfloor$, cut every atom into blocks of size $\ell$, and send its leftover points to $V_0$. Provided $L\ge 2\cdot 4^k$, the new number $k^{\prime }$ of classes satisfiesThe lower bound holds because every old class retains at least one full block: its leftovers total less than $2^k\ell <L$. The discarded fraction is at mostCutting blocks is a further refinement before discarding. Discarding changes only ordered pairs with at least one discarded vertex, so it loses at most $2r$ in the energy. The net increase is therefore at least ${\epsilon }^5/2-2{\epsilon }^6/100\ge {\epsilon }^5/4$.

There can be fewer than $T=\lceil 4{\epsilon }^{-5}\rceil$ such strict energy increases, because the energy stays in $[0,1]$. Through $T$ steps the exceptional fraction is at mostThis also verifies the exceptional-set bound used at every preceding step. Iterating the finite recursion $k\mapsto 2k{4}^k$ at most $T$ times bounds all class counts by some $K=K(\epsilon ,m_0)$. Choosing $N$ sufficiently large in terms of $K$, for instance with $N\ge 4K{4}^K$ and $N\ge 2k_0/\epsilon$, guarantees $L=(N-|V_0|)/k\ge N/(2K)\ge 2\cdot 4^k$ throughout. Thus the process terminates with the desired regular pairs of vertex sets. Choose a finite integer threshold $N_{\ast }$ for these requirements and set $M\ge \max \{K,N_{\ast }\}$. For $m_0\le N<N_{\ast }$, use $N$ singleton sets; every pair is regular because its only admissible nonempty subsets are the whole pair. This completes the Szemerédi regularity lemma for all admissible $N$.

For the half graph, a class of size $s$ has exactly $\lceil s/10\rceil$ elements with fewer than $s/10$ predecessors, and the same number with fewer than $s/10$ successors. Thus its boundary contains at most $2\lceil s/10\rceil$ elements. Let $k_X,k_Y$ be the numbers of classes. The boundary rank obstruction in a half graph gives the union boundbecause $k_X,k_Y\le n/100$. There is therefore an index $m$ away from both boundaries. In its classes $X_i,Y_j$, all four subsets $X^{-}=\{x<m\}$, $X^{+}=\{x>m\}$, $Y^{-}=\{y<m\}$, $Y^{+}=\{y>m\}$ have at least one tenth of their respective class sizes. These subsets are taken within $X_i,Y_j$, which need not be intervals.

In the half graph, $d(X^{-},Y^{+})=1$ and $d(X^{+},Y^{-})=0$. If $(X_i,Y_j)$ were a $1/10$-regular pair of vertex sets, both values would be within $1/10$ of $d(X_i,Y_j)$, implying $1\le 2/10$, a contradiction. At least one class pair is not $1/10$-regular.

Statement and definitions. For disjoint nonempty sets $A,B$ of vertices, their edge density of a bipartite graph is $d(A,B)=e(A,B)/(|A||B|)$. They form a regular pair of vertex sets, or an $\epsilon$-uniform pair, ifwhenever $X\subseteq A$, $Y\subseteq B$, $|X|\ge \epsilon |A|$ and $|Y|\ge \epsilon |B|$.

The Szemerédi regularity lemma states that for every $\epsilon >0$ and positive integer $m_0$, there are integers $M,n_0$ such that every graph of order $n\ge n_0$ has a set partitionwith $m_0\le k\le M$, $|V_0|\le \epsilon n$, equal nonzero sizes $|V_1|=\cdots =|V_k|$, and at most $\epsilon k^2$ unordered pairs $(V_i,V_j)$, $1\le i<j\le k$, that fail to be $\epsilon$-uniform. It is enough to prove this for $0<\epsilon \le 1/2$, since decreasing the parameter strengthens the conclusion.

Energy increment. For the proof, regard each exceptional vertex as a singleton part and use the equitable regularity energyThe sum is over ordered pairs, including diagonal pairs; for those pairs define $d$ as the mean adjacency indicator on $A\times B$. Thus $0\le q\le 1$. Under a refinement, the subrectangle densities have the original density as their weighted mean. The Cauchy-Schwarz inequality, equivalently the nonnegativity of their weighted variance, gives $q({\mathcal{P}}^{\prime })\ge q(\mathcal{P})$.

Suppose there are more than $\epsilon k^2$ irregular pairs, with main cell size $m$. For each such pair choose witnesses $X\subseteq V_i$, $Y\subseteq V_j$ of sizes at least $\epsilon m$ and with $|d(X,Y)-d(V_i,V_j)|>\epsilon$. Refine every main cell by all its witness subsets. Each splits into at most $2^{k-1}$ atoms, and we may use the larger bound $2^k$.

For an ordered irregular rectangle, write its refined densities as $d_{ab}$ and relative areas as $w_{ab}$. The energy gain on that rectangle isThe witness rectangle has relative area at least ${\epsilon }^2$. Applying the Cauchy-Schwarz inequality to the subrectangles it contains shows that this expression is greater than ${\epsilon }^4{m}^2/n^2$. There are two orientations of each irregular pair. Since $km=n-|V_0|\ge n/2$, the total gain is greater than

Restoring equal sizes without losing energy. Put $\ell =\lfloor m/4^k\rfloor$, assuming $m\ge 2\cdot 4^k$. Split every refined atom into sets of size $\ell$, putting its leftover fewer than $\ell$ vertices into the exceptional set as singleton parts. This is a further refinement, so the energy cannot decrease. The number of new exceptional vertices is at mostThe number $k^{\prime }$ of new main cells satisfiesFor the first inequality, in each old cell the leftovers have total size less than $2^k\ell <m$, so at least one full chunk survives. All previous exceptional singleton parts are kept. Treating leftovers as singleton parts is what makes the energy increment survive equitabilization exactly.

Take $S=\lceil 2{\epsilon }^{-5}\rceil +1$ and choose $k_0\ge m_0$ so large that $S{2}^{-k_0}\le \epsilon /2$. Start with $k_0$ equal main cells, with fewer than $k_0$ exceptional vertices. Define a finite bound $M$ by iterating $x\mapsto 2x{4}^x$ a total of $S$ times, starting from $k_0$. Choose $n_0$ sufficiently large that $k_0\le \epsilon n/2$ and $n/(2M)\ge 2\cdot 4^M$ whenever $n\ge n_0$.

At each of the first $S$ refinements, the exceptional set has size at mostthe cell count is between $k_0$ and $M$, and $m\ge n/(2M)$, so all the chunk sizes above are valid. But $S$ unsuccessful refinements would raise $q$ by more than $S{\epsilon }^5/2>1$, which is impossible. Thus the procedure terminates at a partition satisfying the Szemerédi regularity lemma.

Three linearly large uniform sets. Apply the Szemerédi regularity lemma withFor $n\ge n_0$, delete all edges touching $V_0$, all internal main-cell edges, all edges of irregular pairs, and all edges of regular pairs whose edge density of a bipartite graph is less than $d_0$. The respective losses are bounded byThus the total loss is at most $0.00102n^2$, and the remaining graph has at leastedges. Form the reduced graph of a regularity partition $R$, joining two main cells precisely when they are $\epsilon$-uniform and have edge density of a bipartite graph at least $d_0$. If their common size is $m$, the remaining edge count is at most $e(R)m^2$. Since $km\le n$, we obtain $e(R)>k^2/4$. The Mantel theorem gives a triangle in a graph in $R$. Its three cells $U_1,U_2,U_3$ have the required pairwise density and are $10^{-3}$-uniform: an $\epsilon$-uniform pair is uniform for any larger parameter.

Each selected cell has size $m\ge (1-\epsilon )n/M$. ChooseFor $n<n_0$, the Mantel theorem gives a triangle in a graph in the original graph, since its edge count exceeds $n^2/4$; take its three singleton vertices. Their pairs have density one and are uniform for every positive parameter, and $1>cn$. Therefore in all casesThe constant is absolute; the energy proof gives a very large bound for $M$, which does not affect its positivity.

For disjoint nonempty vertex sets $A,B$, writeThe pair $(A,B)$ is $\epsilon$-uniform ifwhenever $X\subseteq A$, $Y\subseteq B$, $|X|\ge \epsilon |A|$, and $|Y|\ge \epsilon |B|$.

The Szemerédi regularity lemma states that for every $\epsilon >0$ and integer $m_0$ there are $M,n_0$ such that every graph on $n\ge n_0$ vertices has a partitionwith $m_0\le m\le M$, $|V_0|\le \epsilon n$, equal sizes $|V_1|=\cdots =|V_m|$, and at most $\epsilon m^2$ pairs $(V_i,V_j)$ that are not $\epsilon$-uniform.

For disjoint nonempty vertex sets $A,B$, writeThe pair $(A,B)$ is a regular pair of vertex sets with parameter $ϵ$ ifwhenever $X\subseteq A$, $Y\subseteq B$, $|X|\ge ϵ|A|$, and $|Y|\ge ϵ|B|$.

The Szemerédi regularity lemma says that for every $ϵ>0$ and $m_0$ there are $M,n_0$ such that every graph on at least $n_0$ vertices has a partitionwhere $m_0\le m\le M$, $|V_0|\le ϵ|V|$, the classes $V_1,\dots ,V_m$ have equal size, and all but at most $ϵm^2$ pairs $(V_i,V_j)$ are $ϵ$-uniform.