A Tight Bound for Hyperaph Regularity

Abstract

The hypergraph regularity lemma—the extension of Szemerédi’s graph regularity lemma to the setting of k-uniform hypergraphs—is one of the most celebrated combinatorial results obtained in the past decade. By now there are several (very different) proofs of this lemma, obtained by Gowers, by Nagle–Rödl–Schacht–Skokan and by Tao. Unfortunately, what all these proofs have in common is that they yield regular partitions whose order is given by the k-th Ackermann function. We show that such Ackermann-type bounds are unavoidable for every $k \ge 2$, thus confirming a prediction of Tao. Prior to our work, the only result of this type was Gowers’ famous lower bound for graph regularity.

Proof of Claim 5.4

1.1 Basic facts.

In order to prove Claim 5.4 we will need several auxiliary results and definitions. We begin with the notion of complexes. Henceforth, the rank of a (not necessarily uniform) hypergraph P is $\max _{e \in P} |e|$. For $r \ge 2$ we denote $P^{(r)} = \left\{ e \in P \,\big \vert \, |e|=r\right\}$ and $P^{(1)}=V(P)$.

Definition A.1

(Complex). A k-complex$(k \ge 2)$ is a k-partite hypergraph P of rank $k-1$ where $P^{(r)} \subseteq \mathcal {K}(P^{(r-1)})$ for every $2 \le r \le k-1$.

Definition A.2

(f-regular complex). Let $f :[0,1]\rightarrow [0,1]$. A k-complex P on vertex classes $(V_1,\ldots ,V_k)$ is $(f,d_2,\ldots ,d_{k-1})$-regular, or simply f-regular, if for every $2 \le r \le k-1$ and every r vertex classes $V_{i_1},\ldots ,V_{i_r}$ we have that $P^{(r)}[V_{i_1},\ldots ,V_{i_r}]$ is $(\epsilon ,d_r)$-regular in $P^{(r-1)}[V_{i_1},\ldots ,V_{i_r}]$, where $\epsilon =f(d_0)$ and $d_0=\min \{d_2,\ldots ,d_{k-1}\}$.

Note that by using the notion of complexes one can equivalently define an f-equitable partition (recall Definition 5.3) as follows; an $(r,a_1,\ldots ,a_r)$-partition $\mathcal {P}$ is f-equitable if $\mathcal {P}^{(1)}$ is equitable and, if $r \ge 2$, every r-complex of $\mathcal {P}$ is $(f,1/a_2,\ldots ,1/a_r)$-regular.

We now state the dense counting lemma of [RS07b] specialized to complexes. We henceforth fix the following notation for $k \ge 3$, $\gamma > 0$;

$$\begin{aligned} F_{k,\gamma }(x) := \frac{\gamma ^3}{12}\left( \frac{x}{2}\right) ^{2^{k+1}} . \end{aligned}$$

(60)

The statement we use below follows from combining Theorem 10 and Corollary 14 in [RS07b], and generalized to the case where the vertex classes are not necessarily of the same size. For a k-polyad F and an edge $e \in F$, we denote the set of k-cliques in F containing e by $\mathcal {K}(F,e)=\{e' \in \mathcal {K}(F) \,\vert \, e \subseteq e'\}$. For a k-complex P we abbreviate $\mathcal {K}(P):=\mathcal {K}(P^{(k-1)})$.

Fact 1

(Dense counting lemma for k-complexes). Let $\gamma > 0$ and let P be a k-complex $(k \ge 3)$ that is $(F_{k,\gamma },d_2,\ldots ,d_{k-1})$-regular with $n_i \ge n_0(\gamma ,d_2,\ldots ,d_{k-1})$ vertices in the $i\text {-th}$ vertex class. Then

$$\begin{aligned} |\mathcal {K}(P)| = (1 \pm \gamma )\prod _{i=2}^{k-1} d_i^{\left( {\begin{array}{c}k\\ i\end{array}}\right) } \cdot \prod _{i=1}^k n_i . \end{aligned}$$

Moreover,¹⁶ let $P^{(k-1)}=(P_1,\ldots ,P_k)$. We have for all edges $e \in P_k$ but at most $\gamma |P_k|$ that¹⁷

$$\begin{aligned} |\mathcal {K}(P,e)| = (1 \pm \gamma )\prod _{i=2}^{k-1} d_i^{\left( {\begin{array}{c}k-1\\ i-1\end{array}}\right) } \cdot n_{k} . \end{aligned}$$

We will also need a slicing lemma for complexes.

Lemma A.3

(Slicing lemma for complexes). Let P be a k-complex $(k\ge 3)$ on vertex classes $(V_1,\ldots ,V_k)$ and let $V_k' \subseteq V_k$ with $|V_k'| \ge \delta |V_k|$. If P is $(f,\,d_2,\ldots ,d_{k-1})$-regular with $f(x) \le \frac{\delta }{2} F_{k-1,\frac{1}{4}}(x)$ and $|V(P)| \ge n_0(d_2,\ldots ,d_{k-1})$ then the induced k-complex $Q=P[V_1,\ldots ,V_{k-1},V_k']$ is $(f^*,d_2,\ldots ,d_{k-1})$-regular with $f^*=\frac{2}{\delta } \cdot f$.

For the proof we will need the notation $P^{(\le i)} = \{e \in P \,\vert \, |e| \le i\}$ where P is any hypergraph.

Proof

We proceed by induction on k. We begin with the induction basis $k=3$. Let $P=(P_1,P_2,P_3)$ be an $(f,d_2)$-regular 3-complex on vertex classes $(V_1,V_2,V_3)$, meaning that each bipartite graph $P_i$ (which is obtained from P by removing $V_i$ and its adjacent edges) is $(\epsilon ,d)$-regular with $d=d_2$ and $\epsilon =f(d)$. Put $Q=(Q_1,Q_2,Q_3)$. We will show that the bipartite graphs $Q_1=P_1[V_2,V_3']$ and $Q_2=P_2[V_1,V_3']$ are each $(\epsilon /\delta ,\,d)$-regular. Since $f(x)/\delta \le f^*(x)$, and since $Q_3=P_3$ is $(\epsilon ,d)$-regular by assumption, this would imply that Q is $(f^*,d_2)$-regular, as needed. To prove that $Q_1$ is $(\epsilon /\delta ,d)$-regular, let $S \subseteq V_2 \cup V_3'$ with $|\mathcal {K}(S)| \ge (\epsilon /\delta )|V_2||V_3'|$. Then $|\mathcal {K}(S)| \ge \epsilon |V_2||V_3|$, hence $d_{Q_1}(S)=d_{P_1}(S) = d \pm \epsilon$, as desired. Similarly, to prove that $Q_{2}$ is $(\epsilon /\delta ,d)$-regular, let $S \subseteq V_1 \cup V_3'$ with $|\mathcal {K}(S)| \ge (\epsilon /\delta )|V_1||V_3'|$. Then $|\mathcal {K}(S)| \ge \epsilon |V_1||V_3|$, hence $d_{Q_2}(S)=d_{P_2}(S) = d \pm \epsilon$. This proves the induction basis.

It remains to prove the induction step. Let P be a $(k+1)$-complex on vertex classes $(V_1,\ldots ,V_{k+1})$ and let $V_{k+1}^{\prime } \subseteq V_{k+1}$ with $|V_{k+1}^{\prime }| \ge \delta |V_{k+1}|$, and suppose P is $(f,d_2,\ldots ,d_k)$-regular with

$$\begin{aligned} f(x) \le \frac{\delta }{2} F_{k,\frac{1}{4}}(x) . \end{aligned}$$

(61)

We need to show that the induced $(k+1)$-complex $Q=P[V_1,\ldots ,V_k,V_{k+1}^{\prime }]$ is $(f^*,d_2,\ldots ,d_k)$-regular. Put $d_0 = \min \{d_2,\ldots ,d_{k-1}\}$, $P^{(k)}=(P_1,\ldots ,P_{k+1})$ and $Q^{(k)}=(Q_1,\ldots ,Q_{k+1})$. Let $i \in [k+1]$, and observe that the regularity assumption on P translates to the following assumptions on $P_i$:

• (i) the k-complex $P_i^{(\le k-1)}$ is $(f,d_2,\ldots ,d_{k-1})$-regular,

• (ii) the k-partite k-graph $P^{(k)}_i$ is $(f(d_0),\,d_k)$-regular in $P^{(k-1)}_i$.

To prove that Q is $(f^*,d_2,\ldots ,d_k)$-regular we need to show that $Q_i$ satisfies the following conditions:

• (i) the k-complex $Q_i^{(\le k-1)}$ is $(f^*,d_2,\ldots ,d_{k-1})$-regular,

• (ii) the k-partite k-graph $Q^{(k)}_i$ is $(f^*(d_0),\,d_k)$-regular in $Q^{(k-1)}_i$.

We henceforth assume $i \ne k+1$, since otherwise $Q_i=P_i$ and so the above conditions follow from the above assumptions together with the fact that $f(x) \le f^*(x)$.

Apply the induction hypothesis with the k-complex $P_i^{(\le k-1)}$ and $V_{k+1}^{\prime }$, using assumption (i), the fact that $f(x) \le \frac{\delta }{2}F_{k-1,\frac{1}{4}}(x)$ by (61) and the statement’s assumption on |V(P)|. It follows that the k-complex $Q_i^{(\le k-1)}=P_i^{(\le k-1)}[V_1,\ldots ,V_{i-1},V_{i+1},\ldots ,V_k,V_{k+1}^{\prime }]$ is $(f^*,d_2,\ldots ,d_k)$-regular, thus proving condition (i).

Apply Fact 1 (dense counting lemma) with $\gamma =1/2$ and the k-complex $P_i^{(\le k-1)}$, using assumption (i), the fact that $f(x) \le F_{k,\frac{1}{2}}(x)$ by (61) and the statement’s assumption on |V(P)|, to deduce that

$$\begin{aligned} |\mathcal {K}(P_i^{(k-1)})| \le \frac{3}{2}\prod _{j=2}^{k-1} d_j^{\left( {\begin{array}{c}k\\ j\end{array}}\right) } \cdot \prod _{\begin{array}{c} 1 \le j \le k+1:\\ j \ne i \end{array}} |V_j| . \end{aligned}$$

On the other hand, applying Fact 1 with $\gamma =1/4$ and the k-complex $Q_i^{(\le k-1)}$, using condition (i), the fact that $f^*(x) = \frac{2}{\delta }f(x) \le F_{k,\frac{1}{4}}(x)$ by (61) and the statement’s assumption on |V(P)|, implies that

$$\begin{aligned} |\mathcal {K}(Q_i^{(k-1)})| \ge \frac{3}{4}\prod _{j=2}^{k-1} d_j^{\left( {\begin{array}{c}k\\ j\end{array}}\right) } \cdot \prod _{\begin{array}{c} 1 \le j \le k:\\ j \ne i \end{array}} |V_j| \cdot \delta |V_{k+1}| \ge \frac{\delta }{2}|\mathcal {K}(P_i^{(k-1)})| . \end{aligned}$$

(62)

We now prove condition (ii). Let $S \subseteq Q_i^{(k-1)}$ satisfy $|\mathcal {K}(S)| \ge f^*(d_0)|\mathcal {K}(Q_i^{(k-1)})|$. Then $|\mathcal {K}(S)| \ge f(d_0)|\mathcal {K}(P_i^{(k-1)})|$ by (62). Therefore $d_{Q_i^{(k)}}(S)=d_{P_i^{(k)}}(S) = d_k \pm f(d_0)$, where the last equality uses assumption (ii). This proves condition (ii), thus completing the induction step and the proof. $\square$

1.2 Proof of Claim 5.4.

Proof

Put $G=G_H^k$, $\delta ^{\prime }=2\sqrt{\delta }$, and let $E \in E_k(\mathcal {P})$ and $V \in V_k(\mathcal {P})$.

Note that E is a $(k-1)$-partite $(k-1)$-graph, and let $(V_1,\ldots ,V_{k-1})$ denote its vertex classes. Thus, $V_j \subseteq \mathbf {V}^j$ for every $1 \le j \le k-1$, and $V \subseteq \mathbf {V}^k$.

Moreover, let $E^{\prime } \subseteq E$, $V^{\prime } \subseteq V$ with $|E^{\prime }| \ge \delta ^{\prime }|E|$, $|V^{\prime }| \ge \delta ^{\prime }|V|$.

To complete the proof our goal is to show that $d_{G}(E^{\prime },V^{\prime }) \ge \frac{1}{2} d_{G}(E,V)$ (recall Definition 2.2).

Consider the following k-partite k-graph and subgraph thereof:

$$\begin{aligned} K= & {} \{ (v_1,\ldots ,v_k) \,\vert \, (v_1,\ldots ,v_{k-1}) \in E \text { and } v_k \in V \} = E \circ V ,\\ K^{\prime }= & {} \{ (v_1,\ldots ,v_k) \,\vert \, (v_1,\ldots ,v_{k-1}) \in E^{\prime } \text { and } v_k \in V^{\prime } \} = E^{\prime } \circ V^{\prime } . \end{aligned}$$

We claim that

$$\begin{aligned} d_{G}(E,V) = \frac{|H \cap K|}{|K|} \quad \text {and}\quad d_{G}(E^{\prime },V^{\prime }) = \frac{|H \cap K^{\prime }|}{|K^{\prime }|} . \end{aligned}$$

(63)

Proving (63) would mean that to complete the proof it suffices to show that

$$\begin{aligned} \frac{|H \cap K^{\prime }|}{|K^{\prime }|} \ge \frac{1}{2} \frac{|H \cap K|}{|K|} . \end{aligned}$$

(64)

To prove (63) first note that

$$\begin{aligned} |K| = |E||V| \quad \text {and}\quad |K^{\prime }|=|E^{\prime }||V^{\prime }| . \end{aligned}$$

(65)

Furthermore,

$$\begin{aligned} e_G(E,V)&= \left| \left\{ \, ((v_1,\ldots ,v_{k-1}),v_k) \in G \,\big \vert \, (v_1,\ldots ,v_{k-1}) \in E,\, v_k \in V \, \right\} \right| \\&= \left| \left\{ \, (v_1,\ldots ,v_{k-1},v_k) \in H \,\big \vert \, (v_1,\ldots ,v_{k-1}) \in E,\, v_k \in V \, \right\} \right| = |H \cap K| , \end{aligned}$$

and similarly, $e_G(E^{\prime },V^{\prime })=|H \cap K^{\prime }|$. Therefore, using (65), we indeed have

$$\begin{aligned} d_G(E,V) = \frac{e_G(E,V)}{|E||V|} = \frac{|H \cap K|}{|K|} \quad \text {and}\quad d_G(E^{\prime },V^{\prime }) = \frac{e_G(E^{\prime },V^{\prime })}{|E^{\prime }||V^{\prime }|} = \frac{|H \cap K^{\prime }|}{|K^{\prime }|} . \end{aligned}$$

Having completed the proof of (63), it remains to prove (64). By Claim 3.8 there is a set of k-polyads $\{P_i\}_i$ of $\mathcal {P}$ on $(V_1,\ldots ,V_{k-1},V)$ such that

$$\begin{aligned} K = \bigcup _i \mathcal {K}(P_i) \,\text { is a partition, with }\, P_i = (P_{i,1},\ldots ,P_{i,k-1},E) . \end{aligned}$$

(66)

For each k-polyad $P_i=(P_{i,1},\ldots ,P_{i,k-1},E)$, let $P_i^{\prime }$ be the induced k-polyad $P_i^{\prime } = P_i[V_1,\ldots ,V_{k-1},V^{\prime }]$. Write $P_i^{\prime }=(P^{\prime }_{i,1},\ldots ,P^{\prime }_{i,k},E)$, and let $Q_i$ be the k-polyad $Q_i=(P^{\prime }_{i,1},\ldots ,P^{\prime }_{i,k-1},E^{\prime })$. Note that $Q_i$ satisfies $\mathcal {K}(Q_i) = \mathcal {K}(P_i) \cap K^{\prime }$. It therefore follows from (66) that

$$\begin{aligned} K^{\prime } = \bigcup _i (\mathcal {K}(P_i) \cap K^{\prime }) = \bigcup _i \mathcal {K}(Q_i) \,\text { is a partition.} \end{aligned}$$

(67)

Suppose $\mathcal {P}$ is a $(k-1,a_1,a_2,\ldots ,a_{k-1})$-partition, and denote $d_j = 1/a_j$ and

$$\begin{aligned} d = \prod _{j=2}^{k-1} d_j^{\left( {\begin{array}{c}k-1\\ j-1\end{array}}\right) } . \end{aligned}$$

Put $\gamma = \frac{1}{8}\delta ^{\prime }$ ($\le \frac{1}{8}$, as otherwise there is nothing to prove). We will next apply the dense counting lemma (Fact 1) to prove the estimates:

$$\begin{aligned} |\mathcal {K}(P_i)| \le (1+\gamma )d|K| \end{aligned}$$

(68)

and

$$\begin{aligned} |\mathcal {K}(Q_i)| \ge \big (1-\gamma )d|K^{\prime }| . \end{aligned}$$

(69)

Note that proving these estimates would in particular imply the bound

$$\begin{aligned} |\mathcal {K}(Q_i)| \ge \delta |\mathcal {K}(P_i)| ; \end{aligned}$$

(70)

indeed, from the assumptions $|E^{\prime }| \ge \delta ^{\prime }|E|$, $|V^{\prime }| \ge \delta ^{\prime }|V|$ and (65) we have that $|K^{\prime }| \ge (\delta ^{\prime })^2|K|$, hence we deduce from (68) and (69) the lower bound

$$\begin{aligned} \frac{|\mathcal {K}(Q_i)|}{|\mathcal {K}(P_i)|} \ge \frac{1-\gamma }{1+\gamma }(\delta ^{\prime })^2 \ge \frac{3}{4} \cdot (2\sqrt{\delta })^2 \ge \delta , \end{aligned}$$

where we used the inequality

$$\begin{aligned} \frac{1-\gamma }{1+\gamma } \ge 1-2\gamma \ge \frac{3}{4} . \end{aligned}$$

(71)

In order to prove (68) and (69) we first need to introduce some notation.

Put $m=n/a_1$ where $n=|V(H)|$ is the size of the vertex set, and put

$$\begin{aligned} \gamma ^{\prime } = \frac{1}{2}\gamma \delta ^{\prime }d \quad \left( =\frac{1}{4}\delta d\right) , \quad d_0 = \min \{d_2,\ldots ,d_{k-1}\} . \end{aligned}$$

Note that $d \ge d_0^{2^k}$. Using the statement’s assumption on f we have (recall (60))

$$\begin{aligned} f(d_0) = \delta ^4\left( \frac{d_0}{2}\right) ^{2^{k+3}} \le \frac{\delta }{2^{6}} \cdot \left( \frac{1}{4}\delta d_0^{2^k}\right) ^3 \left( \frac{d_0}{2}\right) ^{2^{k+1}} \le \frac{\delta }{2}\cdot \frac{\gamma ^{\prime 3}}{12}\left( \frac{d_0}{2}\right) ^{2^{k+1}} = \frac{\delta }{2}F_{k,\gamma ^{\prime }}(d_0).\nonumber \\ \end{aligned}$$

(72)

In particular,

$$\begin{aligned} f(d_0) \le \gamma ^{\prime } d_0 \le \gamma ^{\prime } d_{k-1} . \end{aligned}$$

(73)

Note that if P is an $\ell$-polyad of $\mathcal {P}$, for any $2 \le \ell \le k$, then, since $\mathcal {P}$ is f-equitable, the unique $\ell$-complex of $\mathcal {P}$ containing P is $(f,d_2,\ldots ,d_{\ell -1})$-equitable. Applying Fact 1 (dense counting lemma) with $\gamma ^{\prime }$ implies, using the fact that $f(x) \le F_{k,\gamma ^{\prime }}(x) \le F_{\ell ,\gamma ^{\prime }}(x)$ by (72) and the statement’s assumption on n, that

$$\begin{aligned} |\mathcal {K}(P)| = (1 \pm \gamma ^{\prime })\prod _{j=2}^{\ell -1} d_j^{\left( {\begin{array}{c}\ell \\ j\end{array}}\right) } \cdot m^\ell . \end{aligned}$$

(74)

Let $P_E$ be the unique $(k-1)$-polyad of $\mathcal {P}$ such that $E \subseteq K(P_E)$.

Then $|E|=d_E(P_E)|\mathcal {K}(P_E)|$, and since $\mathcal {P}$ is f-equitable, E is $(d_{k-1},f(d_0))$-regular in $P_E$. In particular, $|E| \ge (d_{k-1}-f(d_0))|\mathcal {K}(P_E)|$.

By (74),

$$\begin{aligned}|\mathcal {K}(P_E)| \ge (1-\gamma ^{\prime })\prod _{j=2}^{k-2} d_j^{\left( {\begin{array}{c}k-1\\ j\end{array}}\right) } \cdot m^{k-1} .\end{aligned}$$

Thus,

$$\begin{aligned} |E| \ge (d_{k-1}-f(d_0))|\mathcal {K}(P_E)| \ge (1-\gamma ^{\prime })d_{k-1}|\mathcal {K}(P_E)| \ge (1-2\gamma ^{\prime })\prod _{j=2}^{k-1} d_j^{\left( {\begin{array}{c}k-1\\ j\end{array}}\right) } \cdot m^{k-1} ,\nonumber \\ \end{aligned}$$

(75)

where the second inequality uses (73). Furthermore, for every $P_i$ as above we have (recall $P_i$ is a k-polyad of $\mathcal {P}$), again by (74), that

$$\begin{aligned} |\mathcal {K}(P_i)|\le & {} (1+\gamma ^{\prime })\prod _{j=2}^{k-1} d_j^{\left( {\begin{array}{c}k\\ j\end{array}}\right) } \cdot m^k = (1+\gamma ^{\prime })d\prod _{j=2}^{k-1} d_j^{\left( {\begin{array}{c}k-1\\ j\end{array}}\right) } \cdot m^k \\\le & {} \frac{1+\gamma ^{\prime }}{1-2\gamma ^{\prime }}d|E||V| \le (1+4\gamma ^{\prime })d|K| , \end{aligned}$$

where the penultimate inequality uses (75), and the last inequality uses (65) and the fact that $\gamma ^{\prime } \le \gamma \le \frac{1}{8}$. This proves (68).

Next we prove (69). Let $\overline{P_i}$ be the unique k-complex of $\mathcal {P}$ containing the k-polyad $P_i$, and let $\overline{P_i}^{\prime }$ be the induced k-complex $\overline{P_i}^{\prime }=\overline{P_i}[V_1,\ldots ,V_{k-1},V^{\prime }]$. Apply Lemma A.3 (slicing lemma) on $\overline{P_i}$, using the fact that $|V^{\prime }| \ge \delta ^{\prime }|V|$ and $f(x) \le \frac{\delta }{2}F_{k-1,\frac{1}{4}}$ by (72), to deduce that $\overline{P_i}^{\prime }$ is $(\frac{2}{\delta }f,\, d_2,\ldots ,d_{k-1})$-regular. Let the k-complex $\overline{Q_i}$ be obtained from the k-complex $\overline{P_i}^{\prime }$ by replacing E$(=P[V_1,\ldots ,V_{k-1}])$ with $E^{\prime }$, and note that the $(k-1)$-uniform hypergraph $\overline{Q_i}^{(k-1)}$ is precisely the k-polyad $Q_i$. Apply Fact 1 (dense counting lemma, the “moreover” part) with $\gamma ^{\prime }$ on $\overline{P_i}^{\prime }$, using the fact that $\frac{2}{\delta }f(x) \le F_{k,\gamma ^{\prime }}(x)$ by (72) and the statement’s assumption on n, to deduce that

$$\begin{aligned} |\mathcal {K}(Q_i)|\ge & {} |E^{\prime }| \cdot (1-\gamma ^{\prime })d|V^{\prime }| - \gamma ^{\prime }|E|\cdot |V^{\prime }| \ge \left( 1-\gamma ^{\prime } - \frac{1}{\delta ^{\prime } d}\gamma ^{\prime }\right) d|E^{\prime }||V^{\prime }| \\\ge & {} \big (1-\gamma )d|K^{\prime }| , \end{aligned}$$

where the second inequality uses the assumption that $|E^{\prime }|\ge \delta ^{\prime }|E|$ and the third inequality uses (65). This proves (69).

Finally, recall that our goal is to prove (64). We have

$$\begin{aligned} |H \cap K|&= \sum _i |H \cap \mathcal {K}(P_i)| = \sum _i d_H(P_i) \cdot |\mathcal {K}(P_i)| \\&\le (1+\gamma )|K| \cdot d\sum _i d_H(P_i) , \end{aligned}$$

where the first equality uses (66) and the inequality is by (68). Put $d^{\prime }=d\sum _i d_H(P_i)$. Then

$$\begin{aligned} \frac{|H \cap K|}{|K|} \le (1 + \gamma )d^{\prime } . \end{aligned}$$

(76)

Observe that for every i, the statement’s assumption on $\mathcal {P}$ implies, together with (70), that

$$\begin{aligned} d_H(Q_i) \ge \frac{2}{3} d_H(P_i) . \end{aligned}$$

(77)

We have

$$\begin{aligned} |H \cap K^{\prime }|&= \sum _i |H \cap \mathcal {K}(Q_i)| = \sum _i d_H(Q_i) \cdot |\mathcal {K}(Q_i)| \\&\ge \sum _i \frac{2}{3} d_H(P_i) \cdot |\mathcal {K}(Q_i)| \ge \frac{2}{3}(1-\gamma )|K^{\prime }| \cdot d\sum _i d_H(P_i) , \end{aligned}$$

where the first equality uses (67), the first inequality uses (77) and the second inequality uses (69). This means that

$$\begin{aligned} \frac{|H \cap K^{\prime }|}{|K^{\prime }|} \ge \frac{2}{3}(1-\gamma )d^{\prime } \ge \frac{2}{3}\cdot \frac{1-\gamma }{1+\gamma }\frac{|H \cap K|}{|K|} \ge \frac{1}{2} \frac{|H \cap K|}{|K|} , \end{aligned}$$

where the second inequality uses (76) and the third inequality uses (71). We have thus proved (64) and are therefore done. $\square$