Part III Extremal Graph Theory

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Part III — Extremal Graph Theory

Theorems with proof

Based on lectures by A. G. Thomason

Notes taken by Dexter Chua

Michaelmas 2017

These notes are not endorsed by the lecturers, and I have modified them (often significantly) after lectures. They are nowhere near accurate representations of what

was actually lectured, and in particular, all errors are almost surely mine.

Tur´an’s theorem, giving the maximum size of a graph that contains no complete

r-vertex subgraph, is an example of an extremal graph theorem. Extremal graph theory is an umbrella title for the study of how graph and hypergraph properties depend on the values of parameters. This course builds on the material introduced in the Part II Graph Theory course, which includes Tur´an’s theorem and also the Erd¨os–Stone theorem.

The first few lectures will cover the Erd¨os–Stone theorem and stability. Then we shall treat Szemer´edi’s Regularity Lemma, with some applications, such as to hered-itary properties. Subsequent material, depending on available time, might include: hypergraph extensions, the flag algebra method of Razborov, graph containers and applications.

Pre-requisites

A knowledge of the basic concepts, techniques and results of graph theory, as afforded by the Part II Graph Theory course, will be assumed. This includes Tur´an’s theorem, Ramsey’s theorem, Hall’s theorem and so on, together with applications of elementary probability.

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Contents III Extremal Graph Theory (Theorems with proof)

1 The Erd¨

os–Stone theorem

Theorem (Tur´an’s theorem). IfGis a graph with|G|=n,e(G)≥tr(n) and

G6⊇Kr+1. ThenG=Tr(n).

Theorem (Erd¨os–Stone, 1946). Let r≥1 be an integer andε >0. Then there existsd=d(r, ε) andn0=n0(r, ε) such that if|G|=n≥n0 and

e(G)≥ 1−1 r+ε n 2 ,

thenG⊇Kr+1(t), wheret=bdlognc.

Lemma. Let c, ε >0. Then there existsn1=n1(c, ε) such that if|G|=n≥n1

and e(G) ≥ (c+ε) n₂

, then G has a subgraph H where δ(H) ≥ c|H| and |H| ≥√εn.

Proof. The idea is that we can keep removing the vertex of smallest degree and then we must eventually get theH we want. Suppose this doesn’t gives us a suitable graph even after we’ve got√εnvertices left. That means we can find a sequence

G=Gn ⊇Gn−1⊇Gn−2⊇ · · ·Gs,

wheres=bε1/2nc, |Gj|=j and the vertex inGj\Gj−1 has degree< cj inGj.

We can then calculate

e(Gs)>(c+ε) n 2 −c n X j=s+1 j = (c+ε) _n 2 −c _n_{+ 1} 2 − _s_{+ 1} 2 ∼εn 2 2

asn gets large (sincecands are fixed numbers). In particular, this is> s₂

. ButGsonly hassvertices, so this is impossible.

Lemma. Letr≥1 be an integer andε >0. Then there exists ad1=d1(r, ε)

andn2=n2(r, ε) such that if|G|=n≥n2and

δ(G)≥ 1−1 r +ε n,

thenG⊇Kr+1(t), wheret=bd1lognc.

Proof of Erd¨os–Stone theorem. Providedn0is large, sayn0> n1 1−1_r+ε₂,ε₂

, we can apply the first lemma toGto obtain a subgraphH⊆Gwhere|H|>pε

2n, andδ(H)≥ 1−1 r+ ε 2 |H|.

We can then apply our final lemma as long aspε

2nis big enough, and obtain

Kr+1(t)⊆H⊆G, witht >

d1(r, ε/2) log pε₂n

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Proof of lemma. We proceed by induction onr. Ifr= 0 or ε≥1

r, the theorem

is trivial. Otherwise, by the induction hypothesis, we may assumeG⊇Kr(T)

for T = ₂_t εr .

Call itK=Kr(T). This is always possible, as long as

d1(r, ε)< εr 2 d1 r−1, 1 r(r−1) .

The exact form is not important. The crucial part is that _r₍_r1₋₁₎ = _r₋1₁ −1

r,

which is how we chose theεto put intod1(r−1, ε).

LetU be the set of vertices inG−K having at least 1−1

r+ ε 2 |K| neigh-bours inK. Calculating 1−1 r+ ε 2 |K|= 1−1 r + ε 2 rT = (r−1)T +εr 2T ≥(r−1)T+t, we see that every element inU is joined to at leastt vertices in each class, and hence is joined to someKr(t)⊆K.

So we have to show that|U| is large. If so, then by a pigeonhole principle argument, it follows that we can always find enough vertices inU so that adding them to Kgives aKr+1(t).

Claim. There is some c >0 (depending onrandε) such that fornsufficiently large, we have

|U| ≥cn.

The argument to establish these kinds of bounds is standard, and will be used repeatedly. We write e(K, G−K) for the number of edges between K

andG−K. By the minimum degree condition, each vertex ofKsends at least 1−1

r+ε

n− |K|edges toG−K. Then we have two inequalities

|K| 1−1 r+ε n− |K| ≤e(K, G−K) ≤ |U||K|+ (n− |U|) 1−1 r+ ε 2 |K|.

Rearranging, this tells us

εn 2 − |K| ≤ |U| ₁ r− ε 2 .

Now note that|K| ∼logn, so for largen, the second term on the left is negligible, and it follows that U ∼n.

We now want to calculate the number ofKr(t)’s inK. To do so, we use the

simple inequality _n k ≤en k k . Then we have #Kr(t) = _T t r ≤ _eT t rt ≤ ₃_e εr rt ≤ ₃_e εr rdlogn =nrdlog(3e/εr)

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Now if we pickdsufficiently small, then this tells us #Kr(t) grows sublinearly

withn. Since|U|grows linearly, and tgrows logarithmically, it follows that for

nlarge enough, we have

|U| ≥t·#Kr(t).

Then by the pigeonhole principle, there must be a setW ⊆U of sizetjoined to the sameKr(t), giving aKr+1(t).

Theorem (Erd¨os–Simonovits). LetF be a fixed graph with chromatic number

χ(F) =r+ 1. Then lim n→∞ ex(n, F) n 2 = 1− 1 r.

Proof. Since χ(F) = r+ 1, we know F cannot be embedded in an r-partite graph. So in particular,F 6⊆Tr(n). So ex(n, F)≥tr(n)≥ 1−1 r n 2 .

On the other hand, given any ε >0, if|G|=nand

e(G)≥ 1−1 r+ε n 2 ,

then by the Erd¨os–Stone theorem, we haveG⊇Kr+1(|F|)⊇F providedn is

large. So we know that for everyε >0, we have lim supex(n, F_n )

2

≤1−

1

r +ε.

So we are done.

Theorem. Givenr∈_N, there exists εr>0 such that if 0< ε < εr, then there

existsn3(r, ε) so that ifn > n3, there exists a graphGof ordernsuch that

e(G)≥ 1−1 r+ε n 2 but Kr+1(t)6⊆G, where t= _{3 log}_n log 1/ε .

Proof. Start with the Tur´an graphTr(n). Letm=n_r. Then there is a class

W of Tr(n) of sizem. The strategy is to addε n₂

edges inside W to obtainG

such thatG[W] (the subgraph ofGformed by W) does not contain aK2(t). It

then follows thatG6⊇Kr+1(t), but also

e(G)≥tr(n) +ε _n 2 ≥ 1−1 r +ε n 2 , as desired.

To see that such an addition is possible, we choose edges insideW indepen-dently with probabilityp, to be determined later. LetX be the number of edges chosen, andY be the number ofK2(t) created. IfE[X−Y]> ε n2

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means there is an actual choice of edges withX−Y > ε n₂

. We then remove an edge from eachK2(t) to leave aK2(t)-free graph with at leastX−Y > ε n₂

edges.

So we want to actually computeE[X−Y]. Seeing that asymptotically,mis

much larger thant, we have

E[X−Y] =E[X]−E[Y] =p _m 2 −1 2 _m t _m₋_t t pt2 ∼ 1 2pm 2₋1 2m 2t_pt2 = 1 2pm 2₍₁₋_m2t−2_pt2₋₁ ) = 1 2pm 2₍₁₋₍_m2_pt+1₎t−1₎_. We pick p= 3εr2_and _ε r= (3r2)−6. Thenp < ε5/6, and m2pt+1< m2ε5/6(t+1)<(m2m−5/2)t−1< 1 2. Hence, we find that

E[x−y]≥ 1 4pm 2_{> ε} _n 2 .

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2 Stability III Extremal Graph Theory (Theorems with proof)

2 Stability

Theorem. Lett, r≥2 be fixed, and suppose|G|=nandG6⊇Kr+1(t). If

e(G) = 1−1 r +o(1) n 2 . Then

(i) There existsTr(n) onV(G) with|E(G)∆E(Tr(n))|=o(n2).

(ii) Gcontains an r-partite subgraph with 1−1

r+o(1)

n

2

edges. (iii) Gcontains an r-partite subgraph with minimum degree 1−1

r+o(1)

n.

Proof. We first prove (ii), which looks the weakest. We will then argue that the others follow from this (and in fact they are equivalent).

By Erd¨os–Stone, we can always find some Kr(s) = K ⊆ Gfor some s =

s(n)→ ∞. We can, of course, always chooses(n) so thats(n)≤logn, which will be helpful for our future analysis. We letCi be theith class ofK.

– By throwing awayo(n) vertices, we may assume that

δ(G)≥ 1−1 r +o(1) n

– LetX be the set of all vertices joined to at leasttvertices in eachCi. Note

that there are s_tr

many copies of Kr(t)’s in K. So by the pigeonhole

principle, we must have|X|< t s_tr =o(n), or else we can construct a

Kr+1(t) insideG.

– LetY be the set of vertices joined to fewer than (r−1)s− s

2t+tother

vertices. By our bound onδ(G), we certainly have

e(K, G−K)≥s(r−1 +o(1))n.

On the other hand, every element of G is certainly joined to at most (r−1)s+tedges, sinceX was thrown away. So we have

((r−1)s+t)(n− |Y|) +|Y|(r−1)s− s 2t +t

≥e(K, G−K).

So we deduce that|Y|=o(n). ThrowY away.

Let Vi be the set of vertices in G\K joined to fewer thant of Ci. It is now

enough to show thate(G[Vi]) =o(n2). We can then throw away the edges in

eachVi to obtain anr-partite graph.

Suppose on the contrary thate(G[Vj])≥εn2, say, for somej. Then Erd¨os–

Stone with r = 1 says we haveG[Vj] ⊇ K2(t). Each vertex of K2(t) has at

leasts− s

2t+ 1 in each other Ci, fori6=j. So we see thatK2(t) has at least

s−2t ₂s_t−1> t common neighbours in eachCi, givingKr+1(t)⊆G.

It now remains to show that (i) and (iii) follows from (ii). The details are left for the example sheet, but we sketch out the rough argument. (ii)⇒(i) is

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straightforward, since anr-partite graph with that many edges is already pretty close to being a Tur´an graph.

To deduce (iii), since we can assumeδ(G)≥ 1−1

r+o(1)

n, we note that if (iii) did not hold, then we haveεnvertices of degree ≤ 1−1

r−ε

n, and their removal leaves a graph of order (1−ε)nwith at least

1−1 r+o(1) n 2 −εn 1−1 r −ε n > 1−1 r+ε 2 (1−ε)n 2 ,

which by Erd¨os–Stone would containKr+1(t). So we are done.

Corollary. Let χ(F) = r+ 1, and let G be extremal for F, i.e. G6⊇ F and

e(G) = ex(|G|, F). Thenδ(G) = 1−1

r+o(1)

n.

Proof. By our asymptotic bound on ex(n, F), it cannot be thatδ(G) is greater than 1−1

r+o(1)

n. So there must be a vertexv∈Gwithd(v)≤ 1−1

r−ε

n. We now apply version (iii) of the stability theorem to obtain an r-partite subgraph H of G. We can certainly pick |F| vertices in the same part of H, which are joined to m = 1−1

r+o(1)

common neighbours. FormG∗ from

G−v by adding a new vertexujoined to thesem vertices. Thene(G∗)> e(G), and so the maximality ofGentailsG∗_⊇_F_.

Pick a copy ofF in G∗_{. This must involve}_u_{, since}_G_{, and hence} _G₋_v _does

not contain a copy ofF. Thus, there must be somexamongst the|F| vertices mentioned. But then we can replaceuwithxand we are done.

Theorem (Simonovits). LetF be (r+ 1)-edge-critical, i.e. χ(F) =r+ 1 but

χ(F)\e) =rfor every edgeeofF. Then for largen, ex(n, F) =tr(n).

and the only extremal graph isTr(n).

Proof. LetGbe an extremal graph forF and letH be anr-partite subgraph with δ(H) = 1−1 r +o(1) n.

Note thatH necessarily haverparts of size 1_r+o(1)

neach. Assign each of theo(n) vertices inV(G)\V(H) to a class where it has fewest neighbours.

Suppose some vertexvhas εnneighbours in its own class. Then it has≥εn

in each class. Pickεnneighbours ofv in each class ofH. These neighbours span a graph with at least 1−1

r+o(1)

rεn

2

edges. So by Erd¨os–Stone (or arguing directly), they span F−wfor some vertex w (contained in Kr(|F|)). Hence

G⊇F, contradiction.

Thus, each vertex ofGhas onlyo(n) vertices in its own class. So it is joined to all but o(n) vertices in every other class.

Suppose some class ofGcontains an edgexy. Pick a setZ of|F| vertices in that class with{x, y} ∈Z. NowZ has 1_r+o(1)ncommon neighbours in each class, so (by Erd¨os–Stone or directly) these common neighbours span a

Kr−1(|F|). But together withZ, we have a Kr(|F|) but with an edge added

inside a class. But by our condition thatF\eisr-partite for anye, this subgraph contains an F. This is a contradiction.

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So we conclude that no class ofGcontains an edge. SoGitself isr-partite, but ther-partite graph with most edges isTr(n), which does not containF. So

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3 Supersaturation III Extremal Graph Theory (Theorems with proof)

3 Supersaturation

Theorem (Erd¨os–Simonovits). Let H be some `-uniform hypergraph. Then for allε >0, there existsδ(H, ε) such that every`-uniform hypergraphGwith |G|=nand e(G)>(π(H) +ε) n ` containsbδn|H|ccopies ofH.

Proof. For eachm-setM ⊆V(G), we letG[M] be the sub-hypergraph induced by these vertices. Let the number of subsetsM withe(G[M])> π(H) +₂ε m_` be η _mn

. Then we can estimate

(π(H) +ε) n ` ≤e(G) = P Me(G[M]) n−` m−` ≤η n m m ` + (1−η) _mn π(H) +₂ε m ` n−` m−` . So if n > m, then π(H) +ε≤η+ (1 +η)π(H) +ε 2 . So η≥ ε 2 1−π(H)−ε 2 >0.

The point is that it is positive, and that’s all we care about. We pickmlarge enough so that

ex(m, H)<π(H) +ε 2 m ` .

ThenH⊆G[M] for at leastη _mn

choices ofM. HenceGcontains at least

η _mn n−|H| m−|H| = η _|_Hn_| m |H|

copies ofH, and we are done. (Our results hold whennis large enough, but we can chooseδsmall enough so thatδn|H|_<_{1 when} _n_{is small)}

Theorem (Lorden, 1961). LetGhave degree sequenced1, . . . , dn. Then

k3(G) +k3( ¯G) = n 3 −(n−2)e(G) + n X i=1 di 2 .

Proof. The number of paths of length 2 inGand ¯Gis precisely

n X i=1 di 2 + n−1−di 2 = 2 n X i=1 di 2 −2(n−2)e(G) + 3 n 3 ,

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since to find a path of length 2, we pick the middle vertex and then pick the two edges. A complete or emptyK3contains 3 such paths; Other sets of three

vertices contain exactly 1 such path. Hence

n

3

+ 2(k3(G) +k3( ¯G)) = number of paths of length 2.

Corollary (Goodman, 1959). We have

k3(G) +k3( ¯G)≥

1

24n(n−1)(n−5).

Proof. Letm=e(G). Then

k3(G) +k3( ¯G)≥ n 3 −(n−2)m+n 2m/n 2 .

Then minimize overm.

Corollary. Form=e(G) andn=|G|, we have

k3(G)≥

m

3n(4m−n

2₎_.

Proof. By Lorden’s theorem, we know

k3(G) +k3( ¯G) = _n 3 −(n−2)e( ¯G) +X _d¯ i 2 ,

where ¯di is the degree sequence in ¯G. But

3k3( ¯G)≤

Xd¯i

2

,

since the sum is counting the number of paths of length 2. So we find that

k3(G)≥ _n 3 −(n−2) ¯m−2 3n _{2 ¯}_m/n 2 , and ¯m= n₂ −m.

Theorem. Let G be a graph. For any graphF, letiF(G) be the number of

induced copies of F in G, i.e. the number of subsets M ⊆ V(G) such that

G[M]∼=F. So, for example,iKp(G) =kp(G). Define

f(G) =X

F

αFiF(G),

with the sum being over a finite collection of graphsF, each being complete multipartite, with αF ∈ R andαF ≥ 0 if F is not complete. Then amongst

graphs of given order, f(G) is maximized on a complete multi-partite graph. Moreover, ifα_K¯₃ >0, then there are no other maxima.

Proof. We may supposeα_K¯₃ >0, because the case ofα_K¯₃ = 0 follows from a

limit argument. Choose a graphGmaximizingf and supposeGis not complete multipartite. Then there exist non-adjacent verticesx, ywhose neighbourhoods

X, Y differ.

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(i) F’s that contain both xandy; (ii) F’s that containy but notx; (iii) F’s that containxbut noty; (iv) F’s that contain neitherxnory.

We may assume that the contribution from (iii) ≥(ii), and if they are equal, then|X| ≤ |Y|.

Consider what happens when we remove all edges betweeny andY, and add edges fromy to everything in X. Clearly (iii) and (iv) are unaffected.

– If (iii)>(ii), then after this move, the contribution of (ii) becomes equal to the contribution of (iii) (which didn’t change), and hence strictly increased. The graphsF that contribute to (i) are not complete, soαF ≥0. Moreover,

sinceF is complete multi-partite, it cannot contain a vertex inX∆Y. So making the move can only increase the number of graphs contributing to (i), and each contribution is non-negative.

– If (iii) = (ii) and |X| ≤ |Y|, then we make similar arguments. The contribution to (ii) is unchanged this time, and we know the contribution of (i) strictly increased, because the number of ¯K3’s contributing to (i) is

the number of points not connected toxandy. In both cases, the total sum increased.

Theorem (Bollob´as, 1976). Let 1≤p≤r, and for 0≤x≤ n p

, letψ(x) be a maximal convex function lying below the points

{(kp(Tq(n)), kr(Tq(n))) :q=r−1, r, . . .} ∪ {(0,0)}.

LetGbe a graph of ordern. Then

kr(G)≥ψ(kp(G)).

Proof. Letf(G) =kp(G)−ckr(G), wherec >0.

Claim. It is enough to show thatf is maximized on a Tur´an graph for any c. Indeed, suppose we plot out the values of (kp(Tq(n)), kr(Tq(n)):

kp(G) kr(G) kp(Tr−1(n)) kp(Tr(n)) kr(Tr(n)) kp(Tr+1(n)) kr(Tr+1(n))

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If the theorem doesn’t hold, then we can pick aGsuch that (kp(G), kr(G))

lies belowψ. Draw a straight line through the point with slope parallel toψ. This has slope 1_c >0 for somec. The intercept on thex-axis is thenkp(G)−ckr(G),

which would be greater thanf(any Tur´an graph) by convexity, a contradiction. Now the previous theorem immediately tells us f is maximized on some complete multi-partite graph. Suppose this hasqclass, say of sizes a1≤a2≤

· · · ≤aq. It is easy to verifyq≥r−1. In fact, we may assumeq≥r, else the

maximum is on a Tur´an graphTr−1(n).

Then we can write

f(G) =a1aqA−caqaqB+C,

whereA, B, C are rationals depending only ona2, . . . , aq−1 anda1+aq (Aand

B count the number of ways to pick aKp andKr respectively in a way that

involves terms in both the first and last classes).

We wlog assumecis irrational. HenceaqaqA−ca1aq=a1aq(A−cB)6= 0.

– IfA−cB <0, replacea1andaqby 0 anda1+aq. This would then increase

f, which is impossible.

– IfA−cB >0 anda1≤aq−2, then we can replacea1, aq bya1+ 1, aq−1

to increasef.

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4 Szemer´edi’s regularity lemmaIII Extremal Graph Theory (Theorems with proof)

4 Szemer´

edi’s regularity lemma

Lemma. Let (U, W) be anε-uniform pair withd(U, W) =d. Then |{u∈U :|Γ(u)∩W|>(d−ε)|W|}| ≥(1−ε)|U| |{u∈U :|Γ(u)∩W|<(d+ε)|W|}| ≥(1−ε)|U|,

where Γ(u) is the set of neighbours ofu.

Proof. Let

X ={u∈U :|Γ(u)∩W| ≤(d−ε)|W|}.

Thene(X, W)≤(d−ε)|X||W|. So

d(X, W)≤d−ε=d(U, W)−ε.

So it fails the uniformity condition. Since W is definitely not small, we must have |X|< ε|U|.

The other case is similar, or observe that the complementary bipartite graph betweenU andW has density 1−dand isε-uniform.

Lemma (Graph building lemma). LetGbe a graph containing distinct vertex subsetsV1, . . . , Vrwith|Vi|=u, such that (Vi, Vj) isε-uniform andd(Vi, Vj)≥λ

for all 1≤i≤j≤r.

Let H be a graph with maximum degree ∆(H)≤ ∆. SupposeH has an

r-colouring in which no colour is used more thanstimes, i.e.H ⊆Kr(s), and

suppose (∆ + 1)ε≤λ∆_and_s_{≤ b}_εu_{c. Then}_H _⊆_G_.

Proof. We wlog assumeV(H) ={1, . . . , k}, and letc:V(H)→ {1, . . . , r} be a colouring ofV(H) using no colour more thanstimes. We want to pick vertices

x1, . . . , xk inGso thatxixj ∈E(G) ifij∈E(H).

We claim that, for 0≤`≤k, we can choose distinct verticesx1, . . . , x` so

that xj ∈Cc(j), and for` < j ≤k, a setXj` ofcandidates forxj such that

(i) X_j`⊆Vc(j);

(ii) xiyj ∈E(G) for all yj ∈Xj` andi≤`such thatij ∈E(H).

(iii) |X`

j| ≥(λ−ε)|N(j,`)||Vc(j)|, where

N(j, `) ={xi: 1≤i≤`andij ∈E(H)}.

The claim holds for`= 0 by takingX0

j =Vc(j).

By induction, suppose it holds for`. To pickx`+1, of course we should pick it

from our candidate setX`

`+1. Then the first condition is automatically satisfied.

Define the set

T ={j > `+ 1 : (`+ 1)j∈E(H)}.

Then eacht ∈T presents an obstruction to (ii) and (iii) being satisfied. To satisfy (ii), for eacht∈T, we should set

Xt`+1=X `

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Thus, to satisfy (iii), we want to exclude thosex`+1that make this set too small.

We define

Yt=

n

y∈X_``₊₁:|Γ(y)∩X_t`| ≤(λ−ε)|X_t`|o.

So we want to find something inX` `+1\

S

t∈TYt. We also cannot choose one of

thexi already used. So our goal is to show that

X_``₊₁−[ t∈T Yt > s−1.

This will follow simply from counting the sizes of|X`

`+1|and|Yt|. We already

have a bound on the size of |X`

`+1|, and we shall show that if|Yt|is too large,

then it violates ε-uniformity.

Indeed, by definition ofYt, we have

d(Yt, X``+1)≤λ−ε≤d(Vc(t), Vc(`+1))−ε.

So either|X`

`+1|< ε|Vc(t)|or|Yt|< ε|Vc(`+1)|. But the first cannot occur.

Indeed, writem=|N(`+ 1, `)|. Thenm+|T| ≤∆. In particular, since the |T|= 0 case is trivial, we may assumem≤∆−1. So we can easily bound

|X`+1| ≥(λ−ε)∆−1|Vc(t)| ≥(λ∆−1−(∆−1)ε)|Vc(t)|> ε|Vc(t)|.

Thus, byε-uniformity, it must be the case that |Yt| ≤ε|Vc(`+1)|.

Therefore, we can bound

X_``₊₁−[ t∈T Yt ≥(λ−ε)m|Vc(`+1)| −(∆−m)ε|Vc(`+1)| ≥(λm−mε−(∆−m)ε)u≥εu > s−1. So we are done.

Corollary. Let H be a graph with vertex set {v1, . . . , vr}. Let 0 < λ, ε < 1

satisfyrε≤λr−1_.

LetGbe a graph with disjoint vertex subsetsV1, . . . , Vr, each of size u≥1.

Suppose each pair (Vi, Vj) isε uniform, andd(Vi, Vj)≥λifvivj ∈E(H), and

d(Vi, Vj) ≤1−λ if vivj 6∈E(H). Then there existsxi ∈ Vi so that the map

vi →xi is an isomorphismH →G[{x1, . . . , xr}].

Proof. By replacing theVi-Vj edges by the complementary set whenevervivj 6∈

E(H), we may assumed(Vi, Vj)≥λfor alli, j, andH is a complete graph.

We then apply the previous lemma with ∆ =r−1 ands= 1.

Theorem (Szemer´edi’s regularity lemma). Let 0< ε < 1 and let` be some natural number. Then there exists someL=L(`, ε) such that every graph has anε-uniform equipartition intomparts for some`≤m≤L, depending on the graph.

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Lemma. Let U0 ⊆ U and W0 ⊆ W, where |U0| ≥ (1−δ)|U| and |W0| ≥ (1−δ)|W|. Then

|d(U0, W0)−d(U, W)| ≤2δ.

Proof. Letd=d(U, W) andd0=d(U0, W0). Then

d= e(U, W) |U||W| ≥ e(U0_{, W}0₎ |U||W| =d 0|U0||W0| |U||W| ≥d 0₍₁₋_δ₎2 . Thus, d0−d≤d0(1−(1−δ)2)≤2δd0 ≤2δ.

The other inequality follows from considering the complementary graph, which tells us

(1−d0)−(1−d)≤2δ.

Lemma. Letx1, . . . , xn be real numbers with

X = 1 n n X i=1 xi, and let x= 1 m m X i=1 xi. Then 1 n n X i=1 x2_i ≥X2+ m n−m(x−X) 2_≥_X2₊m n(x−X) 2_. Proof. We have 1 n n X i=1 x2_i = 1 n m X i=1 x2_i + 1 n n X i=m+1 x2_i ≥m nx 2₊n−m n _nX₋_mx n−m 2 ≥X2+ m n−m(x−X) 2

by two applications of Cauchy–Schwarz.

Proof. Define the index ind(P) of an equipartitionP intok partsVito be

ind(P) = 1

k2

X

i<j

d2(Vi, Vj).

We show that if P is notε-uniform, then there is a refinement equipartition Q into k4k parts, with ind(Q)≥ind(P) +ε₈5.

This is enough to prove the theorem. For choose t ≥ ` with 4tε5 ≥ 100. Define recursively a functionf by

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Let

N =f(d4ε−5e),

and pickL=N16N_.

Then, ifn≤L, then just take an equipartition into single vertices. Otherwise, begin with a partition intotparts. As long as the current partition intokparts is not ε uniform, replace it a refinement into 4k4 _{parts. The point is that}

ind(P)≤ 1

2 for any partition. So we can’t do this more than 4ε

−5 _{times, at}

which point we have partitioned intoN ≤Lparts.

Note that the reason we had to setL=N16N is that in our proof, we want to assume we have many vertices lying around.

The proof really is just one line, but students tend to complain about such short proofs, so let’s try to explain it in a bit more detail. If the partition is notε-uniform, this means we can further partition each part into uneven pieces. Then our previous lemma tells us this discrepancy allows us to push up 1

n

P_x2

i.

So given an equipartitionP that is notε-uniform, for each non-uniform pair (Vi, Vj) ofP, we pick witness sets

Xij ⊆Vi, Xji⊆Vj

with|Xij| ≥ε|Vi|,|Xji| ≥ |Vj|and|d(Xij, Xji)−D(Vi, Vj)| ≥ε.

Fixi. Then the setsXijpartitionViinto at most 2k−1atoms. Letm=b_kn₄kc, and letn=k4k_m₊_ak₊_b_{, where 0}_≤_a_≤₄k _and_b_≤_k_{. Then we see that}

bn/kc= 4km+a

and the parts ofP have size 4km+aor 4km+a+ 1, withb of the larger size. Partition each part of P into 4k sets, of size m orm+ 1. The smallerVi

havingaparts of sizem+ 1, and thelargrer havinga+ 1 such pairs.

We see that any such partition is an equipartition intok4k _{parts of size}_m_or

m+ 1, withak+bparts of larger sizem+ 1.

Let’s choose such an equipartitionQwith parts as nearly as possible inside atoms, so each atom is a union of parts ofQwith at mostmextra vertices.

All that remains is to check that ind(Q)≥ind(P) +ε₈5. Let the sets ofQ withinVi beVi(s), where 1≤s≤4k≡q. So

Vi= q [ s=1 Vi(s). Now X 1≤s,t≤q e(Vi(s), Vj(t)) =e(Vi, Vj).

We’d like to divide by some numbers and convert these to densities, but this is where we have to watch out. But this is still quite easy to handle. We have

m

m+ 1 ≤q|Vi(s)| ≤ |Vi| ≤

m+ 1

m q|Vi(s)|

for alls. So we wantmto be large for this to not hurt us too much. So m m+ 1 d(Vi, Vj)≤ 1 q2 X s,t d(Vi(s), Vj(t))≤ m+ 1 m 2 d(Vi, Vj).

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Usingn≥k16k_{, and hence}

_m m+ 1 2 ≥1− 2 m ≥1− 2 4k ≥1− ε5 50, we have 1 q2 X s,t d(Vi(s), Vj(t))−d(Vi, Vj) ≤ ε 5 49+d(Vi, Vj). In particular, 1 q2 X d2(Vi(s), Vj(t))≥d2(Vi, Vj)− ε5 25. The lower bound can be improved if (Vi, Vj) is not ε-uniform.

LetX∗

ij be the largest subset ofXij that is the union of parts ofQ. We may

assume

X_ij∗ = [

1≤s≤ri

Vi(s).

By an argument similar to the above, we have 1 rirj X 1≤s≤ri 1≤t≤rj d(Vi(s), dj(t))≤d(Xij∗, X ∗ ji) + ε5 49.

By the choice of parts of Q within atoms, and because|Vi| ≥qm = 4km, we

have |X_ij∗| ≥ |Xij| −2k−1m ≥ |Xij| 1−2 k_m ε|Vi| ≥ |Xij| 1− 1 2k_ε ≥ |Xij| 1− ε 10 .

So by the lemma, we know

|d(X_ij∗, X_ji∗)−d(Xij, Xji)|< ε 5. Recalling that |d(Xij, Xji)−d(Vi, Vj)|> ε, we have 1 rirj X 1≤s≤ri 1≤t≤rj d(Vi(s), Vj(t))−d(Vi, Vj) >3 4ε.

We can now allow our Cauchy–Schwarz inequality with n=q2 _and_m₌_r

irj gives 1 q2 X 1≤s,t≤q d2(Vi(s), Vj(t))≥d2(Vi, Vj)− ε5 25+ rirj q2 · 9ε2 16 ≥d2(Vi, Vj)− ε5 25+ ε4 3,

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using the fact that

ri q ≥ 1− 1 m _r i q m+ 1 m ≥ 1− 1 m ≥|X ∗ ij| |Vi| ≥ 1− 1 m 1− ε 10 |X_ij| |Vi| >4ε 5 . Therefore ind(Q) = 1 k2_q2 X 1≤i<j<k 1≤s,t≤q d2(Vi(s), Vj(t)) ≥ 1 k2 X 1≤i<j≤k d2(Vi, Vj)− ε5 25+ε k 2 ε4 3 ≥ind(P) +ε 5 8.

Theorem. Given an integer d, there existsc(d) such that

r(G)≤c|G| for every graphGwith ∆(G)≤d.

Proof. Let t = R(d+ 1). Pick ε ≤ minn1_t,₂d₍_d1₊₁₎

o

. Let ` ≥ t2_{, and let}

L=L(`, ε). We show thatc= L_ε works.

Indeed, letGbe a graph. Colour the edges ofKn by red and blue, where

n ≥ c|G|. Apply Szemir´edi to the red graph with `, ε as above. Let H be the graph whose vertices are {V1, . . . , Vm}, whereV1, . . . , Vm is the partition

of the red graph. Let Vi, Vj ∈ E(H) if (Vi, Vj) is ε-uniform. Notice that

m=|H| ≥`≥t2_{, and}_e_{( ¯}_H₎_≤_ε m

2

. SoH ⊇Kt, or else by Tur´an’s theroem,

there are integersd1, . . . , dt−1 andPdimand

e( ¯H)≥ di 2 ≥(t−1) m/(t−1) 2 ≥ε m 2

by our choice ofεandm.

We may as well assume all pairsVi, Vj for 1≤i < j≤t areε-uniform. We

colour the edge ofKtgreen ifd(Vi, Vj)≥1₂ (in the red graph), or white if< 1₂

(i.e. density >1₂ in the blue graph).

By Ramsey’s theorem, we may assume all pairsViVj for 1≤i < j≤d+ 1

are the same colour. We may wlog assume the colour is green, and we shall find a redG(a similar argument gives a blueGif the colour is white).

Indeed, take a vertex colouring of G with at most d+ 1 colours (using ∆(G)≤d), with no colour used more than|G|times. By the building lemma withH (in the lemma) beingG(in this proof), andG(in the lemma) equal to the subgrpah of the red graph spanned byV1, . . . , Vd+1 (here),

u=|Vi| ≥ n L ≥ c|G| L ≥ |G| ε , r=d+ 1,λ=1₂, and we are done.

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Theorem (Triangle removal lemma). Givenε >0, there existsδ >0 such that if |G| =nandGcontains at mostδn3 _{triangles, then there exists a set of at}

mostεn2_{edges whose removal leaves no triangles.}

Proof. Exercise. (See example sheet)

Corollary (Roth, 1950’s). Letε >0. Then ifnis large enough, andA⊆[n] = {1,2, . . . , n} with|A| ≥εn, thenA contains a 3-term arithmetic progression.

Proof. Define

B={(x, y)∈[2n]2:x−y∈A}.

Then certainly|B| ≥εn2. We form a 3-partite graph with disjoint vertex classes

X = [2n] andY = [2n], Z= [4n]. If we havex∈X, y∈Y andz∈Z, we joinx

toy if (x, y)∈B; joinxtoz if (x, z−x)∈B and joiny tozif (z−y, y)∈B. A triangle inG is a triple (x, y, z) where (x, y),(x, y+w),(x+w, y)∈ B

wherew=z−x−y. Note thatw <0 is okay. A 0-triangle is one with w= 0. There are at least εn2 _{of these, one for each (}_{x, y}₎ _∈ _B_{, and these are edge}

disjoint, becausez=x+y.

Hence the triangles cannot be killed by removing ≤εn2_/_{2 edges. By the}

triangle removal lemma, we must have≥δn3 _{triangles for some}_δ_{. In particular,}

fornlarge enough, there is some triangle that is not a 0-triangle. But then we are done, since

x−y−w, x−y, x−y+w∈A

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5 Subcontraction, subdivision and linkingIII Extremal Graph Theory (Theorems with proof)

5 Subcontraction, subdivision and linking

Theorem (Menger’s theorem). LetGbe a graph and s1, . . . , sk, t1, . . . , tk be

distinct vertices. If κ(G)≥k, then there exists k vertex disjoint paths from {s1, . . . , sk}to {t1, . . . , tk}.

Lemma. Ifκ(G)≥2kandG⊇T K5k, thenGisk-linked.

Proof. Let B be the set of “branch vertices” of T K5k, i.e. the 5k vertices

joined by paths. By Menger’s theorem, there exists 2k vertex disjoint paths joining {s1, . . . , sk, t1, . . . , tk} toB (note that these sets might intersect). We

say one of our 2k pathsimpinges on a path inT K5k if it meets that path and

subsequently leaves it. Choose a set of 2k paths impinging minimally (counting 1 per impingement).

Let these join{s1, . . . , tk} to{v1, . . . , v2k}, where B ={v1, . . . , v5k}. Then

no path impinges on a path in T K5k from {v1, . . . , v2k} to {v2k+1, . . . , v5k}.

Otherwise, pick the path impinging closest to v2k+j, and reroute it to v2k+j

rather than to something in{v1, . . . , v2k}, reducing the number of impingements.

Hence each path (of any 2k) meet at most one path in {v1, . . . , v2k} to

{v2k+1, . . . , v5k} (once we hit it, we must stay there).

Thus, we may assume the paths from{v1, . . . , v2k} to{v4k+1, . . . , v5k}are

not met at all. Then we are done, since we can link upvj tovk+j viav4k+j to

join sj totk.

Lemma. Ife(G)≥k|G|, then there exists someH with|H| ≤2kandδ(H)≥k

such thatGK1+H.

Proof. Consider the minimal subcontractionG0 _of_G_{among those that satisfy} e(G0)≥k|G0|. Then we must in fact have e(G0) =k|G0| , or else we can just throw away an edge.

Since e(G0) = k|G0|, it must be the case that δ(G0) ≤ 2k. Let v be of minimum degree in G0. and setH =G0[Γ(v)]. ThenGK1+H and|H| ≤2v.

To see thatδ(H)≥k, supposeu∈V(H) = Γ(v). Then by minimality ofG0, we have

e(G0/uv)≤k|G0| −k−1.

But the number of edges we kill when performing this contraction is exactly 1 plus the number of triangles containinguv. Souv lies in at leastk triangles of

G0. In other words,δ(H)≥k.

Theorem. Ift≥3 ande(G)≥2t−3|G|, thenGKt.

Proof. Ift= 3, thenGcontains a cycle. SoGK3. Ift >3, thenGK1+H

whereδ(H)≥2t−3_{. So}_e₍_H₎_≥₂t−4_|_H_|_{and (by induction)}_H _K

t−1.

Lemma. If δ(G)≥2k, then Gcontains vertex disjoint subgraphs H, J with

δ(H)≥k,J connected, and every vertex inH has a neighbour in J.

Proof. We may assumeGis connected, or else we can replaceGby a component. Now pick a subgraphJ maximal such that J is connected and e(G/J)≥

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Let H be the subgraph spanned by the vertices having a neighbour in J. Note that ifv∈V(H), thenv has at leastkneighbours inH. Indeed, when we contractJ∪ {v}, maximality tells us

e(G/(J∪ {v}))≤k(|G| − |J|+ 1)−k−1,

and the number of edges we lose in the contraction is 1 plus the number of neighbours ofv(it is easier to think of this as a two-step process — first contract

J, so that we get an H+K1, then contract v with the vertex coming from

K1).

Theorem. LetF be a graph withnedges and no isolated vertices. Ifδ(G)≥2n, thenG⊇T F.

Proof. Ifn= 1, then this is immediate. IfF consists ofnisolated edges, then

F ⊆G(in fact, δ(G)≥2n−1 is enough for this). Otherwise, pick an edgee

which is not isolated. ThenF −e has at most 1 isolated vertex. Apply the previous lemma to Gto obtain H with δ(H)≥2n−1_{. Find a copy of}_T₍_F₋_e₎

in H (apart from the isolated vertex, if exists).

Ifewas just a leaf, then just add an edge (say toJ). If not, just construct a path going throughJ to act ase, sinceJ is connected.

Lemma. We have

c(t)≥(α+o(1))tplogt.

To determineα, letλ <1 be the root to 1−λ+ 2λlogλ= 0. In factλ≈0.284. Then

α= 1−λ

2plog 1/λ ≈0.319.

Proof. Consider a random graph G(n, p), where pis a constant to be chosen later. So G(n, p) has nvertices and edges are chosen independently at random, with probabilityp. Here wefix at, and then try to find the best combination of

nandpto give the desired result.

A partition V1, . . . , Vt of V(G) is said to be complete if there is an edge

betweenVi andVj for alli andj. Note that having a complete partition is a

necessary, but not sufficient condition for having aKtminor. So it suffices to

show that with probability>0, there is no complete partition.

Writingq= 1−p, a given partition with|Vi|=niis complete with probability

Y i,j (1−qninj₎_≤_exp  − X i<j qninj   ≤exp − _t 2 Y qninj/(t₂) (AM-GM) ≤exp − _t 2 qn2/t2 .

The expected number of complete partitions is then

≤tnexp − _t 2 qn2/t2 ,

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As long as we restrict to the choices ofnandq such that

t > n

s

log(1/q) logn ,

we can bound this by

≤exp nlogt− _t 2 ₁ n =o(1) in the limitt→ ∞. We set

q=λ, n= t √

logt

p

log 1/λ.

Then with probability>0, we can find a graph with onlyo(1) many complete partitions, and by removingo(n) many edges, we have a graph with

p

_n

2

−o(n) = (α+o(1))tplogt·n

many edges with no Kt minor.

Lemma. Letk∈NandGbe a graph withe(G)≥11k|G|. Then there exists

some H with

|H| ≤11k+ 2, 2δ(H)≥ |H|+ 4k−1 such thatGH.

Proof. We use our previous lemma as a starting point. Letting`= 11k, we know that we can findH1 such that

GK1+H1,

with|H1| ≤2`andδ(H1)≥`. We shall improve the connectivity of thisH1 at

the cost of throwing away some elements.

By divine inspiration, consider the constantβ≈0.37 defined by the equation 1 =β 1 +log 2 β ,

and the function φdefined by

φ(F) =β`|F| 2 log|F| β` + 1 .

Now consider the set of graphs

C={F :|F| ≥β`, e(F)≥φ(F)}.

Observe thatH1∈ C, sinceδ(H1)≥`and|H1| ≤2`.

LetH2 be a subcontraction ofH1, minimal with respect toC.

Since the complete graph of order dβ`eis not in C, as φ > β`₂

, the only reason we are minimal is that we have hit the bound on the number of edges. Thus, we must have

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and

e(H2/uv)< φ(H2/uv)

for all edgesuvofH2.

Choose a vertexu∈H2of minimum degree, and put H3=H2[Γ(u)]. Then

we have |H3|=δ(H2)≤ 2dφ(H2)e |H2| ≤ β` log _|_H 2| β` + 1 + 2 |H2| ≤`, sinceβ`≤ |H2| ≤2`.

Let’s writeb=β`and h=|H2|. Then by the usual argument, we have

2δ(H3)− |H3| ≥2(φ(H2)−φ(H2/uv)−1)− |H3| ≥bh logh β + 1 −b(h−1) logh−1 b + 1 −b logh b + 1 −3 =b(h−1) log h h−1−3 > b−4,

becauseh≥b+ 1 andxlog 1 +_x1

>1−1

x for realx >1. So

2δ(H3)− |H3| ≥β`−4>4k−4.

If we put H=K2+H3, thenGH, |H| ≤11k+ 2 and

2δ(H)− |H| ≥2δ(H3)− |H3|+ 2≥4k−1.

Theorem. We have

c(t)≤7tplogt

iftis large.

Proof. For larget, we can choosek∈Nso that

5.8tplog₂t≤11k≤7tplogt.

Let ` = d1.01plog₂te. Then k ≥ t`/2. We want to show that G Kt if

e(G)≥11k|G|.

We already know that G H where h = |H| ≤ 11k+ 2, and 2δ(H) ≥ |H|+ 4k−1. We show that H Kt.

From now on, we work entirely inH. Note that any two adjacent vertices have≥4k+ 1 common neighbours. Randomly select 2tdisjoint`setsU1, . . . , U2t

in V(H). This is possible since we need 2t`vertices, and 2t`≤4k.

Of course, we only needt many of those` sets, and we want to select the ones that have some chance of being useful to us.

Fix a partU. For any vertexv (not necessarily inU), its degree is at least

h/2. So the probability thatvhas no neighbour inU is at most h/_`2

/ h_`

≤2−`_.

WriteX(U) for the vertices having no neighbour inU, then_E|X(U)| ≤2−`_h_.

We sayU is bad if |X(U)| > 8·2−`_h_{. Then by Markov’s inequality, the}

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U1, . . . , U2t is≤₄t. By Markov again, the probability that there are more than t

2 bad sets is< 1 2.

We say a pair of sets (U, U0) is bad if there is no U −U0 edge. Now the probability thatU, U0 is bad is_P(U0 ⊆X(U)). If we condition on the event that

U is good (i.e. not bad), then this probability is bounded by

8·2−`h ` / h−` ` ≤8`2−`2 1 + ` h−` ` ≤8`2−`2e`2/(h−`)≤9`2−`2

ift is large (hence`is slightly large andhis very large). We can then bound this by ₈1_t.

Hence, the expected number of bad pairs (where one of them is good) is at most 1 8t 2t 2 ≤ t 4.

So the probability that there are more than t₂ such bad pairs is <1₂.

Now with positive probability, U1, . . . , U2t has at most t₂ bad sets and at

most ₂t bad pairs amongst the good sets. Then we can findtmany sets that are pairwise good. We may wlog assume they areU1, . . . , Ut.

Fixing such a choiceU1, . . . , Ut, we now work deterministically. We would

be done if each Ui were connected, but they are not in general. However, we

can findW1, . . . , WtinV(H)\(U1∪ · · ·Ut) such thatUi∪Wi is connected for

1≤i≤t.

Indeed, ifUi={u1, . . . , u`}, we pick a common neighbour ofuj−1anduj if

uj−1uj6∈E(H), and put it inWi. In total, this requires us to pick≤t`distinct

vertices inV(H)−(U1∪ · · · ∪Ut), and we can do this becauseuj−1anduj have

at least 4k−t`≥t`common neighbours in this set.

Lemma. Letd≥0,k≥2 andh≥d+b3k/2cbe integers.

LetGbe a graph, S ={s1, . . . , sk} ⊆V(G). Suppose there exists disjoint

subgraphsC1, . . . , Ch ofGsuch that

(∗) EachCi is either connected, or each of its component meetsS. Moreover,

eachCi is adjacent to all but at mostdof theCj, j6=i not meetingS.

(†) Moreover, noS-cut of order< kavoidsd+ 1 of C1, . . . , Ch.

ThenGcontains disjoint non-empty connected subgraphsD1, . . . , Dm, where

m=h− bk/2c,

such that for 1≤ i≤ k, si ∈Di, andDi is adjacent to all but at mostdof

Dk+1, . . . , Dm.

Proof. Suppose the theorem is not true. Then there is a minimal counterexample

G(with minimality defined with respect to proper subgraphs).

We first show that we may assumeG has no isolated vertices. If v is an isolated vertex, and v 6∈S, then we can simply apply the result to G−v. If

v∈S, then theS-cut (S, V(G)\ {v}) of orderk−1 avoidsh−k≥d−1 of the

Ci’s.

Claim. If (A, B) is anS-cut of order kavoidingd+ 1 many of theCi’s, then

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Indeed, given such anS-cut, we define

S0=A∩B G0=G[B]−E(S0)

C_i0=Ci∩G0.

We now make a further claim:

Claim. (G0, S0,{C_i0}) satisfies the hypothesis of the lemma.

We shall assume this claim, and then come back to establishing the claim. Assuming these claims, we show thatGisn’t a countrexample after all. Since (G0, S0,{C_i0}) satisfies the hypothesis of the lemma, by minimality, we can find subgraphsD0₁, . . . , D0_minG0 that satisfy the conclusion of the lemma forG0 and

S0. Of course, these do not necessarily contain oursiAssuming these claims, we

can construct the desiredDi. However, we can just add paths from S0 to thesi.

Indeed,G[A] has noS-cut (A00, B00) of order less thankwithS0 ⊆B00, else (A00, B00∪B) is an S-cut of G of the forbidden kind, sinceA, and hence A00

avoids too many of theCi. Hence by Menger’s theorem, there are vertex disjoint

pathsP1, . . . , Pk inG[A] joiningsi tos0i (for some labelling ofS0). Then simply

take

Di =D0i∪Pi.

To check that (G0_{, S}0_,_{_C0

i}) satisfy the hypothesis of the lemma, we first

need to show that theC0

i are non-empty.

If (A, B) avoids Cj, then by definition Cj ∩A = ∅. So Cj = Cj0. By

assumption, there are d+ 1 manyCj’s for which this holds. Also, since these

don’t meetS, we know eachCi is adjacent to at least one of theseCj’s. So in

particular,C_i0 is non-empty since C_i0 ⊇Ci∩Cj0 =Ci∩Cj.

Let’s now check the conditions:

(∗) For each C_i0, consider its components. If there is some component that does not meetS0, then this component is contained inB\A. So this must also be a component ofCi. ButCi is connected. So this component isCi.

SoC_i0=Ci. Thus, eitherCi0 is connected, or all components meetS0.

Moreover, since anyC0

j not meetingS0 equalsCj, it follows that eachCi

and so eachC_i0 is adjacent to all but at mostdof these C_j0s. Therefore (∗) holds.

(†) If (A0, B0) is an S0-cut in G0 avoiding some C_i0, then (A∪A0, B0) is an

S-cut ofGof the same order, and it avoidsCi (sinceCi0 doesn’t meetS0,

and we have argued this impliesC_i0=Ci, and soCi∩A=∅). In particular,

no S0_{-cut (}_A0_{, B}0_{) of} _G0 _{has order less than} _k _{and still avoids} _d_{+ 1 of} C0

1, . . . , Ch0.

We can now use our original claim. We know whatS-cuts of orderkloop like, so we see that if there is an edge that does not join two Ci’s, then we

can contract the eedge to get a smaller counterexample. Recalling thatGhas no isolated vertices, we see that in fact V(G) =∪V(Ci), and |Ci| = 1 unless

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Let

C=[{V(Ci) :|Ci| ≥2}.

We claim that there is a setIof|C|independent edges meetingC. If not, by Hall’s theorem, we can find X⊆C whose neighbourhoodY (inV(G)−S) such that |Y|<|X|. Then (S∪Y, V(G)−X) is anS-cut of order|S| − |X|+|Y|<|S|=k

avoiding ≥ |G| − |S| − |Y| manyCi’s, and this is

≥ |G| − |S| − |X|+ 1≥ |G| − |C| −k+ 1.

But h≤ |G| − |C|+|C₂|, since|G| − |C|is the number ofCi’s of size 1, so we

can bound the above by ≥h−|C|

2 −k+ 1≥h− 3k

2 + 1≥d+ 1. This contradiction shows thatI exists.

Now we can just write down theDi’s. For 1≤i≤k, setDi to besi ifsi is

a C` with|C`|= 1, i.e.si 6∈C. Ifsi∈C, letDi be the edge ofI meeting Si.

LetDk+1, . . . , Dmeach be single vertices of G−S−(ends ofI). Note those

exist because≥ |G| −k− |C| ≥h− b3k

2c=m−kvertices are avaiable. Note

that each Di contains aC` with|C`|= 1. So eachDi is joined to all but≤d

manyDj’s.

Theorem. Let Gbe a graph with κ(G)≥2kande(G)≥11k|G|. ThenG is

k-linked. In particular,f(k)≤22k.

Proof. We know that under these conditions, GH with|H| ≤11k+ 2 and 2δ(H)≥ |H|+ 4k−1. Leth=|H|, and d=h−1−δ(H). Note that

h= 2d−2 + 2δ(H)−h≥2d+ 4k.

LetC1, . . . , Ck be connected subgraphs ofGwhich contract to formH. Then

clearly each Ci is joined to all but at most h−1−δ(H) = d other Cj’s.

Now let s1, . . . , sk, t1, . . . , tk be the vertices we want to link up. Let S =

{s1, . . . , sk, t1, . . . , tk}. Observe that Ghas no S-cut of order<2kat all since

κ(G)≥2k. So the conditions are satisfied, but with 2kinstead ofk.

So there are subgraphsD1, . . . , Dm, wherem=h−k≥2d+ 3kas described.

We may assumesi∈Diandti∈Dk+i. Note that for each pair (Di, Dj), we can

find≥m−2d≥3kotherD` such thatD` is joined to bothDi andDj. So for

eachi= 1, . . . , k, we can find an unusedD`not amongD1, . . . , D2k that we can

use to connect up Di andDk+i, hencesi andti.

Lemma. Ifδ(a)≥t2_and_G_is t+2 3

-linked, then G⊇T Kt.

Proof. Since δ(G) ≥t2_{, we may pick vertices} _v

1, . . . , vt and sets U1, . . . Ut all

disjoint so that|Ui|=t−1 andvi is joined toUi. ThenG− {v1, . . . , vk}is still t+2

2

−t= t₂-linked. SoU1, . . . , Ut−1may be joined with paths to form aT Kt

in G.

Lemma. Letk, d∈_Nwithk≤ d+1

2 , and supposee(G)≥d|G|. ThenGcontains

a subgraph H with

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Proof. Define

Ed,k={F ⊆G:|F| ≥d, e(F)> d|F| −kd}.

We observe thatD∈ Ed,k, butKd6∈ Ed,k. Hence|F|> dfor allF ∈ Ed,k. letH

be a subgraph ofGminimal with respect toH ∈ Ed,k. Thene(H) =d|H|−kd+1,

andδ(H)≥d+ 1, elseH−v∈ Ed,k for somev∈H.

We only have to worry about the connectivity condition. SupposeS is a cutset of H. Let C be a component of G−S. Since δ(h)≥ d+ 1, we have |C∪S| ≥d+ 2 and|H−C| ≥d+ 2.

Since neitherCsupS norH−C lies in that class, we have

e(C∪S)≤d|C∪S| −kd≤d|C|+d|S| −kd

and

e(H−C)≤d|H| −d|C| −kd.

So we find that

e(H)≤d|H|+d|S| −2kd.

But we know this is equal tod|H| −kd+ 1. So we must have |S| ≥k.

Theorem. t2 16 ≤t(t)≤13 t+ 1 2 .

Proof. The lower bound comes from (disjoint copies of) Kt2_/₈_,t2_/₈. For the uppper bound, write

d= 13

_t_{+ 1}

2

, k=t·(t+ 1).

Then we can find a subgraphH withδ(H)≥d+ 1 and κ(H)≥k+ 1. Certainly|H| ≥d+ 2, and we know

e(H)≥d|H| −kd >(d−k)|H| ≥11 _t_{+ 1} 2 |H|. So we knowH is t+1₂ -linked, and soH⊇T Kt.

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6 Extremal hypergraphs III Extremal Graph Theory (Theorems with proof)

6 Extremal hypergraphs

Theorem (Erd¨os). Let G be `-uniform of order n with p n_`

edges, where

p≥2n−1/t`−1_{. Then}_G_contains_K`

`(t) (providednis large).

Proof. Assume thatt≥2. We proceed by induction on`≥2. For each (`−1)-set

σ⊆V(G), let

N(σ) ={v:σ∪ {v} ∈E(G)}.

Then the average degree is

_n `−1 −1 X |N(σ)|= _n `−1 −1 `|E(G)|=p(n−`+ 1).

For each of the n_t

manyt-setsT ⊆V(G), let

D(T) ={σ:T ⊆N(σ)}. Then X T |D(T)|=X σ _|_N₍_σ_)| t ≥ _n `−1 _p₍_n₋_`_{+ 1)} t ,

where the inequality follows from convexity, since we can assume thatp(n−`+ 1)≥t−1 ifnis large.

In particular, there exists aT with

|D(T)| ≥ n t −1 n `−1 p(n−`+ 1) t ≥1 2p t n `−1 .

whennis large. By our choice oft, we can write this as

|D(T)| ≥2t−1n−1/t`−2 n `−1 .

If`= 2, then this is≥2t−1≥t. SoT is joined to aT-set givingKt,t.

If`≥3, then|D(T)| ≥2n−1/t`−2 _n

`−1

. So, by induction, the (`−1)-uniform hypergraph induced by the σ withT ⊆N(σ) contains K_``₋−₁1(t), giving K`

`(t)