Combinatorics Lectured by I. B. Leader Michaelmas Term 2008, 2010 & 2012 Chapter 1 Set Systems 1 Chapter 2 Isoperimetric Inequalities 10 Chapter 3 Intersecting Families 23 Chapter 3 ′ Projections 28 Examples Sheets Note. Chapter 3 was lectured in 2008 and 2012. Chapter 3 ′ was lectured in 2010. Books (for chapter 1) – Bollob´ as, Combinatorics (CUP, 1986) (“Bedtime reading”) – Anderson, Combinatorics and Finite Sets (OUP, 1987) (“Simple and clear”) Pre-requisites – basic concepts of graph theory (graph, path, Hall’s Theorem) – integers mod p – vector spaces Last updated: Tue 23 rd Apr, 2013 Please let me know of any corrections: [email protected]
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Combinatorics
Lectured by I. B. Leader
Michaelmas Term 2008, 2010 & 2012
Chapter 1 Set Systems 1
Chapter 2 Isoperimetric Inequalities 10
Chapter 3 Intersecting Families 23
Chapter 3′ Projections 28
Examples Sheets
Note. Chapter 3 was lectured in 2008 and 2012. Chapter 3′ was lectured in 2010.
Books (for chapter 1)
– Bollobas, Combinatorics (CUP, 1986) (“Bedtime reading”)
– Anderson, Combinatorics and Finite Sets (OUP, 1987) (“Simple and clear”)
Pre-requisites
– basic concepts of graph theory (graph, path, Hall’s Theorem)– integers mod p– vector spaces
Last updated: Tue 23rd Apr, 2013
Please let me know of any corrections: [email protected]
Course description (from the Part III booklet)
The flavour of the course is similar to that of the Part II Graph Theory course, although weshall not rely on many of the results from that course.
We shall study collections of subsets of a finite set, with special emphasis on size, intersectionand containment. There are many very natural and fundamental questions to ask aboutfamilies of subsets; although many of these remain unsolved, several have been answeredusing a great variety of elegant techniques.
We shall cover a number of ‘classical’ extremal theorems, such as those of Erdos-Ko-Radoand Kruskal-Katona, together with more recent results concerning isoperimetric inequalitiesand intersecting families. The aim of the course is to give an introduction to a very activearea of mathematics.
We hope to cover the following material.
Set Systems
Definitions. Antichains; Sperner’s lemma and related results. Shadows. Compression opera-tors and the Kruskal-Katona theorem. Intersecting families; the Erdos-Ko-Rado theorem.
Isoperimetric Inequalities
Harper’s theorem and the edge-isoperimetric inequality in the cube. Inequalities in thegrid. The classical isoperimetric inequality on the sphere. The ‘concentration of measure’phenomenon. Applications.
Intersecting Families (2008 & 2012)
Katona’s t-intersecting theorem. The Ahlswede-Khachatrian theorem. Restricted intersec-tions. The Kahn-Kalai counterexample to Borsuk’s conjecture.
Projections (2010)
The trace of a set system; the Sauer-Shelah lemma. Bounds on projections: the Bollobas-Thomason box theorem. Hereditary properties. Intersecting families of graphs.
Desirable Previous Knowledge
The only prerequisites are the very basic concepts of graph theory.
Introductory Reading
1. Bollobas, B., Combinatorics, C.U.P. 1986.
Chapter 1 : Set Systems
Let X be a set. A set system on X (or family of subsets of X) is a family A ⊂ P(X).
E.g., X(r) = {A ⊂ X : |A| = r}.
Unless otherwise stated, X = [n] = {1, . . ., n}. So, e.g.,∣∣X(r)
∣∣ =
(n
r
).
For example, [4](2) = {12, 13, 14, 23, 24, 34}, where “12” = {1, 2}.
Often, we make P(X) into a graph by joining A to B if |A△B| = 1. I.e., if A = B ∪ {i} forsome i /∈ B, or vice versa. This graph is the discrete cube Qn.
E.g., Q3 Qn (n even) Qn (n odd)
@@
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@@
���
�@
@
��
@@
∅
1 2 3
12 13 23
123������ ���� ��...�� ���� ��...�� ������
X(n)
X(n−1)
X(n−2)
X(n/2)
X(n/2−1)
X(1)
X(0)
������ ��...�� ���� ��...�� ������
X(n)
X(n−1)
X(⌈n/2⌉)
X(⌊n/2⌋)
X(1)
X(0)
If we identify a point A ∈ P(X) with a 0–1 sequence of length n(e.g., {1, 3} 7→ 10100. . .00, via the indicator function A↔ IA),then Qn is precisely the unit cube in Rn. q
∅q1
q3
q13q
2
q23
q12
q123
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-6��� 1
32
Chains and Antichains
A ⊂ P(X) is a chain if for all A, B ∈ A, either A ⊂ B or B ⊂ A.
A ⊂ P(X) is an antichain if for all A, B ∈ A, A 6= B, we have A 6⊂ B.
E.g., {12, 1257, 12357} is a chain, and {1, 346, 2589} is an antichain.
How large can a chain be? We can have |A| = n + 1, e.g. A = {∅, 1, 12, 123, . . ., [n]}. Wecannot beat this, as a chain can meet a “level” X(r) in at most one point.
How large can an antichain be? We can have |A| = n, e.g. A = {1, 2, 3, . . ., n}. Indeed, X(r)
is always an antichain, so we can achieve size(
nn/2
)(for n even) or
(n
⌊n/2⌋)
(for n odd).
Can we beat that?
1
Theorem 1 (Sperner’s Lemma). Let A ⊂ P(X) be an antichain. Then |A| 6(
n
⌊n/2⌋
).
Idea. Inspired by “chains having 6 1 in each layer”, let’s try to decompose cube into chains.
Proof. We’ll decompose P(X) into
(n
⌊n/2⌋
)chains, and then we’re done.
To do this, it’s sufficient to show that:
(i) for each r < n/2, there is a matching (a set of disjoint edges) from X(r) to X(r+1),(ii) for each r > n/2, there is a matching from X(r) to X(r−1).
Then just put together these matchings to form our chains. Taking complements, it’ssufficient to prove (i).
Consider the induced subgraph of Qn spanned by X(r) ∪X(r+1). This is bipartite, andhas d(A) = n − r for all A ∈ X(r) (the things that can be added to the r-set), andd(A) = r + 1 for all A ∈ X(r+1) (the ways to throw out an element of an (r + 1)-set).
Given S ⊂ X(r), the number of S–Γ(S) edges equals |S|(n− r) (counting from below),and is 6 |Γ(S)|(r + 1) (counting from above).
Thus, |Γ(S)| > |S|(
n− r
r + 1
)> |S|, as r < n/2.
So there exists a matching, by Hall’s Theorem. �
Remarks. 1. We can achieve |A| =(
n
⌊n/2⌋
). E.g., A = X(⌊n/2⌋).
2. Uniqueness? When can |A| =(
n
⌊n/2⌋
)? The above proof tells us nothing.
Aim. If A is an antichain, then
n∑
r=0
|A ∩X(r)|(nr
) 6 1.
“Sum of percentages of levels filled is at most 1.”
This trivially implies Sperner. (Exercise: make sure this is clear.)
For A ⊂ X(r), 1 6 r 6 n, the shadow or lower shadow of A is the set system:
∂A = ∂−A = {B ∈ X(r−1) : B ∪ {i} ∈ A for some i /∈ B} ⊂ X(r−1).
E.g., if A = {123, 124, 234, 135}, then ∂A = {12, 13, 23, 14, 24, 34, 15, 35}.
Proposition 2 (Local LYM). Let 1 6 r 6 n, and let A ⊂ X(r). Then|∂A|(
nr−1
) >|A|(nr
) .
“The fraction occupied by ∂A is > that for A.”
(“LYM” = Lubell, Yamamoto, Meshalkin.)
2
Proof. The number of edges from A to ∂A is equal to |A|r (counting from A), and is6 |∂A|(n− r + 1) (counting from ∂A).
Thus |A|r 6 |∂A|(n− r + 1), so|∂A||A| >
r
n− r + 1. But
(n
r−1
)(nr
) =r
n− r + 1. �
Equality in Local LYM? Must have that: ∀A ∈ A, ∀i ∈ A, ∀j /∈ A, we have (A− i) ∪ j ∈ A.(“Add something and remove something, and you’re still in A.”) Thus A = ∅ or X(r).
Theorem 3 (LYM inequality). Let A ⊂ P(X) be an antichain. Then
n∑
r=0
|A ∩X(r)|(nr
) 6 1.
Proof 1. “Bubble down using Local LYM.”
Write Ar for A ∩X(r). Firstly, we have|An|(
nn
) 6 1.
Now, ∂An and An−1 are disjoint subsets of X(n−1) (as we are in an antichain).
So|∂An|(
nn−1
) +|An−1|(
nn−1
) =|∂An ∪ An−1|(
nn−1
) 6 1. So by Local LYM,|An|(
nn
) +|An−1|(
nn−1
) 6 1.
Also, ∂(∂An ∪ An−1) and An−2 are disjoint, so|∂(∂An ∪ An−1)|(
nn−2
) +|An−2|(
nn−2
) 6 1.
So by Local LYM,|∂An ∪ An−1|(
nn−1
) +|An−2|(
nn−2
) 6 1, and thus|An|(
nn
) +|An−1|(
nn−1
) +|An−2|(
nn−2
) 6 1.
Keep going. We get|An|(
nn
) + . . . +|A0|(
n0
) 6 1. �
Equality in LYM? We must have had equality in each application of Local LYM. Hence, forthe greatest r with Ar 6= ∅, we must have Ar = X(r), whence A = X(r) as A is an antichain.
Conclusion: equality in LYM ⇔ A = X(r), some r.
In particular, equality in Sperner ⇔{A = X(n/2) (n even)A = X(⌊n/2⌋) or X(⌈n/2⌉) (n odd)
.
Proof 2. Choose, uniformly at random, a maximal chain C (i.e., C0 ⊂ C1 ⊂ . . . ⊂ Cn, with|Ci| = i for all i).
For a fixed r-set A, we have P (A ∈ C) =1(nr
) (as all r-sets equally likely to be in C).
So P (C meets Ar) =|Ar|(
nr
) (as events are disjoint).
So P (C meets A) =
n∑
r=0
|Ar|(nr
) (as events are disjoint). Thus
n∑
r=0
|Ar |(nr
) 6 1. �
Equivalently, the number of maximal chains = n!, and the number of maximal chains con-taining a given r-set = r!(n− r)!. So
∑nr=0 |Ar| r!(n− r)! 6 n!.
3
Shadows
If A ⊂ X(r), we know |∂A| > |A| r
n− r + 1, but equality is rare (only for A = ∅ or X(r)).
What happens in between?
How would we choose A ⊂ X(r), with |A| given, to minimise |∂A|? It’s believable that for|A| =
(kr
)for some k, we should take A = [k](r), yielding ∂A = [k](r−1).
What if(kr
)< |A| <
(k+1
r
)? It’s believable that we would take A to be [k](r) and some extra
sets from [k + 1](r).
E.g., for |A| =(
7
3
)+
(4
2
)in X(3), we would try A = [7](3) ∪
{A ∪ {8} : A ∈ [4](2)
}.
Two total orderings on X(r)
We’re given A, B ∈ X(r), say A = {a1, . . ., ar} where a1 < . . . < ar, and B = {b1, . . ., br}where b1 < . . . < br.
Say that A < B in the lexicographic or lex order if there exists i with ai < bi and aj = bj
for all j < i. Equivalently, A < B if ai < bi where i = min{j : aj 6= bj}.
“Use small elements if possible (like a dictionary).”
E.g., lex on [4](2): 12, 13, 14, 23, 24, 34.
lex on [6](3): 123, 124, 125, 126, 134, 135, 136, 145, 146, 156,234, 235, 236, 245, 246, 256, 345, 346, 356, 456.
Say that A < B in the colexicographic or colex order if there exists i with ai < bi andaj = bj for all j > i. Equivalently, A < B is ai < bi where i = max{j : aj 6= bj}.
“Don’t use large elements.”
Equivalently, A < B if∑
i∈A 2i <∑
i∈B 2i.
E.g., colex on [4](2): 12, 13, 23, 14, 24, 34.
colex on [6](3): 123, 124, 134, 234, 125, 135, 235, 145, 245, 345,126, 136, 236, 146, 246, 346, 156, 256, 356, 456.
Note that [m](r) is an initial segment (i.e., first t elements, for some t) of [m + 1](r) in colex.Thus we could view colex as an enumeration of N(r). (This is false for lex.)
Aim. Initial segments of colex have smallest shadow ∂.
I.e., if A ⊂ X(r) and C ⊂ X(r) is the initial segment of colex with |C| = |A|, then |∂C| 6 |∂A|.
In particular, |A| =(
k
r
)⇒ |∂A| >
(k
r − 1
).
4
Compressions
Given A ⊂ X(r), we would like to replace A by some A′ ⊂ X(r), where:
(i) |A′| = |A|, (ii) |∂A′| 6 |∂A|, (iii) A′ “looks more like” C than A did.
We would like to find several such “compression” operations: A → A′ → A′′ → · · · → B, forwhich either B = C, or B is so similar to C that we can see directly that |∂B| > |∂C|.
“Colex prefers 1 to 2” inspires the following.
Fix 1 6 i < j 6 n. The ij-compression is defined as follows.
For A ∈ X(r), let Cij(A) =
{A ∪ i− j if j ∈ A, i /∈ AA otherwise
.
And for A ⊂ X(r), let Cij(A) = {Cij(A) : A ∈ A} ∪ {A ∈ A : Cij(A) ∈ A}.
I.e., “replace j by i if possible”.
E.g., if A = {123, 124, 135, 235, 245, 367} then C12(A) = {123, 124, 135, 235, 145, 367}.
We clearly have |Cij(A)| = |A|. Say that A is ij-compressed if Cij(A) = A.
Lemma 4. Let A ⊂ X(r). Then for any 1 6 i < j 6 n, we have |∂Cij(A)| 6 |∂A|.
Proof. Write A′ for Cij(A). We’ll show that for each B ∈ ∂A′− ∂A, we have j /∈ B, i ∈ B,and B ∪ j − i ∈ ∂A− ∂A′. (Then done.)
We have B∪x ∈ A′ for some x, and B∪x /∈ A, so i ∈ B∪x, j /∈ B∪x. We cannot havei = x, as then B∪i ∈ A′, so either B∪i or B∪j is in A, so B ∈ ∂A, contradiction. Thuswe know i ∈ B and j /∈ B. Since B ∪ x ∪ j − i ∈ A, we certainly have B ∪ j − i ∈ ∂A.
Claim. B ∪ j − i /∈ ∂A′.
Proof of claim. Suppose (B ∪ j− i)∪ y ∈ A′ for some y. Then we cannot have y = i,or else B ∪ j ∈ A′, whence B ∪ j ∈ A and so B ∈ ∂A, contradiction.
So i /∈ (B ∪ j − i) ∪ y and j ∈ (B ∪ j − i) ∪ y. And so (B ∪ j − i) ∪ y and B ∪ yare both in A (by definition of A′). So B ∈ ∂A, contradiction �
Remark. We actually showed ∂Cij(A) ⊂ Cij(∂A).
Say that A is left-compressed if Cij(A) = A for all i < j.
Corollary 5. Let A ⊂ X(r). Then there exists B ⊂ X(r) with |B| = |A|, |∂B| 6 |∂A|, andB is left-compressed.
Proof. Define a sequence A0,A1,A2, . . . ⊂ X(r) as follows. Set A0 = A. Having chosenA0, . . .,Ak, if Ak is left-compressed, stop the sequence with Ak. If not, choose i < jwith Ak not ij-compressed, and set Ak+1 = Cij(Ak). This must terminate – e.g., asthe function
∑A∈Ak
∑x∈A x is decreasing in k. The system B = Ak has |B| = |A| and
|∂B| 6 |∂A| by Lemma 4. �
5
Remarks. 1. Alternatively, among all B ⊂ X(r) with |B| = |A| and |∂B| 6 |∂A|, choose onewith minimal
∑A∈B
∑x∈A x.
2. It is possible to apply each Cij at most once if we choose the order sensibly.
Any initial segment of colex is left-compressed. However, the converse is false, for exampleA = {123, 124, 125, 126, 127}.
“Colex prefers 23 to 14” inspires the following.
For U, V ⊂ X with |U | = |V | and U ∩ V = ∅, define the UV -compression as follows.
For A ∈ X(r), let CUV (A) =
{A ∪ U − V if V ⊂ A, A ∩ U = ∅A otherwise
.
“Replace V s with Us.”
And for A ⊂ X(r), let CUV (A) = {CUV (A) : A ∈ A} ∪ {A ∈ A : CUV (A) ∈ A}.
E.g., if A = {123, 145, 147, 234, 235, 267} then C23,14(A) = {123, 145, 237, 234, 235, 267}.
Note that |CUV (A)| = |A|, and that C{i},{j}(A) = Cij(A).
Unfortunately, we can have |∂CUV (A)| > |∂A|. E.g., let A = {147, 478}. Then |∂A| = 5,but C23,14(A) = {237, 478} with |∂C23,14(A)| = 6.
Say A is UV -compressed if CUV (A) = A.
Lemma 6. Let U, V ⊂ X be disjoint, with |U | = |V |. Let A ⊂ X(r). Suppose that for allu ∈ U , there is v ∈ V such thatA is (U−u, V−v)-compressed. Then |∂CUV (A)| 6 |∂A|.
Proof. Write A′ for CUV (A). For any B ∈ ∂A′ − ∂A, we’ll show that U ⊂ B, V ∩ B = ∅,and B ∪ V − U ∈ ∂A− ∂A′. (Then done.)
We have B ∪ x ∈ A′ for some x, and B ∪ x /∈ A. Thus U ⊂ B ∪ x, V ∩ (B ∪ x) = ∅,and B ∪ x ∪ V − U ∈ A. So certainly V ∩B = ∅.
Also, x /∈ U . For if x ∈ U , then there is y ∈ V with A being (U −x, V −y)-compressed.But B ∪ x ∪ V − U ∈ A, so then B ∪ y ∈ A, contradicting B /∈ ∂A. Thus U ⊂ B.
And we have B ∪ V − U ∈ ∂A, because B ∪ x ∪ V − U ∈ A.
Suppose B ∪ V − U ∈ ∂A′. So (B ∪ V − U) ∪ w ∈ A′ for some w.
If w /∈ U , then because (B∪V −U)∪w ∈ A′, we must have (B∪V −U)∪w and B∪win A (by definition of CUV ), contradicting B /∈ ∂A.
If w ∈ U , then we have that A is (U − w, V − z)-compressed for some z ∈ V . So from(B ∪V −U)∪w ∈ A (which is true as it contains V , so cannot have moved), we obtainB ∪ z ∈ A, a contradiction. �
Remark. We actually showed that ∂CUV (A) ⊂ CUV (∂A).
6
Theorem 7 (Kruskal-Katona Theorem). Let A ⊂ X(r) (1 6 r 6 n), and let C be theinitial segment of colex on X(r) with |C| = |A|. Then |∂A| > |∂C|.
In particular, if |A| =(
k
r
)then |∂A| >
(k
r − 1
).
Proof. Let Γ ={
(U, V ) : U, V ⊂ X, |U | = |V | > 0, U ∩ V = ∅, maxV > max U}.
Define a sequence A0,A1, . . . as follows. Set A0 = A. Having chosenA0, . . .,Ak, ifAk isUV -compressed for all (U, V ) ∈ Γ, stop the sequence with Ak. If not, choose (U, V ) ∈ Γ,with |U | minimal, such that Ak is not UV -compressed, and set Ak+1 = CUV (Ak).
For each u ∈ U , setting v = min V , we have (U − u, V − v) ∈ Γ ∪ {(∅, ∅)}. So Ak is(U − u, V − v)-compressed. Thus by Lemma 6, we have |∂Ak+1| 6 |∂Ak|. Continue.
This sequence must terminate, as∑
A∈Ak
∑i∈A 2i is decreasing. The final term,
B = Ak, satisfies: |B| = |A|, |∂B| 6 |∂A|, and B is UV -compressed for all (U, V ) ∈ Γ.
Claim. B = C.
Proof of claim. Suppose not. Then there exists A, B ∈ X(r) with A < B in colex,and A /∈ B, B ∈ B. Set U = A − B, V = B − A. We have maxV > max U(as A < B in colex). Thus (U, V ) ∈ Γ. So B ∈ B ⇒ A ∈ B. Contradiction. �
Remarks. 1. Equivalently, if |A| =(kr
r
)+(kr−1
r−1
)+ . . . +
(ks
s
), where kr > kr−1 > . . . > ks
and s > 0, then |∂A| >(
kr
r−1
)+(kr−1
r−2
)+ . . . +
(ks
s−1
).
2. The proof used only Lemma 6, and not Lemma 4 or Corollary 5.
3. Equality? We can check that if |A| =(kr
)and we have equality in Kruskal-Katona
(i.e., |∂A| =(
kr−1
)), then A = Y (r) for some Y ⊂ X with |Y | = k. But it is false in
general that if equality holds (i.e., |∂A| = |∂C|) then A is isomorphic to C.
(A, B are isomorphic if there is a bijection f : X → X sending A to B.)
For A ⊂ X(r) (0 6 r 6 n − 1), the upper shadow of A is ∂+A ⊂ X(r+1), given by∂+A = {A ∪ x : A ∈ A, x ∈ X, x /∈ A}.
Note A < B in colex on X(r) ⇔ Ac < Bc in lex on X(n−r) with the ground set order reversed.
Corollary 8. Let A ⊂ X(r) (0 6 r 6 n− 1) and let C be the initial segment of lex on X(r)
with |C| = |A|. Then |∂+C| 6 |∂+A|.
Proof. Take complements. �
The shadow of an initial segment of colex on X(r) is an initial segment of colex on X(r−1)
(for if C = {A ∈ X(r) : A 6 a1. . .ar} then ∂C = {B ∈ X(r−1) : B 6 a2. . .ar}), so we have:
Corollary 9. Let A ⊂ X(r), and let C be the initial segment of colex on X(r) with |C| = |A|.Then |∂tA| > |∂tC| for all 1 6 t 6 r. In particular, if |A| =
(kr
)then |∂tA| >
(k
r−t
).
Proof. If |∂tA| > |∂tC| then |∂t+1A| > |∂t+1C| by Kruskal-Katona. �
7
Intersecting Families
Say that A ⊂ P(X) is intersecting if A ∩B 6= ∅ for all A, B ∈ A.
How large can an intersecting family be?
We can have |A| = 2n−1, e.g. A = {A ⊂ X : 1 ∈ A}.
Proposition 10. Let A ⊂ P(X) be intersecting. Then |A| 6 2n−1.
Proof. For any A ∈ P(X), at most one of A, Ac can belong to A. �
Remark. There are many extremal systems. E.g., {A ∈ P(X) : |A| > n/2} for n odd.
What if we restrict to A ⊂ X(r)?
If r > n/2, silly: we can take A = X(r).
If r = n/2, take one of each complementary pair A, Ac – gives 12
(nr
). (Optimal, as we can
never take both of A and Ac.)
So assume r < n/2.
Obvious guess: A = {A ∈ X(r) : 1 ∈ A}. This has |A| =(
n− 1
r − 1
)=
r
n
(n
r
).
We could also try, e.g., B ={A ∈ X(r) : |A ∩ {1, 2, 3}| > 2
}.
E.g., in [8](3), |A| = 21 and |B| = 1 + 3.5 = 16 < 21.
Theorem 11 (Erdos-Ko-Rado Theorem). Let r < n/2, and A ⊂ X(r) be intersecting.
Then |A| 6(
n− 1
r − 1
).
Proof 1. “Bubble down with Kruskal-Katona.”
For any A, B ∈ A, A∩B 6= ∅. I.e., A 6⊂ Bc. Thus, writing A = {Bc : B ∈ A} ⊂ X(n−r),we have that A and ∂n−2rA are disjoint subsets of X(r).
Suppose that |A| >(
n− 1
r − 1
).
Then |A| >(
n− 1
r − 1
)=
(n− 1
n− r
). So |∂n−2rA| >
(n− 1
r
), by Corollary 9.
Thus |A|+ |∂n−2rA| >(
n− 1
r − 1
)+
(n− 1
r
)=
(n
r
). Contradiction. �
Note. Numbers had to work out, given that if A = {A ∈ X(r) : 1 ∈ A}, we have equality,and A, ∂n−2rA partition X(r).
8
Proof 2. Consider a cyclic ordering of X , i.e. a bijection c : Zn → X . How many A ∈ Aare intervals (blocks of r consecutive elements) in this ordering? Answer: at most r.Indeed, suppose {c1, c2, . . ., cr} ∈ A. Then for each 1 6 i 6 r − 1, at most one of{ci−r+1, . . ., ci} and {ci+1, . . ., ci+r} belongs to A.
Also, each r-set is an interval in precisely n r! (n− r)! of the n! cyclic orderings.ր ↑ տ
where insideinterval
outsideinterval
Thus, |A|n r! (n− r)! 6 n! r, and so |A| 6 r
n
(n
r
). �.
Notes. 1. Equivalently, we are double-counting the edges in the bipartite graph with vertexclasses A and the cyclic orderings, where we join A ∈ A to a cyclic ordering c if A isan interval in c.
2. This method is called averaging, or Katona’s method.
Equality in Erdos-Ko-Rado
Want: if A ⊂ X(r) is intersecting (r < n/2) and |A| =(n−1r−1
), then A = {A ∈ X(r) : i ∈ A},
some i.
From proof 2, for each cyclic ordering c, we have r intervals in A. These must be all ofthe r intervals containing a point x(c), say. Our task is to show that x(c) = x(c′) forall c, c′.
Fix a cyclic ordering c, and say x(c) = c0.
Thus {c0, . . ., cr−1}, {c−r+1, . . ., c0} ∈ A, and {c1, . . ., cr}, {c−r, . . ., c−1} /∈ A.
Let c′ be obtained from c by swapping two adjacent elements 6= c0. Say we swap ci andci+1. Wlog (else reflect) i > (n−1)/2. (Not n/2, in case i = (n−1)/2, i+1 = (n+1)/2.)
Then {c0, . . ., cr−1} is an interval of c′, and {c1, . . ., cr} is an interval of c′ (unlessr = (n− 1)/2 and i = (n− 1)/2, in which case {c1, . . ., cr−1, cr+1} is an interval of c′).
But {c0, . . ., cr−1} ∈ A and {c1, . . ., cr} /∈ A (and {c1, . . ., cr−1, cr+1} /∈ A, as disjointfrom {c−r+1, . . ., c0} ∈ A).
Thus x(c′) = c0. Hence x(c) = x(c′) for all c, c′, as required. �
9
Chapter 2 : Isoperimetric Inequalities
“How small can the boundary of a set of given size be?”
E.g., – among subsets of R2 of given area, the disc has smallest perimeter.– among subsets of R3 of given volume, solid sphere has smallest surface area.– among subsets of S2 of given area, circular cap has smallest perimeter.
For a graph G and A ⊂ V (G), the boundary of A is:
b(A) = {x ∈ G : x /∈ A, xy ∈ E(G) for some y ∈ A}.
E.g.,
��
AAs s s s
s s s1 2 3
4 5 6 7
– if A = {1, 2, 5} then b(A) = {3, 4}.
An isoperimetric inequality on G is an inequality of the form: |b(A)| > f(|A|) for allA ⊂ V (G).
Equivalently, minimise the neighbourhood of A, N(A) = A ∪ b(A) = {x : d(x, A) 6 1},where d = usual graph distance. Often a good guess is B(x, r) = {y : d(x, y) 6 r}.
What happens in Qn?
E.g., |A| = 4 in Q3.
�� ��
�� ��
s ss s|b(A)| = 3
�� ��
�� ��
s ss s|b(A)| = 4
Guess: B(∅, r) = X(6r) = X(0) ∪ . . . ∪X(r) are best.
What if |X(6r)| < |A| < |X(6r+1)|? Guess: take A = X(6r) ∪ B, some B ⊂ X(r+1). Thenb(A) = (X(r+1) −B) ∪ ∂+B, so we’d take B an initial segment of lex, by Kruskal-Katona.
This suggests the following. The simplicial ordering on P(X) is defined by: x < y if either|x| < |y|, or |x| = |y| and x < y in lex.
Aim. Initial segments of the simplicial order are best.
Let A ⊂ Qn, and let 1 6 i 6 n.
The i-sections of A are the sets A(i)+ , A
(i)− ⊂ P(X − i) given by
A(i)− = {x ∈ A : i /∈ x}, A
(i)+ = {x− i : x ∈ A, i ∈ x}.
�� ��
�� ��
�� ��
�� ��
↑ i
←− A(i)−
←− A(i)+
The i-compression Ci(A) of A is defined by giving its i-sections:
(Ci(A)
)(i)+
is the initial segment of the simplicial order on P(X − i) of size |A(i)+ |.
(Ci(A)
)(i)− is the initial segment of the simplicial order on P(X − i) of size |A(i)
− |.
10
E.g.,
i ↑ −→Ci
�� ��
�� ��
�� ��
�� ��
�� ��
�� ��
�� ��
�� ��
ss s
s
ss ss
s ss s s s ss
s s
Note that |Ci(A)| = |A|, and also that Ci(A) “looks more like” a Hamming ball than A did.(A Hamming ball is a set A ⊂ Qn with X(6r) ⊂ A ⊂ X(6r+1), some r.)
Say A is i-compressed if Ci(A) = A.
Theorem 1 (Harper’s Theorem). Let A ⊂ Qn and let C be the initial segment of thesimplicial order with |C| = |A|. Then |N(A)| > |N(C)|.
In particular, if |A| >r∑
i=0
(n
i
)then |N(A)| >
r+1∑
i=0
(n
i
).
Remarks. 1. If we knew that A was a Hamming ball, we would be done by Kruskal-Katona.
2. Conversely, Theorem 1 implies Kruskal-Katona: given some B ⊂ X(r), apply Theo-rem 1 to A = X(<r) ∪B.
Proof. Induction on n. Done if n = 1. Given A ⊂ Qn (n > 1), 1 6 i 6 n, claim:
Claim. |N(Ci(A))| 6 |N(A)|.
Proof of claim. Write B for Ci(A). We have |N(A)| = |N(A−)∪A+|+ |N(A+)∪A−|and |N(B)| = |N(B−) ∪B+|+ |N(B+) ∪B−|.
Now, |B+| = |A+|, by definition of B, and |N(B−)| 6 |N(A−)|, by induction.But B+ is an initial segment of the simplicial order, and so is N(B−) (as theneighbourhood of an initial segment is an initial segment). Thus B+ and N(B−)are nested (i.e., one is contained in the other).
And so certainly |N(B−) ∪B+| 6 |N(A−) ∪A+|.
Similarly, we have |N(B+) ∪B−| 6 |N(A+) ∪A−|. Thus |N(B)| 6 |N(A)|.
Define A0, A1, . . . as follows. Set A0 = A. Having chosen A0, . . ., Ak, if Ak is i-compressed for all i then stop the sequence with Ak.
If not, choose i with Ci(Ak) 6= Ak, set Ak+1 = Ci(Ak) and continue. This mustterminate, as
∑x∈Ak
f(x) is decreasing, where f(x) denotes the position of x in thesimplicial order.
Then B = Ak satisfies: |B| = |A|, |N(B)| 6 |N(A)|, and B is i-compressed for all i.
Does B being i-compressed for all i imply that B is an initial segment of the simplicialorder? (If so then we are done, and B = C.)
11
Answer: no, e.g. {∅, 1, 2, 12} ⊂ Q3. However, we are done by Lemma 2 below.
Lemma 2. Let B ⊂ Qn be i-compressed for all i, but not an initial segment of the simplicialorder. Then for n odd, say n = 2k + 1, we have
B = X(6k) − {k + 2, k + 3, . . ., 2k + 1} ∪ {1, 2, 3, . . ., k + 1},
and for n even, say n = 2k, we have
B = X(<k) ∪ {x ∈ X(k) : 1 ∈ x} − {1, k + 2, k + 3, . . ., 2k} ∪ {2, 3, . . ., k + 1}.
(Then done, as in each case we have |N(B)| > |N(C)|.)
Proof. We have some x < y with x /∈ B, y ∈ B. For each i, we cannot have i ∈ x and i ∈ y,as B is i-compressed. Similarly, we cannot have i /∈ x and i /∈ y. Thus x = yc.
So for each x /∈ B, we have at most one later point y ∈ B (namely xc), and for eachy ∈ B, we have at most one earlier point x /∈ B (namely yc).
So B = {z : z 6 y} − {x}, where x is the predecessor of y, and x = yc.
If n is odd, we must have x the last n−12 set.
If n is even, we must have x the last n2 set containing 1. So done. �
Notes. 1. Can also prove Harper’s Theorem by UV -compressions.
2. Can also use these “codimension-1” compressions to prove Kruskal-Katona directly.
For A ⊂ Qn, the t-neighbourhood of A is N t(A) = A(t) = {x ∈ Qn : d(x, A) 6 t}.
Corollary 3. Let A ⊂ Qn with |A| >r∑
i=0
(n
i
). Then for 1 6 t 6 n− r, |N t(A)| >
r+t∑
i=0
(n
i
).
Proof. Theorem 1, plus induction. �
To get a feel for the strength of Corollary 3, we’ll needsome estimates on
∑ri=0
(ni
), etc.
-
6(ni
)
in/2↑( 1
2− ǫ)n
Proposition 4. Let 0 < ǫ < 1/4. Then
⌊( 12−ǫ)n⌋∑
i=0
(n
i
)6
1
ǫe−ǫ2n/2 2n.
Note. This is an exponentially small fraction of 2n (for ǫ fixed, n→∞).
(“This is ∼ ǫ√
n standard deviations from the mean n/2.”)
12
Proof.
(n
i− 1
)=
(n
i
)i
n− i + 1. So for i 6 ⌊(1
2 − ǫ)n⌋, we have
(n
i−1
)(ni
) =i
n− i + 16
(12 − ǫ)n
(12 + ǫ)n
=12 − ǫ12 + ǫ
= 1− 2ǫ12 + ǫ
6 1− 2ǫ.
So,
⌊( 12−ǫ)n⌋∑
i=0
(n
i
)6
1
2ǫ
(n
⌊(12 − ǫ)n⌋
)(sum of a GP).
Similarly,
(n
⌊(12 − ǫ)n⌋
)6
(n
⌊(12 − ǫ
2 )n⌋
)(1− 2
ǫ
2
)ǫn/2−1
(same argument, ǫ→ ǫ/2)
which is 6 2n 2 (1− ǫ)ǫn/2 6 2n 2 e−ǫ2n/2, as 1− ǫ 6 e−ǫ.
Thus,
⌊( 12−ǫ)n⌋∑
i=0
(n
i
)6
1
2ǫ2 e−ǫ2n/2 2n. �
Theorem 5. Let A ⊂ Qn, 0 < ǫ < 1/4. Then|A|2n
>1
2⇒ |A(ǫn)|
2n> 1− 2
ǫe−ǫ2n/2.
“ 12 -sized sets have exponentially large ǫn-neighbourhoods.”
Proof. It is enough to show that, assuming ǫn is an integer, we have|A(ǫn)|
2n> 1− 1
ǫe−ǫ2n/2.
We have |A| >⌈n/2−1⌉∑
i=0
(n
i
), so by Harper we have |A(ǫn)| >
⌈n/2+ǫn−1⌉∑
i=0
(n
i
).
So |(A(ǫn))c| 6
n∑
i=⌈n/2+ǫn⌉
(n
i
)=
⌊n/2−ǫn⌋∑
i=0
(n
i
)6
1
ǫe−ǫ2n/2 2n. �
Remark. The above is concerned with 12 -sized sets, but the same argument would show
that|A|2n
>1
ǫe−ǫ2n/2 ⇒ |A(2ǫn)|
2n> 1− 1
ǫe−ǫ2n/2.
Concentration of Measure
Say f : Qn → R is Lipschitz if |f(x)− f(y)| 6 1 for all x, y adjacent.
A real number M is a median or Levy mean for f if |{x : f(x) 6 M}| > 2n−1 and|{x : f(x) > M}| > 2n−1.
We are now ready to show that “every well-behaved function on Qn is roughly constantnearly everywhere”.
13
Theorem 6. Let f be a Lipschitz function on Qn, with median M . Then
|{x : |f(x)−M | 6 ǫn}|2n
> 1− 4
ǫe−ǫ2n/2 (0 < ǫ < 1/4).
“If you look at the cube, all you see is the ‘middle’ layer.”
Note. This is the “concentration of measure” phenomenon.
Proof. Let A = {x : f(x) 6 M}. Then|A|2n
>1
2, so|A(ǫn)|
2n> 1− 2
ǫe−ǫ2n/2.
But x ∈ A(ǫn) ⇒ f(x) 6 M + ǫn (as f Lipschitz).
And so|{x : f(x) 6 M + ǫn}|
2n> 1− 2
ǫe−ǫ2n/2.
Similarly,|{x : f(x) > M − ǫn}|
2n> 1− 2
ǫe−ǫ2n/2. �
Let G be a graph of diameter D. (I.e., D = max{d(x, y) : x, y ∈ G}.)
Define α(G, ǫ) = max
{1− |A(ǫD)|
|G| : A ⊂ G,|A||G| >
1
2
}.
So “α(G, ǫ) small” says: 12 -sized sets have big ǫD-neighbourhoods.
A sequence G1, G2, . . . of graphs is a Levy family if α(Gn, ǫ)→ 0 as n→∞, for each ǫ > 0.
So, e.g., Theorem 5 tells us that (Qn)∞n=1 is a Levy family, and even a normal Levy family(meaning α(Gn, ǫ) is exponentially small, for any ǫ).
So we again have concentration of measure (“Lipschitz functions on Gn are almost constantnearly everywhere”) for any Levy family. It turns out that many natural families of graphsare Levy families. For example, the permutation groups Sn (made into a graph by: σ adjacentto τ if σ−1τ is a transposition) form a Levy family.
Similarly, we can define α(S, ǫ) for any metric measure space S (of finite diameter and finitemeasure), so we can again define Levy families. It turns out that many natural families ofmetric spaces form Levy families.
Example. The sphere Sn. Two ingredients.
1. Isoperimetric inequality in Sn: |A| = |C| ⇒ |A(ǫ)| > |C(ǫ)| where C is a circularcap. To prove this, we can use compressions, e.g. analogue of ij-compressions.
����q
q?“Stamp on your set” – i.e., replace by the bottom point, if possible.(2-point symmetrisation)
2. Estimate: 12 -sized circular cap has angle π
2 , so its ǫ-neighbourhood is a circular cap
of angle π2 + ǫ. But the surface area of everything else,
∫ π/2
ǫ cosn t dt → 0 as n → ∞,for any fixed ǫ.
We deduced concentration of measure from isoperimetric estimates. Conversely:
14
Proposition 7. Let G be a graph such that for every Lipschitz f : G → R of median M ,we have
|{x ∈ G : |f(x)−M | > t}||G| 6 α (some fixed t, α).
Then|A||G| >
1
2⇒ |A(t)|
|G| > 1− α.
Proof. Let f(x) = d(x, A). Then f is Lipschitz, and has 0 as a median (as |A| > 12 |G|). �
Edge-Isoperimetric Inequalities
For a graph G, A ⊂ V (G), the edge-boundary of A is
∂eA = ∂A = {xy ∈ E(G) : x ∈ A, y /∈ A}.
E.g.,
��
AAs s s s
s s s1 2 3
4 5 6 7
– if A = {1, 2, 5} then ∂A = {14, 23, 45}.
An edge-isoperimetric inequality on G is an inequality of the form A ⊂ G, |A| = m ⇒|∂A| > |f(m)|.
In the cube
E.g., |A| = 4 in Q3.
��
��
s ss s|∂A| = 6
s ss s|∂A| = 4
This suggests that subcubes are best.
The binary ordering on Qn is given by: x < y if max(x△y) ∈ y.
Equivalently, if∑
i∈x 2i <∑
i∈y 2i. “Go up in subcubes.”
Aim. Initial segments of binary minimise ∂.
For A ⊂ Qn and 1 6 i 6 n, the i-binary-compression Bi(A) is defined by giving itsi-sections:
(Bi(A)
)(i)+
is the initial segment of binary on P(X − i) of size A(i)+ .
(Bi(A)
)(i)− is the initial segment of binary on P(X − i) of size A
(i)− .
E.g.,
i ↑ −→Ci
�� ��
�� ��
�� ��
�� ��
�� ��
�� ��
�� ��
�� ��
sss sss
s ss s
s s s s
15
Clearly |Bi(A)| = |A|. Say A is i-binary-compressed if Bi(A) = A.
Theorem 8 (Edge-isoperimetric inequality in the cube). Let A ⊂ Qn, and let C bethe initial segment of binary with |C| = |A|. Then |∂C| 6 |∂A|. In particular, |A| =2k ⇒ |∂A| > (n− k)2k.
Remark. Sometimes called “the theorem of Harper, Lindsey, Bernstein and Hart”.
Proof. Induction on n. Done if n = 1. Given A ⊂ Qn, 1 6 i 6 n.
Claim. |∂Bi(A)| 6 |∂A|.
Proof of claim. Write B for Bi(A).
Have |∂A| = |∂(A−)|+ |∂(A+)|+ |A+△A−|.↑ ↑ ↑
edges inthe bottom
edges inthe top
edgesacross
Similarly, |∂B| = |∂(B−)|+ |∂(B+)|+ |B+△B−|.
Now, |∂(B−)| 6 |∂(A−)| and |∂(B+)| 6 |∂(A+)|, by induction.
Also, |B+△B−| 6 |A+△A−|, because |B+| = |A+|, |B−| = |A−|, and the sets B+,B− are nested, as each is an initial segment of the binary order. Thus |∂B| 6 |∂A|.
Define A0, A1, . . . as follows. Set A0 = A. Having chosen A0, . . ., Ak, if Ak is i-binary-compressed for all i, then stop. If not, choose i with Bi(Ak) 6= Ak and setAk+1 = Bi(Ak). This must terminate, e.g. because
∑x∈Ak
(position of x in binary) isdecreasing.
The final set B = Ak satisfies: |B| = |A|, |∂B| 6 |∂A|, and B is i-binary-compressedfor all i. As before, B need not be an initial segment, e.g. {∅, 1, 2, 3} ⊂ Q3. But weare done by Lemma 9 below.
Lemma 9. Let B ⊂ Qn be i-binary-compressed for all i, but not an initial segment of binary.Then B = P(X − n) ∪ {n} − {1, 2, 3, . . ., n− 1}.
(Then done, as certainly |∂B| > |∂C| in this case.)
Proof. Have some x < y with x /∈ B, y ∈ B. Then for each i, cannot have i ∈ x, i ∈ y, ori /∈ x, i /∈ y, as B is i-binary-compressed. So x = yc.
So for each x /∈ B, at most one y > x has y ∈ B (namely xc), and for each y ∈ B, atmost one x < y has x /∈ B (namely yc).
Thus B = {z : z 6 y} − {x}, where x is the predecessor of y, and x = yc.
Hence y = {n}, as required. (“As n changes hands only once.”) �
Remark. Vital, in the above proof, that the extremal sets in dimension n − 1 were nested
(i.e., given by initial segments of some ordering).
16
For a graph G, the isoperimetric number of G is: i(G) = min
{ |∂A||A| : A ⊂ G,
|A||G| 6
1
2
}.
“How small can the average out-degree be?”
Corollary 10. i(Qn) = 1.
Proof. The set A = P([n − 1]) show i(Qn) 6 1. To show i(Qn) > 1, let C be any initialsegment of binary with |C| 6 2n−1. Then C ⊂ P([n− 1]), so certainly |∂C| > |C|. �
Inequalities in the Grid
The grid is the graph on [k]n ={(x1, . . ., xn) : xi ∈ {1, . . ., k} ∀ i
}, in
which x = (x1, . . ., xn) is joined to y = (y1, . . ., yn) if for some i we have|xi − yi| = 1 and xj = yj for all j 6= i. (“l1-norm”)
Note. For k = 2, this is exactly the graph Qn.[4]2
Do Theorem 1 and Theorem 8 extend to the grid?
Vertex-isoperimetric inequality in the grid
@@@d
d
b(A) ∼ d
∼√
2|A|q q q q q qq q q q qq q q qq q qq qq versus
d
d
b(A) ∼ 2d
∼ 2√|A|q q q q q qq q q q q qq q q q q qq q q q q qq q q q q qq q q q q q
Good guess: sets of the form {x : |x| 6 r} are best, where |x| = x1 + . . . + xn.
What if |{x : |x| 6 r}| < |A| < |{x : |x| 6 r +1}|? We’d take A of the form {x : |x| 6 r}∪B,some B ⊂ {x : |x| = r + 1}.
Define the simplicial ordering on [k]n by: x < y if either |x| < |y|,or |x| = |y| and xi > yi where i = min{j : xj 6= yj}.
��
��
@@@D
D`↓x1 small
ւ x1 big
E.g., on [3]2: (1, 1), (2, 1), (1, 2), (3, 1), (2, 2), (1, 3), (3, 2), (2, 3), (3, 3).
on [4]3: (1, 1, 1), (2, 1, 1), (1, 2, 1), (1, 1, 2), (3, 1, 1), (2, 2, 1),(2, 1, 2), (1, 3, 1), (1, 2, 2), (1, 1, 3), (4, 1, 1), . . . r r r
r r rr r r
-@@IHHHHj@@I
@@IHHHHj@@I
-
Note. This agrees with the previous definition of the simplicial ordering for k = 2.
Aim. Initial segments of simplicial are best for vertex-isoperimetric.
Let A ⊂ [k]n. For 1 6 i 6 n, the i-sections of A are the sets A1, . . ., Ak (or A(i)1 , . . ., A
(i)k ) in
[k]n−1 given by:
At ={x = (x1, . . ., xn−1) ∈ [k]n−1 : (x1, . . ., xi−1, t, xi, xi+1, . . ., xn−1) ∈ A
}.
17
The i-compression of A is the set Ci(A) ⊂ [k]n defined by:
(Ci(A)
)t
is the initial segment of simplicial on [k]n−1 of size |At|.
Say A is i-compressed if Ci(A) = A.
Theorem 11. Let A ⊂ [k]n, and let C be the initial segment of simplicial with |C| = |A|.
Then |N(C)| 6 |N(A)|.
In particular, |A| >∣∣{x : |x| 6 r}
∣∣ ⇒ |N(A)| >∣∣{x : |x| 6 r + 1}
∣∣.
(Called the “vertex-isoperimetric inequality in the grid”.)
Proof. Induction on n. For n = 1: for any A ⊂ [k]1 with A 6= ∅, [k]1, we have |N(A)| >
|A|+ 1 = |N(C)|.
Given A ⊂ [k]n, fix 1 6 i 6 n.
Claim. |N(Ci(A))| 6 |N(A)|.
Proof of claim. Write B for Ci(A).
For any 1 6 t 6 k, have N(A)t = N(At) ∪At−1 ∪At+1 (taking A0 = Ak+1 = ∅).↑ ↑ ↑
ownlayer
below above
So |N(A)| = ∑t
∣∣N(At) ∪At−1 ∪At+1
∣∣.
Similarly, |N(B)| = ∑t
∣∣N(Bt) ∪Bt−1 ∪Bt+1
∣∣.
But |Bt−1| = |At−1|, |Bt+1| = |At+1|, and |N(Bt)| 6 |N(At)|, by induction.
Also, the sets N(Bt), Bt−1, Bt+1 are nested, as each is an initial segment ofsimplicial on [k]n−1.
Thus∣∣N(At) ∪At−1 ∪At+1
∣∣ >∣∣N(Bt) ∪Bt−1 ∪Bt+1
∣∣, and so |N(A)| > |N(B)|.
Among all B ⊂ [k]n with |B| = |A| and |N(B)| 6 |N(A)|, choose one for which∑x∈B (position of x in simplicial) is minimal. Then B is i-compressed for all i (else
Ci(B) contradicts the minimality of B).
So it remains to show that |N(B)| > |N(C)|.
Case 1: n = 2.
B is i-compressed for all i iff B is a down-set. (I.e., if x ∈ B and y has yi 6 xi for alli, then y ∈ B. “Closed under going left and going down.”)
Let s = max{|x| : x ∈ B} and r = min{|x| : x /∈ B}.
18
Then r 6 s (else B = C).
If r = s, then {x : |x| < r} ⊂ B ⊂ {x : |x| 6 r},so certainly |N(B)| > |N(C)|. @
@@q q q q qq q q qq q q qq q qqq
B
r = s
@@@q q q q q qq q q q qq q q qq qq
C
If r < s, cannot have {x : |x| = s} ⊂ B, as B a down-set(and ∃x, |x| = r, x /∈ B). Hence there are x, x′ such that|x| = |x′| = s, x ∈ B, x′ /∈ B, and x = x′ ± (e1 − e2), whereei = (0, . . ., 0, 1, 0, . . ., 0). (I.e., x, x′ are adjacent.)
Similarly, cannot have {y : |y| = r} ⊂ Bc, so again there arey, y′ with |y| = |y′| = r, y ∈ B, y′ /∈ B, and y = y′ ± (e1 − e2).
@@
@ @@
@q a a qyy′
xx′
r
s
qz
But now let B′ = B ∪ y′ − x. Then |N(B′)| 6 |N(B)| (as we have lost > 1 pointfrom the neighbourhood in level s + 1 (e.g. z) and gained 6 1 point in level r + 1),contradicting the choice of B.
Case 2: n > 3.
For x ∈ B with xn > 2, have x − en + ei ∈ B for all i with xi < k (as B is j-compressed for any j 6= n, i). Thus N(Bt) ⊂ Bt−1 for all t = 2, . . ., k. We hadN(B)t = N(Bt) ∪Bt+1 ∪Bt−1. So we actually have N(B)t = Bt−1.
So |N(B)| = |Bk−1|+ |Bk−2|+ . . . + |B1|+ |N(B1)| = |B| − |Bk|+ |N(B1)|.↑ ↑ ↑ ↑
height k height k−1 height 2 height 1
Similarly, have |N(C)| = |C| − |Ck|+ |N(C1)|.
Thus, to complete the proof, it is sufficient to show that |Bk| 6 |Ck| and |B1| > |C1|.
• |Bk| 6 |Ck|
Define a set D ⊂ [k]n by: Dk = Bk, and Dt = N(Dt+1) for t = k − 1, k − 2, . . ., 1.
Then D is an initial segment of simplicial, and D ⊂ B. So |D| 6 |B| = |C|, so D ⊂ C(as D, C are initial segments of simplicial). Thus Dk ⊂ Ck, so |Dk| 6 |Ck|.
• |B1| > |C1|
Define a set E ⊂ [k]n by: E1 = B1, and Et = {x ∈ [k]n−1 : N({x}) ⊂ Et−1} fort = 2, 3, . . ., k.
Then E is an initial segment of simplicial, and E ⊃ B. So |E| > |B| = |C|, so E ⊃ C.Thus E1 ⊃ C1, so |E1| > |C1|. �
Corollary 12. Let A ⊂ [k]n with |A| >∣∣{x : |x| 6 r}
∣∣. Then |A(t)| >∣∣{x : |x| 6 r + t}
∣∣. �
Remark. Can check from this that for any fixed k, the sequence ([k]n)∞n=1 is a normal Levy
family.
19
Edge-isoperimetric inequality in the grid
To minimise |∂A| for |A| given.
n = 2.
rq q q q q qq q q q q qq q q q q qq q q q q qq q q q q qq q q q q q versus
@@@
rq q q q q qq q q q qq q q qq q qq qq
First has size r2, second has size ∼ 12r2,
and both have boundary 2r.
This suggests that squares are best. But:
q q q qq q q qq q q qq q q q →
12k
q q q q q q qq q q q q q qq q q q q q qq q q q q q qq q q q q q qq q q q q q qq q q q q q q =
14k
q q q qq q q qq q q qq q q qq q q qq q q qq q q qq q q qq q q qq q q qq q q qq q q qq q q q
→
34k
q q q q q q q q q qq q q q q q q q q qq q q q q q q q q qq q q q q q q q q qq q q q q q q q q qq q q q q q q q q qq q q q q q q q q qq q q q q q q q q qq q q q q q q q q qq q q q q q q q q qq q q q q q q q q qq q q q q q q q q qq q q q q q q q q q
=
12k
q q q q q q qq q q q q q qq q q q q q qq q q q q q qq q q q q q qq q q q q q q
q q q q q q q q q q q q qq q q q q q q q q q q q qq q q q q q q q q q q q qq q q q q q q q q q q q qq q q q q q q q q q q q qq q q q q q q q q q q q qq q q q q q q q q q q q q→
q q q q q q q q q qq q q q q q q q q qq q q q q q q q q q
q q q q q q q q q q q q qq q q q q q q q q q q q qq q q q q q q q q q q q qq q q q q q q q q q q q qq q q q q q q q q q q q qq q q q q q q q q q q q qq q q q q q q q q q q q qq q q q q q q q q q q q qq q q q q q q q q q q q qq q q q q q q q q q q q q
Sadly, the extremal sets are not nested: there are “phase transitions” at size k2/4 and 3k2/4.(So we can’t order them, which seems to rule out compressions.)
n = 3. Get: cube [a]3 −→ square column [a]2 × [k] −→ halfspace [a]× [k]2
−→ complement of square column −→ complement of cube.
Aim. Best to take [a]d × [k]n−d (some d) or complements.
Observe that if A = [a]d × [k]n−d then |∂A| = dad−1kn−d = d|A|1−1/dkn/d−1.
Theorem 13. Let A ⊂ [k]n with |A| 6 kn/2.
Then |∂A| > min{d|A|1−1/dkn/d−1 : d = 1, 2, . . ., n
}.
(The “edge-isoperimetric inequality in the grid”.)
** Non-examinable section **
Proof (sketch). Wlog, A is a down-set in [k]n. (Stamp on A – i.e., apply 1-dimensionalcompressions.)
For 1 6 i 6 n, define Ci(A) by giving its i-sections:
Ci(A)t is extremal in [k]n−1, with |Ci(A)t| = |At|.↑
i.e., of form [a]d × [k]n−1−d, or complement
Write B for Ci(A).
Then |∂A| = ∑ |∂At|+ |A1| − |Ak|.↑ տ ր
horizontal vertical, as A is a down-set
And |∂B| =∑ |∂Bt| + ? – as extremal sets in [k]n−1 are not nested.
20
Indeed, we can clearly have |∂B| > |∂A|.
Try to introduce a ‘fake boundary’, ∂′. We would want ∂′ 6 ∂, with equality for ourextremal sets, and ∂′Ci(A) 6 ∂′A for all A.
Obvious guess: ∂′A =∑ |∂At| + |A1| − |Ak|. (Note ∂′ = ∂ on down-sets.) So we do
have ∂′B 6 ∂′A, where B = Ci(A). But it fails for Cj , j 6= i.
(Could try ∂′′A =∑n
i=1 |A(i)1 | − |A
(i)k | – cannot work, e.g. because the ‘outside shell’ of
[k]n would have ∂′′ = 0.)
We know: |∂A| = ∂′A > ∂′B =∑ |∂Bt|+ |B1| − |Bk| =
∑f(|Bt|) + |B1| − |Bk|, where
f is the extremal function in dimension n− 1.
Now, f(x) is minimum of some functions of theform c.x1−1/d and c(kn−1−x)1/1−d. Each of theseis concave, so their pointwise minimum f is alsoconcave.
x1−1/d
(kn−1 − x)1−1/d
Consider varying |B2|, . . ., |Bk−1|, keeping |B2| + . . . + |Bk−1| fixed, and such that|B1| > |B2| > . . . > |Bk−1| > |Bk|.
By concavity of f , have ∂′B > ∂′C, where for some λ we have Ct =
{B1 if t 6 λBk it t > λ
.
We have |∂A| = ∂′A > ∂′B > ∂′C, with ∂′C = λf(|B1|) + (k− λ)f(|Bk|) + |B1| − |Bk|(but C still not a down-set).
Now consider varying |B1| and |Bk| (λ fixed), keeping λ|B1| + (k − λ)|Bk| fixed,and keeping |B1| > |Bk|. Again, this is a concave function of |B1| (as it is con-cave+concave+linear), so by concavity of f , we have ∂′C > ∂′D, where either
Dt = D1 ∀ t, or Dt =
{D1 ∀ t 6 λ∅ ∀ t > λ
, or Dt =
{[k]t−1 ∀ t 6 λDk ∀ t > λ
.
D1
D1
D1
D1
D1
D1
{λ
D1
D1
D1
∅{
λ
Dk
Dk
Dk
[k]n−1
[k]n−1
[k]n−1
Thus |∂A| = ∂′A > ∂′B > ∂′C > ∂′D = |∂D|. (Miraculously, D is a down-set.)
So, take as our i-compression the map A 7→ D.
Then finish proof as usual. �
Remark. To make the argument precise, work in the continuous cube [0, 1]n instead, andthen pass to discrete at the end.
** End of non-examinable section **
21
Very few isoperimetric inequalities are known (even approximately).
For example, what about r-sets? Take the graph on X(r) with x, y joined if |x ∩ y| = r − 1(i.e., d(x, y) = 2 in Qn). No good isoperimetric inequality is known.
Most interesting case is r = n/2. It’s conjectured that balls are best – i.e., sets of the form{x ∈ X(r) : d(x, x0) 6 d}, i.e., of the form
{x ⊂ [2r] : |x ∩ [r]| > k
}.
Unknown!
22
Chapter 3 : Intersecting Families
t-intersecting families
Say A ⊂ P(X) is t-intersecting if |x ∩ y| > t for all x, y ∈ A.
How large can A be?
E.g., t = 2. Could take {x : 1, 2 ∈ x} – size 14 .2n. Beaten by {x : |x| > n
2 + 1} – size ∼ 12 .2n.
Theorem 1 (Katona’s t-intersecting theorem). Let A ⊂ P(X) be t-intersecting, where
n + t is even. Then |A| 6∣∣∣X(> n+t
2 )∣∣∣.
Proof. For any x, y ∈ A, have d(x, yc) > t. Thus writing A = {xc : x ∈ A}, we haved(A, A) > t. I.e., A(t−1) is disjoint from A.
Suppose |A| >∣∣∣X(> n+t
2 )∣∣∣. Then |A(t−1)| >
∣∣∣X(>n+t2
−(t−1))∣∣∣ =
∣∣∣X(> n−t2
+1)∣∣∣.
↑by Harper
But∣∣A∣∣ >
∣∣∣X(6 n−t2
)∣∣∣, contradicting A(t−1) ∩A = ∅. �
What about A ⊂ X(r) being t-intersecting? (1 6 t 6 r)
Could guess A0 = {x ∈ X(r) : 1, 2, . . ., t ∈ x}.
Could also try, for 1 6 α 6 r − t, the set Aα ={x ∈ X(r) : |x ∩ {1, 2, . . ., t + 2α}| > t + α
}.
E.g., t = 2, in [7](4): |A0| =(
5
2
)= 10, |A1| = 1 +
(4
3
)(3
1
)= 13, |A2| =
(6
4
)= 15.
in [8](4): |A0| =(
6
2
)= 15, |A1| = 1 +
(4
3
)(4
1
)= 17, |A2| =
(6
4
)= 15.
in [9](4): |A0| =(
7
2
)= 21, |A1| = 1 +
(4
3
)(5
1
)= 21, |A2| =
(6
4
)= 15.
As |A0| is quadratic (in n), |A1| is linear, and |A2| is constant, we will have A0 winning if nis large.
Theorem 2. Let A ⊂ X(r) be t-intersecting, where r, t are fixed, 1 6 t 6 r.
Then |A| 6 |A0| =(
n− t
r − t
), if n is sufficiently large.
Remarks. 1. Sometimes called the “second Erdos-Ko-Rado Theorem”.
2. Bound on n is (16r)r (crude) or 2tr3 (careful).
23
Proof. “Idea: A0 has r − t degrees of freedom.”
Wlog A is maximal t-intersecting, and so there are x, y ∈ A with |x ∩ y| = t (or elsefor all x ∈ A and all i ∈ x, j /∈ x, have x ∪ j − i ∈ A by maximality, hence A = X(r) –contradiction).
Cannot have x ∩ y ⊂ z for all z ∈ A (else |A| 6 |A0|, so done). So choose z ∈ A withz 6⊃ x ∩ y. Then for any w ∈ A, we must have |w ∩ (x ∪ y ∪ z)| > t + 1.
So |A| 6 23r
((n
r − t− 1
)+
(n
r − t− 2
)+ . . . +
(n
0
)),
↑ տ ↑ րw on x ∪ y ∪ z w off x ∪ y ∪ z
which is a polynomial of degree r − t− 1, hence < |A0| for n large. �
Remark. Frankl Conjecture was: A ⊂ X(r) t-intersecting ⇒ |A| 6 max(|A0|, . . ., |Ar−t|)(any n). Proved by Ahslwede and Khachatrian in 1998.
Modular intersections
So far, we have banned |x ∩ y| = 0. What if we instead banned |x ∩ y| ≡ 0 mod something?
E.g., suppose r is odd, and we want A ⊂ X(r) with |x∩y| odd for all x, y ∈ A. We can achieve
|A| =(⌊n−1
2 ⌋r−12
)by taking all sets x of the form: 1, together with r−1
2 of the pairs 23, 45, . . .
What if (still for r odd) we wanted |x ∩ y| even for all x, y ∈ A (x 6= y). We can achieven− r + 1 by taking all x containing 1, 2, . . .r − 1. Amazingly:
Proposition 3. Let r be odd and let A ⊂ X(r) have |x ∩ y| even for all x, y ∈ A (x 6= y).Then |A| 6 n.
Idea. Try to find |A| linearly independent vectors in an n-dimensional vector space.
Proof. View each point x ∈ Qn as a point x in Zn2 , the n-dimensional vector space over the
field Z2. (E.g., {1, 3, 4} ↔ 10110. . .0.) Consider {x : x ∈ A}.
Have 〈x, x〉 = 1 for all x ∈ A (where 〈, 〉 is the usual dot product), and 〈x, y〉 = 0 for allx, y ∈ A, x 6= y. Hence {x : x ∈ A} is linearly independent. (If
∑λixi = 0, dot with
xi to get λi = 0.)
So∣∣{x : x ∈ A}
∣∣ 6 n. �
For r even? If |x ∩ y| even for all distinct x, y ∈ A ⊂ X(r), we can have |A| >(⌊n/2⌋
r/2
), by
taking r/2 of the pairs 12, 34, . . .
If |x ∩ y| is odd for all distinct x, y ∈ A ⊂ X(r), then we must have |A| 6 n + 1 – else addthe point n + 1 to each x ∈ A to contradict Proposition 3.
24
So banning |x ∩ y| ≡ r (mod 2) forces |A| to be small. Does this generalise?
We now show that “s allowed intersections mod p ⇒ |A| 6(ns
)”.
Theorem 4 (Frankl-Wilson Theorem). Let p be prime, and let A ⊂ X(r) be such thatfor all distinct x, y ∈ A we have |x ∩ y| ≡ λi (mod p), some i, where λ1, . . ., λs (s 6 r)are integers, with none ≡ r (mod p).
Then |A| 6(
n
s
).
Remarks. 1. The bound is a polynomial in n, independent of r (!)
2. Bound is essentially best possible. Take A ={x ∈ X(r) : [r − s] ⊂ x
}.
This has |A| =(
n− r + s
s
)∼(
n
s
). (Ratio → 1 as n→∞.)
3. No polynomial bound if we allow |x ∩ y| ≡ r (mod p).
Indeed, write r = a + λp (0 6 a 6 p− 1). (← a→)| {z }
(← p→)(← p→)(← p→). . .| {z }
take this take λ of these
Can have |x ∩ y| ≡ r (mod p) for all x, y ∈ A,
with |A| =(⌊n−a
p ⌋λ
). But the degree grows as λ grows.
Idea. Try to find |A| linearly independent points in a vector space of dimension(
ns
), by
somehow “applying the polynomial (t− λ1). . .(t− λs) to the values of |x ∩ y|”.
Proof. For i 6 j, let M(i, j) be the(ni
)×(nj
)matrix, with rows indexed by X(i) and columns
indexed by X(j), with M(i, j)xy =
{1 if x ⊂ y0 if x 6⊂ y
.
The matrix M(s, r) has(ns
)rows. Let V be the vector space (over R) spanned by those
rows. Thus dimV 6(ns
).
Consider the matrix M(i, s)M(s, r), where i 6 s. For x ∈ X(i) and y ∈ X(r),(M(i, s)M(s, r)
)xy
= the number of z ∈ X(s) with x ⊂ z ⊂ y
=
{ (r−is−i
)if x ⊂ y
0 otherwise
=
(r − i
s− i
)M(i, r)xy.
Thus M(i, s)M(s, r) =(r−is−i
)M(i, r), and the rows of M(i, r) belong to V (as pre-
multiplying is just taking linear combinations of the rows).
Consider M(i) = M(i, r)T M(i, r) – again, must have all rows in V . For x, y ∈ X(r),
M(i)xy = the number of z ∈ X(i) with z ⊂ x, z ⊂ y
=
(|x ∩ y|i
).
25
Write the integer polynomial (t − λ1). . .(t − λs) as∑s
i=0 ai
(ti
), for some integers ai.
(This is possible, as t(t− 1). . .(t− i + 1) = i!(ti
).)
Let M =∑s
i=0 aiM(i). All rows are in V . Then
Mxy =
s∑
i=0
ai
(|x ∩ y|i
)= (|x ∩ y| − λ1). . .(|x ∩ y| − λs)
{≡ 0 (mod p) if |x ∩ y| ≡ some λi (mod p)6≡ 0 (mod p) otherwise
.
Now look at the submatrix whose rows and columns are indexed by A.
It is
6≡ 0 O
. . .
O 6≡ 0
.
Rows are integer-valued, and are linearly independent over Zp, so also over Z, so alsoover Q, so also over R.
Thus |A| 6 dimV 6
(n
s
). �
Remark. We do need p prime. Grolmusz constructed, for each n, a value of r ≡ 0 (mod 6)and A ⊂ X(r) with |x ∩ y| 6≡ 0 (mod 6) for all distinct x, y ∈ A, with |A| > nc log n/ log log n,some c > 0. (Not a polynomial in n.)
Corollary 5. Let A ⊂ X(r) with |x ∩ y| 6≡ r (mod p) for all distinct x, y ∈ A (some prime
p < r). Then |A| 6(
n
p− 1
).
Proof. We are allowed p− 1 values of |x ∩ y| (mod p). �
Two n2 -sets from [n] intersect (on average) in about n/4 points. But an intersection size of
exactly n/4 is very unlikely. However, amazingly:
Corollary 6. Let p be prime, and let A ⊂ [4p](2p) be such that |x ∩ y| 6= p for all x, y ∈ A.
Then |A| 6 2
(4p
p− 1
).
(“Not much of a constraint.”)
Note.(
4pp−1
)is extremely small: have
(n
n/4
)6 2 e−n/32 2n, so
(n
n/4
)is an exponentially small
fraction of(
nn/2
)∼ c√
22n.
Proof. Halving |A| if necessary, we may assume that we never have x, xc ∈ A. Hence, fordistinct x, y ∈ A, we have |x ∩ y| 6= 0, p, so |x ∩ y| 6≡ 0 (mod p).
Thus |A| 6(
4p
p− 1
)by Corollary 5. �
26
Borsuk’s Conjecture
Suppose S is a bounded set in Rn. How many pieces do we need to break S into such thateach piece has smaller diameter than S? Taking S ⊂ Rn to be a regular simplex (i.e., n + 1equally-spaced points), it’s clear that we may need as many as n + 1 pieces.
Borsuk’s Conjecture. n + 1 pieces is always possible.
Known for n = 1, 2, 3, and also for S ⊂ Rn being a smooth convex body or a symmetricconvex body (i.e., x ∈ S ⇒ −x ∈ S).
In fact, Borsuk’s Conjecture is massively false.
Theorem 7 (Kahn, Kalai). For all n, there exists a bounded set S ⊂ Rn such that tobreak S into pieces of smaller diameter, we need > c
√n pieces, for some constant c > 1.
Notes. 1. Our proof will give Borsuk false for n > 2000.
2. We’ll prove Theorem 7 for n of the form(4p2
), p prime – then it follows (with a
different c) for all n, e.g. because there exists p with n/2 6 p 6 n.
Proof. We’ll consider S ⊂ Qn ⊂ Rn. In fact S ⊂ [n](r). (“A genuine idea.”)
For x, y ∈ S, we have ||x−y||2 = 2(r−|x∩y|) (where ||·|| is the usual Euclidean distance).Thus we seek S with min |x ∩ y| = k say, but any subset of S with min |x ∩ y| > k ismuch smaller than S.
Let n =(4p2
), where p is prime. Identify [n] with E(K4p) – the edge-set of the complete
graph on {1, . . ., 4p}. For each x ∈ [4p](2p), let Gx be the complete bipartite graph onvertex classes x, xc.
Let S = {Gx : x ∈ [4p](2p)}. So S ⊂ [n](4p2), and |S| = 12
(4p2p
).
We have |Gx ∩Gy| = |x ∩ y|2 + (2p− |x ∩ y|)2= d2 + (2p− d)2, where d = |x ∩ y|.
This is minimal when d = p, i.e. when |x ∩ y| = p.
��
���
���
�
�
�
�
�
�
�
�x xc
Gx
y
yc
Gy
,,
,,
%%
%%l
ll
l
ee
ee
Now let S′ ⊂ S have smaller diameter than S. Say S′ = {Gx : x ∈ A}.
Then we must have |x ∩ y| 6= p for all x, y ∈ A (else diameter of S′ = diameter of S).
So |A| 6 2(
4p2p−1
), by Corollary 6. Thus the number of pieces needed is
>
12
(4p2p
)
2(
4pp−1
) >c 24p/
√p
4e−p/8 24p> (c′)p (some c′ > 1)
> (c′′)√
n (some c′′ > 1). �
27
Chapter 3′ : Projections
“If a set has small projections, must it be small?”
Let A ⊂ P(X) and let Y ⊂ X . The projection or trace of A on Y is A|Y = {x∩Y : x ∈ A}.Thus A|Y ⊂ P(Y ) – ‘project A onto the coordinates corresponding to Y ’.
E.g., if A = {1, 12, 13, 124, 34, 345} and Y = {1, 3}, then A|Y = {1, 13, 3}.
Say A covers or shatters Y if A|Y = P(Y ).
The trace number (or VC-dimension) of A is tr A = max{|Y | : Y shattered by A}.
Given |A|, how small can trA be? Equivalently, how large can |A| be given tr A < k?
We could take A = X(<k). This clearly does not shatter any k-set (as if |Y | = k thenY /∈ A|Y ). Our aim is to show that we cannot do better than |X(<k)|.
The main idea is that this is trivial if A is a down-set (i.e., if whenever x ∈ A and y ⊂ x thenalso y ∈ A), because then if |A| > |X(<k)|, A must contain a set of size > k, and we’re nowdone because A is a down-set.
Next idea: try to compress an arbitrary A into a down-set.
For 1 6 i 6 n, the i-down-compression of A is defined as follows. For x ∈ P(X), let
Di(x) =
{x− {i} if i ∈ xx otherwise
,
and for A ⊂ P(X), let
Di(A) = {Di(x) : x ∈ A} ∪ {x ∈ A : Di(x) ∈ A}.
I.e., we ‘compress A downwards in direction i’. Note that |Di(A)| = |A|. We say that A isi-down-compressed if Di(A) = A.
Remark. Di is a 1-dimensional compression.
Theorem 1 (Sauer-Shelah Lemma). If A ⊂ P(X) with |A| > |X(<k)| then tr A > k.
Proof. Claim. For any A ⊂ P(X) and 1 6 i 6 n, we have tr Di(A) 6 tr A.
Proof of claim. Write A′ for Di(A). Suppose A′ shatters Y . We shall show that Aalso shatters Y . If i /∈ Y then A′|Y = A|Y , and so we are done.
So suppose i ∈ Y . Then for z ⊂ Y with i /∈ z we have z ∪ {i} ∈ A′|Y , so thereexists x ∈ A′ with x∩Y = z∪{i}. But then i ∈ x, so x, x−{i} ∈ A (by definitionof A′). Thus z, z ∪ {i} ∈ A|Y . Hence A|Y = P(Y ).
Set B = Dn(Dn−1(Dn−2(. . .(D1(A)). . .))). Then |B| = |A|, trB 6 tr A, and B is adown-set. But |B| > |X(<k)|, so B contains some k-set, so tr B > k. �
In general, do upper bounds on some projections∣∣A|Yi
∣∣ give us upper bounds on |A|?
28
For example, Sauer-Shelah says that if∣∣A|Y
∣∣ 6 2k − 1 for all k-sets Y , then |A| 6 |X(<k)|.
A brick or box in Rn is a set of the form [a1, b1]× [a2, b2]× · · · × [an, bn], where ai 6 bi forall i. A body S ⊂ Rn is a finite union of bricks. The volume of S is written |S| or m(S).
Remarks. 1. In fact, everything will go through for a general compact S ⊂ Rn.
2. A set system A ⊂ Qn gives a body A ⊂ Rn, namely A =⋃
x∈A Bx, where Bx =
[x1, x1 + 1]× · · · × [xn, xn + 1]. We have |A| = m(A), where for a body S, m(S) = |S|is the volume of S.
For a body S ⊂ Rn and Y ⊂ [n], the projection of S onto the span of {ei : i ∈ Y } is denotedby SY . For example, if S ⊂ R3, then S1 is the projection of S onto the x-axis:
S1 = {x1 ∈ R : (x1, x2, x3) ∈ S for some x2, x3 ∈ R},
and S12 is the projection of S onto the xy-plane:
S12 = {(x1, x2) ∈ R2 : (x1, x2, x3) ∈ S for some x3 ∈ R}.
We have that SA ⊂ R|A|.
What bounds on |S| do we get given bounds on some SY ?
For example, let S be a body in R3. Then trivially |S| 6 |S1||S2||S3| as S ⊂ S1 × S2 × S3.Similarly, |S| 6 |S12||S3| as S ⊂ S12 × S3.
What if |S12| and |S13| are known? This tells us nothing – e.g., S = [0, 1/n]× [0, n]× [0, n].
What if |S12|, |S13| and |S23| are known?
Proposition 2. Let S be a body in R3. Then |S|2 6 |S12||S13||S23|.
Remark. We have equality if S is a brick.
For S ⊂ Rn, the n-sections are the sets S(x) ⊂ Rn−1 for each x ∈ R defined by
S(x) = {(x1, x2, . . ., xn−1) ∈ Rn−1 : (x1, x2, . . ., xn−1, x) ∈ S}.
Proof of Proposition 2. Consider first the case when each 3-section is a square, i.e. whenS(x) = [0, f(x)] × [0, f(x)]. Then |S12| = M2, where M = maxx∈R f(x). Also |S13| =|S23| =
∫f(x) dx, and |S| =
∫f(x)2 dx. Thus we want:
(∫f(x)2 dx
)2
6 M2
(∫f(x) dx
)2
.
But∫
f(x)2 dx 6 M∫
f(x) dx as f(x) 6 M for all x, so this indeed holds.
For the general case, define a body T ⊂ R3 by
T (x) =[0,√|S(x)|
]×[0,√|S(x)|
].
Then |T | = |S| and |T12| 6 |S12|, as |T12| = maxx∈R |T (x)|.
29
Let g(x) = |S(x)1| and h(x) = |S(x)2|, so |S(x)| 6 g(x)h(x). Then
|T13| = |T23| =∫ √
|S(x)| dx 6
∫ √g(x)h(x) dx.
Also, |S13| =∫
g(x) dx and |S23| =∫
h(x) dx. So we will be done (by reduction to thecase of T ) if (∫ √
g(x)h(x) dx
)2
6
(∫g(x) dx
) (∫h(x) dx
),
i.e. ∫ √g(x)h(x) dx 6
(∫g(x) dx
)1/2 (∫h(x) dx
)1/2
,
which is just the Cauchy-Schwarz inequality on the functions√
g(x) and√
h(x). �
We say that sets Y1, Y2, . . ., Yr cover [n] if⋃r
j=1 Yj = [n]. They are a k-uniform cover ifeach i ∈ [n] belong to exactly k of the Yj .
For example, for n = 3: {1}, {2}, {3} is a 1-uniform cover; {1}, {2, 3} is a 1-uniform cover;{1, 2}, {1, 3}, {2, 3} is a 2-uniform cover; {1, 2}, {1, 3} is not uniform.
Our aim is to show that if Y1, Y2, . . ., Yr form a k-uniform cover then |S|k 6 |SY1||SY2
| · · · |SYr |.
Let C = {Y1, Y2, . . ., Yr} be a k-uniform cover of [r]. Note that C is a multiset, i.e. repetitionsare allowed – for example, {1, 1, 2, 3, 23} is a 2-uniform cover of [3].
Put C− = {Yi : n /∈ Yi} and C+ = {Yi − n : n ∈ Yi} as usual, so |C+| = k, and C− ∪ C+ is ak-uniform cover of [n− 1].
Note that if n ∈ Y then |SY | =∫|S(x)Y −n| dx (e.g. if S ⊂ R3 then |S13| =
∫|S(x)1| dx), and
this holds even if Y = [n]. Also, if n /∈ Y then |SY | > |S(x)Y | for all x (e.g. |S1| > |S(x)1|for all x).
In the proof of Proposition 2, we used Cauchy-Schwarz:
∫fg 6
(∫f2
)1/2(∫g2
)1/2
.
Here, we will need Holder’s inequality:
∫fg 6
(∫fp
)1/p (∫gq
)1/q
, for1
p+
1
q= 1.
Whence, iterating, we get:
∫f1. . .fk 6
(∫fk1
)1/k
. . .
(∫fk
k
)1/k
.
Theorem 3 (Uniform Covers Theorem). Let S be a body in Rn, and let C be a k-uniform cover of [n].
Then |S|k 6∏
Y ∈C|SY |.
Proof. The proof is by induction on n. The case n = 1 is trivial. Given a body S ⊂ Rn forn > 2, we have
30
|S| =
∫
x
|S(x)| dx
6
∫
x
∏
Y ∈C+
|S(x)Y |1/k∏
Y ∈C−
|S(x)Y |1/k dx (induction)
6∏
Y ∈C−
|SY |1/k
∫
x
∏
Y ∈C+
|S(x)Y |1/k dx
6∏
Y ∈C−
|SY |1/k∏
Y ∈C+
(∫
x
|S(x)Y | dx
)1/k
(Holder)
=∏
Y ∈C−
|SY |1/k∏
Y ∈C+
|SY ∪n|1/k
=∏
Y ∈C|SY |1/k .
�
Corollary 4 (Loomis-Whitney Theorem). Let S be a body in Rn.
Then |S|n−1 6
n∏
i=1
|S[n]−i|.
Proof. The family [n]− 1, [n]− 2, . . . , [n]− n is an (n− 1)-uniform cover of [n]. �
Remark. The case n = 3 of the Loomis-Whitney theorem is Proposition 2.
Corollary 5. Let A ⊂ Qn, and let C be a k-uniform cover of [n].
Then |A|k 6∏
Y ∈C
∣∣A|Y∣∣.
In particular, if C is a uniform cover with∣∣A|Y
∣∣ 6(2|Y |)c for all y ∈ C then |A| 6 (2n)c.
Proof. For the first part, apply Theorem 3 to the body
A =⋃
x∈A
[x1, x1 + 1]× · · · × [xn, xn + 1].
Then m(A) = |A| and m(A|Y ) =∣∣A|Y
∣∣ for all Y .
For the second part, suppose that C is a k-cover. Then
|A|k 6∏
Y ∈C
∣∣A|Y∣∣ 6
∏
Y ∈C
(2|Y |)c =
(2
P
Y ∈C|Y |)c
=(2kn)c
.�
Our next aim is the ‘Bollobas-Thomason box theorem’, that for any body S ⊂ Rn there is abox B ⊂ Rn with |B| = |S| and |BY | 6 |SY | for all Y ⊂ [n]. This has no right to be true! Forexample, we can then read off all possible projection theorems – just check them for boxes.
A uniform cover C of [n] is irreducible if we cannot write C = C′ ⊔ C′′ where C′ and C′′ areuniform covers. For example, if n = 3 then 12, 13, 23 form an irreducible cover, but 1, 23,13, 2 do not.
31
Lemma 6. There are only finitely many irreducible uniform covers of [n].
Proof. Suppose C1, C2, C3, . . . are distinct irreducible covers. List P([n]) as E1, E2, . . ., E2n .Choose a subsequence Ci1 , Ci2 , . . . on which the number of copies of E1 is increasing (notnecessarily strictly). Then choose a subsequence of Ci1 , Ci2 , . . . on which the number ofcopies of E2 is increasing. Repeating for E3, then E4, then . . . , then E2n , we obtain asubsequence Cj1 , Cj2 , Cj3 , . . ., on which the number of copies of Ei is increasing for alli. But then Cj2 is not irreducible, as Cj2 ⊃ Cj1 . Contradiction. �
Theorem 7 (Bollobas-Thomason Box Theorem). Let S be a (non-empty) body in Rn.Then there is a box B ∈ Rn with |B| = |S| and |BY | 6 |SY | for all Y ⊂ [n].
Proof. We may assume without loss of generality that |S| > 0 and n > 2. Take real variablesxY for each Y ∈ P([n]) with Y 6= ∅ or [n], with constraints:
(i) 0 6 xY 6 |SY | for all Y ;
(ii) xY 6∏
i∈Y xi for all Y with |Y | > 2; and
(iii) |S|k 6∏
Y ∈C xY for each k-uniform irreducible cover C 6= {[n]}.
If (iii) is satisfied for all irreducible covers, then it is satisfied for all uniform covers. Wedenote the condition (iii) for all uniform covers by (iii)′. We ‘want a minimal solution’.
We have a solution, namely xY = |SY | for all Y . The solution set is compact, so thereexists a solution with minimal
∑Y xY . We must have xY > 0 for all Y , because every
Y occurs in some uniform cover, whence (iii)′ gives |xY | > 0 (as |S| > 0).
Claim. For 1 6 i 6 n, xi appears on the RHS of an inequality from (iii) in whichequality holds.
Proof of claim. We must have xi on the RHS of some constraint for which equalityholds, or else we could decrease xi (as the set of constraints is finite). It is not aninequality from (i), as xi > 0. If it is an inequality from (iii) then we are done.
If it is an inequality from (ii), then xY =∏
j∈Y xj for some Y with {i} ∈ Y . Wemust have xY on the RHS of an inequality that is an equality (by minimality ofxY ), which must be of type (iii). So |S|k =
∏Z∈C xZ , for some irreducible cover
C with Y ∈ C.
Then C − {Y } ∪{{j} : j ∈ Y
}is also a uniform cover with equality in (iii)′, and
{i} belongs to the cover. Now take any irreducible cover C′ from this cover whichincludes {i}.
Thus for each i, we have a uniform cover Ci with equality in (iii) and with {i} ∈Ci. Consider C =
⋃ni=1 Ci. Then C is a uniform cover with equality in (iii)’, and
{1}, {2}, . . ., {n} ∈ C. Put C′ = C −{{1}, {2}, . . ., {n}
}. Then C′ is also a uniform
cover, say a k-cover, and we have |S|k 6∏
Y ∈C′ xY and |S|k+1 =∏
Y ∈C′ xY
∏ni=1 xi.
Thus |S| =∏n
i=1 xi. Now for any Y , consider the uniform cover {Y, Y c} of [n]. Wehave
|S| 6 xY xY c 6
(∏
i∈Y
xi
)(∏
i∈Y c
xi
)= |S|,
so xY =∏
i∈Y xi. Thus B = [0, x1]× [0, x2]× · · · × [0, xn]. �
32
Intersecting families of graphs
What happens to intersecting families if we have more structure in our ground set?
One natural example is to take our ground set to be [n](2), the edges of the complete graph
on [n]. There are a total of 2(n2) graphs on [n].
How many graphs can we find such that any two intersect in something containing P2, thepath of length 2? We want to find max |A| subject to G, H ∈ A⇒ G ∩H ⊃ P2.
Clearly |A| 6 12 2(n
2), as we cannot have both G ∈ A and Gc ∈ A for any graph G.
We can get |A| ∼ 12 2(n
2) by fixing x ∈ [n] and taking
A ={
G : dG(x) >n
2+ 1}
.
This has
|A| ∼(
1
2− c√
n
)2(n
2).
Similarly, we can get |A| ∼ 12 2(n
2) for G ∩H containing a star.
Conjecture 8. If G, H ∈ A⇒ G ∩H contains a triangle, then |A| 6 18 2(n
2).
Note that we can obtain |A| = 18 2(n
2) by taking A to consist of all graphs G which containsome fixed triangle.
Theorem 9. Let A ⊂ P([n](2)
)be such that if G, H ∈ A then G ∩H contains a triangle.
Then |A| 6 14 2(n
2).
Proof. We want |A| 6 2(n2)−2 = 2(n
2)(1−2/(n2)), so it is enough to find a uniform cover C of
[n](2) such that for all Y ∈ C we have∣∣A|Y
∣∣ 6 2c|Y |, where c = 1− 4/n(n− 1).
For n even, take all Y of the form B(2) ∪ (Bc)(2) with |B| = 12 |A|. This is clearly a
uniform cover. Now for any such Y , G ∩H is not bipartite and so G and H meet onY . Thus A|Y is intersecting, whence
∣∣A|Y∣∣ 6 1
2 2|Y | = 22(n/2
2 )−1 = 22(n/2
2 )(
1− 1
2(n/22 )
)
,
so we need
1− 1
2(n/22
) 6 1− 4
n(n− 1).
For n odd, we do the same thing but with |B| = (n− 1)/2. �
33
Mich. 2008 COMBINATORICS – EXAMPLES 1 IBL
1. Write down two antichains of size 10 in P ([5]). Write down an antichain of size 8 in
P ([5]) whose members do not all have the same size.
2. What are the 99th, 100th and 101st elements in the colex order on N(4)? For which
A ∈ N(4) is it true that A and the successor of A (in colex) have the same sum?
3. Let A ⊂ [9](3) with |A| = 28. How small can the lower shadow of A be? And the upper
shadow?
4. Let n be even, and let A ⊂ P(X) be a set system that contains no chain of length 3.
Prove that |A| ≤(
nn/2
)
+(
nn/2 − 1
)
.
5. Let A ⊂ P(X) be an antichain not of the form X (r), 0 ≤ r ≤ n. Must there exist a
maximal chain that is disjoint from A?
6. A set system A ⊂ P(X) is called a cross-cut if for every B ∈ P(X) there exists A ∈ A
with B ⊂ A or A ⊂ B. Prove that every cross-cut contains a cross-cut of size at most(
nbn/2c
)
. Does every cross-cut contain a cross-cut that is an antichain?
7. Let A ⊂ X(r), and let U, V ⊂ X with |U | = |V |, U ∩ V = ∅ and maxU < max V . If A
is left-compressed, can we have |∂CUV (A)| > |∂A|?
8. Find a set system A for which equality holds in the Kruskal-Katona theorem but which
is not isomorphic to an initial segment of colex.
9. Let x1, . . . , xn be non-zero real numbers, and let a be real. Show that at most(
nbn/2c
)
of the sums∑
i∈A xi , A ⊂ [n], can be equal to a.
+10. For n = 2r + 1, give an explicit bijection f : X (r) → X(r+1) such that A ⊂ f(A) for
every A ∈ X(r).
+11. Let A ⊂ X(r) and B ⊂ X(r+1) be initial segments of colex with |A| = |B|. Do we
always have |∂A| ≤ |∂B|?
Mich. 2008 COMBINATORICS – EXAMPLES 2 IBL
1. Let r < n/2. What is the largest intersecting family contained in X (≤r)?
2. A set system A ⊂ P(X) is an up-set if whenever x ∈ A and x ⊂ y then also y ∈ A.
Explain why every maximal intersecting family is an up-set of size 2n−1. Conversely, if A
is an up-set with |A| = 2n−1, must A be intersecting?
3. Let A ⊂ P(X) with |A| = 2n−1 + 1, so that some pair x, y ∈ A must be disjoint. What
is the smallest number of such disjoint pairs that A can have? And what if |A| = 2n−1+2?
4. Let A ⊂ Q6 with |A| = 26. How small can the vertex-boundary of A be? And the
edge-boundary?
5. Suppose that we try to use codimension-1 compressions to prove the (false) result that
initial segments of the simplicial order on Qn minimise the edge-boundary. Where does
the proof go wrong?
6. Write down a direct proof of the Kruskal-Katona theorem using codimension-1 com-
pressions.
7. Does the sequence of paths P1, P2, P3, . . . form a Levy family? What about the sequence
of two-dimensional grids [1]2, [2]2, [3]2, . . .?
8. Give an example of a connected graph G for which the extremal sets for the vertex-
isoperimetric inequality do not form a nested family: in other words, there is no ordering
of the vertices of G such that to minimise the neighbourhood of a set of m vertices it is
best to take the first m vertices in that ordering.
9. Let f1, . . . , fn2 be Lipschitz functions on Qn. Prove that, for n sufficiently large, there
exists a point x ∈ Qn such that |fi(x)− fi(xc)| < n/100 for all i.
10. Let A ⊂ Qn with |A| =∣
∣X(≤r)∣
∣, where r < n/2. Prove that we can always find a set
of(
nr
)
disjoint edges between A and Ac.
11. Let A1, A2, . . . , Ad ⊂ P(X) be intersecting families. Prove that |A1 ∪A2 ∪ . . . ∪Ad| ≤
2n − 2n−d.
12. Let A ⊂ P(N) be an intersecting family of finite sets. Must there exist a finite set
F ⊂ N such that the family {A ∩ F : A ∈ A} is intersecting? And what if A ⊂ N(r)?
Mich. 2008 COMBINATORICS – EXAMPLES 3 IBL
1. Let A ⊂ [4]3 with |A| = 23. How small can the vertex-boundary of A be?
2. The distance between two (non-empty) subsets A and B of [k]n is d(A,B) =
min {d(x, y) : x ∈ A, y ∈ B}. Show that if d(A,B) ≥ t then the distance between the
first |A| and the last |B| points of [k]n in the simplicial order is also at least t.
3. Suppose that we wish to remove edges from [k]n in such a way that the resulting graph
has all components of size at most kn/2. Show that we must remove at least kn−1 edges.
4. The diameter of a set A ⊂ Qn is max {d(x, y) : x, y ∈ A}. Deduce from Harper’s
theorem that if A ⊂ Qn has diameter d, where d < n and d is even, then |A| ≤∣
∣X(≤d/2)∣
∣.
5. Let r be an odd positive integer. Show that there are infinitely many values of n for
which there exists a set system A ⊂ [n](r) satisfying |x ∩ y| even for all distinct x, y ∈ A
and having size |A| = n.
6. Does there exist an even positive integer r and a value of n greater than r+1 for which
there exists a set system A ⊂ [n](r) satisfying |x ∩ y| odd for all distinct x, y ∈ A and
having size |A| = n?
7. Let A ⊂ [n](r) be such that the number of values taken by |x ∩ y|, over all distinct
x, y ∈ A, is s. Prove that |A| ≤(
ns
)
.
8. Give a constructive proof that the Ramsey number R(t) grows faster than any polyno-
mial in t, by considering the graph on [p3](p2−1), where p is prime, in which x is joined to
y if |x ∩ y| is congruent to −1 mod p.
9. Consider the graph on Rn in which x is joined to y if ‖x− y‖ = 1, where ‖.‖ denotes
the usual Euclidean distance. Show that this graph has chromatic number at least cn, for
some constant c > 1.
+10. Let A be a subset of [k]n of diameter d, where k is odd and d is even. How large can
A be?
Mich. 2010 COMBINATORICS – EXAMPLES 1 IBL
1. Write down two antichains of size 10 in P ([5]). Write down an antichain of size 8 in
P ([5]) whose members do not all have the same size.
2. What are the 99th, 100th and 101st elements in the colex order on N(4)? For which
A ∈ N(4) is it true that A and the successor of A (in colex) have the same sum?
3. Let A ⊂ [9](3) with |A| = 28. How small can the lower shadow of A be? And the upper
shadow?
4. Let n be even, and let A ⊂ P(X) be a set system that contains no chain of length 3.
Prove that |A| ≤(
nn/2
)
+(
nn/2 − 1
)
.
5. Let A ⊂ P(X) be an antichain not of the form X(r), 0 ≤ r ≤ n. Must there exist a
maximal chain that is disjoint from A?
6. A set system A ⊂ P(X) is called a cross-cut if for every B ∈ P(X) there exists A ∈ A
with B ⊂ A or A ⊂ B. Prove that every cross-cut contains a cross-cut of size at most(
n⌊n/2⌋
)
. Does every cross-cut contain a cross-cut that is an antichain?
7. Let A ⊂ X(r), and let U, V ⊂ X with |U | = |V |, U ∩ V = ∅ and maxU < maxV . If A
is left-compressed, can we have |∂CUV (A)| > |∂A|?
8. Find a set system A for which equality holds in the Kruskal-Katona theorem but which
is not isomorphic to an initial segment of colex.
9. Let x1, . . . , xn be non-zero real numbers, and let a be real. Show that at most(
n⌊n/2⌋
)
of the sums∑
i∈A xi , A ⊂ [n], can be equal to a.
+10. For n = 2r + 1, give an explicit bijection f : X(r) → X(r+1) such that A ⊂ f(A) for
every A ∈ X(r).
+11. Let A ⊂ X(r) and B ⊂ X(r+1) be initial segments of colex with |A| = |B|. Do we
always have |∂A| ≤ |∂B|?
Mich. 2010 COMBINATORICS – EXAMPLES 2 IBL
1. Let r < n/2. What is the largest intersecting family contained in X(≤r)?
2. A set system A ⊂ P(X) is an up-set if whenever x ∈ A and x ⊂ y then also y ∈ A.
Explain why every maximal intersecting family is an up-set of size 2n−1. Conversely, if A
is an up-set with |A| = 2n−1, must A be intersecting?
3. Let A ⊂ P(X) with |A| = 2n−1 + 1, so that some pair x, y ∈ A must be disjoint. What
is the smallest number of such disjoint pairs that A can have? And what if |A| = 2n−1 +2?
4. Let A ⊂ Q6 with |A| = 26. How small can the vertex-boundary of A be?
5. Suppose that we try to use codimension-1 compressions to prove the (false) result that
initial segments of the simplicial order on Qn minimise the edge-boundary. Where does
the proof go wrong?
6. Write down a direct proof of the Kruskal-Katona theorem using codimension-1 com-
pressions.
7. Does the sequence of paths P1, P2, P3, . . . form a Levy family? What about the sequence
of two-dimensional grids [1]2, [2]2, [3]2, . . .?
8. Give an example of a connected graph G for which the extremal sets for the vertex-
isoperimetric inequality do not form a nested family: in other words, there is no ordering
of the vertices of G such that to minimise the neighbourhood of a set of m vertices it is
best to take the first m vertices in that ordering.
9. Let f1, . . . , fn2 be Lipschitz functions on Qn. Prove that, for n sufficiently large, there
exists a point x ∈ Qn such that |fi(x) − fi(xc)| < n/100 for all i.
10. Let A ⊂ Qn with |A| =∣
∣X(≤r)∣
∣, where r < n/2. Prove that we can always find a set
of(
n
r
)
disjoint edges between A and Ac.
11. Let A1, A2, . . . , Ad ⊂ P(X) be intersecting families. Prove that |A1 ∪ A2 ∪ . . . ∪ Ad| ≤
2n − 2n−d.
12. Let A ⊂ P(N) be an intersecting family of finite sets. Must there exist a finite set
F ⊂ N such that the family {A ∩ F : A ∈ A} is intersecting? And what if A ⊂ N(r)?
Mich. 2010 COMBINATORICS – EXAMPLES 3 IBL
1. Let A ⊂ [4]3 with |A| = 23. How small can the vertex-boundary of A be?
2. The distance between two (non-empty) subsets A and B of [k]n is d(A, B) =
min {d(x, y) : x ∈ A, y ∈ B}. Show that if d(A, B) ≥ t then the distance between the
first |A| and the last |B| points of [k]n in the simplicial order is also at least t.
3. The diameter of a set A ⊂ Qn is max {d(x, y) : x, y ∈ A}. Deduce from Harper’s
theorem that if A ⊂ Qn has diameter d, where d < n and d is even, then |A| ≤∣
∣X(≤d/2)∣
∣.
4. Prove that, for any A ⊂ P(X), the number of sets shattered by A is at least |A|.
5. What is wrong with the following ‘proof’ that if S is a body in R3 then |S|
2≤
|S12||S23||S31|: if we let T = S×S ⊂ R6, then |S|
2= |T | ≤ |T12||T34||T56| = |S12||S31||S23|.
6. Let A ⊂ P(X) be a set system that does not shatter any of the sets
{1, 2, 3}, {2, 3, 4}, . . . , {n − 2, n − 1, n}, {n − 1, n, 1}, {n, 1, 2}. Explain why when n is a
multiple of 3 it is trivial that |A| ≤ 7n/3, and then prove that in fact this holds for every
value of n.
7. Let Y1, Y2, . . . , Yr ⊂ [n] be sets that do not form a uniform cover of [n]. Show that
knowledge of |SY1||SY2
| . . . |SYr| does not imply any upper bound on |S|.
8. Let S and T be bodies in R2, and let B and C be boxes verifying the Bollobas-Thomason
box theorem for S and T respectively. Show that if S ⊂ T then it is always possible to
choose B and C such that B ⊂ C. What happens in R3?
+9. Let A be a subset of [k]n of diameter d, where k is odd and d is even. How large can
A be?
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