Tải bản đầy đủ - 0trang
6



2. Operations on sets and set systems



where A, B ⊂ X. For a family F ⊂ 2X the complement family is

defined by F c = {F c : F ∈ F}.

Now we introduce two operations on families of sets that will turn

out to be very useful in establishing various upper bounds.

Let F ⊂ 2X and let x ∈ X be a fixed element. The operation

squash at an element x is defined by

Sx (F) = {Fx : F ∈ F},

where

Fx =



F \ {x} if x ∈ F ∈ F and F \ {x} ∈ F,

F



otherwise.



Let us explain the operation squash at x in words as well. All the

members F of F with x ∈ F are left unchanged. For each F with

x ∈ F ∈ F we check whether F \ {x} is in F or not (so Fx depends

on F). In the first case, in order to avoid duplication, we leave it

unchanged. In the latter case we replace F by F \ {x}. For example,

if

F = {{x, y, z}, {x, y}, {y, z}, {z}},



(2.1)

then



Sx (F) = {{x, y, z}, {y}, {y, z}, {z}}.

By definition we have |Sx (F)| = |F|. The following trivial observation is essential in proving that certain properties are not altered by

squashing.

Observation 2.2. Let F ∈ Sx (F) and x ∈ F . Then both F and

F \ {x} are members of F ∩ Sx (F).

Exercise 2.3. Let F ⊂ 2X and x ∈ X. Recall from (0.1) and (0.2)

that

F(x) = {F \ {x} : x ∈ F ∈ F},

F(¯

x) = {F : x ∈ F ∈ F}.

Prove that

x) and (Sx (F))(¯

x) = F(x) ∪ F(¯

x).

(Sx (F))(x) = F(x) ∩ F(¯



2. Operations on sets and set systems



7



Let us define the diameter diam(F) of F by

diam(F) = max |A B|.

A,B∈F



Proposition 2.4 ([84]). diam(Sx (F)) ≤ diam(F).

Proof. Consider the squash of F at x. Take two arbitrary sets A, B ∈

F. As long as Ax Bx ⊂ A B there is no problem. The only other

possibility is

Ax Bx = (A B) {x}.

In this case we have x ∈ A ∩ B. By symmetry, let x ∈ Ax and

x ∈ Bx . By Observation 2.2, A \ {x} is in F as well. Now Ax Bx =

(A \ {x}) B implies diam(Sx (F)) ≤ diam(F).

Let us remark that applying the operation squash at different

elements x, y ∈ X will produce a family independent of the order of

operations, that is, Sx (Sy (F)) = Sy (Sx (F)). Consequently, applying

the operation squash at all x ∈ X just once will produce a family F˜

satisfying the property that

(2.2)



if x ∈ F ∈ F˜ then F \ {x} ∈ F˜ for all x ∈ X.



We call F˜ a squashed family obtained from F. For the example (2.1),

it follows that F˜ = Sz (Sy (Sx (F))) = {{x}, ∅, {y}, {z}}. We say that

a family F is hereditary (or a downset) if G ⊂ F ∈ F implies G ∈ F.

In other words, F is hereditary if it is a union of power sets.

Proposition 2.5. A squashed family F˜ is hereditary.

Proof. Let G ⊂ F ∈ F˜ . We show that G ∈ F˜ by induction on |F \G|.

For |F \ G| = 1, the statement is simply (2.2). For the induction step

choose H with G H F ; then F ∈ F and the induction hypothesis

yield H ∈ F. Applying the induction hypothesis to the pair (G, H)

yields G ∈ F as desired.

F|T



For a family F ⊂ 2X and a subset T ⊂ X, we define the trace

by

F|T = {F ∩ T : F ∈ F}.



Let us remark that F|T is a usual family, that is, a subset of 2T .

Consequently, |F|T | ≤ 2|T | .



8



2. Operations on sets and set systems



Proposition 2.6 ([48]). Let F ⊂ 2X and x ∈ X. For every T ⊂ X

one has

|Sx (F)|T | ≤ |F|T |.



(2.3)



Proof. If x ∈ T then the operation squash does not affect the trace

on T . Let T = {x} T and T0 ⊂ T . If {x} ∪ T0 ∈ Sx (F), then

{T0 , {x} ∪ T0 } ⊂ Sx (F) ∩ F by Observation 2.2. If {x} ∪ T0 ∈ Sx (F)

and T0 ∈ Sx (F), then T0 ∈ F or {x} ∪ T0 ∈ F by the definition of

the squash operation. Therefore we always have

|Sx (F)|T ∩ {T0 , {x} ∪ T0 }| ≤ |F|T ∩ {T0 , {x} ∪ T0 }|.

Summing this over all 2|T |−1 choices of T0 ⊂ T gives (2.3).

The second operation on families of sets is called shifting. For

convenience let X = {1, 2, . . . , n} (or [n] for short). For 1 ≤ i = j ≤ n

we are going to define the (i, j)-shift Si,j for families F ⊂ 2[n] . The

formal definition that goes back to Erd˝

os, Ko, and Rado is as follows:

Si,j (F) = {si,j (F ) : F ∈ F},

where

si,j (F ) =



(F \ {j}) ∪ {i} if i ∈ F, j ∈ F and (F \ {j}) ∪ {i} ∈ F,

F



otherwise.



In words, we only try to make changes in members of F if they do

not contain i but contain j. In this case we check whether the subset

obtained by replacing j with i is already in F. Only if not do we then

replace the subset with the new one.

Exercise 2.7. Verify the following.

(i) |Si,j (F)| = |F|.

(ii) |si,j (F )| = |F |, so F ⊂



[n]

k



implies Si,j (F) ⊂



[n]

k



.



Exercise 2.8. Let F(a, ¯b) = {F \ {a} : F ∈ F, F ∩ {a, b} = {a}} for

a, b ∈ [n]. Show that

Si,j (F)(i, ¯j) = F(i, ¯j) ∪ F(j, ¯i) and Si,j (F)(j, ¯i) = F(i, ¯j) ∩ F(j, ¯i).

∈ We will present two propositions that show the importance

of shifting: the first one says that shifting preserves the intersecting



2. Operations on sets and set systems



9



property, and the second one claims that shifting does not increase

the size of the shadow. To make the statements precise we need two

definitions. Let t be a positive integer. We say that a family F ⊂ 2X

is t-intersecting if |F ∩ F | ≥ t for all F, F ∈ F. We simply say that

a family is intersecting if it is 1-intersecting. For a positive integer p

with 1 ≤ p ≤ n, define the p-shadow σp (F) by

σp (F) = {P ∈



[n]

p



: ∃ F ∈ F, P ⊂ F }.



Proposition 2.9. If F ⊂ 2X is t-intersecting, then Si,j (F) is tintersecting as well.

Proof. Suppose for contradiction that there exist F, G ∈ F with

|si,j (F ) ∩ si,j (G)| < t ≤ |F ∩ G|.

In this case, by symmetry, we may assume that si,j (F ) = F but

si,j (G) = G and, moreover, that F ∩ {i, j} = G ∩ {i, j} = {j}. When

performing the (i, j)-shift on F , why did not we replace j with i? The

only possible reason is that the set F := (F \{j})∪{i} is already in F.

However, |F ∩G| = |si,j (F )∩si,j (G)|, contradicting the t-intersecting

property of F.

Proposition 2.10. σp (Si,j (F)) ⊂ Si,j (σp (F)).

Proof. Let P ∈ σp (Si,j (F)). If |P ∩ {i, j}| = 0 or 2, then we can

find F ∈ F such that P ⊂ F , and it follows that P ∈ σp (F) and

P ∈ Si,j (σp (F)). So we may assume that |P ∩ {i, j}| = 1. Set

R = P \ {i, j}. Since P ∈ σp (Si,j (F)), there is Q ∈ Si,j (F) such that

P ⊂ Q.

First suppose that P = R {i}. If Q ∈ F then P ∈ σp (F) and

P ∈ Si,j (σp (F)). If Q ∈ F then F := (Q \ {i}) ∪ {j} ∈ F and

R {j} ⊂ F . Thus R {j} ∈ σp (F) and P = R {i} ∈ Si,j (σp (F)).

Next suppose that P = R {j}. Then Q ∈ F and P ∈ σp (F). If

i ∈ Q then R {i} ∈ σp (F). If i ∈ Q then (Q \ {j}) ∪ {i} ∈ F and

R {i} ∈ σp (F) again. Thus si,j (P ) = P and P ∈ Si,j (σa (F)).

We say that a family G ⊂ 2X is s-union if |G ∪ G | ≤ |X| − s for

all G, G ∈ G. We say that a family is union if it is 1-union.

Exercise 2.11. Let s be a positive integer with s ≤ |X|.



10



2. Operations on sets and set systems

(i) Show that G ⊂ 2X is s-union if and only if the complement

family G c is s-intersecting.

(ii) Show that if G ⊂ 2X is s-union then Si,j (G) is s-union.



Exercise 2.12. Show that if a family G ⊂ 2X is union, then |G| ≤

2|X|−1 . (Hint: Use (i) of Theorem 1.3 and (i) of Exercise 2.11.)

If G := si,j (F ) = F then

us the following.



a∈F



a−



a∈G



a = j − i. This gives



Observation 2.13. If 1 ≤ i < j ≤ n and Si,j (F) = F ⊂ 2X , then

a>

F ∈F a∈F



a.

G∈Si,j (F ) a∈G



Since X is finite, the above observation guarantees that after a

finite number of repeated applications of the (i, j)-shift for various 1 ≤

˜ =

i < j ≤ n, eventually we end up with a family F˜ satisfying Si,j (F)

˜

F for all 1 ≤ i < j ≤ n. Such a family is called shifted. From

Proposition 2.9 (resp. Exercise 2.11) it is clear that in establishing

upper bounds for the size of t-intersecting (resp. s-union) families we

can restrict ourselves to shifted families. By Proposition 2.10, the

same is true when trying to prove lower bounds on the size of the

p-shadow.

Finally we establish the following useful bound concerning shadows.

Theorem 2.14 (Kruskal–Katona Theorem (integer version)). Let

0 ≤ p < k ≤ x be integers. Suppose that F is a family of k-element

subsets with |F| ≥ xk . Then |σp (F)| ≥ xp .

The following proof is taken from [51].

Proof. Let p < k be fixed. We prove the statement by induction on

x. The initial step x = k is trivial. We suppose x ≥ k + 1, that is,

x − 1 ≥ k. By Proposition 2.10 we may suppose that F is shifted.

First we consider the case p = k − 1. Recall the definitions of F(x)

and F(¯

x) from (0.1) and (0.2).

1)) ⊂ F(1).

Claim 2.15. σk−1 (F(¯



2. Operations on sets and set systems



11



Proof. Let G ∈ σk−1 (F(¯

1)). Then G∪{j} ∈ F for some j ∈ {1} G.

By shiftedness, G ∪ {1} ∈ F, that is, G ∈ F(1).

Claim 2.16. |F(1)| ≥



x−1

k−1



.



Proof. Suppose the contrary, that is, |F(1)| <

|F(¯

1)| = |F| − |F(1)| >



x

x−1

−

k

k−1



x−1

k−1



. We infer that



=



x−1

.

k



So we may apply the induction hypothesis to F(¯1). Then we obtain

|σk−1 (F(¯

1))| ≥ x−1

k−1 > |F(1)|. This contradicts Claim 2.15.

We note that

σk−1 (F) ⊃ F(1)



{{1}



G : G ∈ σk−2 (F(1))}.



In fact, clearly σk−1 (F) contains F(1), which consists of subsets not

containing 1. If G ∈ σk−2 (F(1)), then there is H ∈ F(1) such that

G ⊂ H, and {1} G ⊂ {1} H ∈ F, yielding that {1} G is in

σk−1 (F). Thus we get

|σk−1 (F)| ≥ |F(1)| + |σk−2 (F(1))|.

Applying the induction hypothesis with Claim 2.16 to F(1), we infer

that |σk−2 (F(1))| ≥ x−1

k−2 . This proves

|σk−1 (F)| ≥



x−1

x−1

+

k−1

k−2



=



x

.

k−1



This completes the proof for the case p = k − 1.

Next we prove the general case p < k by induction on i := k − p,

where we have already proved the case i = 1. Suppose that the

statement is true for the case i, that is, |F| ≥ xk implies |σk−i (F)| ≥

x

k−i . Since σk−(i+1) (F) = σ(k−i)−1 (σk−i (F)), we can apply the

induction hypothesis to the case k = k−i and p = k −1 (that is, i =

x

.

1); we get |σk−(i+1) (F))| = |σk −1 (σk (F))| ≥ k x−1 = k−(i+1)

Remark 2.17. One can define

x

k



=



x

k



for all real numbers x by



x(x − 1) · · · (x − k + 1)

.

k!



Note that xk is a polynomial of degree k. Since the equality xk −

x−1

− x−1

k

k−1 = 0 is true for all integer values of x with x ≥ k, it is



12



2. Operations on sets and set systems



an identity. That is, it is true for all real values of x. By the same

proof one can show that Theorem 2.14 holds for all real x with x ≥ k.

This is called the Lov´

asz version of the Kruskal–Katona Theorem.

If F ⊂ X

k , then the (k − 1)-shadow σk−1 (F) is sometimes just

called the shadow (or immediate shadow) of F and denoted by σ(F).

We will return to the problem of estimating σ(F) in Chapter 6 and

refine the Lov´

asz version.



Chapter 3



Theorems on traces



In the preceding chapter we worked hard to introduce some operations

on families and prove some of their properties. In this chapter our

efforts will be rewarded by some almost trivial proofs of classical

results.

Let us start with the following important result that was proved

independently by three sets of authors: Perles and Shelah, Sauer, and

Vapnik and Chervonensky. The names of the third set of authors are

ordered according to the Russian alphabet.

Theorem 3.1. Let n > k ≥ 0 be integers. If F ⊂ 2[n] satisfies

k

[n]

such that F|T = 2T ,

|F| > i=0 ni , then there exists a T ∈ k+1

where F|T = {F ∩ T : F ∈ F} denotes the trace of F on T .



Proof. Suppose, to the contrary, that F ⊂ 2[n] with F|T = 2T for

X

. In view of Propositions 2.5 and 2.6, after repeatedly

all T ∈ k+1

applying squashing we obtain a hereditary family G ⊂ 2[n] with |G| =

X

. We claim that

|F| and still satisfying |G|T | < 2|T | for all T ∈ k+1

|G| ≤ k for all G ∈ G.

Indeed, if |G| ≥ k + 1 for some G ∈ G, then by the hereditary

G

we have G|H = 2H , a

property 2G ⊂ G, and thus for all H ∈ k+1

contradiction. Consequently, we have proved that |G| ≤ k for all G ∈

13



14



3. Theorems on traces

k



G and |G| ≤ i=0 ni . But this contradicts the fact that |G| = |F|

k

and our assumption that |F| > i=0 ni .

Let us introduce the arrow notation (n, m) → (a, b) with the

following meaning: For every F ⊂ 2[n] with |F| = m there exists

T ⊂ [n] with |T | = a such that |F|T | ≥ b.

Again, by Proposition 2.6, in order to establish the veracity of an

arrow relation, it is sufficient to check it for hereditary families.

Theorem 3.2. The following hold.

(i) (n, m) → (n − 1, m) for m ≤ n.

(ii) (n, m) → (n − 1, m − 1) for m ≤ 1 + n +

(iii) (n, m) → (3, 7) for m > 1 + n +



n−1

2 .



n2

4 .



Proof. (i) If G ⊂ 2[n] is hereditary, then ∅ ∈ G. If, moreover, m =

|G| ≤ n, then not all the 1-element subsets are in G, that is, {x} ∈ G

for some x ∈ [n]. Then, by the hereditary property, G ∈ G for all G

containing x. Therefore G ⊂ 2[n]\{x} , implying |G|[n]\{x} | = |G|.

(ii) If {x} ∈ G for some x ∈ [n], the above proof works. Thus we

may assume that ∅ and all 1-element sets are in G. Since m = |G| ≤

n−1

more subsets in G, there is

1 + n + n−1

2 and there are at most

2

an element y ∈ [n] which is not contained in any 2-element member

of G. By the hereditary property, the only member of G containing y

is {y}. Consequently, G|[n]\{y} = G \ {y}, proving (ii).

(iii) If |G| = 3 for some G ∈ G, then |G|G | = 8 and we have nothing

[n]

[n]

to prove. Consequently, we may assume that G ⊂ [n]

0

1

2 .

[n]

Let G (i) = G ∩ i for i = 0, 1, 2. Then G (2) is a graph on n vertices.

2

The number of edges is at least m−|G (0) |−|G (1) | ≥ m−1−n > n4 by

the assumption. By Exercise 3.4 there are three edges {x, y}, {x, z},

and {y, z} in it (forming a triangle). Now, letting T = {x, y, z}, it

T

T

follows that G|T = T0

1

2 , and thus |G|T | = 1 + 3 + 3 = 7.

Remark 3.3. In the above theorem (i) and (ii) were originally proved

by Bondy and Bollob´as, respectively. As to (iii), it was conjectured

by Lov´

asz and proved by Frankl; see [48].



3. Theorems on traces



15



Exercise 3.4. Let G be a graph on n vertices with more than n2 /4

edges. Show that G contains a triangle.

For any F ⊂ 2X and Y ⊂ X, the arrow relation

(|X|, |F|) → (|Y |, |F|Y | + 1)

is not true. We apply this fact to the following construction.

Construction 3.5. Let n be of the form n = dq, where d and q are

positive integers. Let X = X1 · · · Xq be a partition with |Xi | = d,

1 ≤ i ≤ q. Define a hereditary family F := F(d, q) = 2X1 ∪ · · · ∪ 2Xq .

Note that 2Xi ∩ 2Xj = {∅} and |F| = 1 + (2d − 1) nd . If Y = X \ {x}

for some x ∈ X, then F|Y = F \ {F ∈ F : x ∈ F }. So it follows that

X

|F|Y | = |F| − 2d−1 for all Y ⊂ n−1

. Thus

(dq, 1 + (2d − 1) nd ) → (dq − 1, 1 + (2d − 1) nd − 2d−1 + 1)

is not true.

The next result shows that F(d, q) is an extremal example. Recall

the definitions of F(x) and F(¯

x) from (0.1) and (0.2).

Theorem 3.6. Let n and d be fixed positive integers, with n > d.

Let |X| = n and let F ⊂ 2X be a hereditary family satisfying |F| ≤

d

1 + 2 d−1 n. Then one of the following holds.

(i) |F(x)| < 2d−1 for some x ∈ X. In this case |F|Y | > |F| −

2d−1 for Y = X \ {x}.

(ii) F is isomorphic to F(d, q) for some q (in particular, d

X

.

divides n). In this case |F|Y | = |F|−2d−1 for all Y ∈ n−1

Proof. The proof is based on the following lemma of some independent interest.

Lemma 3.7. Let H be a hereditary family, with |H| = 2t for some

t+1

1

≥ 2 t+1−1 and H attains the minimum if and

t ≥ 0; then H∈H |H|+1

only if H = 2T for some t-element set T .

Let us postpone the proof of the lemma and prove the theorem

first. The following identity is easily established by reversing the order



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3. Theorems on traces



of summation:

|F \ {∅}| =

F ∈F x∈F



1

=

|F |



x∈X



H∈F (x)



1

.

|H| + 1



Supposing that (i) does not hold, we have |F(x)| ≥ 2d−1 for all x ∈ X.

d

By the lemma the quantity in the bracket is at least 2 d−1 , yielding

d

|F| ≥ 1 + 2 d−1 n. Since the assumption of Theorem 3.6 is |F| ≤

d

1 + 2 d−1 n, we have equality. Therefore, we must have equality all the

way, that is, |F(x)| = 2d−1 , and by the lemma F(x) = 2T (x) with

X

some T (x) ∈ d−1

. We infer that every x ∈ X is contained in a

unique d-element set {x} T (x). Consequently, these d-element sets

partition X, that is, F is isomorphic to F(d, q), where q = nd is the

number of these d-element sets. That is, (ii) holds.

Proof of Lemma 3.7. Let us use the hereditary property of H to

prove a simple inequality. Here hi denotes the number of i-element

t

sets in H. Then hi = 0 for i > t and i=0 hi = |H|. Since |H| = 2t

t

t

t

we have that i=0 hi = i=0 i . We will show that

r



r



hi ≥



(3.1)

i=0



i=0



t

i



for all r with t ≥ r ≥ 0. Suppose, to the contrary, that ri=0 hi <

r

t

i=0 i for some 0 ≤ r < t. Since (3.1) holds for r = t, there must

exist some s with r < s ≤ t such that hs > st . Then we can choose

G ⊂ H ∩ Xs with |G| = st . By Theorem 2.14 |σi (G)| ≥ ti for

0 ≤ i < s, and by the hereditary property it follows that σi (G) ⊂ H.

Thus hi ≥ ti for 0 ≤ i < s. Since r < s, it then follows that

r

r

t

i=0 hi ≥

i=0 i . This contradicts our indirect assumption and

thereby proves (3.1).

We label the members of H in size-increasing order, that is, let

H = {H1 , H2 , . . . , H2t }, where |H1 | ≤ |H2 | ≤ · · · ≤ |H2t |. We partition H into H0 H1 · · · Ht with |Hi | = ti by choosing members

starting from H1 according to the above order, that is, H0 = {H1 },

H1 = {H2 , . . . , Ht+1 }, H2 = {Ht+2 , . . . , H(t )+t+1 }, and so on. In

2



view of (3.1), it follows that if H ∈ Hi then |H| ≤ i. Thus, for