Tải bản đầy đủ - 0 (trang)
Chapter 9. The Erdős matching conjecture

Chapter 9. The Erdős matching conjecture

Tải bản đầy đủ - 0trang

48



9. The Erd˝

os matching conjecture



Erd˝os proved that |F| ≤ |B(n, k, s)| for fixed k and s and for n

very large. In this chapter we prove the following.

Theorem 9.2. Let n ≥ (2s + 1)k − s and F ⊂

then |F| ≤ |B(n, k, s)|.



[n]

k



. If ν(F) ≤ s



Proof. It is easy to see that the (i, j)-shift does not increase ν(F).

So we may assume that F is shifted. For a subset A ⊂ [s + 1] and a

family G ⊂ 2[n] , let

G(A) = {G \ [s + 1] : G ∈ G, G ∩ [s + 1] = A}.

. Note

Note that if G is k-uniform and |A| ≤ k, then G(A) ⊂ [n]\[s+1]

k−|A|

also that A ∩ [s] = ∅ for all A ⊂ [s + 1] with |A| ≥ 2. We simply use B

for

for our target family B(n, k, s). Then we have B(A) = [n]\[s+1]

k−|A|

all A ⊂ [s + 1] with |A| ≥ 2. In particular,

(9.2)



|F(A)| ≤ |B(A)| for A ⊂ [s + 1] with |A| ≥ 2.



The equality

|G| =



(9.3)



|G(A)|

A⊂[s+1]



should be obvious. For simplicity, let f0 = |F(∅)| and let fi = |F({i})|

for 1 ≤ i ≤ s + 1. Then (9.1) will follow once we show the following:

s+1



fi ≤ s



f0 +



(9.4)



i=1



n−s−1

.

k−1



Indeed, B(∅) = B({s + 1}) = ∅, showing that the RHS of (9.4) is

s+1

equal to |B(∅)| + i=1 |B({i})|. This, together with (9.2) and (9.3),

yields |F| ≤ |B(n, k, s)|.

We deduce (9.4) from the following two inequalities:

(9.5)



f0 ≤ sfs+1 ,

s



fi + (s + 1)fs+1 ≤ s



(9.6)

i=1



n−s−1

.

k−1



Let us see how these inequalities imply (9.4). In view of (9.5), the

LHS of (9.4) does not exceed the LHS of (9.6). Thus (9.4) follows

from (9.6).



9. The Erd˝

os matching conjecture



49



For the proof of (9.5) and (9.6) we need to use the fact that F is

shifted. Recall that the immediate shadow is defined by

σ(F) =



G∈



[n]

k−1



: G ⊂ F for some F ∈ F



.



We use the following general result to prove (9.5).

Theorem 9.3 ([52]). Let H ⊂



[n]

k



. If ν(H) ≤ s then |H| ≤ s|σ(H)|.



We will prove the above theorem at the end of this chapter. Since

ν(F(∅)) ≤ ν(F) ≤ s, the above theorem yields

(9.7)



f0 = |F(∅)| ≤ s|σ(F(∅))|.



Then (9.5) follows from (9.7) via the following easy observation.

Claim 9.4. σ(F(∅)) ⊂ F({s + 1}).

Proof. Suppose, to the contrary, that there is an H ∈ σ(F(∅)) such

that H ∈ F({s + 1}). Since H ∈ σ(F(∅)), we have H ∈ [n]\[s+1]

and

k−1

H ⊂ F for some F ∈ F. Then we can write F = H {y} for some

y ≥ s + 2. Since F is shifted, it follows that (F \ {y}) {s + 1} =

{s+1} H ∈ F. This means that H ∈ F({s+1}), a contradiction.

Next we show (9.6). For the (k − 1)-uniform families F({i}),

1 ≤ i ≤ s + 1, shiftedness implies

(9.8)



F({s + 1}) ⊂ F({s}) ⊂ · · · ⊂ F({1}).



The assumption n ≥ (2s + 1)k − s is equivalent to n − (s + 1) ≥

(2s + 1)(k − 1). Therefore one can choose 2s + 1 pairwise disjoint

(k − 1)-element subsets H0 , H1 , . . . , H2s ⊂ [n] \ [s + 1].

Let us construct an auxiliary bipartite graph G with vertex partition V (G) = V1 V2 , where V1 = [s + 1], V2 = {H0 , . . . , H2s }, and

there is an edge joining i ∈ V1 and Hj ∈ V2 if and only if Hj ∈ F({i}).

Note that s + 1 pairwise independent edges in G would immediately

yield s + 1 pairwise disjoint edges in F, a contradiction. Now, by

invoking the Kăoning Theorem (see, e.g., Theorem 2.1.1 in [25]), we

can choose an s-element subset C ⊂ V (G) covering all edges of G.

That is, at least one endpoint of every edge of G is contained in C.

We denote by di the degree of i ∈ V1 . Note that d1 + · · · + ds+1 is



50



9. The Erd˝

os matching conjecture



the number of edges |E(G)|. Note also that ds+1 ≤ ds ≤ · · · ≤ d1 by

(9.8).

Claim 9.5. |E(G)| + sds+1 ≤ s(2s + 1).

Proof. Let us define x by |C ∩ V1 | = s − x. Since |C| < s + 1, one

can choose i ∈ V1 \ C. Then |C ∩ V2 | = x implies di ≤ x and, in

particular, ds+1 ≤ x. Noting that |V1 \ C| = x + 1 ≤ s + 1, we have

|E(G)| ≤ |C ∩ V1 ||V2 | + |V1 \ C|x ≤ (s − x)(2s + 1) + (s + 1)x.

Together with sds+1 ≤ sx, we get |E(G)| + sds+1 ≤ s(2s + 1).

We want to deduce (9.6) from Claim 9.5. Choose H0 , . . . , H2s

uniformly at random. Then the probability of Hj ∈ F({i}), or equivalently the probability that i ∈ V1 and Hj ∈ V2 are adjacent, is

fi

n−s−1

k−1



.



Thus the expected value of di is

(2s + 1)fi

n−s−1

k−1



.



Therefore the expected value of

s+1



di + sds+1 = |E(G)| + sds+1

i=1



is exactly

2s + 1

n−s−1

k−1



s+1



fi + sfs+1



.



i=1



By Claim 9.5, this never exceeds s(2s + 1), which proves (9.6).

Finally we prove Theorem 9.3.

Proof. Let H ⊂ [n]

k . We may assume that H is shifted and ν(H) =

s. For fixed s ≥ 2 we prove that

(9.9)



|H| ≤ s|σ(H)|



for all (n, k) with n ≥ k ≥ 1 by double induction on k and n.



9. The Erd˝

os matching conjecture



51



First consider the case k = 1. In this case we may assume that

H = {1, 2, . . . , s}. Then σ(H) = {∅} and |σ(H)| = 1. So we always

have equality in (9.9).

Next we deal with the case n ≤ (s+1)k−1, that is, sk ≥ n−k+1.

Let G be a bipartite graph with vertex partition V (G) = V1 V2 ,

where V1 = H and V2 = σ(H), and H ∈ V1 and K ∈ V2 are adjacent if

and only if H ⊃ K. For H ∈ V1 the degree of H is k, while for K ∈ V2

the degree of K is at most |[n] \ K| = n − (k − 1). Thus by counting

the number of edges from both sides we get |V1 |k ≤ |V2 |(n − k + 1),

that is,

n−k+1

|σ(H)| ≤ s|σ(H)|.

|H| ≤

k

Finally let n ≥ (s + 1)k. We prove (9.9) for (n, k) by assuming

that (9.9) holds for (∗, k − 1) and (n − 1, k). We want to apply the

induction hypothesis to the following two families:

[n − 1]

,

k

[n − 1]

D = {H \ {n} : n ∈ H ∈ H}} ⊂

.

k−1

C = {H : n ∈ H ∈ H} ⊂



Since C ⊂ H we have ν(C) ≤ ν(H) = s, and it follows from the

hypothesis that |C| ≤ s|σ(C)|.

Claim 9.6. ν(D) ≤ s.

Proof. Assume, on the contrary, that we have s + 1 pairwise disjoint

subsets D1 , . . . , Ds+1 ∈ D. Then Di {n} ∈ H for 1 ≤ i ≤ s + 1, and

|[n − 1] \ (D1



···



Ds+1 )| = (n − 1) − (s + 1)(k − 1) ≥ s.



Thus we can find s vertices x1 , . . . , xs in [n − 1] \ (D1 · · · Ds+1 ).

Since H is shifted, we have Di {xi } ∈ H for all 1 ≤ i ≤ s, and

moreover Ds+1 {n} ∈ H. This contradicts ν(H) = s.

By the hypothesis we get |D| ≤ s|σ(D)| and

|H| = |C| + |D| ≤ s(|σ(C)| + |σ(D)|).

Thus it remains to show the following.

Claim 9.7. |σ(C)| + |σ(D)| = |σ(H)|.



52



9. The Erd˝

os matching conjecture



Proof. Let σ(H) = C˜



˜ where

D,



C˜ = {K ∈ σ(H) : n ∈ K},

˜ = {K ∈ σ(H) : n ∈ K}.

D

˜ Moreover, if K {n} ∈ H, then since H

Then clearly σ(C) ⊂ C.

is shifted we have K {i} ∈ H for all i < n and K ∈ σ(C). This

˜ On the other hand, it follows (without using

means that σ(C) = C.

˜ and |σ(D)| = |D|.

˜ Thus

shiftedness) that σ(D) = {K \ {n} : K ∈ D}

˜ + |D|

˜ = |σ(H)|.

we have |σ(C)| + |σ(D)| = |C|

This completes the proof of Theorem 9.3.

Exercise 9.8. In Theorem 9.3 suppose further that H is shifted.

.

Show that if |H| = s|σ(H)| then H = [sk+k−1]

k

In Theorem 9.2 we only proved the inequalities, but one can further prove the uniqueness of the optimal families, that is, if |F| =

|B(n, k, s)| then F ∼

= B := B(n, k, s). The idea of the proof is as

follows. First, it is not so difficult to show this for shifted families.

Second, one can also show that if ν(F) = s and Si,j (F) = B then

F∼

= B. The proof of this part is similar to the proof of Lemma 10.5.



Chapter 10



The

Ahlswede–Khachatrian

Theorem



Let n, k, and t be positive integers with n ≥ k ≥ t. A family F ⊂ [n]

k

is called t-intersecting if |F ∩ F | ≥ t for all F, F ∈ F. Let M (n, k, t)

os, Ko,

denote the maximum size of t-intersecting families in [n]

k . Erd˝

and Rado proved in [36] that if n > n0 (k, t) then M (n, k, t) = n−t

k−t .

In this chapter we present a result by Ahlswede and Khachatrian

which tells us M (n, k, t) for all n ≥ k ≥ t.

Frankl introduced the following t-intersecting families:

F (k) (n, t, r) :=



F ∈



[n]

k



: |F ∩ [t + 2r]| ≥ t + r



for r ≥ 0. Without loss of generality we may assume that n ≥ t + 2r,

that is, r ≤ n−t

2 . He conjectured in [44] that

(10.1)



M (n, k, t) =



max



0≤r≤ n−t

2



|F (k) (n, t, r)|.



Exercise 10.1. Let n, k, and t be fixed, and let Fr = F (k) (n, r, t).

Show that |Fr | − |Fr+1 | is positive, zero, or negative if and only if

n − (k − t + 1) 2 +



t−1

r+1

53



54



10. The Ahlswede–Khachatrian Theorem



is positive, zero, or negative, respectively. For example, |F (5) (9, 3, 1)| =

|F (5) (9, 3, 2)| = 21 as shown int he picture below. (Hint: Compare

|Fr \ Fr+1 | and |Fr+1 \ Fr |.)

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F (5) (9, 3, 1)



F (5) (9, 3, 2)



The above exercise shows that

max |F (k) (n, t, r)| = |F (k) (n, t, 0)| =

r



n−t

k−t



if and only if n ≥ (t + 1)(k − t + 1). So the result by Erd˝

os, Ko, and

Rado confirms (10.1) for n > n0 (k, t). Also, the case n = 4m and

k = 2m of the conjecture already appeared in the paper [36], and

was popularized by Erd˝

os for a number of years (e.g. [33]). Frankl

[45] proved (10.1) (for t ≥ 15) for the exact range of n, that is,

n ≥ (t + 1)(k − t + 1), by using shifting and then counting the lattice



10. The Ahlswede–Khachatrian Theorem



55



paths corresponding to subsets in a shifted family; see Chapter 15.

Then Wilson [115] gave a completely different proof for the same

result (for all t) by studying the spectra of a graph on [n]

k reflecting

the t-intersecting property; see Section 27.3. The case of small n was

more dicult. Frankl and Fă

uredi [56] proved (10.1) for the cases

n > (k − t + 1)c t/ log t, where c is some absolute constant. Then

it was Ahlswede and Khachatrian who finally established (10.1) in

general. They gave two proofs, one based on the method of generating

sets [5] and the other using the so-called pushing-pulling method [7].

Both approaches are purely combinatorial and may be considered

‘dual’ to each other in some sense.

Theorem 10.2 (Alhswede–Khachatrian Theorem). Let n ≥ k ≥ t ≥

1 and 0 ≤ r ≤ n−t

2 be integers.

(i) If

(10.2)



(k − t + 1) 2 +



t−1

r+1



< n < (k − t + 1) 2 +



r−1

r



,



then1

M (n, k, t) = |F (k) (n, t, r)|.

Moreover, F (k) (n, t, r) is the unique extremal configuration up to isomorphism2 .

(ii) If

(10.3)



(k − t + 1) 2 +



t−1

r+1



= n,



then

M (n, k, t) = |F (k) (n, t, r)| = |F (k) (n, t, r + 1)|.

If moreover t ≥ 2, then F (k) (n, t, r) and F (k) (n, t, r + 1) are the only

extremal configurations up to isomorphism.

We will present the second proof from [7]. Here we show a consequence of the pushing-pulling method, which will play a key role in

the proof. To state the result we need some definitions. For integers

1 ≤ i, j ≤ n with i = j and a subset F ⊂ [n], let Fi,j denote the

1

We regard r−1

as being ∞ for r = 0 in (10.2), that is, we put no restriction on

r

the upper bound for n in this case.

2

is t-intersecting and |F | = M (n, k, t), then F ∼

That is, if F ⊂ [n]

=

k

(k)

F (n, t, r).



56



10. The Ahlswede–Khachatrian Theorem



subset obtained by renaming i and j to each other in F ; more precisely, Fi,j = F {i, j} if |F ∩ {i, j}| = 1, and Fi,j = F otherwise. We

say that Fi,j is obtained by (i, j)-exchange from F . For a family of

subsets F ⊂ 2[n] let

Fi,j = {Fi,j : F ∈ F}.

We say that F is exchange stable on [h] if Fi,j = F for all 1 ≤ i, j ≤ h

(i = j). Note that in this case if {F ∩ [h] : F ∈ F} contains an aelement subset, then it contains all a-element subsets of [h]. Suppose

that Fi,j = F for some i, j. Then there exists a unique h such that F

is exchange stable on [h] but not on [h + 1]. This number h is called

a head size (or homogeneous part size) of F and denoted by h(F).

The following easy fact is useful.

Claim 10.3. Let F ⊂ 2[n] be exchange stable on [s]. If A, B ∈ F

satisfy |A ∩ [s]| = a and |B ∩ [s]| = b, then there are A , B ∈ F such

that |A ∩ B ∩ [s]| ≤ max{0, a + b − s}.

Proof. Since F is exchange stable on [s], we may choose A , B from

A, B by changing A ∩ [s], B ∩ [s], respectively, so that A ∩ [s] = [a],

B ∩ [s] = [s − b + 1, n]. If a < s − b + 1 then A ∩ B ∩ [s] = ∅.

Otherwise |A ∩ B ∩ [s]| = |A ∩ [s]| + |B ∩ [s]| − |[s]| = a + b − s.

The following is the key lemma for the proof of Theorem 10.2.

We will prove the lemma in the next chapter.

Lemma 10.4. Let n, k, t and r be integers with k ≥ t ≥ 2, 0 ≤ r ≤

[n]

n−t

be a shifted t-intersecting family

2 , and n > 2k − t. Let F ⊂ k

with |F| = M (n, k, t). If

n < (k − t + 1) 2 +



t−1

r



,



then h(F) ≥ t + 2r. (We put no restriction on the upper bound on n

if r = 0.)

Having this lemma, we can prove Theorem 10.2 relatively easily.

We just recall a simple fact about a complement family. For a family

of subsets F ⊂ 2[n] let F c denote the complement family, that is,

F c = {[n] \ F : F ∈ F}.



10. The Ahlswede–Khachatrian Theorem



57



[n]

c

It is readily verified that if F ⊂ [n]

k is t-intersecting, then F ⊂ k

is t -intersecting, where k = n − k and t = n − 2k + t. Moreover, if F

is shifted, then F c is also shifted but in the opposite direction, that is,

if F ∈ F c , i ∈ F , j ∈ F , and 1 ≤ i < j ≤ n, then (F \ {i}) ∪ {j} ∈ F c .



Proof of Theorem 10.2 for shifted families. Let n ≥ k ≥ t ≥

1. In Chapter 4 we dealt with the case t = 1 already, so we may

be a shifted t-intersecting family

assume that t ≥ 2. Let F ⊂ [n]

k

of maximum size. Recall that shifting preserves the size and the tintersecting property of the family. We will show that F is actually

F (k) (n, t, r) or F (k) (n, t, r + 1). Let k = n − k, t = n − 2k + t, and

r = k − t − r. For F ∈ F let us define its head part F h and tail part

F t by F h = F ∩ [t + 2r] and F t = F \ F h .

Case (i). Assume (10.2). Then, by Lemma 10.4, the head size of

F is such that h(F) ≥ t + 2r, and F is exchange stable on [t + 2r].

Next we will verify that the complement family has the corresponding

exchange property. By direct computation we see that (10.2) can be

rewritten as

t −1

r −1

.

(k − t + 1) 2 +

< n < (k − t + 1) 2 +

r +1

r

is t -intersecting and

Recall that the complement family F c ⊂ [n]

k

reverse shifted. So, again by Lemma 10.4, it follows that h(F c ) ≥

t + 2r = n − (t + 2r). Since the direction of shifting for F c is the

opposite, the head of F c starts from n, so F c is (and of course also

F is) exchange stable on [t + 2r + 1, n].

If |F h | ≥ t + r for all F ∈ F, then F ⊂ F (k) (n, t, r), and by the

maximality, F = F (k) (n, t, r).

So we may assume that |F h | ≤ t + r − 1 for some F ∈ F. By

the shiftedness we may assume that |F h | = t + r − 1. Since F is

exchange stable on [t + 2r], all (t + r − 1)-element subsets of [t + 2r]

are contained in {F ∩ [t + 2r] : F ∈ F}. Thus by Claim 10.3 we can

find F, G ∈ F such that |F h | = |Gh | = t + r − 1 and

|F h ∩ Gh | = |F ∩ G ∩ [t + 2r]| ≤ 2(t + r − 1) − (t + 2r) = t − 2.

We also have |F t | = |Gt | = k − (t + r − 1) and

|F t | + |Gt | − |[t + 2r + 1, n]| = 2k − t − n + 2 ≤ 1.



58



10. The Ahlswede–Khachatrian Theorem



Since F is exchange stable on [t+2r+1, n], we apply Claim 10.3 again,

and we may assume that |F t ∩ Gt | = |F ∩ G ∩ [t + 2r + 1, n]| ≤ 1.

Therefore we have |F ∩ G| = |F h ∩ Gh | + |F t ∩ Gt | ≤ (t − 2) + 1 < t,

which contradicts the t-intersecting property.

Case (ii). Assume (10.3). Then we have n < (k − t + 1) 2 + t−1

r

and h := h(F) ≥ t + 2r by Lemma 10.4. Since (10.3) is equivalent to

n = (k − t + 1) 2 +

we also get n < (k − t + 1)(2 +



t −1

r −1 )



t −1

r



,



and



h := h(F c ) ≥ t + 2(r − 1) = n − (t + 2r + 2).

Thus F is exchange stable both on [t + 2r] and on [t + 2r + 3, n].

If |F h | ≥ t + r for all F ∈ F, then F ⊂ F (k) (n, t, r), and by the

maximality we have F = F (k) (n, t, r).

So we may assume that |F h | ≤ t + r − 1 for some F ∈ F. By the

shiftedness we may assume that |F h | = t + r − 1. Moreover, since F

is exchange stable on [t + 2r], we may assume that F ∩ [t + 2r + 2] =

[r + 2, t + 2r + 2].

If |G∩[t+2r +2]| ≥ t+r +1 for all G ∈ F, then F ⊂ F (k) (n, t, r +

1), and by the maximality we have F = F (k) (n, t, r + 1).

So we may assume that |G ∩ [t + 2r + 2]| ≤ t + r for some G ∈ F.

By the shiftedness we may assume that G ∩ [t + 2r + 2] = [t + r].

Then we have |F ∩G∩[t+2r +2]| = t−1. Let R = [n]\[t+2r +2].

As for the remaining part, we have |F ∩ R| = k − (t + r + 1) and

|G ∩ R| = k − (t + r). Since F is exchange stable on R, we can

use Claim 10.3. Then we may assume that F ∩ G ∩ R = ∅ because

|F ∩ R| + |G ∩ R| − |R| = 2k − t − n + 1 ≤ 0. Thus |F ∩ G| = t − 1,

which contradicts the t-intersecting property.

We have proved Theorem 10.2 for shifted families (assuming the

key lemma which will be shown in the next chapter). To complete

the proof we need to show that if a family can be made to be one of

the extremal configurations after applying shifting operations, then

it was isomorphic to the extremal configuration from the beginning.

In fact the following holds.



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Chapter 9. The Erdős matching conjecture

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