Chapter 9. The Erdős matching conjecture
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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.