Chapter 4. The Erdős–Ko–Rado Theorem via shifting
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20
4. The Erd˝
os–Ko–Rado Theorem via shifting
Now let k > 1 be ﬁxed. Let us suppose that the statement holds
for all (n , k ) with n ≥ 2k and k < k, and for all (n , k) with n >
n ≥ 2k. We will consider the case (n, k). In view of Proposition 2.9,
we may assume that F is shifted. Recall the deﬁnitions of F(n) and
F(¯
n) from (0.1) and (0.2).
Claim 4.2. Both F(¯
n) and F(n) are intersecting.
Proof. Since F(¯
n) ⊂ F, the statement is trivial for F(¯
n). Suppose
indirectly that G, H ∈ F(n) with G ∩ H = ∅. Since |G| + |H| =
2(k − 1) < n − 2, there exists i ∈ [n − 1] such that i ∈ G ∪ H. As F
is shifted, the set H ∪ {i} is also in F. However,
(H ∪ {i}) ∩ (G ∪ {n}) = H ∩ G = ∅,
a contradiction. This completes the proof of the claim.
Applying the induction hypothesis to both F(¯
n) ⊂
[n−1]
F(n) ⊂ k−1 gives us
|F| = |F(¯
n)| + |F(n)| ≤
(n − 1) − 1
(n − 1) − 1
+
k−1
(k − 1) − 1
[n−1]
k
=
and
n−1
,
k−1
which completes the proof of the theorem.
Recall from Exercise 2.12 that if F ⊂ 2[n] is 1-union then |F| ≤
2n−1 . To extend this result to s-union families let us deﬁne
(4.2)
K(n, q) =
{K ⊂ [n] : |K| ≤ p}
if q = 2p,
{K ⊂ [n] : |K ∩ [2, n]| ≤ p} if q = 2p + 1.
Exercise 4.3. Let s = n − q. Show that K(n, q) is s-union, that is,
|K ∪ K | ≤ q for all K, K ∈ K(n, q).
Exercise 4.4. Verify that the size |K(n, q)| can be represented as
if q = 2p + 1. Verify also
(i) pi=0 ni if q = 2p and (ii) 2 pi=0 n−1
i
that |K(n, n − 1)| = 2n−1 .
Theorem 4.5 (Katona union theorem). Let n > s > 0 and q = n−s,
and suppose that G ⊂ 2[n] is s-union, that is,
(4.3)
|G ∪ G | ≤ q
for all G, G ∈ G. Then either (i) or (ii) holds.
4. The Erd˝
os–Ko–Rado Theorem via shifting
(i) q = 2p and |G| ≤
(ii) q = 2p + 1 and |G|
p
n
i=0 i .
p
≤ 2 i=0 n−1
i
21
.
The following simple proof is due to Wang [114].
Proof. Apply induction on n. The initial case is n = s + 1. In this
case, q = 1 and the only maximal s-union family is {∅, {i}} for some
i ∈ [n]. Thus (ii) holds for p = 0. So suppose n ≥ s + 2, that is,
q ≥ 2. Using (ii) of Exercise 2.11, assume that G is shifted. As usual
let G(n) = {G \ {n} : n ∈ G ∈ G} and G(¯
n) = {G ∈ G : n ∈ G}. Since
G(¯
n) ⊂ G, it follows that |F ∪ F | ≤ q for F, F ∈ G(n).
Claim 4.6. For F, F ∈ G(n), |F ∪ F | ≤ q − 2.
Proof. Since F ∪ {n} and F ∪ {n} are in G, (4.3) implies |F ∪ F | ≤
q − 1. Suppose indirectly that equality holds, that is, |F ∪ F | = q − 1.
Since (n − 1) − (q − 1) = s ≥ 1, we can ﬁnd j ∈ [n − 1] \ (F ∪ F ).
By the shiftedness of G, the set F ∪ {j} is also in G. However,
|(F ∪ {n}) ∪ (F ∪ {j})| = |F ∪ F | + 2 = q + 1, a contradiction.
Let us apply the induction hypothesis to G(¯
n) and G(n), both in
2[n−1] . Adding up the two inequalities, we obtain the desired upper
bound in both cases (i) and (ii). The reason that this simple argument
works is the fact that q and q − 2 have the same parity. Let us do the
detailed calculation for case (ii). Note that we have q = 2p + 1 for
G(¯
n) while q − 2 = 2(p − 1) + 1 for G(n). Thus we get
p
|G(¯
n)| ≤ 2
i=0
p−1
|G(n)| ≤ 2
i=0
Noting that
n−2
i
+
p
(n − 1) − 1
i
n−1
n−2
=2
+2
,
0
i
i=1
(n − 1) − 1
i
=2
n−2
i−1
=
n−1
i
p
n−2
.
i−1
i=1
, we ﬁnally obtain
p
|G| = |G(¯
n)| + |G(n)| = 2
i=0
n−1
.
i
The calculation for case (i) is slightly simpler and we leave it to the
reader.
22
4. The Erd˝
os–Ko–Rado Theorem via shifting
It is easy to show that (ii) implies the Erd˝os–Ko–Rado Theorem.
[n]
Let n ≥ 2(p + 1) and q = 2p + 1 < n, and let F ⊂ p+1
be an
arbitrary intersecting family. Deﬁne G = F ∪ {G ⊂ [n] : |G| ≤ p}.
We show that G is (n − q)-union. In fact, if G, G ∈ F then
|G ∪ G | = |G| + |G | − |G ∩ G | ≤ 2(p + 1) − 1 = q,
and if G ∈ G and G ∈ G \ F then |G ∪ G | ≤ |G| + |G | ≤ q. Now
it follows from (ii) of Theorem 4.5 that |G| ≤ |K(n, q)|. Both G and
K(n, q) contain all subsets of size at most p of [n]. In excess K(n, q)
(p+1)-element sets containing 1. Thus |F| ≤ n−1
contains the n−1
p
p
follows.
Since A B ⊂ A∪B for all pairs of subsets, the following theorem
is formally stronger than the Katona union theorem.
Theorem 4.7 (Kleitman diameter theorem). Suppose that n > d > 0
are integers and F ⊂ 2[n] is a family of diameter at most d, that is,
|F F | ≤ d for all F, F ∈ F. Then according to the parity of d
either (i) or (ii) holds.
(i) d = 2p and |F| ≤
(ii) d = 2p + 1 and |F|
p
n
i=0 i .
p
≤ 2 i=0 n−1
i
.
Proof. In view of Proposition 2.4, squashing will not increase the diameter, max |F F |. Thus, by Proposition 2.5 we obtain a hereditary
family G ⊂ 2[n] such that |G| = |F| and diam(G) ≤ d.
Claim 4.8. G is (n − d)-union, that is, |G ∪ G | ≤ d for G, G ∈ G.
Indeed, by the hereditary property for G, G ∈ G, the set G =
G \ G is also in G. Now G ∪ G = G G implies the claim. Now we
get the desired inequalities from Theorem 4.5.
Chapter 5
Katona’s circle
Trying to solve extremal problems by looking straight at the whole of
2[n] or even [n]
k might be too diﬃcult or very complicated. Katona’s
ingenious idea was to look at such problems on a much smaller family
of subsets, that is, the circle. Let us deﬁne this circle.
For an arbitrary ordering of the n elements of [n] on a circle, we
consider only subsets that form an arc on this circle, that is, the elements of this set are consecutive. More formally, let (a1 , . . . , an ) be a
circular permutation of 1, 2, . . . , n. By circular we mean that we consider the element a1 to be the consecutive element of an and we do not
distinguish the n possible permutations (a1 , . . . , an ), (a2 , . . . , an , a1 ),
(a3 , . . . , an , a1 , a2 ), . . . , (an , a1 , a2 , . . . , an−1 ) that yield the same circular arrangement. There are (n − 1)! circular permutations and each
of them contains n subsets of size i, 1 ≤ i < n, forming an arc of the
circle. For a circular permutation π of [n], let C(π) denote the family
of all arcs of π. So |C(π)| = n2 − n because C(π) consists of n arcs of
size i for each 1 ≤ i < n.
We will always imagine that the n elements are arranged on the
circle in clockwise order. This will allow us to speak of the ﬁrst and
last elements of an arc.
Let us illustrate the use of the circle by giving an alternate proof
of the Erd˝os–Ko–Rado Theorem.
23
24
5. Katona’s circle
Theorem 5.1 (Erd˝os–Ko–Rado Theorem on the circle). Let n ≥ 2k
be positive integers and ﬁx a circular permutation of [n]. Let E =
{E1 , . . . , Em } be a collection of circular arcs of k elements such that
Ei ∩ Ej = ∅ for any i, j. Then m ≤ k.
Proof. Without loss of generality let A = (a1 , . . . , ak ) be a member of
E. Since E consists of subsets of size k, no circular arc of size k (except
for A itself) contains A completely. By the intersection property all
the remaining arcs Ei have either their ﬁrst or their last element in A.
The candidates for the ﬁrst element except for A are a2 , . . . , ak . For
1 ≤ i ≤ k − 1, we denote the arc starting at ai+1 by Si and the arc
terminating at ai by Ti . Note that 2k ≤ n implies that Si ∩ Ti = ∅.
Thus for each 1 ≤ i ≤ k − 1 at most one of the two subsets Si and Ti
is in E. Together with A this gives us m ≤ (k − 1) + 1 = k.
Katona deduced the Erd˝
os–Ko–Rado Theorem from the circle
version by simple double counting. Let n ≥ 2k. Note that a ﬁxed
k-element subset is an arc in exactly k!(n − k)! circular permutations.
Let F ⊂ [n]
k be an intersecting family. Let us count, in two diﬀerent
ways, the number N of pairs (F, π), where F ∈ F appears as an arc
of a circular permutation π, that is, F ∈ C(π). The following picture
shows an example for the case n = 4, k = 2, and F = 1, 2, 1, 3, 2, 3.
In this example the members of F are indicated by thick arcs, and
we see that N = 12.
(1234)
(1243)
1
{1, 2} 4
1
2
3
3
4
3
2
4
3
2
1
4
2
1
3
3
4
3
2
1
3
(1432)
1
3
4
1
2
(1423)
1
3
2
1
2
(1342)
1
2
4
1
{1, 3} 4
(1324)
1
4
2
4
3
4
2
4
2
3
1
1
1
1
1
1
{2, 3} 4
2
3
3
2
4
4
3
2
2
3
4
3
4
2
2
4
3
On the one hand, each F ∈ F appears in k!(n − k)! circular
permutations, implying N = |F|k!(n − k)!. On the other hand, there
5. Katona’s circle
25
are (n − 1)! circular permutations and each of them contains at most
k arcs of F by Theorem 5.1. Thus N ≤ (n − 1)!k. Therefore we get
|F|k!(n − k)! ≤ (n − 1)!k,
or equivalently
|F| ≤
n−1
.
k−1
Probably, a simpler way is by comparing F and the family
A1 =
A∈
[n]
k
:1∈A
for each circular permutation π. Since |C(π) ∩ A1 | = k for all π, we
always have
|C(π) ∩ F| ≤ |C(π) ∩ A1 |.
This implies
n−1
k−1
because each k-element subset of [n] occurs with the same multiplicity
k!(n − k)!. This completes an alternative proof of Theorem 4.1.
|F| ≤ |A1 | =
Let us next prove the following important theorem concerning
antichains. Recall that a family S ⊂ 2[n] is an antichain if S contains
no two members S and S such that S S .
Theorem 5.2 (Yamamoto’s inequality1 [112]). Suppose that S ⊂
2[n] is an antichain. Then
1
(5.1)
≤ 1.
n
S∈S
|S|
Noting that ∅ ∈ S (resp. [n] ∈ S) implies S = {∅} (resp. S =
{[n]}), in proving (5.1) we may assume that ∅, [n] ∈ S. Then we shall
consider arcs on a circular permutation π again, that is, we look at
S ∩ C(π).
Proposition 5.3. Suppose that S0 ⊂ C(π) is an antichain. Then
|S0 | ≤ n, and moreover, equality holds only if all members of S0 have
the same size.
1
This inequality has been rediscovered several times by several authors, including
Lubell [90] and Meshalkin [93], and it is also called the LYM inequality.
26
5. Katona’s circle
Proof. Note that if the last element of two arcs is the same, then the
longer one contains the shorter one. Consequently, for the antichain
S0 there can be at most one member of S0 for every last element.
This proves |S0 | ≤ n.
Should we have equality, there must be exactly one member of
S0 for every last element. If not all are of the same size, then we can
ﬁnd two consecutive last elements, say ai and ai+1 , such that the arc
in S0 corresponding to ai+1 is longer. However, this means that this
arc contains the shorter arc ending in ai , a contradiction.
Proof of Theorem 5.2. We count the number N of pairs (S, π),
where S ∈ S appears as an arc on a circular permutation π. Since
each S ∈ S appears in |S|!(n − |S|)! circular permutations, it follows
that N = S∈S |S|!(n − |S|)!. On the other hand, there are (n − 1)!
circular permutations and each of them contains at most n arcs of S
by Proposition 5.3. This yields N ≤ (n − 1)!n = n!. Thus
|S|!(n − |S|)! ≤ n!,
S∈S
and dividing both sides by n! gives (5.1).
Let us next deduce Sperner’s Theorem (Theorem 1.1) from Yamamoto’s inequality. For convenience we restate it here.
Theorem. Let S ⊂ 2[n] be an antichain. Then |S| ≤
over, equality holds if and only if S =
or
[n]
n
2
[n]
n
2
n
n
2
. More-
for n even and S =
[n]
n
2
for n odd.
Proof. Noting that nk attains its maximum (as a function of k) for
n+1
k = n2 in the even case, and for k = n−1
2 , 2 in the odd case, (5.1)
implies
1
1
|S|
=
≤
≤ 1,
n
n
n
n
2
S∈S
n
2
S∈S
n
|S|
that is,
|S| ≤
n
n
2
.
Moreover, in the case of equality one must have |S| = n2 for all S ∈ S
if n is even. If n is odd, we can only deduce that |S| = n2 or
5. Katona’s circle
|S| =
n
2
27
for all S ∈ S. However, by Proposition 5.3, |S| =
only possible if all S ∈ S have the same size. That is, S =
S=
[n]
n
2
n
n
2
[n]
n
2
is
or
.
Our next target is an old theorem of Erd˝
os. A family S ⊂ 2[n]
is called a t-antichain if S contains no t + 1 members S0 , S1 , . . . , St
forming a chain, that is, S0 S1 · · · St .
Let k1 , k2 , . . . , kt be such that kn1 , . . . , knt are the largest t binomial coeﬃcients. Note that for n−t odd, {k1 . . . , kt } = [ n−t+1
, n+t−1
],
2
2
n−t n+t
n−t
and for n−t even, {k1 . . . , kt } is either [ 2 , 2 −1] or [ 2 +1, n+t
2 ].
Proposition 5.4. Suppose that S ⊂ 2[n] is a t-antichain and π is a
circular permutation of [n]. Then we have
(i) |S ∩ C(π)| ≤ tn,
(ii)
S∈S∩C(π)
1
≤
|S|!(n − |S|)!
t
i=1
n
.
ki !(n − ki )!
Proof. The arcs with identical last element form a chain. This proves
that in S ∩ C(π) there can be at most t members with the same last
element, proving (i).
n
1
/n! and thus the t
= |S|
To prove (ii) we note that |S|!(n−|S|)!
largest values are for {k1 , . . . , kt }. By (i), the left-hand side (LHS)
of (ii) has at most tn terms. Each value of |S| can occur at most n
times. These facts imply (ii).
Theorem 5.5 (The Erd˝os t-antichain theorem). Let n ≥ t, and let
S ⊂ 2[n] be a t-antichain. Then we have
t
|S| ≤
i=1
n
.
ki
Proof. Let us sum (ii) of Proposition 5.4 over all (n − 1)! circular
permutations. Since each S ∈ S is counted |S|!(n − |S|)! times, it
follows that
t
|S| ≤ (n − 1)!
i=1
n
=
ki !(n − ki )!
t
i=1
n
,
ki
28
5. Katona’s circle
as needed.
Next we prove a very simple but useful statement about the shadof arcs on a circular permuows. Let k ≥ 2. For a family S ⊂ [n]
k
tation π, let σ π (S) denote the family of arcs of size k − 1 that are
contained in some member of S. If S consists of all n arcs of size k,
then |σ π (S)| = n.
be a family of arcs on
Proposition 5.6. Let k ≥ 2 and let S ⊂ [n]
k
a circular permutation π. Unless S consists of all n arcs of size k, it
follows that
|σ π (S)| ≥ |S| + 1.
Proof. To every S ∈ S associate the arc of size k − 1 obtained by
deleting the last element of S. This shows that |σ π (S)| ≥ |S|. Let
|S| < n. Then there are two consecutive arcs, say (a1 , . . . , ak ) and
(a2 , . . . , ak+1 ), such that the ﬁrst is in S but the second is not. Then
the arc (a2 , . . . , ak ) of size k − 1 is in σ π (S) but has not been counted
yet. This proves |σ π (S)| ≥ |S| + 1.
The next statement is about several — not necessarily distinct —
families. Such statements have proved useful in dealing with problems
concerning one family as well as being interesting in their own right.
Let t ≥ 2 and let F1 , . . . , Fr ⊂ 2[n] . We say that F1 , . . . , Fr
are r-cross union if F1 ∪ · · · ∪ Fr = [n] for all choices of Fi ∈ Fi ,
i = 1, . . . , r.
Proposition 5.7. Let 1 ≤ ki < n for 1 ≤ i ≤ r. Let Fi be a family
of arcs of size ki on a circular permutation π = (a1 , . . . , an ). Suppose
that k1 + · · · + kr ≥ n and F1 , . . . , Fr are r-cross union. Then we
have
r
(|Fi | + ki ) ≤ rn.
(5.2)
i=1
Proof. We use induction on k1 + · · · + kr . First we prove the base
case k1 + · · · + kr = n. The idea is very simple. For every 1 ≤ j ≤ n
(j)
(j)
(j)
consider the r arcs F1 , . . . , Fr , where the ﬁrst element of F1 is
(j)
(j)
aj , the ﬁrst element of F2 is aj+k1 , . . . , and the ﬁrst element of Fr
is aj+k1 +k2 +···+kr−1 . Then these r arcs partition the whole circle.
5. Katona’s circle
29
Therefore not all r can be members of the corresponding families.
(j)
That is, there is some i such that Fi ∈ Fi . For the n choices of j
we get altogether at least n missing sets, yielding
|F1 | + |F2 | + · · · + |Fr | ≤ rn − n = rn − (k1 + · · · + kr ),
which is equivalent to (5.2).
Now suppose that k1 + · · · + kr > n and we are in the induction
step. Suppose that we can choose i, 1 ≤ i ≤ r, such that ki ≥ 2
and |Fi | < n. Then we replace Fi with σ π (Fi ) and leave the other
families unchanged. In view of Proposition 5.6 we have that
|σ π (Fi )| + ki − 1 ≥ |Fi | + ki .
Consequently, (5.2) follows from the induction hypothesis.
The only remaining case is where after reordering the families we
have k1 = · · · = ks = 1 and |Fs+1 | = · · · = |Fr | = n for some s ≥ 0.
We may further assume that |Fs | < n. If ks+1 + · · · + kr ≥ n, then
|Fs+1 | = · · · = |Fr | = n implies that Fs+1 , . . . , Fr are not (r − s)cross union already, and F1 , . . . , Fr cannot be r-cross union. Thus
we may assume that ks+1 + · · · + kr ≤ n − 1 and so s ≥ 1.
Then without loss of generality |F1 | < n and therefore |F1 |+k1 ≤
n. Now consider the (r − 1)-cross union families F2 , . . . , Fr . Recall
that we have assumed k1 + · · · + kr ≥ n + 1 and k1 = 1. Thus
k2 + · · · + kr ≥ (n + 1) − k1 = n and we may apply the induction
r
hypothesis and obtain i=2 (|Fi |+ki ) ≤ (r−1)n. Adding |F1 |+1 ≤ n,
the inequality (5.2) follows.
We say that a family F ⊂ 2[n] is r-wise union if F1 ∪· · ·∪Fr = [n]
for all F1 , . . . , Fr ∈ F.
Theorem 5.8 ([43]). Let 0 < k < n, r ≥ 2 and kr ≥ n. If F ⊂
is r-wise union, then |F| ≤ n−1
k .
[n]
k
Proof. Let π be a circular permutation. Applying Proposition 5.7
to F1 = F2 = · · · = Fr := F ∩ C(π), we get |F ∩ C(π)| ≤ n − k.
Adding up this inequality over all (n − 1)! cyclic permutations gives
k!(n − k)!|F| ≤ (n − k)(n − 1)!, or equivalently |F| ≤ n−1
k .
30
5. Katona’s circle
Let us note that the above theorem is best possible. Indeed, more
if and only if F = Yk for
careful analysis shows that |F| = n−1
k
[n]
some Y ∈ n−1
, except for the case r = 2 and n = 2k; cf. [50].