2 Application: the memory allocation problem
Tải bản đầy đủ  0trang
111
8.3 Sperner’s theorem
{k + 1, . . . , 2k} the set of the last k elements. By Theorem 8.3, both S and T
k
have symmetric chain decompositions of their posets of subsets into m = k/2
symmetric chains: 2S = C1 ∪ · · · ∪ Cm and 2T = D1 ∪ · · · ∪ Dm . Corresponding
to the chain
Ci = {x1 , . . . , xj } ⊂ {x1 , . . . , xj , xj+1 } ⊂ . . . ⊂ {x1 , . . . , xh } (j + h = k)
we associate the sequence (not the set!) Ci = (x1 , x2 , . . . xh ). Then every subset of S occurs as an initial part of one of the sequences C1 , . . . , Cm . Similarly
let D1 , . . . , Dm be sequences corresponding to the chains D1 , . . . , Dm . If we
let Di denote the sequence obtained by writing Di in reverse order, then
every subset of T occurs as a ﬁnal part of one of the Di . Next, consider the
sequence
L = D1 C1 D1 C2 . . . D1 Cm . . . Dm C1 Dm C2 . . . Dm Cm .
We claim that L is a universal sequence for the set {1, . . . , n}. Indeed, each
of its subsets A can be written as A = E ∪ F where E ⊆ S and F ⊆ T .
Now F occurs as the ﬁnal part of some Df and E occurs as the initial part
of some Ce ; hence, the whole set A occurs in the sequence L as the part of
Df Ce . Thus, the sequence L contains every subset of {1, . . . , n}. The length
of the sequence L is at most km2 = k
k
k/2
∼ 2k
2
kπ ,
k 2
k/2 .
Since, by Stirling’s formula,
2
the length of the sequence is km2 ∼ k kπ
· 22k =
2 n
π2 .
8.3 Sperner’s theorem
A set system F is an antichain (or Sperner system) if no set in it contains
another: if A, B ∈ F and A = B then A ⊆ B. It is an antichain in the sense
that this property is the other extreme from that of the chain in which every
pair of sets is comparable.
Simplest examples of antichains over {1, . . . , n} are the families of all sets of
ﬁxed cardinality k, k = 0, 1, . . . , n. Each of these antichains has nk members.
Recognizing that the maximum of nk is achieved for k = n/2 , we conclude
n
. Are these antichains the largest ones?
that there are antichains of size n/2
The positive answer to this question was found by Emanuel Sperner in
1928, and this result is known as Sperner’s Theorem.
Theorem 8.5 (Sperner 1928). Let F be a family of subsets of an n element
n
set. If F is an antichain then F  ≤ n/2
.
A considerably sharper result, Theorem 8.6 below, is due to Lubell (1966).
The same result was discovered by Meshalkin (1963) and (not so explicitly)
by Yamamoto (1954). Although Lubell’s result is also a rather special case
112
8 Chains and Antichains
of an earlier result of Bollobás (see Theorem 8.8 below), inequality (8.1) has
become known as the LYM inequality.
Theorem 8.6 (LYM Inequality). Let F be an antichain over a set X of n
elements. Then
−1
n
≤ 1.
(8.1)
A
A∈F
Note that Sperner’s theorem follows from this bound: recognizing that
is maximized when k = n/2 , we obtain
F  ·
n
n/2
−1
≤
A∈F
n
A
n
k
−1
≤ 1.
We will give an elegant proof of Theorem 8.6 due to Lubell (1966) together
with one of its reformulations which is pregnant with further extensions.
First proof. For each subset A, exactly A!(n − A)! maximal chains over
X contain A. Since none of the n! maximal chains meet F more than once,
we have A∈F A!(n − A)! ≤ n!. Dividing this inequality by n! we get the
desired result.
Second proof. The idea is to associate with each subset A ⊆ X, a permutation
on X, and count their number. For an aelement set A let us say that a
permutation (x1 , x2 , . . . , xn ) of X contains A if {x1 , . . . , xa } = A. Note that
A is contained in precisely a!(n − a)! permutations. Now if F is an antichain,
then each of n! permutations contains at most one A ∈ F. Consequently,
A∈F a!(n − a)! ≤ n!, and the result follows. To recover the ﬁrst proof,
simply identify a permutation (x1 , x2 , . . . , xn ) with the maximal chain {x1 } ⊂
{x1 , x2 } ⊂ . . . ⊂ {x1 , x2 . . . , xn } = X.
8.4 The Bollobás theorem
The following theorem due to B. Bollobás is one of the cornerstones in extremal set theory. Its importance is reﬂected, among other things, by the list
of diﬀerent proofs published as well as the list of diﬀerent generalizations.
In particular, this theorem implies both Sperner’s theorem and the LYM
inequality.
Theorem 8.7 (Bollobás’ theorem). Let A1 , . . . , Am be aelement sets and
B1 , . . . , Bm be belement sets such that Ai ∩ Bj = ∅ if and only if i = j. Then
m ≤ a+b
a .
This is a special case of the following result.
113
8.4 The Bollobás theorem
Theorem 8.8 (Bollobás 1965). Let A1 , . . . , Am and B1 ,. . ., Bm be two sequences of sets such that Ai ∩ Bj = ∅ if and only if i = j. Then
m
i=1
ai + b i
ai
−1
≤ 1,
(8.2)
where ai = Ai  and bi = Bi .
As we already mentioned, due to its importance, there are several diﬀerent
proofs of this theorem. We present two of them.
First proof. Our goal is to prove that (8.2) holds for every family F =
{(Ai , Bi ) : i = 1, . . . , m} of pairs of sets such that Ai ∩ Bj = ∅ precisely
when i = j. Let X be the union of all sets Ai ∪ Bi . We argue by induction
on n = X. For n = 1 the claim is obvious, so assume it holds for n − 1 and
prove it for n. For every point x ∈ X, consider the family of pairs
Fx := {(Ai , Bi \ {x}) : x ∈ Ai }.
Since each of these families Fx has less than n points, we can apply the induction hypothesis for each of them, and sum the corresponding inequalities (8.2).
−1
i
The resulting sum counts n − ai − bi times the term aia+b
, corresponding
i
i −1
to points x ∈ Ai ∪ Bi , and bi times the term ai +b
ai
points x ∈ Bi ; the total is ≤ n. Hence we obtain that
m
(n − ai − bi )
i=1
ai + b i
ai
−1
+ bi
−1
ai + b i − 1
ai
, corresponding to
−1
≤ n.
k
Since k−1
= k−l
l
l , the ith term of this sum is equal to n ·
k
Dividing both sides by n we get the result.
ai +bi −1
.
ai
Second proof. Lubell’s method of counting permutations. Let, as before, X
be the union of all sets Ai ∪ Bi . If A and B are disjoint subsets of X then
we say that a permutation (x1 , x2 , . . . , xn ) of X separates the pair (A, B) if
no element of B precedes an element of A, i.e., if xk ∈ A and xl ∈ B imply
k < l.
Each of the n! permutations can separate at most one of the pairs (Ai , Bi ),
i = 1, . . . , m. Indeed, suppose that (x1 , x2 , . . . , xn ) separates two pairs
(Ai , Bi ) and (Aj , Bj ) with i = j, and assume that max{k : xk ∈ Ai } ≤
max{k : xk ∈ Aj }. Since the permutation separates the pair (Aj , Bj ),
min{l : xl ∈ Bj } > max{k : xk ∈ Aj } ≥ max{k : xk ∈ Ai }
which implies that Ai ∩ Bj = ∅, contradicting the assumption.
We now estimate the number of permutations separating one ﬁxed pair.
If A = a and B = b and A and B are disjoint then the pair (A, B) is
114
8 Chains and Antichains
separated by exactly
n
a+b
a!b!(n − a − b)! = n!
a+b
a
−1
n
counts the number of choices for the positions of
permutations. Here a+b
A ∪ B in the permutation; having chosen these positions, A has to occupy
the ﬁrst a places, giving a! choices for the order of A, and b! choices for the
order of B; the remaining elements can be chosen in (n − a − b)! ways.
Since no permutation can separate two diﬀerent pairs (Ai , Bi ), summing
up over all m pairs we get all permutations at most once
m
n!
i=1
ai + b i
ai
−1
≤ n!
and the desired bound (8.2) follows.
Tuza (1984) observed that Bollobás’s theorem implies both Sperner’s theorem and the LYM inequality. Let A1 , . . . , Am be an antichain over a set X.
Take the complements Bi = X \ Ai and let ai = Ai  for i = 1, . . . , m. Then
bi = n − ai and by (8.2)
m
i=1
n
Ai 
m
−1
=
i=1
ai + b i
ai
−1
≤ 1.
Due to its importance, the theorem of Bollobás was extended in several
ways.
Theorem 8.9 (Tuza 1985). Let A1 , . . . , Am and B1 , . . . , Bm be collections of
sets such that Ai ∩ Bi = ∅ and for all i = j either Ai ∩ Bj = ∅ or Aj ∩ Bi = ∅
(or both) holds. Then for any real number 0 < p < 1, we have
m
pAi  (1 − p)Bi  ≤ 1.
i=1
Proof. Let X be the union of all sets Ai ∪ Bi . Choose a subset Y ⊆ X at
random in such a way that each element x ∈ X is included in Y independently
and with the same probability p. Let Ei be the event that Ai ⊆ Y ⊆ X \
Bi . Then for their probabilities we have Pr [Ei ] = pAi  (1 − p)Bi  for every
i = 1, . . . , m (see Exercise 8.4). We claim that, for i = j, the events Ei
and Ej cannot occur at the same time. Indeed, otherwise we would have
Ai ∪ Aj ⊆ Y ⊆ X \ (Bi ∪ Bj ), implying Ai ∩ Bj = Aj ∩ Bi = ∅, which
contradicts our assumption.
Since the events E1 , . . . , Em are mutually disjoint, we conclude that
Pr [E1 ] + · · · + Pr [Em ] = Pr [E1 ∪ · · · ∪ Em ] ≤ 1, as desired.
8.5 Strong systems of distinct representatives
115
The theorem of Bollobás also has other important extensions. We do not
intend to give a complete account here; we only mention some of these results
without proof. More information about Bollobástype results can be found,
for example, in a survey by Tuza (1994).
A typical generalization of Bollobás’s theorem is its following “skew version.” This result was proved by Frankl (1982) by modifying an argument of
Lovász (1977) and was also proved in an equivalent form by Kalai (1984).
Theorem 8.10. Let A1 , . . . , Am and B1 , . . ., Bm be ﬁnite sets such that
Ai ∩ Bi = ∅ and Ai ∩ Bj = ∅ if i < j. Also suppose that Ai  ≤ a and Bi  ≤ b.
Then m ≤ a+b
a .
We also have the following “threshold version” of Bollobás’s theorem.
Theorem 8.11 (Füredi 1984). Let A1 , . . . , Am be a collection of asets and
B1 , . . . , Bm be a collection of bsets such that Ai ∩ Bi  ≤ s and Ai ∩ Bj  > s
.
for every i = j. Then m ≤ a+b−2s
a−s
8.5 Strong systems of distinct representatives
Recall that a system of distinct representatives for the sets S1 , S2 , . . . , Sk is
a ktuple (x1 , x2 , . . . , xk ) where the elements xi are distinct and xi ∈ Si for
all i = 1, 2, . . . , k. Such a system is strong if we additionally have xi ∈ Sj for
all i = j.
Theorem 8.12 (Füredi–Tuza 1985). In any family of more than r+k
sets
k
of cardinality at most r, at least k + 2 of its members have a strong system
of distinct representatives.
Proof. Let F = {A1 , . . . , Am } be a family of sets, each of cardinality at
most r. Suppose that no k + 2 of these sets have a strong system of distinct
representatives. We will apply the theorem of Bollobás to prove that then
m ≤ r+k
k . Let us make an additional assumption that our sets form an
antichain, i.e., that no of them is a subset of another one. By Theorem 8.8
it is enough to prove that, for every i = 1, . . . , m there exists a set Bi , such
that Bi  ≤ k, Bi ∩ Ai = ∅ and Bi ∩ Aj = ∅ for all j = i.
Fix an i and let Bi = {x1 , . . . , xt } be a minimal set which intersects all
the sets Aj \ Ai , j = 1, . . . , m, j = i. (Such a set exists because none of
these diﬀerences is empty.) By the minimality of Bi , for every ν = 1, . . . , t
there exists a set Sν ∈ F such that Bi ∩ Sν = {xν }. Fix an arbitrary element
yi ∈ Ai . Then (yi , x1 , . . . , xt ) is a strong system of distinct representatives
for t + 1 sets Ai , S1 , . . . , St . By the indirect assumption, we can have at most
k + 1 such sets. Therefore, Bi  = t ≤ k, as desired.
In the case when our family F is not an antichain, it is enough to order
the sets so that Ai ⊆ Aj for i < j, and apply the skew version of Bollobás’s
theorem.
116
8 Chains and Antichains
8.6 Unionfree families
A family of sets F is called runionfree if A0 ⊆ A1 ∪ A2 ∪ · · · ∪ Ar holds for
all distinct A0 , A1 , . . . , Ar ∈ F. Thus, antichains are runionfree for r = 1.
Let T (n, r) denote the maximum cardinality of an runionfree family F
over an nelement underlying set. This notion was introduced by Kautz and
Singleton (1964). They proved that
Ω(1/r2 ) ≤
log2 T (n, r)
≤ O(1/r).
n
This result was rediscovered several times in information theory, in combinatorics by Erdős, Frankl, and Füredi (1985), and in group testing by Hwang
and Sós (1987). Dyachkov and Rykov (1982) obtained, with a rather involved
proof, that
log2 T (n, r)
≤ O(log2 r/r2 ).
n
Recently, Ruszinkó (1994) gave a purely combinatorial proof of this upper
bound. Shortly after, Füredi (1996) found a very elegant argument, and we
present it below.
Theorem 8.13 (Füredi 1996). Let F be a family of subsets of an nelement
underlying set X, and r ≥ 2. If F is runionfree then F  ≤ r + nt where
t := (n − r)
That is,
r+1
2
.
log2 F /n ≤ O log2 r/r2 .
Proof. Let Ft be the family of all members of F having their own tsubset.
That is, Ft contains all those members A ∈ F for which there exists a telement subset T ⊆ A such that T ⊆ A for every other A ∈ F. Let Tt be
the family of these tsubsets; hence Tt  = Ft . Let F0 := {A ∈ F : A < t},
and let T0 be the family of all tsubsets of X containing a member of F0 , i.e.,
T0 := {T : T ⊆ X, T  = t and T ⊃ A for some A ∈ F0 }.
The family F is an antichain. This implies that Tt and T0 are disjoint. The
family F0 is also an antichain, and since t < n/2, we know from Exercise 5.11
that F0  ≤ T0 . Therefore,
F0 ∪ Ft  ≤ Tt  + T0  ≤
It remains to show that the family
F := F \ (F0 ∪ Ft )
n
.
t
(8.3)
117
Exercises
has at most r members. Note that A ∈ F if and only if A ∈ F, A ≥ t and
for every tsubset T ⊆ A there is an A ∈ F such that A = A and A ⊇ T .
We will use this property to prove that A ∈ F , A1 , A2 , . . . , Ai ∈ F (i ≤ r)
imply
A \ (A1 ∪ · · · ∪ Ai ) ≥ t(r − i) + 1.
(8.4)
To show this, assume the opposite. Then the set A \ (A1 ∪ · · · ∪ Ai ) can be
written as the union of some (r − i) telement sets Ti+1 , . . . Tr . Therefore, A
lies entirely in the union of A1 , . . . , Ai and these sets Ti+1 , . . . , Tr . But, by
the choice of A, each of the sets Tj lies in some other set Aj ∈ F diﬀerent
from A. Therefore, A ⊆ A1 ∪ · · · ∪ Ar , a contradiction.
Now suppose that F has more than r members, and take any r + 1 of
them A0 , A1 , . . . , Ar ∈ F . Applying (8.4) we obtain
r

Ai  = A0  + A1 \ A0  + A2 \ (A0 ∪ A1 ) + · · ·
i=0
+Ar \ (A0 ∪ A1 ∪ · · · ∪ Ar−1 )
≥ tr + 1 + t(r − 1) + 1 + t(r − 2) + 1 + · · · + t · 0 + 1
=t·
r+1
r(r + 1)
+r+1=t
+ r + 1.
2
2
By the choice of t, the righthand side exceeds the total number of points n,
which is impossible. Therefore, F cannot have more than r distinct members.
Together with (8.3), this yields the desired upper bound on F .
Exercises
8.1. Let F be an antichain consisting of sets of size at most k ≤ n/2. Show
that F  ≤ nk .
8.2. Derive from Bollobás’s theorem the following weaker version of Theorem 8.11. Let A1 , . . . , Am be a collection of aelement sets and B1 , . . . , Bm be
a collection of belement sets such that Ai ∩Bi  = t for all i, and Ai ∩Bj  > t
for i = j. Then m ≤ a+b−t
a−t .
8.3. Show that the upper bounds in Bollobás’s and Füredi’s theorems (Theorems 8.7 and 8.11) are tight. Hint: Take two disjoint sets X and S of respective
sizes a + b − 2s and s. Arrange the selement subsets of X in any order: Y1 , Y2 , . . .. Let
Ai = S ∪ Yi and Bi = S ∪ (X \ Yi ).
8.4. Use the binomial theorem to prove the following. Let 0 < p < 1 be a real
number, and C ⊂ D be any two ﬁxed subsets of {1, . . . , n}. Then the sum of
pA (1 − p)n−A over all sets A such that C ⊆ A ⊆ D, equals pC (1 − p)n−D .
118
8 Chains and Antichains
8.5. (Frankl 1986). Let F be a kuniform family, and suppose that it is intersection free, i.e., that A∩B ⊂ C for any three sets A, B and C of F . Prove that
k
F  ≤ 1+ k/2
. Hint: Fix a set B0 ∈ F , and observe that {A∩B0 : A ∈ F , A = B0 }
is an antichain over B0 .
8.6. Let A1 , . . . , Am be a family of subsets of an nelement set, and suppose
that it is convex in the following sense: if Ai ⊆ B ⊆ Aj for some i, j, then B
m
belongs to the family. Prove that the absolute value of the sum i=1 (−1)Ai 
n
does not exceed n/2 . Hint: Use the chain decomposition theorem. Observe that
the contribution to the sum from each of the chains is of the form ±(1 − 1 + 1 − 1 . . .),
and so this contribution is 1, −1 or 0.
8.7. Let x1 , . . . , xn be real numbers, xi ≥ 1 for each i, and let S be the set of
all numbers, which can be obtained as a linear combinations α1 x1 +. . .+αn xn
with αi ∈ {−1, +1}. Let I = [a, b) be any interval (in the real line) of length
n
b − a = 2. Show that I ∩ S ≤ n/2
. Hint: Associate with each such sum
ξ = α1 x1 + . . . + αn xn the corresponding set Aξ = {i : αi = +1} of indices i for which
αi = +1. Show that the family of sets Aξ for which ξ ∈ I, forms an antichain and
apply Sperner’s theorem. Note: Erdős (1945) proved a more general result that
if b − a = 2t then I ∩ S is less than or equal to the sum of the t largest
binomial coeﬃcients ni .
8.8. Let P be a ﬁnite poset and suppose that the largest chain in it has size
r. We know (see Theorem 8.1) that P can be partitioned into r antichains.
Show that the following argument also gives the desired decomposition: let
A1 be the set of all maximal elements in P ; remove this set from P , and let
A2 be the set of all maximal elements in the reduced set P \ A1 , etc.
8.9. Let F = {A1 , . . . , Am } and suppose that
Ai ∩ Aj  <
1
min{Ai , Aj } for all i = j.
r
Show that F is runionfree.
8.10. Let F = {A1 , . . . , Am } be an runionfree family. Show that then
i∈I Ai =
j∈J Aj for any two distinct nonempty subsets I, J of size at
most r.
9. Blocking Sets and the Duality
In this chapter we will consider one of the most basic properties of set systems — their duality. The dual of a family F consists of all (minimal under
setinclusion) sets that intersect all members of F . Dual families play an important role in many applications, boolean function complexity being just
one example.
9.1 Duality
A blocking set of a family F is a set T that intersects (blocks) every member
of F . A blocking set of F is minimal if none of its proper subsets is such.
(Minimal blocking sets are also called transversals of F .) The family of all
minimal blocking sets of F is called its dual and is denoted by b (F).
Proposition 9.1. For every family F we have b (b (F )) ⊆ F. Moreover, if F
is an antichain then b (b (F )) = F .
Proof. To prove the ﬁrst claim, take a set B ∈ b (b (F )). Observe that none of
the sets A \ B with A ∈ F can be empty: Since B is a minimal blocking set of
b (F ), it cannot contain any member A of F as a proper subset, just because
each member of F is a blocking set of b (F ). Assume now that B ∈ F. Then,
for each set A ∈ F, there is a point xA ∈ A \ B. The set {xA : A ∈ F } of all
such points is a blocking set of F , and hence, contains at least one minimal
blocking set T ∈ b (F ). But this is impossible, because then B must intersect
the set T which, by it deﬁnition, can contain no element of B.
To prove the second claim, let F be an antichain, and take any A ∈ F.
We want to show A is in b (b (F )). Each element of b (F ) intersects A, so A
is a blocking set for b (F ). Therefore A contains (as a subset) some minimal
blocking set B ∈ b (b (F )). Since b (b (F )) is a subset of F (by the ﬁrst part
of the proof), the set B must belong to F . Hence, A and its subset B are
both in F . But F is an antichain, therefore A = B, so A ∈ b (b (F )).
S. Jukna, Extremal Combinatorics, Texts in Theoretical Computer Science.
An EATCS Series, DOI 10.1007/9783642173646_9,
© SpringerVerlag Berlin Heidelberg 2011
119
120
9 Blocking Sets and the Duality
1
2
. . .
r
Fig. 9.1 Example of a selfdual family.
Let us consider the following problem of “keys of the safe” (Berge 1989).
An administrative council is composed of a set X of individuals. Each of
them carries a certain weight in decisions, and it is required that only subsets
A ⊆ X carrying a total weight greater than some threshold ﬁxed in advance,
should have access to documents kept in a safe with multiply locks. The
minimal “coalitions” which can open the safe constitute an antichain F. The
problem consists in determining the minimal number of locks necessary so
that by giving one or more keys to every individual, the safe can be opened
if and only if at least one of the coalitions of F is present.
Proposition 9.2. For every family F of minimal coalitions, b (F)  locks are
enough.
Proof. Let b (F ) = {T1 , . . . , T }. Then give the key of the ith lock to all the
members of Ti . It is clear that then every coalition A ∈ F will have the keys
to all locks, and hence, will be able to open the safe. On the other hand, if
some set B of individuals does not include a coalition then, by Proposition 9.1,
the set B is not a blocking set of b (F ), that is, B ∩ Ti = ∅ for some i. But
this means that people in B lack the ith key, as desired.
A family F is called selfdual if b (F ) = F .
For example, the family of all kelement subsets of a (2k − 1)element set
is selfdual. Another example is the family of r + 1 sets, one of which has r
elements and the remaining r sets have 2 elements (see Fig. 9.1).
What other families are selfdual? Our nearest goal is to show that a family
is selfdual if and only if it is intersecting and not 2colorable. Let us ﬁrst
recall the deﬁnition of these two concepts.
A family is intersecting if any two of its sets have a nonempty intersection.
The chromatic number χ(F ) of F ⊆ 2X is the smallest number of colors
necessary to color the points in X so that no set of F of cardinality > 1 is
monochromatic. It is clear that χ(F) ≥ 2 (as long as F is nontrivial, i.e.,
contains at least one set with more than one element).
The families with χ(F ) = 2 are of special interest and are called 2colorable.
In other words, F is 2colorable iﬀ there is a subset S such that neither S
nor its complement X \ S contain a member of F . It turns out that χ(F ) > 2
is a necessary condition for a family F to be selfdual.
For families of sets F and G, we write F G if every member of F contains
at least one member of G.
9.2 The blocking number
Proposition 9.3. (i) A family F is intersecting if and only if F
(ii) χ(F ) > 2 if and only if b (F ) F.
121
b (F).
Proof. (i) If F is intersecting then every A ∈ F is also a blocking set of F , and
hence, contains at least one minimal blocking set. Conversely, if F
b (F )
then every set of F contains a blocking set of F, and hence, intersects all
other sets of F .
(ii) Let us prove that χ(F ) > 2 implies b (F) F. If not, then there must
be a blocking set T of F which contains no set of F. But its complement
X \ T also contains no set of F , since otherwise T would not block all the
members of F . Thus (T, X \ T ) is a 2coloring of F with no monochromatic
set, a contradiction with χ(F ) > 2.
For the other direction, assume that b (F)
F but χ(F ) = 2. By the
deﬁnition of χ(F) there exists a set S such that neither S nor X \ S contain
a set of F . This, in particular, means that S is a blocking set of F which
together with b (F) F implies that S ⊇ A for some A ∈ F, a contradiction.
Corollary 9.4. Let F be an antichain. Then the following three conditions
are equivalent:
(1)
(2)
(3)
b (F ) = F;
F is intersecting and χ(F ) > 2;
both F and b (F ) are intersecting.
Proof. Equivalence of (1) and (2) follows directly from Proposition 9.3. Equivalence of (1) and (3) follows from the fact that both F and b (F) are antichains.
9.2 The blocking number
Recall that the blocking number τ (F ) of a family F is the minimum number
of elements in a blocking set of F , that is,
τ (F) := min {T  : T ∩ A = ∅ for every A ∈ F } .
We make two observations concerning this characteristic:
If F contains a kmatching, i.e., k mutually disjoint sets, then τ (F) ≥ k.
If F is intersecting, then τ (F ) ≤ minA∈F A.
A family F can have many smallest blocking sets, i.e., blocking sets of
size τ (F ). The following result says how many. The rank of a family F is the
maximum cardinality of a set in F .
Theorem 9.5 (Gyárfás 1987). Let F be a family of rank r, and let τ = τ (F ).
Then the number of blocking sets of F with τ elements is at most rτ .