We letf(m, k, l) denote the maximum number of subsets of [m] with no trace Kkand no chain of size l+ 1. This was introduced by Frankl in the following conjecture.
Conjecture 3.2.1 (Frankl [Fra89]) Assume m+l ≥2k. Then f(m, k, l)≤
m k−1
+
m k−2
+· · ·+ m
k−l
. (3.16)
Forl≥k we easily deducef(m, k, l) from the bound (2.1). The following lemma was found by R. Israel [Isr] which extends the argument in [Fra89].
Lemma 3.2.2
f(m, k, l)≤f(m−1, k−1, l−1) +f(m−1, k,2l). (3.17) Proof of Lemma 3.2.2 Let A be a simple matrix on m rows with no configuration Kk and no chain of size l+ 1. Decompose A in the standard way like in (2.6)
A=
00· · ·0 11· · ·1 B1B2 B2B3
(3.18) where B2 consists of those columns that are repeated when the first row of Ais removed. We see that [B1B2B3] is simple and has noKk. Also it cannot have a chain of length 2l+ 1 since then either [B1B2] or [B2B3] will have a chain of length l+ 1 contradicting our choice of A. Thus [B1B2B3] has at
most f(m−1, k,2l) columns. We see that B2 has no chain of length l since but to prove Conjecture 3.1.2(in view of Conjecture 3.2.1) we would like a 3 in place of the 4. A quick consequence of this lemma proves Conjecture 3.2.1 in one case.
Proof of Theorem 3.2.3 LetAbe the matrix associated with F. Use the standard decomposition of (2.6) and note that the argument of Lemma 3.2.2 yields
Thus the bound (3.20) follows and so an extremal family has equality in (3.22,3.23). We may assume by induction onkand without loss of generality that
But then B3 has at most one column, the column of all 0’s since a column with between 1 and k−2 1’s violates simplicity of A and any column of at least k−1 1’s yields a chain of sizek+ 1 in A (k−1 from B2 under 0’s and
and by equality in (3.23) we obtain|B3|= 1 and equality in (3.25). Thus by or the (0,1)-complement. The latter is impossible so B1 = [m−1]k−1
and B3 =
[m−1]
0
and soA has the desired form.
Later in this section we see that the case of equality forf(m, k, k) behaves entirely differently.
Theorem 3.2.4 Let F ⊆2[m] be an antichain with no trace K2. Then
|F | ≤m (3.27)
with equality if and only if F = Proof of Theorem 3.2.4 The result is true for m = 2,3 using Theo-rem 3.1.1 alone. Now repeat the standard decomposition (2.6) for the second row but using the fact you have an antichain so that for example, B2 = ∅.
Then
This yields the bound of (3.27) with equality if and only if we have equalities in both (3.30) and (3.31), so in particular |C3| = 1. Thus to avoid K2 in the first two rows we must haveC1 =∅ orC5 = ∅. Assume, without loss of generality, thatC5 =∅. We may decompose A as follows
A=
We have chosenDto consist of the rows ofC1 containing a [01] (which forces α and β in order to avoid a K2) and then we choose E similarly and F is in the remaining rows. It is obtained from |C3|= 1, that α=β.
We allow matrices to have 0 rows in this decomposition whereDisl1×k1, E isl2×k2, F isl3×k3 and each are simple, form an antichain and have no a chain in A, a contradiction. The same happens with E, so assume D is (m−2)×(m−2). Using induction and (3.28) we have that D = [m−2]1 or [m−2]m−3
. In the former case we quickly deduce that α (and β) is all 0’s to avoid a chain of two columns and so (3.28) holds forA. In the latter case we can assume m−2≥3 and so to avoid having aK2 the vectorα cannot have two 0’s but then a chain of two columns will be formed, a contradiction.
Theorem 3.2.5 Let F ⊆ 2[m] be an antichain with no trace K3. Assume with equality if and only if
F = us-ing Theorem 3.1.1. For m ≥ 6 use the decomposition of (3.29). Now [C1C2C3C4C5] is simple and has no K3. Also C3 is simple, an antichain,
This immediately yields the bound of (3.33) with equality if and only if we have equality in (3.35) and (3.36). By Theorem 3.2.4, C3 is [m−2]1
or the (0,1)-complement. Without loss of generality assume the former. Then because A is an antichain, C2, C4 =∅. Also C5 is either a column of 0’s or is empty. Now C1 is simple, an antichain with no K3, and so by induction
|C1| ≤ m−22
. In order to have equality in (3.35) we deduce |C1| = m−22 and|C5|= 1 so eitherC1 = [m−2]2
or the complement and soC5 is a column of 0’s. Thus C1 = [m−2]2 Proof of Theorem 3.2.6 The result is true for m = 5,6,7 using Theo-rem 3.1.1 alone. Assume m≥8 and let A be the matrix associated with F. Use the decomposition of (3.29). Now [C1C2C3C4C5] is simple and has no
If the inequality in (3.39) is strict then the inequality (3.37) holds as desired.
So assume (3.39) holds with equality. Then by Theorem 3.2.5, without loss of generality we may assume C3 = [m]2
. Similarly C1 is an antichain with no K4 and so by induction
|C1| ≤ m−23
. But now (3.37) follows.
3.2.1 A Surprising Construction.
In contrast to the previous results which tend to support the conjecture, the following result perhaps casts doubt on the conjecture.
Theorem 3.2.7 Let m, k, t be given satisfying t+k−1 ≤ m and k ≥ 1.
There exists anm×f(m, k, k)simple matrixA(m, k, t)with no configuration Kk and column sumst, t+ 1, t+ 2, . . . , t+k−1(hence no chain of sizek+ 1).
Proof of Theorem3.2.7 We use induction on m with the additional hy-pothesis for k≥2, t≥1 that
A(m, k−1, t)⊆A(m, k, t−1). (3.40)
We use the base cases:
A(m, k,0) =
[m]
k−1
∪ [m]
k−2
∪ · · · ∪ [m]
0
for k ≥2 (3.41)
A(m,1, t) = ~1t
~0m−t
, (3.42)
where~ap is the vector of p a’s. We can easily handle the case m = 1 since either k= 1 or t= 0. The inductive construction for k ≥2 and t≥1 is
A(m, k, t) =
11· · ·1 00· · ·0 A(m−1, k, t−1) A(m−1, k−1, t)
(3.43) Note thatk+(t−1) = (k−1)+t≤m. Now considerm >1. By construction (3.43) A(m, k, t) has f(m −1, k, k) +f(m −1, k −1, k −1) = f(m, k, k) columns and is simple with column sums t, t+ 1, . . . , t+k−1. If A(m, k, t) has a Kk using the first row then it must have a Kk−1 under the 0’s in A(m−1, k−1, t), a contradiction. If A(m, k, t) has a Kk not using the first row then, sinceA(m−1, k−1, t)⊆A(m−1, k, t−1), we must have aKk in A(m−1, k, t−1), a contradiction.
To verify (3.40), first consider t = 1. Then (3.41) ensures that A(m, k− 1,1)⊆A(m, k,0). For k= 2 note that
A(m,2, t) =
11· · ·1 00· · ·0 A(m−1,2, t−1) A(m−1,1, t)
(3.44)
and A(m,1, t+ 1) =
11· · ·1 A(m−1,1, t)
(3.45) We use induction to obtain A(m−1,1, t) ⊆ A(m−1,2, t−1), then (3.40) follows fork = 2. Fork >2 and t >1 we use the inductive constructions for A(m, k−1, t+1), A(m, k, t). By induction,A(m−1, k−1, t)⊆A(m−1, k, t−1) and A(m−1, k−2, t+ 1)⊆ A(m−1, k−1, t) so that A(m, k−1, t+ 1) ⊆
A(m, k, t).
This parallels the inductive construction of Anstee and Murty [AM85] for matrices of the same size with no trace [k]t
. Its interest is in showing that matrices achieving the bound for f(m, k, k) are hardly unique which is in considerable contrast to Theorems 3.1.1,3.2.4,3.2.5.
An alternative construction follows from the tower
A(p,1, k)⊆A(p,2, k−1)⊆ · · · ⊆A(p, k,1) (3.46)
withp=m−k−t. Taking the (0,1)-complement of the tower withpreplaced We obtain a new construction achieving equality for f(m,2k + 1,2k + 1) wherep=m−k−t as follows: Here we use the cross product construction defined in Definition 2.2.1. It is easy to verify that the matrix is simple, has no K2k+1 and column sums t, t+ 1, . . . , t+ 2k. The number of columns is
Let us first give a proof of the most remarkable property of osh.
Proof of Theorem 3.1.6 Use induction onmand secondarily on|F |, the result being easy for m = 1 as long as you remember the empty set. Now consider a family F of subsets of [m]. If no member of F contains m, then F ⊂2[m−1]and use induction. If all elements ofF containmthen if we define F0 ={E−m : E ∈ F }, we find osh(F) =osh(F0) and can use induction on m. Otherwise split F into two new families of subsets of [m]:
F00 ={E : E ∈ F, m 6∈ F }, (3.50)