Journal ArticleDOI

# Partition-free families of sets

Peter Frankl
01 Aug 2019-Proceedings of The London Mathematical Society (Oxford University Press (OUP))-Vol. 119, Iss: 2, pp 440-468
TL;DR: In this paper, it was shown that the extremal families are not unique, and that they can be shown to be monotone if n = 3m+1 and n=3m+2.
Abstract: Let $m(n)$ denote the maximum size of a family of subsets which does not contain two disjoint sets along with their union. In 1968 Kleitman proved that $m(n) = {n\choose m+1}+\ldots +{n\choose 2m+1}$ if $n=3m+1$. Confirming the conjecture of Kleitman, we establish the same equality for the cases $n=3m$ and $n=3m+2$, and also determine all extremal families. Unlike the case $n=3m+1$, the extremal families are not unique. This is a plausible reason behind the relative difficulty of our proofs. We completely settle the case of several families as well.

### 1 Introduction

• This famous result served as the starting point of the presently burgeoning field of extremal set theory.
• The authors have a few results concerning this and some related questions that will appear in [10].

### 2 Basic tools

• The following lemma is a generalization of the main lemma from Kleitman’s paper [16].
• Let us choose the pairwise disjoint sets from the claim randomly with uniform distribution.

### 6 The proof of Theorem 4

• For a change, in this section the authors give a proof with a somewhat different and hopefully simpler analysis.
• Exactly three are present (and three are missing).
• Exactly two of the m- and 2m-sets are missing.
• Therefore, the equality cannot hold in this case.
• The authors conclude that they are in the situation c1 and in each triple there in exactly one present m-set and (2m+1)-set, moreover, both belong to the same family.

### 7 Discussion

• One natural direction to extend these results is to study r-partition-free families, defined in the introduction, as well as to study their r-partite analogues.
• Another natural generalization of partition-free families, that was overlooked so far, are the r-box-free families (also defined in the introduction).
• More generally, the authors may ask the following question.
• The following sharp result may be proved using a direct generalization of Kleitman’s argument [16].
• The authors sum up all the obtained inequalities and multiply them by the corresponding ( n s1 ) (except for the first one, which they multiply by 1 2 ( n m ) ).

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arXiv:1706.00215v1 [math.CO] 1 Jun 2017
Partition-free families of sets
Peter Frankl, Andrey Kupavskii
Abstract
Let m(n) denote the maximum size of a family of subsets which does not contain
two disjoint sets along with their union. In 1968 Kleitman proved that m(n) =
n
m+1
+
. . . +
n
2m+1
if n = 3m + 1. Conﬁrming the conjecture of Kleitman, we establish the
same equality for the cases n = 3m and n = 3m + 2, and also determine all extremal
families. Unlike the case n = 3m + 1, the extremal families are not unique. This is a
plausible reason behind the relative diﬃculty of our proofs. We completely s ettle the
case of several families as well.
1 Introduction
For a positive integer n let [n] := {1, 2, . . . , n} be the standard n-element set and 2
[n]
its
power set. Subsets of 2
[n]
are called families
.
In 1928 Sperner [19] proved that if a family has size greater than
n
n/2
, then it must
contain two subsets F, G, such that F ( G. This famous result served as the starting point
of the presently burgeoning ﬁeld of extremal set theory.
Paul Erdős was behind many of the early developments. In connection with an ana lytic
problem of Littlewood and Oﬀord he proved [4] that if |F| is larger than the sum of the l
largest binomial coeﬃcients, then F contains a chain
F
0
( F
1
( . . . ( F
l
.
As much as by his results, Erdős also contributed to the development of extremal set
theory by his many problems. Under the inﬂuence of Erdős, the young and promising
physicist Daniel Kleitman switched t o mathematics and went on to solve lots of beautiful
problems. Many of these result and proofs are presented in the long chapter [11]. For an
introduction to the topic the reader is advised to consult the books [1], [2], [3], [1 3].
The generic extremal set theory problem is as follows. Suppose that F does not contain
a certain type of conﬁgurations. Determine or estimate the maximum of |F|. Let us give as
an example the problem which is the main topic of the present paper.
The f amily F 2
[n]
is called partition-free
if t here are no F
0
, F
1
, F
2
F satisfying
F
1
F
2
= and F
0
= F
1
F
2
. How large can |F| be?
Moscow Institute of Physics and Technology, Ecole Polytechnique Fédérale de Lausanne; Email:
kupavskii@yandex.ru Research supported in part by Swiss National Science Foundation grants no. 200020-
162884 and 200021-175977 and by the grant N 15-01-03530 of the Russian Foundation for Basic Research.
1

This problem was proposed to Kleitman by Erdős. Half a century ago Kleitman [16]
proved the following beautiful result.
Theorem 1 (Kleitman [16]). Suppose that n = 3m + 1 for some positive integer m. Let
F 2
[n]
be partition-free. Then
|F|
2m+1
X
t=m+1
n
t
. (1)
Example 1. Let n = 3m + l, 0 l 2 and deﬁne K(n) := {K [n] : m + 1 |K|
2m + 1}. It is evident that K(n) is partition-free. This shows that (1) is best possible.
It is conjectured in [16] t hat (1) holds for n = 3m and n = 3m + 2 as well. However,
for nearly half a century no progress was made on this problem. The main purpose of the
present paper is to conﬁrm Kleitman’s conjecture.
Let us mentions that Kleitman’s proof is elegant and short. Unfortunately, our proo f is
much more technical. A r eason that suggests that no easy proof exists might be that while
for n = 3 m + 1 K(n) from Example 1 is the unique
family attaining equality in (1), it is no
longer true for n = 3m + 2 and n = 3m.
Example 2. Let F 2
[n]
be partition-free and deﬁne F
d
:= {F [n + 1] : F [n] F}.
It is easy to see that F
d
is partition-free and satisﬁes |F
d
| = 2| F|. We call F
d
the double
of
F.
Note the identity
3m + 1
m + 1
+
2m + 1
m + 2
+ . . . +
3m + 1
2m + 1
=
3m + 1
m
+ . . . +
3m + 1
2m
,
implying 2
2m+1
X
t=m+1
3m + 1
t
=
2m+1
X
t=m+1
3m + 2
t
.
Consequently, |K(3m + 1)
d
| = |K(3m + 2)|.
Example 3. Fix an element x [n] and deﬁne
˜
K
x
(n) :=
n
F
[n]
m
: x F
o
K(n) \
n
G
[n]
2m + 1
: x G
o
.
Since
3m1
m1
=
3m1
2m
, one has |
˜
K
x
(3m)| = |K(3m)|. It can be checked easily that
˜
K(n) is
partition-free.
Theorem 2. Suppose that m 6 and n = 3m + 2 or n = 3m. If F 2
[n]
is partition-free,
then (1) holds. Moreover, fo r n = 3m+2 the equality in (1) is possible (up to the permutation
of the g round set) only when F = K( 3m + 2) or F = K( 3m + 1)
d
. For n = 3m the equality
in (1) is possi b l e only when F = K(3m) or F =
˜
K
x
(3m) for some x [n].
2

Let us remark also that in view of Example 2 t he inequality (1) for n = 3m + 1 follows
from the case n = 3m + 2.
Deﬁnition 1. Three families F
1
, F
2
, F
3
2
[n]
are called cross partition-free
, if there is no
possible choice of A F
1
, B F
2
, C F
3
such that one of those sets is equal to the disjoi nt
union of the other two.
For the case n = 3m + 1 and n = 3m + 2 one can extend (1) to this situation, although
in the case n = 3m + 2 we get a new extremal example.
Theorem 3. Suppose that F
1
, F
2
, F
3
2
[n]
are cross partition-free, n = 3m + 1 or n =
3m + 2, m 6. Then
|F
1
| + |F
2
| + |F
3
| 3
2m+1
X
t=m+1
n
t
. (2)
Moreover, for n = 3m + 1 the equality hol ds only when F
1
= F
2
= F
3
= K(3m + 1). For
n = 3m + 2 the equality up to the permutation of the indices of the families and the elements
of the ground set holds only in the following three cases:
F
1
= F
2
= F
3
= K(3m + 2) ,
F
1
= F
2
= F
3
= K(3m + 1)
d
,
F
1
= {F 2
[n]
: m+2 |F | 2m+1} , F
2
= F
3
= {F 2
[n]
: m+1 |F | 2m+2} .
Note that (2) implies (1 ) f or n = 3m + 1, and also gives the uniqueness of the extremal
family. At the same time, the n = 3m + 2 case of Theorem 3 implies the n = 3m + 2 case
of Theorem 2, along with the characterization of the extremal fa milies.
For n = 3 m one can do better.
Example 4. Let n = 3m and deﬁne
A :={A [n] : m |A| 2m + 1} (3)
B := C :={B [n] : m + 1 |B| 2m}. (4)
It is easy to check that A, B, C are cross partition-f r ee. Using
3m
m
=
2m+1
m
3m
2m+1
, it
follows that
|A| + |B| + |C| = 3|K(3m)| +
1
m
3m
2m + 1
.
Theorem 4. Suppose that F
1
, F
2
, F
3
2
[n]
are cross partition-free, n = 3m 18. Then
|F
1
| + |F
2
| + |F
3
| 3|K(3m)| +
1
m
3m
2m + 1
. (5)
Moreover, the equality holds only for the f amilies F
i
of the form as in Example 4.
3

It is natural to extend the notion of partition-free to more sets. L et r 2 be an integer.
A family F 2
[n]
said to be r-partition-free if there are no pairwise disjoint members
F
1
, . . . , F
r
F such that F
1
. . . F
r
F as well.
For n = rm + q, 0 q < r the most nat ura l construction of an r-partition-free family is:
K(n, r) := {K [n] : m + 1 | K| rm + r 1} .
In [5] it was proven that for n = rm + r 2 the unique optimal family is K(n, r) .
However, for r 3 the general situation is complex. It seems to be diﬃcult to ﬁnd
a plausible conjecture covering all congruence classes modulo r. We have a few results
concerning this and some r elated questions that will appear in [10]. Let us just state one of
them.
An r-box
is a conﬁguration consisting of 2
r
1 subsets, namely, r pairwise disjoint sets
B
1
, . . . , B
r
along with all possible non-empty unions of them.
Theorem 5 ([10]). Suppose that n = rm + r 2, m > r
2
and F 2
[n]
contains no r-box.
Then
|F| < |K(n, r)| or F = K(n, r) hold.
In the papers [7], [8], [9] the authors advanced in related problems of Erdős and Kleitman
on families t hat contain no s pa irwise disjoint sets.
Kleitman [17] considered the following related problem. What is the maximum size u(n)
of a family F 2
[n]
without three distinct members satisfying A B = C. The diﬀerence
with partition-free families is that one does not require A and B to be disjoint. Kleitman
proves u(n)
n
n/2
(1 +
c
n
) for some absolute constant c.
An “abstract” version of this problem was solved by Katona and Tarja n [15]. Let v(n)
denote t he maximum size of a family F without three distinct members A, B, C such that
A C and B C. Katona and Tarja n proved that v(2m + 1) = 2
2m
m
.
This result was the starting point of a lot of research. The central problem might be
stated as to determine t he largest size of subsets of the boo lean lattice without a certain
subposet. We refer the reader to the survey [12]. One of the important recent advancements
in the topic was the result of [18], where the authors showed that for any ﬁnite poset there
exists a constant C, such that the largest size o f a family without an induced copy of this
poset has size at most C
n
n/2
. However, the va lue of C is unknown in most cases, including
the “diamond” poset, and we hope that the methods developed in t he present paper would
be helpful to attack these problems.
Suppose that F 2
[n]
has no three sets A, B, C, such that |A B| s a nd A B = C.
How large a family F can be? A natural generalization of Example 1 suggests the family
K
s
(n) := { K [n] : m |K| < 2m s} for some m < n, where m is chosen so that the
cardinality of K
s
(n) is maximized. In the discussion section we speak about how much we
can advance in this problem using our methods.
The structure of the remaining part of the paper is as follows. In the next section we
develop some of the basic tools we use. In Section 3 we prove the n = 3m + 1 case of
4

Theorem 3, which is the easiest result and which allows t he reader to get familiar with some
of the methods. In Section 4 we prove the n = 3m + 2 case of Theorem 3, which also implies
the n = 3m + 2 case of Theorem 2. In Section 5 we prove the n = 3m case of Theorem 2,
which is the hardest proo f in the paper. Finally, in Section 6 we prove Theorem 4. In
Section 7 we discuss our results and related questions.
2 Basic tools
For a family F
i
2
[n]
and an integer t, 0 t n, we deﬁne F
(t)
i
:= {F F : |F | = t} and
f
t
i
= |F
(t)
i
|. Let y
t
i
:=
n
t
f
t
i
denote the number of t-sets missing from F
i
. For a single
family F we use the nota tion f
t
, y
t
.
The following lemma is a generalization of the main lemma fr om Kleitman’s paper [16].
We use the following notation: for i [3], let i
+
= i + 1, i
= i 1, with 3
+
= 1 and 1
= 3
(so that we always have {i, i
+
, i
} = [3]).
Lemma 6. Suppose that F
1
, F
2
, F
3
2
[n]
are cross partition-free. Let s
1
, s
2
, s
3
be nonnega-
tive integers satisfying s
1
+ s
2
+ s
3
n. Then the following inequality hol ds.
3
X
i=1
y
s
i
i
n
s
i
+
y
s
i
+
+s
i
i
n
s
i
+
+s
i
2. (6)
We deduce (6) using the following claim.
Claim 7. Let S
1
, S
2
, S
3
[n] be pairwise disjoint sets satisfying |S
i
| = s
i
, i = 1, 2, 3. Suppose
that F
1
, F
2
, F
3
2
[n]
are cross partition-free. Th en
3
X
i=1
|F
i
{S
i
, S
i
+
S
i
}| 4. (7)
Proof. If S
i
F
i
holds for each i = 1, 2, 3, then S
i
+
S
i
/ F
i
for each i, and (7 ) holds.
Now, by symmet r y, we may assume that S
3
/ F
3
. By the cross par tition-free property, one
of the relations S
1
/ F
1
, S
2
/ F
2
, S
1
S
2
/ F
3
holds, completing the proof of (7).
Proof of Lemma 6. Let us choose the pairwise disjoint sets from the claim randomly with
uniform distribution. Then for S
i
[n], |S
i
| = s
i
, the probability of S
i
/ F
i
is
y
s
i
i
(
n
s
i
)
. That is,
the LHS of (6) counts the expected number of missing sets among the 6 sets S
i
, S
i
S
j
, i, j
[3]. On the other hand, by Claim 7 , this number is always at least 2 , concluding the proof.
For a partition-free family F 2
[n]
we can set F
i
:= F and infer:
3
X
i=1
y
s
i
n
s
i
+
y
s
i
+
+s
i
n
s
i
+
+s
i
2. (8)
5

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### "Partition-free families of sets" refers background in this paper

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TL;DR: In this paper, it was shown that if all the Xi are not less than 1, then the interval (1, + 1 ) contains Cn,m s u m s s l, n even.
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519 citations

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