0 7 8 9
6 1 7 8
5 0 2 7
1+ 6 1 3
0
7
8
9
6
1
7
8
1 9
3 2
5 1+
2 3
1+ 5
6 0
5 1+ 9 8
0 6 5 9
2 1 0 6
7 3 2 1
8 7 1+ 3
9 8 7 5
3 9 8 7
1+ 5 6 0
6 0 1 2
1 2 3 4
1
9
8
7
7 1 2
3 5 2 4 6
2 4 3 5 0
9 3 4 6 1
8 9 5 0 2
8 6 1 3
3
7 0 2 4
1+
6 1 3 5
5
0 7 8 9
6
1 9 7 8
0
2 8 9 7
1
2
9
8 2 3
9 3 1+
0 4 5
2 5 6
1+ 6 0
6 0 1
1 7 8
3 8 9 7
5 9 7 8
COM BIN A TOR 1 A L
MAT HEM A TIC S
YEA R
Febr-ual'y 1969 - June 1970
SPERNER TYPE THEOREMS
1
by
2
G. KATONA
Mathematical Institute of the Hungarian Academy of Sciences, Budapest
Department of Statistics, University of North Carolina at Chapel Hill
Institute of Statistics Mimeo Series No. 600.17
JULY 1969
This research was partially sponsored by the National Science Foundation
under Grant GU-2059.
2
This work was done while the author was at the Department of Statistics,
University of North Carolina at Chapel Hill.
Sperner Type Theorems
1
by
G. Katona
2
Mathematical Institute of the
Hungarian Academy of Sciences, Budapest
First, we should like to list four problems from several branches of
mathematics.
A logical or truth function is an n-dimensional function de-
1.
fined on the n-dimensional 0-1 vectors and taking on the values
A truth function
f
0, 1.
is said to be monotonically increasing if
If we fix the values of the variables
a. ,a. , .•• ,a. ,
11
12
1r
say,
a. *,a. *, •.. ,a. *, then we call the set {a. *} an implicant of
11
12
1r
1j
f(a ,a , •.• ,a.*, ••• ,a. *, ... ,a ) = 1 for all the values of a.'s
l 2
n
11
1r
J
(j#i ,
l
if
l=1,2, ... ,r).
For example, let us consider a ballot with
at least
f
k
n
people voting, and if
of them vote "yes", suppose that the proposal is accepted.
It is obvious that this vote function is monotonically increasing, and we
obtain an implicant if we fix the votes of at least
k
people as "yes"
and the votes of some other people as "no".
From this example, it is obvious that an important class of implicants
are those which cease to be implicants when some of their fixed values are
1
This research was partially sponsored by the National Science
Foundation under Grant GU-2059.
2
This work was done while the author was at the Department of Statistics,
University of North Carolina at Chapel Hill.
2
These are called prime-implicants.
omitted.
the votes of exactly
k
In our example, if we fix
people as "yes", we obtain a prime-implicant.
Returning to the general case, a prime-implicant can not have a
fixed value equal to zero (e.g., a. * = 0) if the function is monotonically
1
increasing.
For by changing this zero to
f(al, •.• ,a. *, ••• ,a
11
a. 'so
r
That is,
1
a
ir
i1
*+l, ••• ,a)
n
*, .•. ,a.
1
r -l
*
I
also equals
by the monotonicity
1
for all values of the
would also be an implicant, contradict-
ing the definition of prime implicants.
That means we may consider the
prime implicants as subsets of the variables (the elements of which are
fixed to be
subset.
1), and such a subset can not be contained in any other such
The question arises:
What is the maximum number of prime im-
plicants of a monotonically increasing truth function?
It is then clear
that the problem is equivalent to the following:
Given a set of n
(p)
elements, what is the size of a maximal family of
subsets with the property that none of its members contains any
other one?
2. [2].
Given a positive square-free integer m,
what is the size
of a maximal subset of divisors of m with the property that none of
these divisors is a multiple of any other one.
{2·3, 2·5, 3·S}
divisor of
is such a set.
E.g., if
m=30=2·3·S,
In general, we can associate with every
m a subset of its prime factors, and, therefore, if a set of
divisors has the above property then the family of the associated subsets
has the property described under (P).
problem (P).
Thus, this problem also leads to
3
Given a set of real nwnbers,
3. [8].
a.1,
property
n
2
of the
E.
1
E
1
1
(1
E.
for some
I
E
and
1
n),
E1,.a.1,
and
n
E.1= 1 E.1 I a.1
i.
I = (b, b+1).
and an interval
L:.
1.-=1
I,
lie in
Denote by
1,
I
i
$
n
n
L1= 1 E.a.
1 1
i
$
nwnbers
Let
1
i
~
with the
where
E.
1.-
= 0 or
be two numbers lying in
A and
A
respectively.
::>
A'
How many
+1?
where
I,
the set of indices for which
A'
cannot hold, since this would
n
n
Ei=l Eia i - Ei=l Ei'a i = EiEA a i - EiEA , a i = LiEA- A, a i ~ 1,
imply
which is a contradiction, for two numbers of difference at least
not lie in an interval
I
1
could
I
and
(b, b+l).
That means, if we take all the numbers
n
LIE.
1=
1 a.1
if we consider the subsets of indices for which
lying in
E. = 1,
the family of
1
these subsets possesses the property described under (P).
Thus, this
problem also leads to problem (P).
3A.
We can formulate this problem in the language of the prob-
ability theory, too.
on the values
of
0
and
Let
1
E.
1
(1
$
i
$
n)
with probability
be a random variable taking
~.
What is the maximum value
~
P(b < E~
1= lE.a
1 i < b+l)?
4.
Finally,we give a problem connected with finite search theory.
Let a set
S
of
an unknown element of
n
S,
elements be given and suppose we are looking for
having information as to whether certain sub-
sets contain the unknown element or not.
It is reasonable to assume that
any two of these subsets are such that if we received the information concerning the first one, we can never predict the information concerning
the second one.
non-void parts.
In other words, any two subsets divide the set into four
4
(Following Renyi, two such subsets are called qualitatively independent .)
The problem is the following:
S
~
(i
j)
What is the size of a maximal family
A.1.
having the property that
and
A.
J
are qualitatively independent?
Clearly,
and
none of the subsets
are qualitatively independent if and only if
A.
J
Ai,A. ,A.,A.
J
1.
(A
J
means
S-A)
contains any other one.
again lead to problem (P).
The answer to problem (P) is contained in the following theorem:
I.
A
A.
1.,
SPERNERS THEOREM [2].
Let
S
be a set of n
{A ,A , .•. ,A } be a family of subsets of S
1 2
m
=
~
A.
J
(i
F j).
elements, and let
with the Spemer-property:
Then
m S
We now give three proofs of this theorem.
I.
PROOF (Sperner, [ 2],
of subsets of
S
such that
the family of all subsets
i.
lB.
1.
B
Let
I =v
(1
C satisfying
S
1
{B ·B
l
i
S
r) .
ci = v-I
be a family
2 ,··· ,B r }
Then
and
c(B)
C
c
denotes
B.
1.
for some
We need a simple lemma:
~1.
n
1928) •
If
B
elements, and if
=
{B1,···,B)
IB.I
=v
1.,
I c(B) I
is a family of subsets of a set
s i s
(l
~
r •
r)
then
v
n-v+l .
S
of
5
PROOF:
Count the number of the pairs
Ici = v-I;
B., C where
and
1.
then
r·v $ !c(B) I·(n-v+l)
because a set
C can be contained in at most
n-v+l
B. 'so
1.
of the
The
proof is complete.
Let us now return to the proof of the theorem.
and let
B = {A.1.1 , •.• ,A.1.
}
is the family of maximal elements of
r
Now we 'prove that i f
(1)
= maxI$i:O;m IAi I ,
v
be those indices for which the maximum is attained,
i l ,i 2 ,···,i r
that is,
Put
A
is an optimal family and
n
A.
is even, then
<.!!.
- 2 '
V
Assume the opposite,
r«n/2 + 1)/(n/2))
>
v >
n
2.
Then
Ic(B)! ~ r·v/(n-v+l) ~
r.
A. , ••• ,A.
by the elements of c(B), we obtain a new
1.1
1. r
family having more than m elements and possessing the Sperner property
Replacing
(or briefly, which is a Sperner-family).
A.
of
In the case when
v $ (n+I)/2,
Ic(B)1
v
This contradicts the maxima1ity
:0;
~
r,
n
however, if
is odd, similar considerations lead only to
v
= (n+I)/2,
then
v/(n-v+l)
= 1 and
that is, there always exists an optimal family with
(n-I)/2.
If
so does
A=
{AI, •.• ,A }
m
possesses the Sperner-property, then obviously
- ••. ,A- }.
A* = {AI,
m
Thus, for an optimal family
Comparing this with (1), we obtain
A,
$
IA.I
1.
IAil
=
2n
¥
must hold if
(1 $ i
:0;
m).
is even.
n
If
n
is odd,
6
we have
IAil
of
IAil
$
(n+l)/2
and thus, there is an optimal family with
= (n-l)/2 = [n/2].
S
It is trivial that the family of all [n/2] tuples
is a Sperner-family.
The proof is finished, but the uniqueness of the optimal family was
proved only for even
n's.
If
n
is odd, we proved that
and in an optimal family all the subsets have
ments.
(n-l)/2
m
or
$
([n/2])
(n+l)/2
ele-
We will prove later, that there exist only two optimal families:
the family of all the
(n-l)/2
tuples and the family of all the
(n+l)/2
tuples.
2. PROOF
(De Bruijn, [1], 1952).
2.
We say that a family
1
3.
IBll + JBrl
respect to
=n
(that is, the chain is symmetrical with
n/2).
l6MMA 2.
It is possible to divide all the subsets of S into dis-
joint syrronetrioal ohains.
PROOF
trivial.
We shall use induction on
Assume now, that for
n-l
From
struct two (if
r
Bi n
i
= 1,
=
For
n.
the statement is
n=l,
we have constructed disjoint sym-
B.1 = {B.1 l ,B.1 2 , ... ,B.1ri }
only one) new chains:
{B.
1l
U
{x }, ... , B.
n
1,r i
1
U
{x }}
n
we con-
7
where
X
n
is the new, n-th element.
It is obvious that the new chains
are symmetrical, further that every subset is contained in at least one
new chain and that the new chains are disjoint.
Now we return to the proof of the theorem.
be contained in at least one symmetrical chain.
Every [n/2]-tuple must
On the other hand, every
symmetrical chain contains exactly one [n/2]-tuple because of the symmetry.
Thus, the number of symmetrical chains is exactly
Finally, a
symmetrical chain can contain at most one element from a Sperner-family
and the theorem follows.
3. PROOF
subsets.
A maximal chain is a chain of
(Lubell, [18], 1966).
Count all the maximal chains of a set of
n
elements.
imal chain is determined uniquely by a permutation of the
(in which order we add them to the preceding subset) .
maximal chains is
containing
A=
nl
Now fix a subset
A as an element.
{Al, .•• ,A }
m
more than one
We obtain
A max-
elements
Thus the number of
A and count the maximal chains
IAII(n-IAI)!
If
is a Sperner-family, then no maximal chain can contain
A..
1
Thus,
m
IA. I !(n-IA.I)!
1
1
n!.
$
i=l
Hence, we obtain
m
1
i=l
or
n
n+l
1
8
1,
(2)
where
is the number of
A. 's
l.
eralization of the Sperner theorem.
m
with
IA.I
l. = k.
Indeed, by
~
(2) is already a gen-
(~) ~ ([n/2])
we obtain
1.
and the proof is complete.
REMARK.
The inequality (2) was first observed by L.n. Meshalkin [4]
(1963) .
Now we will consider several generalizations of Sperner's theorem.
The four problems mentioned at the beginning will serve as a basis for
new generalizations.
Let us first return to Problem 1.
For practical
purposes, it is very important to write a truth function in a well-determined form, for example, in a disjunctive-normal form.
This is the basis
of the synthesis of a truth function in electrical engineering.
The
disjunctive-normal form is a disjunction of different terms, where a term
is a conjunction of the variables
of
and
a. 's
l.
and
~.' s
l.
such that at most one
occurs in the same term:
f(al,a "" ,a )=(a
&a. &••• &ai )v ••• v(~. &a. &••• &a. ).
i 11 l.12
Z
n
lr
l.tl l.t2
l.ts
(aVO = 0,
OVI = IVa = IVI = 1,
0&0 = 0&1 = 1&0 = 0,
1&1 = 1).
It is a well-known fact in the theory of truth functions that the disjunction of all the prime-implicants has minimal length (minimal number
of operations) among the disjunctive-normal forms of a monotonically increasing truth function.
That means the number of needed electrical
9
components will be minimal if we make the synthesis on a basis of this
disjunctive-normal form instead of others.
Thus, the real problem, in
this case, is not to determine the maximum number of prime-imp1icants but
the maximum of the sum of variables of the prime-imp1icants.
language of subsets, we have to determine
A=
{A , ..• ,A }
m
1
is a Sperner-fami1y.
n
Lk=Ok~
maximlUll of
max L~=lIAil,
Or, in the
where
Here we can use (2):
What is the
if
We can write
n
L
(3)
.k=O
where
{x}
denotes the least integer not less than
x.
Here, if
k ~ {~}
then
n(n-1) .•. (n-{n/2}+1)
1. .. ({n/2}-1)
(4)
and if
k
<
{n/2},
then
n-k
=
{n/2}
n-k+1
~
{n/2},
{n/2}+1
n-k+2
1 • 2 ... k
n ••• (n-k+1)
k-1
n-{n/2} • k ~ k
thus
(5)
Using (4) and (5) in (3), we obtain
n
L k~ ~
{n/2}({n/2}) .
k=O
Thus we can state:
THEOREM
2.
If
A = {AI' ..... Am}
m
IA.I
1
i=l
is a Bpemer-famiZy .. then
10
Let us now consider Problem 3.
n
mum number of sums
~i=lEiai
Previously we proved that the maxi(b ,b+l)
lying in an open interval
is less
than or equal to the solution of the Sperner-problem, that is,
If we choose
[n/2],
a. = 1
1.
(1
~
~
i
n),
i
~,~
E. = 0,1,
1.
instead of
uniquely by
sums have the value
E = -1,1
then the sums
1
does not change.
Multiplying by 2, we obtain
and an interval of length 2.
i
a 's
i
tive
It is clear that if we use
thus the maximum number of sums lying in an open
interval of length
I ai I
with
~
the set of values of sums
the
([n/2])
that is, our estimation obtained by Sperner's theorem is sharp.
Now let us generalize the problem.
E =
n
then
a 's
i
by
1
Here we can already allow nega-
(instead of
n
1.= IE 1..a.1.
~.
a.
1.
~
Indeed, in this case,
1) •
is unchanged if we replace some of
Hence we obtain the following result:
-a .•
1.
If
a 1....... an
are real numbers with
la.1
1-
n
then the maximum number of sums
n
interval of length 2 is ([n/2]) •
1- 1-
But what can we say about longer intervals, say, of length
~.
the subset of indices with
1.
(b ,b+2h)
sets there does not exist a pair
lA-A'
n
Let us again associate with each sum
I
~
h
A, A'
2h?
lying in an interval
1.= IE 1..a.1.
E.
1.
and
lying in an open
Eo 1E.a.
1-=
~ 1..
Clearly, among these sub-
for which
A
~
A'
and
because of
n
n
E.a.
1. 1.
E
~ a.
1. 1.
( L
i=l
iEA
2
L
a.
1.
L
iEA-A'
a. )
1.
~
(
L
iEA'
HA
a.
1.
-
2h.
a.
1.
- L
HA'
a.)
1.
11
Thus, the following theorem will be useful.
THEOREM 3.
[8].
Let
A
= {A 1,A 2, ... ,Am} be a family of subsets of
a set S of n elements, and h
A.
].
(6)
::>
A.
J
IA.-A.I ~ h
]. J
a natural number.
(1
:0;
i, j
:0;
If
m)
never holdS, then m is less than or equal to the sum of the
h
largest
binomial coefficients of order n.
This theorem will be a special case of the following theorem:
THEOREM 3A.
(ErdHs, 1945, [8]).
Let
A = {A1, ... ,Am} be a family
of different subsets of a set S of n elements and h a natural numbel'.
If the family has the property that a sequence of different sub-
sets
(7)
does not exist, then m is less than
h binomial coefficients of order n,
or'
equaZ to the sum of the largest
and this estimate is the best
possible.
PROOF OF THEOREM 3A.
chain at most
h
The proof is based on Lemma 2.
of the above subsets can occur.
the chain is less than
h,
In a symmetrical
then this number is obviously at most
only have to count the number of chains having length at least
Consider a subset of
t
If the length
OCn+~2]
elements.
it is contained in a chain of length at least
of
t.
We
t.
Because of the symmetricity,
t.
On the other hand, each
12
l
chain of length at least
must contain a subset of
l
Thus, the number of chains of length at least
n
([(n+l)/2])'
[~+~2)
elements.
is exactly
Thus, the desired estimate is:
h-l
(L l·
(the number of symmetrical chains with length
l)J
l=l
+ (h • (the number of symmetrical chains with length at least
h»
h
L
(the number of symmetrical chains with length at least
l)
l=l
[(n+h-l)/2]
h
=
\
L ([(n+l)/2])
n
(.)
/
l=l
i
1.
[(n-h+l) /2]
This latter quantity is the sum of the
The family of all the
[(n+h-l)/2]-tuples
.
h
largest binomial coefficients.
[(n-h+l)/2]-tuples,
[(n-h+l)/2]+1-tuples,
... ,
shows that the estimate of the theorem is the best
possible.
PROOF OF THEOREM 3.
Sperner.
(Erdgs' original proof followed the proof of
Here we follow De Bruijn's proof.)
If a family has no subsets
of type (6), then it cannot have subsets of type (7), thus the same estimate is valid.
... ,
The family of all [(n-h+l)/2]-tuples, [(n-h+l)/2]+1-tuples,
[(n+h-l)/2]-tuples shows again that the estimate is the best
possible.
We can formulate Theorem 3A in another form, too:
13
THEOREM 3B.
Ah =
Let
{Ah1,···,Ah~}
elements.
If each
A1
= {A11,···,A1m1}'
AZ
= {A21, ... ,A2m2}'
be disjoint families of subsets of a set
A.1.-
(1 ::;
i ::;
h)
• • • ..J
S of n
is a Spemer-family, then
h
m.
1
i=l
is less than or equal to the sum of the
of order n,
h
largest binomial coefficients
and this estimate is the best possible.
PROOF.
The family
U.h1= lAo1
A
3A, thus the same estimate
satisfy the conditions of Theorem
is valid.
Returning now to Problem 3A, can we generalize it to
vectors
a.
with
1
la. I
1
~
I?
two-dimensional
The answer is yes, but not trivial.
It is
clear that in this case we can reduce the problem to the case when the
first coordinate of the
a. 's
1
is non-negative.
However, the Sperner-
theorem does not help, because the sum of vectors
with
la. I
1
~
1
can have absolute value less than
a.
in the half plane
1.
(In the half-line
1
this cannot happen, which is the reason for the applicability of the
Sperner-theorem in the one-dimensional case.)
a 's
sum of the
i
But, in a quadrant, the
have again absolute value greater than
Thus, in-
1.
stead of the Sperner-theorem, we can use a theorem where the elements
are divided into two disjoint parts:
THEOREM
Let
n
1
and
::; n 2 .
neither
4.
S2
(Kleitman [11] and Katona [19], independently (1965».
be disjoint sets of n 1 and
elements respectively
If A = {A 1, ... ,Am} is a family of subsets of Sl
U
S2
andif
14
(i ~ j)
(8)
nor
(9)
can
occur~
then
where
PROOF,
n
By Lemma 2, we can divide the subsets of
symmetrical chains.
B ,B , ... ,B
l 2
h
Let
be such a chain.
the following family of intersections of
a common intersection
(i F k)
Ai and Ak
and
since
(8).
AE~
B
i
with
8
2
with those
are disjoint because of (9).
Ai (1
Moreover,
1
into disjoint
Ai
Denote by
A.
J
which have
8 :
1
were true, then the sets
Bi~Bk.
8
BiuA
~ i ~ h)
and
Namely, if
BkUA
AEA
i
would satisfy (9),
is a 8perner-family because of
Thus, we can use Theorem 3B:
[(n+h-l)/2]
\
/
i=[ (n-h+l) /2]
Hence we obtain
[(n +h-l)/2]
2
(10)
m
~
\
_/-
\
/
where the first sum runs over all the symmetrical chains given in the decomposition of
8 .
1
15
From the proof of Theorem 3A, we know that the number of symmetrical
chains of length at least
h
in
Sl
is
It follows from
(10) that
nl
nZ
([(nl+h)/Z]) ([(nz-h+l)/Z])
i=O
( [ (n
nl+n Z
+n ) / Z] ) .
l
Z
The proof is finished.
This theorem is a stronger form of Sperner's theorem since the condition is weaker, but the result is the same.
the optimal family is not unique.
family of different
A's
The only difference is that
An example of an optimal family is the
satisfying
IAnsll = u
and
IAnSZI =
Using Theorem 4, it is not difficult to verify the statement of
Problem 3A in the two-dimensional case.
dimensional case is not yet proved.
3 parts does not help.
The statement for the three-
The generalization of Theorem 4 for
There are two possible generalizations.
If we
exclude
AinSZ"'A/S Z
AinS3~AjnS3'
Ai nSl::J Aj nS l
AinS Z = AjnS Z
AinS3~AjnS3'
AinSl~AjnSl
AinSZ~AjnSZ
A ns = A ns ,
i 3
j 3
(i)
Ains l = AjnS l
(ii)
(iii)
then for
Sl
and
SZUS ,
3
the conditions of Theorem 4 are satisfied, and
the same result is valid, but it is not useful.
16
If we exclude
(i)
(11) (ii)
(iii)
AinS l
A nS
Ains l
AjnS l
j
Ains Z
l
= AjnS Z
AinSZ:JAjnS Z
AinS Z
Ainsl:JAjnSl
= AjnS Z
it would be useful, but the estimate of Theorem 4 is not true in this
case.
For example, if
family of all subsets
Isil = n-Z,
ISzl = Is , = 1
3
(n
is odd), the
A satisfying one of
(i)
n-3
IA n Sll = - Z
IA n Szl
0
IA n S3'
0,
(ii)
n-3
IA n Sl' = -Z-
IA n Szl
1
IA n S3 1
1,
(iii)
n-l
IA n Sll = - Z
IA n Szl = 0
(iv)
n-l
IA n Sll = -Z-
IA n Szl
1
does not contain subsets of type (11).
IA n S31
0,
Nevertheless, it has
n
(> «n-l)/Z))
OPEN PROBLEM.
IA n S3 1 = 1,
elements.
What is the maximum number of elements of a family
A
of subsets having the property that none of (11) can hold?
Let us consider again Problem Z.
trary non-square-free natural number
sets (which are
0-1
Is its statement true for an arbim,
too?
Obviously, instead of sub-
valued functions), we have to consider non-negative
integer-valued functions which have an upper bound (the exponent of
in
m) for the values at each point.
THEOREM
5.
(De Bruijn, T~bergen, Kruyswijk [1], 195Z).
Let
Pi
17
satisfying the condition
(12)
where the
ak's
age given positive integers.
f.
::;
f.
J
l.
does not hold for any pair
where
~s
M
[(L~ la.)/2]
~=
~
If
(i:l j)
then
the number of functions satisfying
= [a/2].
In the case of
1
a.
J
::; n),
(1::; j
the theorem reduces to the
Sperner-theorem.
PROOF.
First we prove a generalization of Lemma 2, and give the
generalizations of the needed concepts.
We say that a set
of functions satisfying (12) is a chain, if
f.
l.
(r
~
i
~
1)
there exists an
differs from
f.l.- 1
for every
xk .
l.
f. (x
l.
k•
)
l.
We say that a chain
l
::; f
2
such that
f.
l.
=
and
fi_l(X)
F
n
n
k=l
k=l
::;
only in one place by
~
f. (x)
f
if
F = {f ,···, f }
l
r
::;
1,
f
r
and each
that is
18
l6MMA 3.
[1].
It is possible to divide all the functions satisfying
(12) into disjoint symmetrical chains.
PROOF.
trivial.
chains
fY
F.1.
We shall use induction over
Assume now that for
Fl ,F2 , ... ,F u '
If
f
fO
i,r.-2
1.
f?1.,r.-l
1.
Clearly,
.
F1.~ =
For
n=l,
we have the disjoint symmetrical
fY(x) = y.
n
n-l
x's,
We construct for a chain
new chains in the following manner:
an
£1
i,r.-2
1.
f2
i,r.-2
1.
f.1.,r.- 2
fl
i,r.-l
1.
f~1.,r.-l
an
£.1.,r.- 1
1.
fl
i,r.
1.
f2
i,r.
1.
1.
1.
{j
j
j
j+l
an
£.1 f. 2' ... , f .
. ,f .
. , ... , f .
.}
1. , 1.
1.,ri-J 1.,ri-J
1.,ri-J
min(a , r.-l))
n
1.
the statement is
is a function defined on the first
denotes its extension with
{f.l, ... ,f. }
1.
1.ri
n-l,
n.
(0:<; j :<;
will be a symmetrical chain again.
Indeed,
n
n
L
fil(X k ) +
L
k=l
k=l
an
f.
. (~)
1.,r -J
i
n-l
L
k=l
fil(X k ) +
[7
k=l
n-l
L
n-l
fil(~) +
L
f.
. (~) + j + a
n
1.,r - J
i
k=l
k=l
n
f.
(x )
1. , r
k
i
j
+ j + a
n
L
k=l
a
k
.
19
Further, it is easy to see that:
1.
Every function is contained in at least one new chain.
2.
The new chains are disjoint.
Now, we can return to the proof of the theorem.
Every function having
the value-sum
n
[a/2]
(13)
must be contained in a symmetrical chain.
On the other hand, every sym-
metrical chain contains exactly one function of type (13).
Thus, the
number of symmetrical chains is exactly the same as the number of functions
satisfying (13), that is,
M.
The set of all functions satisfying (13) shows that our estimate is
the best possible.
The theorem is proved.
Recently, Schonheim [9] proved a common generalization of Theorems
3A and 5, further of Theorems 4 and 5.
Now we give a generalization of all three theorems in a more general
language.
We will say that a directed graph
G
is a symmetrical chain-graph
if
1.
There exists a partition of its vertices into disjoint subsets
and all the directed edges connect a vertex of
vertex of
2.
k
O
k.
1
::;
k
l
::;
= k n-i
(0 ::; i
...
:::;
k[n/2]
(0 ::; i ::; n)
< n).
K.
1
with a
20
3.
There exists a partition of its vertices into disjoint symmetrical
chains, where a symmetrical chain is a set of vertices such that
the vertices are connected by a directed path, and further, if the
starting point of this path is in
K
1
.•
n-l
Consider now a set
tices of a graph
IB-AI
then the endpoint is in
K. ,
= 1.
S
of
n
elements.
Let its subsets be the ver-
G and connect two vertices
A
and
B if
B~A
and
It is easy to see that this graph satisfies the conditions 1
and 2, and by Lemma 2, condition 3 also.
Similarly, if we consider the graph of functions of type (12) and
we connect two vertices
where
f(x) + 1
= g(x),
f
and
g
if
f
=g
except in one place
x
then by Lemma 3, it will also be a symmetrical
chain-graph.
The direct sum
be the following.
hEH,
and
G+H
of two symmetrical chain-graphs
Its vertices will be the ordered pairs
is joined to
and
If
subsets of
SluS .
2
(g,h)
and
and
S2'
H will
gEG,
hI
and
are joined.
G is the graph of the subsets of a set
of the subsets of a (disjoint) set
G and
then
G+H
Sl
and
H is the graph
is the graph of the
The situation is similar in the case of finite functions
of type (12); the direct sum produces a larger graph of similar type.
The generalization of the Sperner (and also of De Bruijn-TenbergenKruyswijk) theorem in this language is the following.
a ,a , .•• ,a
n
l 2
If we have a set
of vertices of a symmetrical chain-graph and no two of them
are connected by a directed path, then
m
k[n/2] .
21
The generalization of Theorem 3A would then be:
(14)
I f no
h+l
of
chain, then
(0 ~ i
k.
1
(a. fa.)
1
J
m
~
~
the sum of the
h
lie in a directed
largest numbers of type
n).
In a direct sum graph, we will use a weaker condition instead of
(14).
THEOREM
6.
KO,K1,···,Kn
Let
(of
G and
H be syrronetrical chain-graphs with levels
ko, .. ·,kn
elements), Pespectively.
elements) and LO,L1,
If we have a set
tices of the direct swn graph
G+H
(gl,hl),
such that no
h+l
,Lp
(of
,(g~hm)
lo, ... ,lp
of ver-
different ones of
then satisfy the condition
g.
1
= ••• = gi
3
w
lie 1.-n a directed path in
H (in this order)
(15)
h.
1
W
g. , ••. ,g.
1
for some w
a
of type
(l
lie in a directed path in
G (in this order)
~
w
~
h+l),
G
and
then
m
~
the sum of the
h
largest numbers
Ok.l
..
1.- a-1.-
I:.
1.-=
REMARK I.
nand
+l
1h
W
p
If
H
are subsets-graphs of sets
8
1
and
8
2
elements, respectively, then
k.
1
and
On the other hand, (15) means that there are no
h+l
different subsets
of
22
Al,A2' ...
'~+1
in the given family of subsets of
SluS 2
such that
AlnS2cA2nS2c •.. cAwnS l
(16)
... = ~+l nS 2
holds for some
w
(1
~
w
~
h+l).
It is clear if (16) holds, then
also holds, that is, in this case we have a weaker condition than the one
in Theorem 3A, but we obtain the same result.
The connection between this
case of Theorem 6 and Theorem 3 is the same as the connection between
Theorem 4 and 5perner's theorem.
REMARK 2.
If we put
REMARK 3.
Let us consider now another important special case.
h=l
in the above example, we obtain Theorem
4.
51
f
be a one-element set, and let the vertices of
defined on
Sl'
such that
is a directed edge from
f
directed path of length
n+l.
Let
to
H be the same with
rectangular
(n+l)
x
(p+l)
0
~
g
p
f
~
nand
only if
instead of
lattice.
f
G be the "functions"
is an integer.
g = f+l.
n.
Let
G+H
Thus
There
G will be a
is in this case a
23
"' "-
"-
"-
H
p+l
"-
"
"-
"-
".
n+l
G
In this special case, Theorem 6 would state that if we have a set of
points of this rectangle, no two of which are connected by a directed
path, then the maximal number of these points is the length of a maximal
diagonal (a diagonal is a set of vertices with the same coordinate-sums),
that is
min(n+l,p+l).
Schonheim's generalization of Theorem 3A in this special case would
state that if we have a set of points in this rectangle with the property
that
(17)
no
h+l
different members of this set lie in a directed path,
then the maximal number of these points is the sum of the lengths of the
h largest different diagonals.
Theorem 6 says that if we exclude the existence of
h+l
different
points lying in a directed path which consists of two straight lines
(instead of (17», we obtain the same maximum.
First we prove this special case.
lEMMA 4.
o~
j
(i~j)
~
b;
1.-~J
onZy to
Let
R be a graph with vertices
are integers)
(i~
j+l)
and
(i~j)
(0 ~ i ~ a;
where there are directed edges from
(i+l~
j).
If we have a set of m vertices
24
such that there are no
h+1
(i1~j1)~···~ (ih+1~jh+1)
different vertices
with the property
i
(18)
i
for some
< •
(1 ~
w
+l
<
1w
w
w
<
w
< j
•••
w
•••
~ h+1)~
then
m
~
sum of the Zengths of the
h
Zargest different diagonaZs.
PROOF,
o
~
Let
j
~
b
The set of vertices
(iO,j)
is called the iO-th coZumn.
where
i
O
is fixed and
The rows are similarly defined.
V be a set of vertices satisfying the conditions of the lemma and
denote by
c
the number of columns with exactly
t
Obviously, by (18)
c
t
=0
if
t
>
h.
t
vertices from
V.
Thus
h
(19)
=
a
+ 1.
t=o
Let us count in two different ways the number of points which are at
least u-th elements of
V in any column starting from below.
umn where the number of elements of
0,
V is less than
u,
In a col-
we have to count
so counting column by column we obtain
C
u + 2c u+ l + 3c u+2 + ..• +
(h-u+l)~
.
On the other hand, counting row by row, we obtain that this number is at
most
(h-u+l) (b-u+2)
because we do not have to consider the first
rows, and in the other rows we can have at most
condition (18).
(20)
C
u
h-u+l
u-l
such points by
Thus we obtain the inequality
+ 2cu+l + 3c u+2 + ... +
(h-u+l)~ ~
(h-u+l)(b-u+2),
(l~u~h).
25
We have to maximize
If
a+l
~-l
~
under the conditions (19) and (20).
b-h+2, then obviously the optimal solution is
= c
=
= O.
l
Assume now that
a+l
that there is an optimal solution with
~', ~-l'
solution
c.
n-l
= c.
n-l
, •.. ,c l ',
' - 2(b-h+2-c. '),
n
with
>
b-h+2,
~
then
Li=lic i
= b-h+2,
~-3 = ~-3' , ... ,
",h . ,
=
L..
1=
llC.
,
1
satisfy (20), it follows that (20) holds for the
"s
C.
1
If we have an optimal
~-2 = ~-2 '+b-h+2-~ , ,
h
is also an optimal solution because
and since the
First we will show
= b-h+2.
~
~'<
b-h+2.
= a+l,
~
too.
= h-I:
u
~-l + 2~
ch-l - 2(b-h+2-ch) + 2(b-h+2)
ch-l + 2ch ~ 2 (b-h+2)
~
u
h-2:
C
(h-u-l)~_2
u + ••. +
=
C
+
(h-u)~_l
(h-u+l)~
+
u + ... + (h-u-l) (ch_2+b-h+l-~)
+ (h-u) (ch-l - 2(b-h+2-ch» + (h-u+l)(b-h+2)
c~
~
+ ... + (h-u-l)ch_2 + (h-u)c
h_l
+ (h-u+l)ch
(h-u+l) (b-u+2).
Thus, for this special optimal solution we have
(21)
C
u + 2c u+ l + ..•
instead of (20).
~-l
~-2 =
=
2 or 1,
•••
If
a+l
~
+ (h-u) ~-l
b-h+4,
according to whether
= c l = o.
~
(h-u+l) (h-u)
,
cl '
b-h+2
a+l = b-h+4
Let us assume that
with
is fixed) .
~ -1
<
2,
h-l)
then the optimal solution is:
a+l
>
that there exists an optimal solution for which
~=
(l ~ u ~
or
b-h+4.
~-l
b-h+3,
It is easy to see
= 2 (naturally
If we have another optimal solution
,
~-l'
then we can change to another one through
... ,
~-l
2,
26
which satisfies (21).
Following this procedure, we obtain that there
o
b-h+2, c.
= 2,
n-l
exists an optimal solution of the form
0 0 0
= 2, ... , cv +l = 2, Cv
~-2
determined by
... = c ol =
1 or 0,
(i,j)
satisfying
i+j
diagonal of the rectangle and is denoted by
D
y' Dy + l '
... ,
where
Dy + h - l '
that the number of points of the
and hence maximal.
because any
v
where
is
(19).
The set of points
are
0,
h
y
h
=
=k
Dk ·
is called the
The
middle diagonals
h
[ (a+b-h+l) /2] .
I t is easy to see
L'lC.a ,
middle diagonals is just
This means that they are also the
k-th
h
1
largest diagonals
arbitrary diagonals satisfy the conditions of the lemma.
The lemma is proved.
PROOF OF THEOREM 6.
By part 3 of the definition of symmetrical
chain graphs, the vertices of
metrical chains.
Denote by
along these chains.
respectively.
Thus,
G
G'
G'
It follows that
and
and
and
H
H'
H'
G'+H'
sufficient to prove the theorem for
can be partitioned into symthe graphs which have edges only
are subgraphs of
is a subgraph of
G'+H'
instead of
G
and
G+H.
G+H.
H
So it is
However,
consists of rectangular lattices, and condition (17) of the theorem
G'+H'
simply means condition (18) in every such rectangular lattice.
We know
that an optimal set of points in every rectangle is the union of the
middle diagonals.
(g,h)
Define the levels of
E
M.
J
iff
G'+H'
in the following manner:
for some
i.
By the definition of the direct sum, it is easy to see that the
M. 's
J
satisfy the first part of the definition of a symmetrical chain-graph.
The
z
=
h
middle levels of
[(n+p-h+l)/2].
G'+H'
h
where
27
We will show that the union of the
rectangles is just the union of the
h
h
middle diagonals for all the
middle levels in
First we verify that an element of the
rectangle is an element of the
h
h
G'+H'.
middle diagonals in a
middle levels in
G'+H'.
Let us con-
sider a fixed rectangle which is a direct sum of two symmetrical chains
from
G'
and
spectively.
i+a
=
n-i,
(22)
H'
If
with vertices
go
K ,
i
E
gO,gl, ..• ,ga
and
then by the symmetry
ga
hO,hl, ... ,h
E
K n i
re-
b
and thus
or
n-a
2
i
(Obviously,
n
and
a
have the same parity.)
Similarly, i f
h
O
E
L.,
. J
then
(23)
j
p-b
2
D,
is in
I f a point
r
one of the
h
middle diagonals of the
rectangle, then
k +,e
(24)
=
r
thus
(25)
E
Since
y
=
h
Mi+j+k+,e'
middle levels of
or using (22), (23),(24)
n+p-a-b + r
2
[(a+b-h+l)/2] ~ r ~ y + h - 1,
Conversely, let
(g,h)
E
M
(n+p-a-b)/2 + r ~ z + h - 1,
the
(gk,h,e)
we have
and (25) means that
z
=
[(n+p-h)/2] ~
(gk,h,e)
is in one of
G'+H'.
(g,h)
be an element of
M, where z ~
s
s~z+h-l.
Then
is contained in a rectangle which is a direct sum of two symmetrical
28
K
E
L
E
n-a
2
If
r
(gk,h ),
l
(g,h)
= k+l =
then
and thus
(n-a)/2 + (p-b)/2 + k + l = s,
s - (n-a)/2 - (p-b)/2
z -
~
2
the following inequality holds:
n+p-a-b
-2
r
~
~
(g,h) is an element of one of the
h
middle diagonals.
Thus we have proved that the points of the
optimal set.
For the union of
h
h
middle levels form an
arbitrarily chosen levels of
the conditions of the theorem are satisfied; so the
h
must be the
h
M
L~ Ok.l
thus the optimal number is the sum of the
1=
1
.;
a-1
largest.
these numbers.
that is for
The number of elements in
G+H,
middle levels
is obviously
a
h
largest of
The proof is complete.
Now we will prove another generalization of a theorem of Erdgs.
We
intend to give a class of problems where the maximum is a sum of binomial
coefficients.
We will state the theorem only for subsets of a finite set,
but everything is the same for any symmetrical chain graph.
Let
be a set of
S
of subsets of
S,
axioms
~
1.
(1
a.
1
i
n
elements and let
where
Al,A2' ... '~
~
where the
s),
satisfy a finite system
a. 's
1
I
of
satisfy several conditions:
a.
1
is a finite formula containing the variables
and the signs
E
(is an element of),
(number of elements),
2.
Al,A2' ... '~ be families
If
=
"',
Cj
c C
k
or
,c 2 , ... ,C
ki
,
(is contained by),
~.
satisfy
Cl, ... ,C .,
k1
either
<,
c
C
l
F,
then for every pair
holds (or both).
29
Let
DO
D
l
C
c
•••
ID.I
=
1
the property
C
be a fixed sequence of subsets of
(0 :s; i
i
I
of families satisfying
l
D
n
l
n l
An + l
A1+ ' An+l
2 ,,,.,ob
If
but having elements only among
l
A1 ,A Z, .•. ,Ah
then denote by
:s; n).
(l = n+l,n-l,n-3, •.. ; l
S
with
is a system
DO,D ,··· ,D ,
l
n
~ 1)
the families
defined by
Ali
2
J
and having elements only from
I
tems of families satisfying
:s; j
n l
A1+ , ... '0An+l
b
There exists a system
3.
L
n-l+l
{D. :
:s;
of families satisfying
DO,Dl, ••• ,D ,
n
such that among the sys-
and having elements only from
D(n-l+l)/2,···,D(n+l-l)/2'
h
is maximal for all
l
(l
n+l,n-l,n-3, ••• ;
l
~
1).
i=l
THEOREM
of n
and
7.
elements,
Al'A Z'··· ,Ah
If
I
A1,A 2, . .. ,Ah are families of subsets of a set S
is a system of axioms satisfying
satisfy
L
1, 2,
and
3,
then
h
I
IA.I
1
:s;
I
(~)
J
i=l
where
j
runs over the number of elements of the subsets lying in
A7+1, ... ,~+1 which are defined under 3.
PROOF.
By Lemma 2, we can divide the set of all subsets into dis-
joint symmetrical chains.
E(n+l-l)/2·
This estimate is best possible.
Consider a fixed chain
E(n-l+l)/2' ... '
We will show that the maximal number of elements that
30
can have in
is
h
t
IA.I
.
1
i=l
In order to show this, it is sufficient to verify that if we have a sys-
I,
tern of families satisfying
of the elements of
S
will also satisfy
E(n-t+l)/Z, ..• ,E(n+t-l)/Z
D(n+t-l)/Z.
then the system obtained by permutation
S
from
However, the above property follows from 1.
Cl'C Z' ••• ,C k .
1
D(n-t+l)/Z' ..• '
The proof of
If after the permutation
we would have
then the re-permuted sets
F. ,
satisfying
would satisfy
1
(in contradiction to our hypothesis) because our
F.
1
E, c, u, n, I I,
signs
Indeed, we can obtain
by a permutation of
the above statement is trivial.
C '.
k1
I.
=,
~
are invariant relative to the permutation.
A more detailed proof would use induction over the number of signs in
So, in every chain of length
t
F .•
1
we determined an upper bound
h
t
IA.I·
1
(26)
i=l
In order to obtain an upper bound for
(Z6) for all chains.
~~=lIAil,
we only have to sum
The best method to accomplish this is to form an
optimal system in every chain and to number the sets in all the chains.
Let
l* t*
A ,A , .••
Z
1
,Abt*
be an optimal system in a chain of length
where the star denotes the appropriate permutation.
The permutation does
not change the number of elements of a set in
then an arbitrary set
length
t*
Ak,
t
«n-.t+l) /Z
t*
DEAk·
~
D of
j
~
j
DjiA
If
elements is an element of a chain of
(n+.t-l) /Z),
Similarly, if
t,
n+l
k
be an element of the optimal system.
and by definition of
then no set
D of
j
~
and
elements can
31
The upper bound is established.
We only have to prove that the
I.
optimal systems defined in this way satisfy
we would have an axiom
a.
1.
and sets
ffttrls were false then
satisfying
Cl ,C Z ,··· ,C k ·
1.
However, by property Z i t follows that
(perhaps with equal terms) which is
F..
1.
form a chain
Cl ,C Z" " ,C k .
1.
n+l.
part of a chain of length
From that we can form, by an appropriate permutation of elements of
the chain
DO
D c ...
l
C
D .
n
C
Thus, the sets
Ci,Ci, ... ,C
k.
S,
thus ob-
1.
tained would be elements of this chain
DO
D c ...
l
C
tation does not change the number of elements, so
C
D .
n
The permu-
Ci,Ci, •.. ,C
k.
would
1.
remain elements of the corresponding families
would satisfy
F.
n+l
An+l
A] , .•. , h
which contradicts our hypothesis.
1.
and they
The proof is
finished.
ExAMPLEs.
1. SPERNER'S THEOREM.
,3
I
contain only one axiom:
(CIEA], CZEA], clc z , ClcC Z)·
(Cl,C Z):
D(n+l-l)/Z
Let
we can only choose a family
if we choose
In any chain
A]
D(n-l+l)/Z"'"
with a single element, and
it satisfies the condition 3, and so, by Theorem 7
D[n/Z]'
II
z. ERDOS' THEOREM
,3
(Cl,CZ'···'C h+ l ):
ClcCZc ... cC h+ l ),
(Theorem 3A).
(ClEA], ... ,Ch+1EA], cl/cZ""'Cl/Ch+l"",Ch/ch+l'
where
we can choose a family
choose for
l
n+l
be maximal for every
h~n+l.
A]
with at most
the family
l.
In a chain
D(n-l+l)/Z, .•. ,D(n+l-l)/Z
min(l,h)
elements, and if we
D[(n-h+l)/Z], ... ,D[(n+h-l)/Z]
Thus, by Theorem 7:
i t will
32
[(n+h-l)/2]
\
n
/
(i)
.
i=[(n-h+l)/2]
13(Cl ,C 2 ,···,Ch + l ):
3.
(C1EA]""'~+lEA], C/C2 , ... ,C l "St+l'
... , Ch"Ch + l , ClcC2c ... c~+1)
where
~
h
n+l
and
[(n-h+l)/2]
D(n-l+l) /2".' . ,D (n+l-l) /2
min«(n+l-l)/2)- k, h)
family
<
k
<
[(n+h-l)/2].
we can choose a family
In a chain
A]
elements, and if we choose for
Dk+l,Dk+2, ... ,Dk+h'
with at most
l
it will be maximal for every
n+l
l.
the
Thus, by
Theorem 7:
k+h
max
n
IA]I
(.)
1
.
i=k+l
4.
The previous example with
IC11=k
instead of
[(n+h)/2]
n
\
(.)
1
/
i=[(n-h)/2]
i"k
IC11~k.
.
[n/2] + 1) .
In a chain
D(n-l+l)/2"" ,Dn+l - l )/2
choosing every other one, leaving out
we choose
we can form a maximal family
D[n/2]+1'
... D[n/2]_2,D[n/2],D[n/2]+2"'"
Thus, if for
A]
l = n+l
this family will be maximal
33
for every
t.
By Theorem 7,
"\
max JA]I
=
/
n-l
Z
(-l)i=~l) [n/Z]
6.
A version of ErdHs' theorem (Theorem 3B)
13
(C ,C ) :
l z
(C ICC 2' C1E~ , CZEAR)
,3
Cl :
ClEA] , Cl EA Z
13
Cl :
C EA], Cl EA
l
3
..,3
C1 :
ClE~_l' ClE~ •
The solution is the same as in the example 2.
7.
l3
(C l ,C Z,C 3) (ClEA], C2EA], C3EA], Cl'fC2, Cl #C 3 ,
ClC 3 ,
Cl CCZcC 3)
13
(C l ,C Z ,C 3) (ClEA z' CzEA z, C3EA z' CI#C Z' C!"C 3 · CiC 3 ,
ClcCZcC 3)
For
t
= n+l
an optimal solution is
AZ = {D[n/Z]' D[n/Z]+l}'
since the optimum for
A] = {D[n/Z]-l,D[n/Z]}'
and it is an optimal solution for each
t=2 is
3 and for
t=l is
t,
2.
n+Z
([n/2]+1) .
34
8.
Let
for every
hand
(1
w
w
~
H be integers with the property
~
h+l)
and
< i
Let us construct a graph with vertices
~
(n-l+l)/2
from
(n+l-l)/2
only to
(i,j)~
the vertex
4:
~
'i
and
1
j
~
~
(i~j)
H.
and
w
~
~
H.
There are directed edges
(18)~
A.
J
correspond to
so we can use Lemma
the maximal number of vertices is the sum of lengths of the
diagonals of this rectangle.
So if for
l=n+l
n+l.
where
Let
the condition (27) is just
h+H-l
we choose the
h
h
largest
largest
diagonals in the "middle":
11
A+
{D[(n+H-h)/2]~
A~+l
{D[(n+H-h)/2]-1' ... , D[(n+H-h)/2]+h-2}
.•.
~
D[(n+H-h)/2]+h-l}
An+1 = {D[(n+H-h)/2]-(H-l)~
'~
l
A1,
then
""
rectangles~
max
1
i=l
~
since the corresponding vertices will also form the
Thus, we may apply Theorem 7:
H-l [(n+H-h)/2]-j+h+l
L IA. t
}
l
~ will have the optimal property in all the smaller
"middle" diagonals.
H
... ~ D[(n+H-h)/2]+h-H
L )
-'-------~--
j=O i=[(n+H-h)/2]-j
h
35
n
n
n
([(n+H-h)/Z]-H+l) + Z([(n+H-h)/Z]-H+Z) + 3([(n+H-h)/Z]-H+3) + ...
n
n
+ Z([(n+H-h)/Z]+h-Z) + ([(n+H-h)/Z]+h-l)
Obviously, conditions Z and 3 are strong.
ample 5 we omit the second axiom and
n
For instance, if in ex-
is even, then for
l=n+l
the
In this case,
{D ,D , ... ,D _ } does not form an optimal solution for l=n-l because
n Z
Z 4
{D ,D , .•• ,D _ } has more elements. Thus, in this case Theorem 7 does
l 3
n l
not work, although it is easy to see that in this case
n-l
Z
max IA.I
1.
holds, too.
A and
Au{e}
Let
A]
e
be a fixed element of
can contain at most one.
S.
If
AcS-{e},
That means
S·, IA]I ::; (Zn/ Z) = Zn-l •
only half of the subsets of
A]
then among
can contain
The construction
of example 5 shows that this estimate is best possible.
Now, we prove a generalization of this statement.
THEOREM 8. If A = {Al~ ... ~Am} is a family of subsets of a set S
of n
elements and no f:wo different subsets satisfy the condition
A. c A.,
1.
IA.-A.I
J
J
1.
::;
k-l
(k
then
m
::;
\
(~)
1.
/
i::[n/Z] (mod k)
and this estimate is best possible.
2::
Z)
36
First we prove the following lemma:
LEMMA 5.
set S
or
If
of n
k~2,
then it is possible to divide the subsets of a
elements into disjoint chains which are
(a)
of length
(b)
of length less than
PROOF OF
k
lEMMA 5.
k
and symmetrical.
We use induction on
chain of length 2 which is symmetrical.
propriate chains for
n-l.
we form two chains
x •
n
{BI, ... ,B }
k
have property (a) again.
For
n=l,
we have one
Assume we have constructed ap-
From these chains, we will constuct new
chains using" the n-th element,
k
n.
From a chain
and
of length
{BIU{xn}, ••. ,BkU{x }}.
n
{BI, ... ,B }
i
From a chain
{B , ••. ,B }
I
k
(i < k)
They
we form the
following two chains:
{B , ... , B., B. U {x }},
I
1.
1.
n
{B
I
U
{x },
n
... , B. I
1.-
The lengths of the new chains of this type will be
be symmetrical;
IBII+IB. u{x }I = IBII+IB. I + 1
1.
n
1.
IB U{xn }! + lB.1.- l U{xn }! = IB11+IB.1.- 11 + 2 = n.
1
~k,
U
{x }}.
n
and they will
= n-l+l = nand
It is clear, further,
that the new chains are disjoint and that they contain every subset of
s.
The proof of the lemma is finished.
PROOF OF THEOREM 8.
by Lemma 5.
Let us consider a set of disjoint chains given
A chain of length at most
by the condition of the theorem.
(28)
IAI
m
~
k
cannot have two elements of
Thus,
number of chains.
By symmetry, every chain of type (b) has an element of size
[n/2],
and
A
37
every chain of type (a) has an element of size
chain cannot contain two elements of size
= [n/2] (mod k).
= [n/2] (mod k),
Since a
the number
of chains is exactly
\
/
n
(i)
,
i:=: [n/2 Hmod k)
which by (28) yields the desired inequality.
having
i=[n/2] (mod k)
The family of all sets
elements shows that the estimate is best possible.
The proof is complete.
Brief List of Further Results.
We have seen in Lemma 1 that
question is the following:
B = {B l ,··· ,B r }
r =
(~),
then
and
r
minlc(B)
1,
a.
J
~
j
(t
~
r • v/(n-v+l).
!c(B)1
An interesting
if
The natural conjecture that if
N
I
(v-I) is correct.
More generally, every
r>O
(av) + (av-ll) + ..• +
v
v-
in a uniquely determined manner if
~
~
"What is the minimum of
is fixed?"
r
t
Ic(B)1
~
j
minlc(B)!
v).
v
is fixed and
a
v
> ••• >a,
t
J. B. Kruskal [15] proved that
av )
(av - l )
(at)
(v-l + v-2 + ... + t-l .
A simpler proof is given by Katona [16].
A similar problem is the following.
family of subsets of
S
(I S I =n)
Ic(B) I
If
B = {B ,B , .•• ,B }
l 2
r
I B.\
=v,
1
and
v
v-k+l
• r
The exact minimum is still an open problem.
.
is a
38
Using the result mentioned above, Milner [13] proved the following
theorem:
A
If
= {Al, .•• ,A }
m
is a family of subsets of
different ones are in the relation
m :5
A,cA"
such that no two
and such that for all pairs
J
1
S
n
([(n+k+l)/2])
and the estimate is best possible.
Sperner's Theorem says if we have a family
m
there is a pair satisfying
>
A=
{Al, ..• ,A },
m
A,cA, ,
Kleitman [7] con-
J
1
where
sidered the question of the least number of such pairs we must have:
if
m
= ([n/2])+x,
then the number of such pairs is at least
x:5 ([n/~]+l)' but for larger
This is exact if
x
x([n/2]+1).
the exact estimate
is not yet known.
The most interesting application of Sperner's theorem is given by
Kolmogorov [17].
~
The concentration function of a random variable
is
defined as follows:
Q~(v)
~1,~2' •.• '~
If
+
~2
are independent,
n
x
~:5
x+v) .
identically distributed random var-
Q~(v) will denote the concentration function of the sum
iables,
~l
sup P (x <
n
+ .•• +
~.
n
Kolmogorov proved
Q ~(v)
n
c(v)
1
In
In the proof, he used Sperner's theorem through the sums
mentioned
in
the third example in the introduction.
LE,a,
1
1
which were
39
For the more-dimensional generalization of this probability theory
theorem, Sazonov [5] needed a new generalization of Sperner's theorem.
This generalization is given by Meshalkin [4]:
be a family of partitions of a set
notes the partition
ferent partitions
of
S
have subsets
elements, where
is fixed).
(r
A·1 l ,A·1 2 ,···,A.1r
A., A.
1
J
n
A = {Al, ••. ,Am}
Let
A , A
ik
jk
Ai
de-
Assume no two dif-
with the property
Then
m
~
n!
max - - , - - - n .
n !
L:n.=n l
r
1
and this estimate is best possible.
or
if
If we exclude two partitions
for each
k,
A. ,
1
then the problem is still
open.
Renyi [3] asked how many subsets we can choose having the Sperner
property if we choose every subset independently and with probability
n
1/2 •
The answer is
<[n/2]) ~ 2n/~
(2/I3)n
which is definitely less than
12/n. The other generalizations of Sperner's theorem
are not yet investigated in the random case.
An old problem is:
how many Sperner families are there?
to give a lower bound for the number
S
of
n
elements.
The family of all
w(n)
It is easy
of Sperner families on a set
[n/2]-tuples is a Sperner family.
Similarly, an arbitrary subfamily of this is again a Sperner family.
So
40
The newest result was obtained by Kleitman [20]:
)
2
~(n)
2
REFERENCES.
[1].
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On the set of divisors of a number. Nieuw Archief voor
Wiskunde, 2, 23 (1952), 191-193.
[2].
Sperner, E.: Ein Satz Uber Untermengen einer endlichen Menge,
Math. Z. 27 (1928), 544-548.
[3].
Renyi, A.: On random subsets of a finite set,
3 (26) (1961), 355-362.
[4].
Meshalkin, L.D.: Generalization of Sperner's theorem on the number of subsets of a finite set, Theory of Probability and
its Application, 8(1963), 203-204. (English translation)
[5].
Sazonov, V.V.: On multi-dimensional concentrations, Theory of
Probability and its Applications, 11 (1966) (English
translation).
[6].
Sevast'yanov, V.A.: On multi-dimensional concentrations, Theory
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[7].
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[8].
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[9].
Schonheim, J.: A generalization of results of P. Erdos and
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Theory, 1970 (to appear).
[10].
Anderson, I.: On the divisors of a number,
Soc. 43 (1968), 410-418.
[11].
Kleitman, D.: On a lemma of Littlewood and Offord on the distribution of certain sums, Math. Z. 90 (1965), 251-259.
[12].
Hilton, A.J. and Milner, E.C.: Some intersection theorems for systems of finite sets, Quart. J. Math. 18 (1967) 369-384.
Mathematica,
J. London Math.
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[13].
Milner, E.C.: A combinatorial theorem on systems of sets,
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[14].
Katona, G: Intersection theorems for systems of finite sets,
Acta Math., Acad. Sci. Hung. 15 (1964), 329-337.
[15].
Kruskal, F.B.: The number of simplices in a comples, Mathematical Optimization Techniques, Edited by R. Bellman,
1963, 251-278.
[16].
Katona, G.: A theorem on finite sets, Theory of I Graphs, Proc.
Colloq. held at Tihany, Hungary. 1966, Akademiai Kiad6,
1968, 187-207.
[17].
Kolmogorov, A.: Sur les proprietes des fonctions de concentrations
de M. P. Levy, Ann. Ins. II. Poincare 16, 1 (1958) 27-34.
[18].
Lubell, D.: A short proof of Sperner's Lemma,
Theory 1 (1966),299.
[19].
Katona, G.: On a conjecture of Erdos and a stronger form of
Sperner's Theorem, Studia Sci. Math. Hungary 1 (1966),
59-63.
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Kleitman, D.: On Dedekind's problem: The number of monotone
Boolean functions, (to appear in Proc. Amer. Math. Soc.).
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