1. This research was partially supported by the Air Force Office of Scientific
Research under Contract No. AFOSR-68-l4l5, University of North Carolina at
Chapel Hill, Department of Statistics.
2.
Department of Mathematics, University of North Carolina at Chapel Hill.
DISJOINING PERMUTATIONS IN FINITE BOOLEAN ALGEBRAS
Douglas G. Kellyl,2
Department of Statistics
University of North CaroZina at ChapeZ HiZZ
Institute of Statistics Mimeo Series No. 824
May, 1972
DISJOINING PERMUTATIONS IN FINITE BOOLEAN ALGEBRAS
Douglas G. Kellyl
Departments of ~ffithematics and Statistics
University of North Carolina at Chapel Hill
ABSTRACT
The existence of a map that permutes the members of a family of finite
sets so that every set is mapped into a disjoint set is shown to be equivalent
to a certain set of inequalities involving order-ideals of sets.
themselves are shown to admit such permutations.
Order-ideals
DISJOINING PERMUTATIONS IN FINITE BOOLEAN ALGEBRAS
Douglas G. Kellyl
Departments of Mathematics and Statistics
University of North Carolina at Chapel Hill
1.
Let
B
= B(X)
INTRODUCTION
be a finite Boolean algebra, which we will regard as the
family of all subsets of a finite set
X.
We will reserve the name 'set' for
B;
subsets of
X;
'family' will be used to refer to subsets of
subsets of
X.
We will use Roman capitals for sets and script capitals for
families.
We will denote the operations of union, intersection, and contain-
B as well as among subsets of X,
ment, among subfamilies of
S,
respectively.
B,
H: H*
= {A*:
A
E
H*
H}.
H - A is not
dinality of a set
The expression
H n A*,
G and
A - B,
if
H
for sets or families, de-
A which are not members of
although
A or of a family
for later use that if
u, nand
is defined as the family of complements of members
notes the set of members of
families,
by
B will be denoted by A ~ A*;
Complementation in
is a subfamily of
of
i.e., sets of
A-B = AnB*
A is denoted by
H are subfamilies of
IHI
(GnH)*
=
G* n
IGnHI
=
!G*nH*I,
H*,
=
B.
for sets.
IAI
B,
Note that for
or
The car-
IAI.
We note
then
IH*I,
(GuH)*
=
G* u
H*,
and thus
IGuHI
=
IG*uH*I·
Since we will not use the ring-theoretic notion of ideal, we define an
~
ideaZ to be what is usually called an 'order-ideal':
a subfamily
I
of
B
2
for which
A ~ BEl
implies
A E I.
An ideal in a fi.nite Boolean algebra is
uniquely determined by its generators, i.e., its maximal elements.
B,
any subfamily of
leA)
we denote by
I
A is
A,
the smallest ideal containing
which is the family of all subsets of members of
ideal with one generator; we write
If
for
A
A.
I({A}).
A principal ideal is an
If
I
is an ideal,
1*
is a filter; this may be taken as the definition.
H
F. J. Dyson has noted (unpublished communication) that if
ideals in a finite Boolean algebra, then
any ideal
(1)
IHnII ~ IHnI*I.
and
I
are
In other words,
H has the property
For any ideal
IHnII ~ IHnI*I.
I.
Dyson's proof is by induction on
Ixl;
we will give a proof below (Proposition
6 and Corollary 7) which, though by induction on
lxi,
is different from
Dyson's.
It is easy to see (and is proved as part of Theorem 1 below) that property (1) is enjoyed by any subfamily
tation; that is, a bijection ¢
joint for every
A in
H.
of
H of
H to
B which has a disjoining permuH such that
¢A
and
A are dis-
\Vhat is perhaps more surprising is the converse,
that (1) implies the existence of a disjoining permutation of
H.
This is
easily obtained, however, by the application of P. Hall's 'marriage' theorem
to the 'disjointness relation' of
Theorem 1.
H (defined below), in the proof of
We will refer to (1) as 'Dyson's condition'.
We recall first some results from matching theory.
The reader is re-
ferred to the books by C. L. Liu [4] and L. Mirsky [5] for elaboration and
proofs.
If
R
~
S
is an injection of
in
R.
If
A
~
S,
x
is any relation between sets, a matching of Sunder R
T
S
to
R(A)
T which is contained (as a set of ordered pairs)
denotes
{bET: (a,b) E R for some
a
in
A}.
If
3
Sand
under
(If
T are finite, then according to
R exists if and only if
Sand
P,
Hall's theorem a matching of
IAI s IR(A)I
for every subset
T are not finite, then according to
R,
A of
S
S.
Rado's extension of
Hall's theorem, which can be found in Mirsky's book, we need the additional
hypothesis that
iff
R({a})
IAI ~ IR(A)I
is finite for each
for each finite subset
a
in
A of
S;
then a matching exists
S.)
A corollary to Hall's theorem, found in Liu's book, provides in the
finite case that if there is a positive integer
R-I({b})
for all
a
In particular, if
S,
in
Sand
RS S
then a matching of
x
S
S
b
in
T,
Sunder
such that
R({a})
then a matching of
is symmetric and
to
k
R
R({a}) = k > 0
~
for each
a
exists.
D S H x H
H
H of a Boolean algebra, defined by
on a subfamily
(A,B)
2.
THEOREM 1:
H
k
Sexists.
We will be applying these results to the symmetric relation
Then
~
€
H iff A and B are disjoint members of H.
D
DISJOINING PERMUTATIONS AND DYSON'S CONDITION
Let
H
be any subfamily of a finite Boolean algebra
has a disjoining permutation if and only if
H
B.
satisfies Dyson's
condition (1).
Proof:
let
Suppose first that
1 be any ideal.
disjoining; since
t
Hnl* = 0,
If
Otherwise, the image of
t
is a disjoining permutation of
and
then the desired inequality is tribial.
Hnl* under t
is injective,
H,
is contained in
IHnII
~
Hnl since t
is
IHnI*1
For the converse, we notice that a disjoining permutation is just a
matching of
H under
D ;
H
so by Hall's theorem,
H has a disjoining permu-
in
4
tation if and only if
(2)
IAI ~ IDH(A)I
If a subfamily
Hn I
(since
for every subfamily
H is given, let I
A of
An B
A of
=~
iff
AS
=
H.
I(A*).
It is clear that
ASH
B*), and that
n 1*.
D(A) =
Property (2)
0
follows.
(A somewhat artificial transfinite version of this theorem runs as follows:
If
H is a set of members of a Boolean algebra (ring of sets) such
H intersects all but finitely many members of H,
that each member of
H has a disjoining permutation iff
erated order-ideal
I.
IHn11
~
IHnI*1
then
for every finitely-gen-
The proof is accomplished by inserting the word
'finite' before each of the two occurrences of the word 'subfamily' in the
above proof.)
The notion of a disjoining permutation has two obvious grapn-theoretical
formulations.
(See the books by Harary [3] and Liu [4] for definitions of
terms used in this and the next three paragraphs.)
The use of Hall's theorem
suggests the natural bipartite graph associated with the relation
vertex set comprises two disjoint copies of
sider the graph
G(H)
H.
~
whose vertices are the members of
is in
tex is adjacent to all others.
whose
On the other hand, we can con-
adjacent if and only if they are disjoint as sets.
multiple edges; if
DH,
G(H)
H,
any pair being
is a graph without
H, then G(H) has a single loop, and this verExcept for this possibility,
the complement of the intersection graph of
H.
G(H)
is just
Since every finite graph is
the intersection graph of some family of nonempty subsets of a finite set, it
follows that every finite graph without loops or multiple edges (or with a
single loop whose vertex is adjacent to all others) is
G(H)
for some
H.
5
A disjoining permutation then corresponds to a spanning subgraph of
G(H)
~l1hich
is the vertex-disjoint union of circuits and paths of length 1.
The paths of length I correspond to cycles of length 2 in the permutation.
Using this formulation we can easily prove
PROPOSITION 2:
If
H
has a disjoining permutation, then it has a dis-
joining permutation in whicl! all cycles are odd, and such that in any cycle,
each set is disjoint only from its neighbors.
Proof:
G'
Let
be the spanning subgraph of
given disjoining permutation.
If
G'
corresponding to the
G(H)
has an even circuit, its vertices can
be covered by a set of paths of length 1, viz. either of the two sets of alternate edges in the circuit.
If
G'
has an odd circuit with a chord, then
one of the two smaller circuits formed by the chord is odd; the remaining vertices are even in number and hence can be covered by a set of paths of length
1.
Repeating this process will produce a spanning subgraph consisting of
chordless odd circuits and paths of length 1.
This subgraph corresponds to a
0
disjoining permutation of the desired description.
The following three propositions are easy consequences of Theorem 1.
COROLLARY 3:
If
H
disjoining permutation if and only if
Proof:
I
If
=
0
A*
=
A,
H
then
H - A does.
ICH-A)nII - ICH-A)nI*1 + IAnII - IAnI*I.
IAnI*1 = IA*nII = IAnII.
I (H-A)nhl.
with
is any ideal, then
IHnII - IHnI*1
But
A
has a subfamily
SO
IHf'lII
~
IHnI*1
iff
ICH-A)nII
~
has a
6
COROLLARY 4:
Hand
H*
both have disjoining permutations if and only
H = H*.
if
Proof:
H = H*,
If
then complementation is a disjoining permutation of
H (in fact, a disjoining invoZution, all cycles being of length 2).
Conversely, if both
any ideal
so that
I
H*
have disjoining permutationa, then for
we have
IHnII =
A',
generated by
and
Hand
A is in
IHnhl.
Il = I A - A.
let
H*
Now for any
iff
A in
Then
B,
I
A is in
1HnIl*1 = IHnIA*I-l.
But
is the principal ideal
A
H iff
1HnIli = IHnIAI-l
H iff A is in H*.
IHnI~1 = IHnI~*1 ,
so
A is in
COROLLARY 5:
If
H satisfies Dyson's condition and H'
an injection
u:
H'
+
and
IHnI A' = IHnIA*1
0
is a family with
H such that A ~ uA for each A in H',
then
H'
also satisfies Dyson's condition.
Proof:
H has a disjoining permutation, say,
H' ,•
disjoining permutation of
for
Anu
-1
~,
~uA ~ uAn~uA
But then
=
~.
u
-1
~u
is a
o
We conclude this section by noting that for Dyson's condition to hold it
is not sufficient that
IHnIAI·.~
counterexample is furnished for
and the three 2-subsets of
X,
IHnI *'
A
for every principal ideal
Ixl = 3
by the family
which satisfies
lA'
A
H consisting of 0
Dyson's condition for every
principal ideal, but has no disjoining permutation.
B.
Moreover, if we let
(j
= 1,2, .. "n =
I HnB.1
J
~
I HnB
.1
n-J
IXI),
when
J
denote the family of j-subsets of
X
it is not sufficient for Dyson's condition that
j
:o;~.
Any saturated chain of sets (maximal family
7
of sets which is totally ordered by inclusion) satisfies this condition but
~
not Dyson's.
3.
EX~lPLES
Here we give a few examples of types of families that have disjoining
permutations.
The only example in which the proof presents any difficulty is
the first one:
all ideals have disjoining permutations.
PROPOSITION 6:
Proof:
the sets
Write
A.-A.
A
B,
A is any subfamily of
{Al~
c
= l,
(i,j
IU~Ail.
duction on
~
J
1
If
,An};
,n)
IAI ~ ID (A)(A)I.
then
1
we need to show that among subsets of
are at least
If this number is
0
or
n
distinct sets.
1,
then
We use in-
A contains only one
or two sets; in either case the assertion is easily seen to be true.
then that the assertion is true when
Let
a
be any member of
If
Al-a, ... ,An-a
there are at least
Denote
~ 1,
= k-l
Ai-{a}
by
and let
Ai-a.
are distinct sets, then by the inductive hypothesis
distinct sets among subsets of the differences
These are obviously subsets of the differences
(A.-a) - (A.-a).
J
1
n
n
UlA •
i
IU~Ail
Suppose
Ai-A
as
j
well.
So we need to consider the case in which the sets
all different.
Notice that if
is the union of the other with
B-a
= C-a
while
{a}.
= B-a
while
(i
A
1
B,
are not
A, B
A-a
=
So we can renumber the sets:
= l, ... ,m); ~+i = A.u{a}
1
are all distinct.
then one of
Moreover, it is impossible that
A, B, C are all different.
in such a way that
Am+l-a, ..• ,An-a
A-a
Al-a, ... ,An-a
(i
= l, ... ,m),
and
8
Thus, by the inductive hypothesis, there are at least
sets of the sets
(Ai-a) - (Aj-a)
sets of
for
A. - A.
1.
J
for
i,j = m+l, ... ,n.
i,j = m+l, •. "n,
n-m
These are also sub-
and none of them contains
Also, by the inductive hypothesis, there are at least
sets of the sets
A. - A.
produces at least
for
J
1.
i,j = l, •.. ,m;
the
n-m
And each of these sets contains
sets obtained earlier.
sets of the sets
COROLLARY 7:
a,
to each of these
for
J
i = m+l, •• ,2m,
so is distinct from each of
The result is at least
n
distinct sub-
0
A. - A..
J
1.
a
A. - A.
m distinct subsets of the sets
a.
m distinct sub-
adjoining
1.
j = l, ..• ,m.
distinct sub-
Every order ideal in a finite Boolean algebra admits a
disjoining permutation.
Proof:
For any subset
A
H,
of an order ideal
lCA) s H,
An apparent strengthening of Proposition 6 is as follows.
are distinct sets, a region of the Venn diagram for
n
2
(not necessarily distinct) sets
A.
or
1.
A.*.
1.
B n
I
...
COROLLARY 8:
If
AI, .•. ,A
n
AI"" ,An
is one of the
AI' ..• ,An
where each
B
i
is either
PROPOSITION 9:
j+k ~ n.
B.
Bj
U
Bj +l
U ••• U
which are
AI, ••• ,A .
n
denotes the family of sets of size
J
we obtain
(i,j=l, ••• ,n)
unions of regions of the Venn diagram for
Recall now that
AI, ... ,A ,
n
are distinct sets, then there are at least
n distinct subsets of the differences
only if
n'
If
If we apply Proposition 6 to the finite Boolean algebra whose
points are the regions of the Venn diagram for
~
n B
and thus
j
in
B.
Bk has a disjoining permutation if and
9
Proof:
H = B. u Bj +l u
J
Let
I =B0 u Bl U
HnI = B. U Bj +l
J
U
U
B
n-j
...
U
...
u Bk ·
HnI* = H,
we have
B ... ~ H
if
n-J
j+k > n,
If
but
j ~ ~n.
then for
HnI = f/J
So
and
j >J§n
if
H has no disjoining per-
mutation.
If
j+k:;; n,
H = Bj
then
BuB
u
n-k
n-k-l
(or
U
Bj +l
II = f/J
U • •• U
if
k <
II'
mentation is a disjoining permutation for
Bn - k- l
I)'
11 *
And if
self has a disjoining permutation, for each member of
exactly
n-i
( . ) other members of
where
II'
=
i:;;~,
then
U •••
=
Bi
it-
is disjoint from
~
As a consequence ot Proposition 9, we observe that for any
G = BO u Bl
1
so that comple-
11'
B
i
I
o
B.•
~
U
u Bk u Bn has a disjoining permutation.
G is a aombina-
toriaZ geometry (see Crapo and Rota [1]) of a special type:
a free geometry (i.e. Boolean algebra).
k:;; n-l,
the truncation of
We conjecture that the family of flats
of any finite combinatorial geometry has a disjoining permutation.
We note
that Greene [2] has proved what would seem to be a partial result in this direction:
that in any combinatorial geometry there is an injection
points to the copoints which is disjoining in that
copoint
¢p
for any
p
E
p
¢
from the
is not a member of the
X.
If the family of flats of a combinatorial geometry has a disjoining per-
mutation, it follows (since
BO IJ
...
u
there are at least as many flats of size
1, 2, ... ,n.
B.
is an ideal for any
J
:;;
j
as of size
j) that
n-j,
~
for
j
=
This is in contrast to (but certainly does not contradict) Rota's
celebrated conjecture that there are no more flats of rank
etry) than of rank
n-j,
for
j:;;
j
(in the geom-
~n.
We conclude with a sufficient condition for the existence of a di.sjoining
~
permutation for a lattice of sets which is not a boolean algebra.
Let
G be
10
an arbitrary subfamily of
~
sub1attice.
0
If
€
G,
then
complemented, and also that
A~
A,
and for a subfamily
ments of members of
H.
which forms, under intersection, an infimum-
B(X)
G is also a lattice.
X is in
H of
G.
G let
IHnII ~ IHnTI
IRI
~
IHI,
so the hypothesis
A G-ideal is the intersection
B.
If
for every
H is a subfamily of
G-idea1
I,
then
G with the property that
H has a disjoining permutation.
Proof: For any A S H, let I be the G-idea1 generated by A.
A s DH(A)
A S HnI
and
and
DH(A)
HnI
S
G by
R be the family of all G-comp1e-
In general, of course,
G with an ideal of
PROPOSITION 10:
G is
Denote complementation in
of the following proposition is quite strong.
of
Suppose that
is a G-idea1, so
DH(A).
Thus
IAI
:s;
I S DH(A).
I DH(A) I •
Moreover,
-
A S I.
Then
So
0
REFERENCES
1.
On the Foundations of Combinatorial
Theory: Combinatorial Geometries (preliminary edition), M.I.T.
Henry H. Crapo and Gian-Car10 Rota,
Press, Cambridge, Mass., 1970.
2.
Curtis Greene,
A Rank Inequality for Finite Geometric Lattices.
Theory~ 9 (1970) 357-364.
J. Combinatorial
3.
Frank Harary, Graph Theory.
Reading, Mass., 1969.
4.
C. L. Liu, Introduction to Combinatorial Mathematics.
Company, New York, 1968.
5.
L. Minsky,
Addison-Wesley Publishing Company,
Transversal Theory.
McGraw-Hill Book
Academic Press, New York, 1971.
FOOTNOTES
AMS subject classifications:
Primary 05.04, 05.15~ 06.60;
Secondary 05.35
Key words and phrases: Finite Boolean algebra; disjointness relation on
finite sets; order ideal.
1. This research was partially supported by the Air Force Office of
Scientific Research under Contract No. AFOSR-68-1415, University of North
Carolina at Chapel Hill, Department of Statistics.