EXTREMAL PROBLEMS IN DIMENSION THEORY
FOR PARTIALLY ORDERED SETS
by
Robert Joseph Kimble, Jr.
B.S. United States Naval Academy
(1970)
SUBMITTED IN PARTIAL FULFILLMENT
OF THE REQUIREMENTS FOR THE
DEGREE OF
DOCTOR OF PHILOSOPHY
at the
MASSACHUSETTS INSTITUTE OF TECHNOLOGY
Signature
.Qepartment of Mathematics, May 4, 1973
Certified
b..
. . ..
t (
Thesis Supervisor
Accepted by......
Chairman, Departmental Committee
on Graduate Students
Archives
(AUG
8 1973
2.
EXTREMAL
PROBLEMS IN DIMENSION THEORY
FOR PARTIALLY ORDERED SETS
by
Robert J. Kimble, Jr.
Submitted to the Department of Mathematics on May 4, 1973
in partial fulfillment of the requirement for the degree
of Doctor of Philosophy.
ABSTRACT
This thesis is concerned with answering the following
"Given a finite poset, P, how many lists of
question:
the elements of P are needed to satisfy the condition
that x is below y in P if and only if it is before
y in each of the lists?" Chapter 1 gives a new proof of
a theorem by Hiraguchi which states tha& for a poset
containing n > 4 elements, at most [N] lists are
needed. In Chapter 2 we use the ideas of Chapter 1 to
characterize those posets on 2n > 6 elements which
need n such lists. The last two Ehapters solve the
same problem for posets on 2n + 1 > 7 elements which
need n such lists.
Thesis Supervisor: Curtis Greene
Title: Assistant Professor of Mathematics
3.
Acknowledgements
The author would like to thank Curtis Greene for
serving as thesis advisor and for investing much
taluable time and effort in the preparation of this
thesis and in the general education of the author.
Special thanks are also due to W. Thomas Trotter and
Kenneth P. Bogart for several interesting and
enlightening conversations on the subject.
4.
Table of Contents
Introduc tion
Chapter 1
-
Hiraguchi's Theorem
Chapter 2
-
Characterization of 2n element Posets
with Dimension n for
Chapter 3
-
Characterization of 2n+1 element Posets
with Dimension n for
Chapter 4
-
n > 3
n > 5
Characterization of 2n+1 element Posets
with Dimension n for
Appendix
References
Biographical Sketch
n < 4
5.
Introduction
The idea of dimension pervades many branches of
mathematics, extending far beyond its geometric origins.
In many cases the introduction of an intrinsic dimension
has been a fruitful source of new ideas.
As a purely
combinatorial invariant, however, the dimension of a
discrete structure is often extremely hard to calculate
or even characterize in any simple way.
The case of
Kuratowski's theorem and the general problem of testing
graphs E6r planarity is an outstanding exception.
The theory of partially ordered sets (posets)
provides a natural but difficult problem of this type:
given a poset P, find the smallest integer
that
P
can be imbedded in
Fk .
componentwise ordering of
dimension of
P.
It
R k,
k
such
preserving the natural
We define
k
to be the
then turns out that this invariant
has another important combinatorial interpretation:
it is the smallest number of lists of the elements of
P
such that
before
y
x < y
in
P
if and only if
x
is
in every list.
Despite the elementary nature of the problem,
6.
very few results were known until recently.
papers
of Szpilrajn [8], Dushnik and Miller [6],
-
etc.
The first
gave simple classes of examples and proved
the equivalence of the two definitions given above.
The main result was that a product of k chains itself
has dimension k.
More recently, a paper of Baker,
Fishburn and Roberts [1] gives a thorough analysis of
the case
k = 2.
The starting point for this paper,
however, is the work of Hiraguchi [71 in 1951, the
subject of which is primarily concerned with bounds on
the dimension of P.
In his paper, Hiraguchi proves a basic inequality:
if
P
is a poset on a set
is at most
1.XI.
X, then the dimension of
P
His proof was extremely long and
complex, and was shortened somewhat by Bogart [3].
The
first main result of this thesis it a strengthening of
Hiraguchi's theorem, together with a new proof which is
far shorter and more illuminating than either of the
previous ones.
This proof is contained in chapter 1.
In recent work, Bogart and Trotter were able to
characterize those partially ordered sets
the dimension is
-7iXI
provided that
JXj
P
for which
is even, and
7.
k > 3.
k = 3,
With a single exception when
they
are obtained by taking the elements of ranks 1 and
k-1 in a Boolean algebra with k atoms.
In chapter 2
we show how the methods of section 1 lead naturally to
a short proof of Bogart and Trotter's theorem.
The next two chapters are concerned with solving
the same problem when
JXI
i
characterize those posets with
dimension k.
odd.
That is, we
2k + 1
elements and
The final solution is that, aside for
13 exceptions when
k = 3, every such case is obtained
by adding a single point arbitrarily to one of the
Bogart-Trotter examples.
The proof of this result is
much more difficult than the preceding ones, although
most of the difficulty lies in handling the small cases.
In fact, it is relatively easy to give a short,
self-contained proof for the case
k > 5.
This is the
content of chapter 3.
In chapter 4 we continue the much more difficult
case
k = 4.
These arguments in turn are based on a
list of exceptional posets for
k = 3
obtained by
exhaustive search of all posets on 7 points.
One of the most difficult problems in this area -
8.
still largely unsolved -
is to find good methods fdr
computing the dimension of explicit posets.
Even when
the number of points is small, the problem can be
prohibitively difficult.
In the course of proving the
main results of this thesis, it was necessary to develop
a wide variety of computational tools.
of good algorithms,
it
In the absence
is resillts of this type which may
be the most useful for developing a full understanding of
the subject.
9.
Chapter 1
Hiraguchi ' s Theorem
A poset, P, on a set, X, is a partial order relation
P C X
X.
x
A linear extension, L, of
order
L C X
write
x <P-y
in
or
P
x
y
X
such that
or
y >
P C L.
and
y
of
is above
X
x
in
P.
any
P
x
x < y
or
y > x.
is below
An antichain, A, in
P
is a subset of
X
is a subset of
P.
P
if
X
whose
A chain, C,
linearly ordered by
x c X, [x] + = {y s Xjx > y}
P.
For
and is called the
closed principal ideal determined by x; dually,
(x) + = {y s Xix < y}
and is called the open principal
filter determined by x.
linearly ordered by
L
will represent this as
sets
X
and
Y
If
and
X = {x , x 2
x2 <
x1 <
L = x1x
2
''
X
n.
are linearly ordered by
is
x'I}
.''
se <'
n-, we
If two disjoint
K
and
L,
respectively, then concatenhbion of the two strings,
KL will represent
K
U
L
U
(X
x
y
Two elements
(and we write
P
is in
<y,x>
we
If the partial order
elements are mutually incomparable in
in
<x,y> e P
are said to be incomparable in
nor
<x,y>
either
x 1% y).
If
is a linear
-x, and we say that
is obvious we simply write
x
P
Y).
Finally, let
P
10.
be a poset on
X
and let
Now, let
P
be a poset on
Y C X.
X.
of all linear extensions of
P.
Szpilrajn [8], we have that
P =
PY =P
Then
Let
t
n
Y x
Y.
be the set
Then by a theorem of
L.
With this we
L ej
can make the following
Definition:
The dimension of
P,
d(P), is the smallest
size of set of linear extensions of
section is
P
whose inter-
P.
We conclude this preliminary section with one last
Definition:
The width of
size of an antichain of
P,
w(P),
is the greatest
P.
For the rest of the paper we assume that
In this chapter we assume
on a set X.
P
is a poset
IXI = n.
We begin with some basic elementary results concerning the calculation of
d(P).
Lemma 1:
(i)
(ii)
If
If
Y C X
P
disjoint union of
then
d[PIY] < d(P).
is a poset on
PlY
and
Y
U
Z
PIZ, then
and
P
is the
li.
21.
d(P) = max {d(PIY), d(PIZ),
If
(iii)
P
has a maximal element x, then
= d[PIX\{x}].
d(P)
If
(iv)
other
z > x
and
then
z c X),
have the same comparabilities
y
and
z < y
if f
(z
x
d(P)
= d(PIX\{xl)
d(P) = max {d(PIX\{x}), 21
(v)
X
if
x
%
for all
x 96y
if
and
y.
d(P)
is a chain, then
X
If
z > y
iff
= 1
and if
d(P) = 2.
is an antichain,
Proof:
(i)
P = L in
If
...
n
Lk, then
p|y
=
L1 Yno.. .eL LkIY
(ii)
PlY = L in ... nLk
Let
Piz = min .. a.
n
where
(iii)
P
=
j < k.
..MJ,\LJ
P = L M1
00
PIX\{x} = L C)
Let
and let
Then
M Lk
)
..
Lk
Then
L1 xn...' Lkx.
(iv)
for some
Let
M1
PIX\{x} = L1n...
and
N
f or
f
0Lk
1 < i < k.
Then
If
L i = M iyNI
y < X,
then
12.
n
P = M 1yxN (..
P
=
M 1 y xN 1
(v)
r
MkyxNk.
M2 xyN2
If
X
Proof:
let
Li
let
Li
=
If
Let
y
n... O MkxyNk
iti
= 1.
P = (x x
2
' 'x
PIX\{xl = Lj \ ...
*
Lk
=
For
Lk.
be a linear extension of
Let
If
X
is
2 x1 )
d(P) < 1 + d[PIX\{x}].
then
x C X,
*
LitX\{x}.
then
x,
'
is a chain then
an antichain, then
Lemma 2:
If
P
1 <i
such that
[Lk! (x)+]x[LkiX\Ex]+]
Lk+l = [Lk IX\[xl+]x[LkI(x)+].
and
Then
P = L *o.,..Lk+l.
Corollary 3: If
Lemma 4:
Le t
C
Y C X
then
d(P ) <
be a chain in
a linear extension of
P
P.
jX\Yj + d(PIY).
Then there exists
in which every element of
is above everything with which it is incomparable.
Proof:
Let
Let
C = {xl,... xm}
X0 = XVX 1]I,
let
X
=
with
x 1 < ... < xm.
= [x il+\[ i+l1I
for
C
13.
1 < i < m
L
XM
and let
=
[x m+.
be a linear extension of
L = L0 Li
...
LM
0 < i < m, let
For
PjXi.
Then
is the required linear extension.
We are now in a positibn to prove two basic
inequalities for
Hiraguchi
[7].
The first was known to
d(P).
The second is new.
Theorem 5:
(i)
d(P) < w(P)
(ii) d(P) < max {2, n-w(P)}
Proof:
(i)
By Dilworth's Theorem [5] there is a set of
w(P) chains in
P
whose union is
For each chain of
X.
P
such a set construct a linear extension of
Lemma 4.
Then
P
as in
is the intersection of these linear
extensions.
(ii)
w(P)
Let
A C X
and assume that
any two elements of
be an antichain of
P
|X\AI > 2.
and
X\A.
Let
x
of size
y
Then by Corollary 3,
d(P) < d(PJA U {x,y}) + n-w(P) -
2.
But by the
reduction of Lemma 1, it is clear that
PIA U
{x,y}
be
has the same dimension as one of the four following
posets:
/
/
a1
a2 a3
A
Ix
x
x
A
y
Y
A
ar
A
It is easily verified that all of
y
these posets have dimension 2; for instance, the
first is
a 1 a 2 xa 3 y
a 3 a 2 ya 1 x.
d(P) < 2 + n - w(P) n - w(P) > 2.
2 = n -
Thus,
w(P)
provided that
d(P) < 2.
Otherwise
We obtain as an immediate corollary the following
Corollary 6:
containing
(Hiraguchi)
n > 4
Let
elements.
P
Then
be a poset on a set
d(P) < [n].
This was the main result of Hiraguchi's 1951 paper [71.
His proof was extremely long and Bogart recently gave a
shortened version [31.
In a sense, Theorem 5 can be
regarded as a refinement of Hiraguchi's theorem.
The
proof is much simpler and more direct than those of
Bogart and Hiraguchi and it contains more information.
X
15.
In fact the technique used here is the basis for our
attack on the next two problems of this paper.
16.
Chapter 2
Characterization
of 2n element Posets with Dimension n.
In 1941 Dushnik and Miller [6] gave an example
of a poset, P, on a set, X, where
d(P) = n; namely, let
X
lxi
= 2n
be the set of n-i element
{l,...,n}
subsets and 1 element subsets of the set
and let
P
and
be set inclusion.
We call this poset
S 2 n*
Furthermore, there is a six element poset,
, of dimension 3.
C6 =
showed that for
n > 3
Bogart and Trotter [4]
these are the only examples.
In
this chapter we show how the methods of Chapter 1 can be
used to give a relatively short proof of this result.
The main idea is to find conditions under which
the basic inequality
d(P) < n - w(P)
can be made strict.
In Lemmas 9, 12 and 13 we give three important conditions
of this type.
By Corollary 3 it is sufficient to give
conditions under which
IX\Al = 3
and
d(P) < 2
adding more elements preserves the relation
d(P) <
|XI-|AI.
since
17.
But first we need some ground work.
will be a maximal antichain in
A C X
element of
A
and we define
Given A, we
P.
X\A
to be the elements of
A
define
In this chapter,
above some
dually.
A*
Thus,
X
is
*
the disjoint union of A, A
and A*.
In Lemma 1 we gave some conditions whereby we could
remove an elemet from a poset without lowering its
In such a case we say, after Hiraguchi [71,
dimension.
that
We now give another such
is d-removable.
x
condition
Definition:
is no
z e X
Let
y
incomparable with
element.
x >- y
x < y
iff
Then unless
x.
x
and there
Furthermore, we
as a cover.
be a maximal element of
precisely one element
X
x)
x < z < y.
such that
refer to such a pair,
Lemma 7:
covers
(y
x < y
P
covering
Suppose that the element of
have a minimum or a maximum
X\{y}
y
is a chain,
is
d-removable.
Proof:
Let
PIX\{y} = L ()...ILk,
that the elements incomparable with
element, z.
We may assume that in
k > 2.
x
Ll,
Suppose
have a minimum
x
is below z.
18.
Let
L1 '
be
t
P = L
Then
with
L
placed immediately above
y
We may assume that in
2 < i < k, let
above
x.
Suppose then that the
l L 2yf...CLky.
elements incomparable with
Li'
Then
Corollary 8:
x
L1
be
have a maximum element z.
x is above z.
Li
x.
with
y
Then for
placed immediately
P = Lly 0 L 2 'fl ... G\Lk *
y
Let
that the elements of
and
x
incomparable with
X
Then if
k minimal elements.
Suppose
be as in Lemma 7.
x
have
d(PIX\{y}) > k,
y
is
d-removable.
Let
Proof:
assume that
PIX\{y} = L 1
x
O.
m >
OVLm,
k
is below each of the
k.
We may
minimally
incomparable elements at least once in the first k linear
extensions.
For
1 < i < k, let
placed immediately above
P =
...
L
Lemma 9:
nLk'
Suppose
....
A*
x.
Lk+ly
L '
be
d(P) < 2.
with
y
Then
...
is empty,
Lmy'
w(.PIA*)
does not contain a subposet isomorphic to
Then
Li
*
< 2
and
PjA*
19.
A | = 1, 2.
assertion is true for
and
lA j. By Theorem 5 the
By induction on
Proof:
IAI = n > 3
If
has a maximum element, then by application
PIA
Suppose then
of Lemma 1 we can use the inductive step.
w(PIA*) = 2.
that
Then by Dilworth's Theorem, there
C1
are two maximal chains,
C lU) C2 = A .
that
x, and
Suppose
C1
x
or
y+
and
a
and
a < yi
1
y1 .
}
for
and
a Iyi+1 }
n a1
...
Then
an-l xbn-n-1
assume that
elements.
y > z
P =b
C
Let
and
w > x
...
.*..
C2
Yi
Let
By Lemma 1,
Yi = {bi}
b n-a
yn-1 ...
for
1
1
We can therefore
both contain at least two
be the top two of
be the top two of
{A c Ala 4x
and
b y1 .
a < yi
and
1 < i < n.
for
Xi = {a i}
we can assume that
1 < i < n.
1 < i < n.
By
is below
A
Xi = {a e Ala < x
Let
such
> yn-1*
...
Lemma 1 we can assume that everything in
either
P A*
in
has one element,
y1 >
has n-i elements,
C2
C2 ,
and
C2 .
Lemma 1 and we can assume that
Now, by hypothesis we must have
C1
We can assume
x # z
w > z
and
w # y
by
by Lemmas 1 and 7.
or
y > x.
If
both, we can again use Lemmas 1 and 7 to reduce to the
20.
inductive step.
Thus, we can assume
w > z
and
By Lemmas 1 and 7 we can assume everything in
below both
w and y.
PIX\{wl = L1
elements,
fn
and
x
P
minimal in
Proof:
Since
and
Suppose
PIX\{w}
we have
and
Suppose
A
"'
x.
is
By the inductive step
y,
M
for suitable
Lemma 10:
L2 .
y
M2 .
x
contains two maximal
L1 = M 1 x
Then
P = M xw
is maximal in
x nv y.
Then
and
P
L 2 = M 2y
M 2 wy.
and
y
is
d(P) < 1 + d(PIX\{x,y}).
PjX\{x,y} = 1L., ...
Then
Lk.
P = yL1x...flyLkx 0 (LkI(x)+)x(LkIX\[xl+\[yl+)y(Lkt(y)+).
x < y, is said to be a cover of
A cover,
Definition:
rank k iff there are precisely k pairs,
x < a, b < y
that
and
<ab>, such
a ev b.
The following very useful lemma is due to Hiraguchi [7].
Lemma 11:
d(P)
If
x < y
is a cover of rank 0 or 1, then
< 1 + d(PIX\{x,y}).
Proof:
Let
PIX\{x,y} = LCn.fLk.
cover of rank 1, let
<a,b>
If
x -< y
is a
be the pair of incomparable
21.
x < a
elements such that
a linear extension of
let
PIX\{x,y}
1 <
(LklX\[x]+)x(Lk
i < k
that
L
let
=
Example:
Proof:
x -< y
Then
P I(A
U
A
A*)
Then
are possible:
x*y
A* = {zl.
and
\ z; x > z
P
such
n+L'
x
is covered
x,z
* or
d(P)
PJA
1
< 2.
Assume then that
Then the following cases
x > y > z; x > y
and
and let
is empty, then the
A*
or
assertion follows from Lemma 9.
= {x,y}
Lin ...
Now,
is a cover of rank 0.
has one element.
If either
=
be a cover in which
x < y
Then
P
and
' y; or x > y,z and y \ z.
In the first case we can assume by Lemmas 1 and 7 that
everything is below x.
the result holds.
Then by Lemma 1 and Theorem 5,
The second case follows immediately
from Lemma 1 and Theorem 5.
is
For
be a linear extension of
Suppose
Lemma 12:
A*
Li'
Lk
b < a.
in which
x)+\[y]+)y(LkI(y)+).
Li' X\{x,y}.
Let
only by y.
A
and assume
Lk' = (Lk1(x)+)x(Lk1y)+Ex+)y(T kIX\[yl+)
L'+1 =
and
b < y
and
In the third case we can
assume, using Lemma 7, that everything in A is either
22.
between
x
and
z
we can assume that
or below both
A
x
and
y.
has two elements.
By Lemma 1
The resulting
poset has 5 elements, and hence its dimension is 2.
In the last case
P
has the same dimension as
x
We may remove the circled
z
element by Lemma 7 and we may remove
A*
|A
Suppose
is empty.
Proof:
by Lemma 1.
d(P) < 2.
Thus we have
Lemma 13:
y
Then
U A*=
4
and neither
A
nor
d(P) < 3.
By use of the elementary reductions from the
preceding lemmas (incomparable minimal and maximal
elements or covers of rank 0) we-reduce the cases to the
following two:
an antichain and
A* = {c,d}
A
= {a,bc}
a,b,c > d
A* = {d}
and
or
= {a,b}
A
with both being antichains and
Using Lemma 1 we can assume that
A
with
A
and
a,b > c,d.
has at most
15 elements - a 1 , comparable only to a; a 2 , comparable
only to b; a 3 comparable only to c; a 4 comparable only
to d; a 5 comparable only to a and b; a 6 , comparable
23.
only to a and c; a7 , comparable only to a and d;
a 8 , comparable only to b and c; a9 ; comparable only
to b and d; alo, comparable only to c and d;
a1 1 , comparable to all but d; a 1 2 , comparable to all but
c; a 1 3 , comparable to all but b; a 1 , comparable to all
but a; and a 1 5 , comparable to all four.
first case,
P =
a 1 a 3 a 6 a 8 a 1 da 1 5a
a2
13 apa 1 0ca
2 agaa 5 aa 2 ba
5 da 7 a 6 a1 3 al1 a1 2 a1 5 alaaga 8 a1 4ba 1 0
n daa
Then in the
1 0 a 4 a 8a 9 a1 5 a1 2alla 5 a 2 ba1 3 a 6 a7
f...
a 3ca
. .
a 3 ca1 a
In the second case, P =
da 4aa a
7
0a
1 2 ca1 0 a 1 3 a 6 a1 5 a1 1 a 5 aa1
a 8 a9 a 2 ba 3
2 da 9 ca 3 a 1 0 aa 1 4 a 5 a 1 1a 15 a 12 ba6
n a a ca
1
5
1 3 a7
*0
a aa 4
3 a 2a 1 1 a 6 a8 daal2a 1 3 a 5 a9abaa 1 0 a4
The above proof is an example of a proof by "brute force".
There appears to be no elegant reduction in view of the
24.
fact that any of the elements of
A U A*
may be
removed and still leave a poset of dimension 3.
Further-
more, there are plenty of examples of posets where
removing two maximal or two comparable extremal
elements lowers the dimension by 2.
The only recourse
then was to find three linear extensions of
intersection yielded P.
P
whose
These were found by the author
in a relatively short time, but they were done mainly
by trial and error.
Thus, if the reader attempts to
find the motivation for these lists, he may encounter
some difficulty.
We can summarize the results of Lemmas 9, 12 and 13
in the following form.
Corollary 14
jX\Al > 3.
Suppose
Then
A
is a maximal antichain and
d(P) < IX\Al
except possibly in the
following cases:
(i)
X\A
is an antichain lying either entirely
above or entirely below A
(ii)
with
of
A
X\A.
*X\Al = 3
and
PIX\A
or
lying between the minimal and maximal elements
25.
Proof:
IX\AI > 4
If
and
A*, then the
A* # $3
corollary follows from Lemma 13.
A* = 0
If
and
X\A
is not an antichain, then the corollary follows from
Lemma 9.
Hence (i) must hold.
If
IX\AI = 3, then
either (i) or (ii) holds by Lemmas 9 and 12.
The main theorem now follows easily:
Theorem 15:
|X| = 2n
(Bogart and Trotter [4]).
and
n > 3.
Then
C6
or
P = S2 n
d(P) = n
Suppose
(or the
dual of C6 .
n = 3, then there are only a few six element
If
Proof:
posets satisfying the restt±ictions of Theorem 5 and
Corollary 14.
If
then we may assume from Theorem 5 and Corollary 14
n > 4
that
Examination of these yields the conclusion.
X = A U A
*
x1 e A
Now, let
for two n element antichains
y1 c A
and
be incomparable.
such pair exists, then
d(P) = 2 < n.
d(P|X\{x ,y1 }) = n -
and by induction
S2n-2.
where
Thus,
xi
".
Yi
1
A = {x,...,xn
for
I and
1 < i < n.
allows us to conclude that
A
A
A
If no
Thus, by Lemma 10,
PIX\{x1 ,y1 } m
= {yl,...,Yn}
Removing
xi < yj for
and
xi
and
yj
2 < i, J < n
*
26.
and
i
that
j.
x
< Yj
Removing
for
Finally, removing
x
< y2
and
x2
x2
y2
and
1 < i, j < n
x3
y1 .
and
Thus
y3
allows us to conclude
and
2 # i /
j # 2'
allows us to conclude
P = S 2 n*
27.
Chapter 3
Characterization of 2n+1 element Posets
with Dimension n,. n > 5
After Bogart and Trotter proved the preceding theorem,
they posed the question of characterizing maximal
The
dimensional posets with an odd number of elements.
techniques they had used to prove Theorem 15 were too
cumbersome to attack this problem.
They had not discovered
Theorem 5 (11), which is extraordinarily useful.
Using
this fact, we conclude that a 2n+l element poset having
dimension n must have width n or n+l.
Armed with this
knowledge, we proceed in the following manner; first,
if the poset has width n+l, we show that it consists of
two antichains, one of size n and the other of size n+l.
If the width is n, we show that the poset has two disjoint
n element antichains.
If
n > 5
we then show that the
poset in fact has a subposet which is isomorphic to S n*
2
This says that there are no really new examples of
posets of dimension n, if we are allowed 2n+l elements
in our poset instead of only 2n.
To make this more
precise, we first make the following
28.
D-efinition:
A poset
if for every
P
on
X
x e X, d(PjX\{xl)
is said to be irreducible
= d(P) -
1.
In other
words, if we remove any element of X, we lower the
dimension.
In this chapter, we will prove the following
Proposition:
For
n > 5
there are no irreducible
2n+1 element- posets of dimension n.
Suppose then that
w(P) = n
We know that
w(P) = n + 1.
Let
element antichain.
Lemma 9, A
lX|
= 2n+l
or
n+l.
n = 3
and
Suppose for now that
A C X
and let
Assume that
d(P) = n > 3.
A*
be a four
is empty.
By
must be a three element antichain.
Examination of the seven element posets satisfying
the above oonditions yields the following seven examples:
and
Note that first four contain a subposet isomorphic to S6.
29.
and the last three are irredudible.
We are now in a
position to state
for
n > 4.
Proof:
lXi
Suppose
Lemia 16:
Then
Let
A
P
= 2n + 1, d(P) = n
A*
be an antichain of size
n + 1.
Then there is at least one pair of
is empty.
n = 4, A = {al, a 2 , a3 , a 4 , a 5 }
{b1 , b2,
By
is also an antichain
A
incomparable minimal and maximal elements.
A
w(P) = n + 1
contains a d-removable element.
Corollary 14 we may assume that
and
and
b 3 , bg}
a1
with
Suppose
and
b1 .
",
By removing them
we must be left with one of the seven posets of
dimension 3 listed above.
irreducible poset:
first
b2
2
If
that
Suppose we are left with the
a5
b,
a
b3
b4
a3
a4
< b 2 , b3 , b 4 .
would be d-removable.
Now removing
a5
then removing them allows us to conclude
a2 , b 3
But then either
So we may assume
must leave either
a
or
a5 < bi.
a5
30.
b
b2
b
b
b2
b4
or
al
a3
a4
ai
a5
In both instances we conclude that
'\ b 2 , then removing
dimension 2.
b
> a
and
a4
a4
a5
By symmetry
Now if
leaves a poset of
Again by symmetry we may thus conclude
a 1 < b2 , b 3 , b .
that
b3
b
b1 > a 2 , a4 .
we are also led to conclude
a
a3
b2
b3
Thus
b4
P
But then removing
a
b 2, a 3
a2
a3
a4
a5
leaves a poset of dimension 2, contradicting the
hypothesis that
d(P) = 4.
Suppose that removing
irreducible poset:
a1 , b 1
leaves the last
31.
b2
b3
b4
a2
a3
a4
a5
As befoDe, we may assume that
a5 < b1 .
allows us to conclude that
Now removing
a4 , b4
a ,< b2 , b
But then removing
a3 , b4
leaves a
poset of dimension 2.
a,
b
leaves the
a4
a5
b3 allows us to deduce that
a1 < b 2 , b4
Finally, suppose that removing
second irreducible poset:
b2
b3
a2
a3
Removing a2
and
a
a5 < b1 .
< b 3.
b4
Removing
Removing
dimension 2.
a 5, b 2
a4 , b3
allows us to deduce that
then leaves a poset of
32.
Thus, by induction removing a pair of incomparable
minimal and maximal elements leaves a poset containing
S 2 n-2*
Proceeding in a manner similar to above allows
us to conclude that
to
P
contains a subposet isomorphic
S 2 n, thus establishing the Lemma.
We now proceed to develop the tools necessary for handling
w(P) = n.
the case
The following lemma gives important
d(P) < n - w(P) -
conditions under which
1.
The proof
is very much like the proof to Lemma 12 and the lists
used here are based on those of Lemma 12.
Due to the
length of the proof and the fact that it has little
expository value, it is given in the Appendix.
Lemma 17.
A
Let
be a maximal antichain of
P
such
*
that neither
A
nor
is not an antichain.
then
Proof:
A*
If
is empty and at least one
A*U A*I
< 5,
d(P) < 3.
Given in the Appendix.
Lemma 18:
Let
A
a poset, P.
Then
Proof:
PtA
Let
be the set of all minimal elements of
d(P) < 1 + w(PIA ).
= C1
U
U0..
Ck
where each
C
is a
33.
Li, 1 < ± < k
Let
chain.
be a linear extension of
C
in which every element of
Let
incomparable to it.
M
P = L
.
IA* I
A
Let
=
If
Proof:
5
and
L1
= C1 U
A
where
Then
be the set of minimal elements of
w(PIA*) < 3, then
d(P)
P.
d(P) < 3.
< 3
by the
w(PIA*) = 3.
we may assume that
Hence,
where these are disjoint
C2 U) C3
we may conclude that
d(P) < 1 + d(PIA
P jC 2 (UC
A.
1 IA*)
By the now legendary pigeon-hole principle
chains.
Then
on
w(PIA*) < 2, then
previous lemma.
Then
(MIA)(L
Lk+l*
Lemma 19:
If
is below everything
Lk+l
reverses the order of
P
3
.
C1
U
has only one element.
C2
U
C3) < 3 by Lemma 9 unless
We can remove either
C1
or
C2
and use Lemma 9 unless neither is a cover of rank 0
which can happen only if
\/
PIA
removing the top element of
C2
.
But then
allows us to use
Lemma 9.
We can summarize the results of Lemmas 17 and 19 in
34.
Corollary 20:
IX\A| > 5.
Suppose
A
d(P) <
Then
is a maximal antichain and
IX\Al - 1
except possibly in
the following cases:
(i)
and
X\A
lies either entirely above or below
IX\Al - 1
w(PIX\A) >
U
X\A = A
(ii)
A
A*
and
A
and
A*
are both
nonempty antichains.
Proof:
(i)
follows from Lemma 19 and
(ii)
follows from Lemma 17.
w(P) = n.
We are now ready to start on the case where
Lemma 21:
Furthermore,
in P, A
*
is an n element set.
*
A
Let
{c,bn I
n > 5.
Then
P
contains a subposet
S 2 n'
A*
= {al,...,an}.
an antichain.
=
suppose that for some n-element antichain, A,
isomorphic to
Proof:
IXI = 2n+l, d(P) = w(P)
Suppose
{c}, A = {bl,***.bn I and
Then Corollary 20 says that
We may assume that
c < bn.
*
A
Removing
by Lemma 11 allows us to conclude that
is
35.
PIA U A \{an,bn}
with
2n-
1 < i < n - 1, let
Now for
as follows.
Place
bi
L
place it above
Now, if
bn > c,
below
b
an
be constructed as
Now
c
bn, keeping in mind that
bi, place it
is incomparable to
but above
is either
otherwise, it must be put
a,;
Do the same for
below ai.
J P
If it is incomparable to
incomparable to or below ai.
ai
l< i<n-1.
bi
aj.
above
for
ai
ai
and
bi.
Otherwise it must
*
be place above
b
and
above
between
elements below
j
some
an
and
and
above
n
A
and place the remaining
bn
ai.
c < ai
P = L
hen
bn
"
c
c
i, then
an > b,
A
arbitrarily, place the elements of
an
c
Now place the rest of
bi.
Note that if
c < b
by hypothesis.
n-.
...
for
Now, if
So we may suppose
an > c, then we use the same linear orders
If
as above with the following modifications.
We find
some
bn nu ai
i
between
or
an % bi.
If
bn
below
\,ai
an
and
n -
1
such that
If none exists, then
and
in
a
1
Lii
and
If
bn
an
and for some
an
, bi, then
PA U A
bn
reverse them.
an > bi, place
j $ i, for which
Z S2n'
is immediately
Then
bn
P = L1
..
Ln-10
immediately above
bn < aj, place
36.
b
immediately above
must exist or else
If
bn < a
below
ai
bn
and
bn
an \,b.,
P
an
=
note that such
Then
then we place an
L
in
L1
...
-
an > b , place
If
an
then follow the above instructions replacing
Again, if this is not possible, then
Ln-l*
immediately
again such
Ln-l.
a
(\...n
P = L
j # i, for which
and for some
Then
L
w(P) = n + 1.
immediately below
must exist.
in
PIX\{bn}
bn
a
C,
-
by
c.
X
Now suppose we have a seven element poset with three
minimal elements and three maximal elements.
Further-
more, suppose that the remaining element is below at
least two of the maximal elements and above at least
two of the minimal elements.
The only two such posets
which don't have a d-removable element by Lemma 1 are
or
The ones which do have a d-removable element all have
dimension 2 and so do these two.
Thus, all such posets
37.
We are now ready for
have dimension 2.
Lemma 22:
lXI
Suppose
more, suppose
P
= 9, d(P) = w(P) = 4; further-
has four maximal elements, four
minimal elements and a ninth element below at least
two of the maximal elements and above at least two of
the minimal elements.
Then
P
restricted to the
maximal and minimal elements is isomorphic to
Proof:
Let
a1 , a 2 , a 3 , a 4
b 1 , b 2 , b3, b 4
S3 .
be the minimal elements, let
be the maximal elements and let
a 1 , a 2 < c < b3 , b .
a3
If either
incomparable to either
b1
or
leaves a poset of dimension 2.
or
a4
is
b 2 , then removing them
Now
a1
and
a2
must
have different comparabilities and so must
a3
and
a4 , b1
and
therefore assume
if
a2
\,b 2 , then
Likewise for
c < b2
a3
b2 ,
b 1 > a 1 n,b 2
b2
b3
b
and
We may
b4 > a 4
and
\ b 3.
Now,
is d-removable by Corollary 8.
a3
if
'v
b 3.
If
a 2 < b1 , then
is a cover of rank 0, and removing it leaves a
poset of dimension 2.
a 3 -v b 4 .
below
and
c,
But then
Thus,
PIX\{c}
then we must have
a 2 '\ b1 .
- S8.
If
Similarly,
a3
a1 , a 2 , a 3
is also
all below
38.
b 1and
b2
or else removing two leaves a poset of
dimension 2.
a1 , a2 or a3
But then
is d.-removable.
Thus, we have the lemma.
Lemma 23:
and
P
Suppose
IXi = 2n + 1, d(P) = w(P) = n > 5.
has n minimal elements.
subposet isomorphic to
Proof:
Let
elements.
B = {b
Then
P
contains a
S 2 ne
A = {a1 ,...,an}
be the set of minimal
Then by Corollary 20, w(PIA*) = n.
,.-.,bn}
the element of
two elements of
CA
be an antichain and let
A*\B.
A
Now if
c
c
be
is not above at least
and below at least two elements of B,
then the lemma is established by Lemma 21.
then that
Let
a 1 , a 2 < c < bn-1,
a 3 ,...,an < bl...,6bn-2.
bno
Then
Suppose
Now suppose that
PIX\{a 1 ,bn}
has
dimension 2 which contradicts the hypothesis that
d(P) = n > 5.
Assume then that
a3
b3 .
Removing them
gives a poset satisfying the hypothesis of Lemma 22.
Thus we conclude that
a
< b; for
i < i < 2
ai
\ bi
2 # i # j # 2.
for
Now, if
1 < i < n
a 3 q bi,
then removing them implies that
the same comparabilities as
bi
and
b3
has
and is therefore
39.
d-removable.
a3 < bl, b2.
Thus
b3 > an-1
a1 < c
Now
1 < i < n
c,
1 < I < n
implies that
bn
± # 3.
and
is a cover of rank 1.
a 3 < bi,
Removing it implies that
i # 3, and removing
Similarly,
and
b3 > a ,
PIA U B = S2n'
Thus
We wrap us the remaining difficulties with
Lemma 24:
Suppose
such that
JA *,
d(P)
A
|Aj
is a maximal antichain in
> 2
jA U Agl = 6.
and
P
Then
< 4.
Proof:
Given in the Appendix.
This immediately strengthens Corollary 20.
Suppose
Corollary 25:
IX\AI > 5.
Then
is a maximal antichain and
A
d(P) <
IX\Aj - 1
except possibly in
the following cases:
(i)
size
either
IX\AI
(ii)
-
1
A*
or
A*
contains an antichain of
or
IX\Al = 5,
IA*I,
IA,
>
2
and both are
4o.
antichains.
Lemmas 16, 21 and 23, and Corollary 25 are summarized in
Theorem 26:
For
n > 5,
there are no irreducible
2n+1 element posets of dimension n.
41.
Chapter 4
CharacterIzation of 2n+1 element *Posets
.with Dimension n, n < 4
In Chapter 3 we were able to show that there were
no irreducible 2n+1 element posets of dimension n for
n > 5.
In this chapter we will show that this same
statement holds for
n = 4,
but not for
n = 3.
The existence of seven element irreducible posets
of dimension 3 has been known for some time.
Baker,
Fishburn and Roberts [1] give an example isomorphic to
and Bogart and Trotter give the following example:
Subsequently Trotter and this author have independently
examined all seven element posets of width 3 and 4 to
discover all those irreducible of dimension 3.
42.
Including the first two mentioned, here - along with
duals of course - is a complete list:
0
073-
I
x
43.
With this knowledge we can ask the question, "are
there any irreducible nine element posets of dimension 4?"
If the width is 5, the answer is no by Lemma 16.
If the
width is four, then by the previous lemmas we can conclude
that such a poset has two disjoint antichains of size 4
*
or it has a four element antichain, A, such that
A*
are antichains of sizes 2 and 3.
A
and
In the first case
we can assume by Lemma 22 that we have a four element
*
antichain A, such that
A
has one element and
A*
is
also an antichain.
Lemma 27:
lXi
Suppose
= 9
Furthermore, suppose that
of size 4.
Proof:
Then
P
and
P
d(P) = w(P) = 4.
has two disjoint antichains
contains a subposet isomorphic to S .
The only case not covered by Lemmas 20 and 21 is
the case in which the remaining element is above both
antichains, and removing it along with an element it
covers, leaves an irreducible poset of dimension 3 in
seven elements.
Let
Then we can assume
A
be a four element antichain.
A = {al, a 2 , a 3 , a 4 },
*
A* = {b1 , b 2 , b 3 , b4}
c is above
and
A*
{c}.
a 1 , a 2 , a 3 and a4, then
Clearly, if
PJA U A4
*
~
8
44.
of
al, c
c > a,, a 2 , a 3 .
a 4'
Assume that
leaves
Suppose the removal
.
Then we can assume
without loss of generaltty that the poset is labelled
a2
as follows:
b
b2
c, a3
b3
b
c, a 2 , we can conclude that
Now, by removing
Removing
a4
a3
b3 , b4 < a .
we can also conclude that
But then removal of
a 4 , b3
Suppose that the removal of
b2 < al.
bh,
leaves a poset of dimension 2.
a1 , c
leaves
04
a2
a3
a
b2
b3
b4
Then it can be labelled either
a2
a4
b
a
or
4
b
b2
b3
In either case, removal of
that
N
a
> b,
, b
b
c, a 2
Then
allows us to conclude
a
or
a2
can be
45.
removed.
Finally, suppose removal of
a1 , c
leaves
a
a2
3
a4
Then it can be labelled either
4
2
b2
b3
b
b2
3
b4
or
1
b4
In either case, removal of
a 4 , b4
leaves a poset of
dimension 2.
Assume that
of
c, a
a1 , a 2 < d
~'j
a 3 , a4 .
Suppose removal
leaves
a2
a3
a4
b3
b4
Again we can assume it is labelled
b1
Removing
or
a
a
or
c, a 2
we can conclude
> b 1 , b2, b3, 1b4.
a2
b2
b1 , b 2 < a1
r,
b 3 , b4
In the first case either
can be removed.
In the second case
a 4 > b4
is a cover of rank 0 and removing it leaves a poset of
dimension 2.
46.
Suppoee that removal of
c, a1
leaves
Then it can be labelled either
a2
a3
a
a
a3
a2
or
bi
b2
b3
b4
In the first case, removing
b1
c, a 2
that
But then removing
b 1 , b3P
b 4 < a1
"
b 2*
leaves a poset of dimension 2.
removing
b
< a1
c, a 2
In the second case,
But then either
Suppose then that removing
a
or
a1
the same comparabilities as
a2
c, a1
For any way that this is labelled, removal of
gives
a4 , b1
allows us to conclude that
k b 2, b 3 , b .
be removed.
b2
b3
b4
allows us to conclude
leaves
c, a 2
a 2 , and hence,
either can be removed.
Finally, assume that
Suppose that removal of
a1 < c
c, a1
'V
a 2 , a3 , a 4 .
leaves
can
47.
that it
Again we can assum
a2
b1
Then
a4 > b
b4 < a
a4 , b4
3
2
and
is labelled
d4
13
-4
is a cover of rank at most 1 unless
b, .
c
In the first case, removing
leaves a poset of dimension 2.
case removing
a4 , b 3
Suppose that removal of
leaves a poset of dimension 2.
c, a1
leaves
Then we can label it
2
3
bi
b2
b3
Randoving
a 3 , b3
we can conclude that
I
a2
b
P =
a4, b4
a2
a
a
b2
b
we conclude
a3
a4
3
4y
b
1
b4
3C
PIX\{a 3 , b3
Then removing
In the second
2
48.
But then removing
leaves a poset of dimension 2.
al, b 1
So we may assume that removing
c, a
which can be labelled
a
b1
But then removing
a
b2
a
b3
b3
leaves
b4
leaves a poset of dimension 2.
The proof of this last statement illustrates the
straightforward technique used to show that certain
types of nine element partially ordered sets have
dimension three or less.
The same technique can be
used to show that for the type of nine element poset
mentioned immediately before Lemma 27, its dimension is
at most 3.
Since the proof is tedious and unenlightening,
consisting essentially as an exhaustion of such posets,
only the statement of this result is given here.
It
has also been verified by Trotter.
In view of Lemma 27, the statement of Theorem 26
holds for
n = 4.
Thus, summarizing the results of
this chapter gives us
49.
Theorem 28:
There are 13 seven element irreducible
posets of dimension 3 up to isomorphism and dual
isomorphism classes.
For
n > 4, there are no
2n+l element irreducible posets of dimension n.
50.
App end ix
We have two possibilities:
Proof to Lemma 17:
Aj = 3 or 4.
then
d(P IA U A ) < 2
Lemma 9.
I=
jA
Suppose
unless
4.
P JA
If
< 2,
w(PIA
~
by
In that case, removal of one of the two
top chains by Lemma 11 and Corollary 14 gives the
conclusion.
Thus we may assume that
= 3.
w(P IA)
This gives us five cases:
a
b
a
b
C
c
C
2)
1)
3)
x
x
x
x
c
6
5)
a
b
or
5)
c
b
a
b
be the element of
A*.
For the rest of this lemma,
Let
d
let
ai, 1 < i < 15
have the same interpretation as in
Lemma 13; also, let
xj, 1 < i < 15, be comparable only
to
x. and the same things to which
ai
is comparable.
51.
Let
be comparable only to
x0
can assume that
x
has only the elements listed.
By removi ng
Case 1:
b, c > d.
If
As in Lemma 13, we
x.
d,
a
a, x
we can assume that
then removing
b
leaves a
a
poset equivalent t 0
x
a3
al
xi
x6
a4
a6
d
Using Lemma 7, we can replace
a
a3.
and
by an element covering
But this poset has dimension 2 by Lemma 9.
Thus, we may assume that
When
a4
a > d.
x > d, P =
xla1 a 3 x 6 a 6 a8 x 1 lalldx 1 5 a1 5 x
3 a1 3 al
a1 0 cx1 2 a1 2 a9 x7 a 7 x 5 xa 5 aa 2 ba 4)...
a 2 x 5 a 5 dx7 x 6x1 3xllxl 2 x1 5x 1 xa7 a 6 a1 3a1 al
1 2 a 1 5 alaaga 8 al
1 ba1 0 a3ca 4
(
da 4 ala g
When
x
"'
x 5 al2 x12
8 a 9a1 '
a3xa
55 a2 ba3
ns...
13 a 6xa 7 x 7 a 3 calxxa
d, we use the same linear orders as above with the
following modifications:
we drop
x7 , x1 2 3 x 1 3 and x 1 5;
and in the second linear order we put
d
immediately above x.
52.
Case 2:
As above we can assume that
a, b, c > d.
We
can also use the same linear orders with the following
modifications:
we drop
x 6 , x7 , X 1 4
the third linear order we put
Case 3:
x
and
x1 5 ;
and in
immediately below b.
As in case I we can assume
a, b, c > d.
We
can also use the same linear orders with the following
modifications:
x11
we drop
first linear order we put
Case 4:
and
x12;
and in the
immediately below
x
Again we can assume that
a, b, c > d.
c.
Then
x1 a 3 x 3x 6 x 8 x1ldx 1 5 xl3x 1 4a1 0 x 1 0 cx 1 2 x 9x 7 x 5 ax 2 bagxgxex (
n
a0 x2 x 5 dx 7 x 6 x 1 3xllxl2x 1 5 xlaxgx 8 x 1 4bxlox 3x4xa1 0 a 3 ca4
n
x0dx 4 agx 1 0 a1 0 x14x 8 x 9 x1 5 x 1 2x1 1 x 5 x 2 bx 1 3 x 6 x 7 x 3 a 3 cslax
Case 5:
Again we can assume that
a, b, c > d.
P =
...
Then we
can use the above linear orders with the following
modifications:
we drop
a3
second linear order we put
Suppose
SAo
I = 3.
If
A*
and
x
a10 ;
and in the
immediately above
c.
is a chain, then removal of
..
53.
A*
antichain.
Let
and
A*= {x > d
A
is an
A
by Lemma 11 gives the result unless
= {a, b,
c}.
a, b, c > d, and. we have the
Then we can assume that
following cases
x < a, b, c.
Case 1:
a a 3 a 6 a 8 aldxx 1 5 a1
a 2 a 5 da 7 a 6 a, 3 xx7
Gn
da 4 a
5
a14aix
1313
1 3 allal2x 12 a
10 a 1 4 a 9 a 1 5 a 1 2 xx 4x 1 0
c ,-x < a, b.
Case 2:
P=
Then
14
1 0 a1 0
cx12 12
5 x 1 5 alaa 9
99 77'5aa 2x4a4n...
xga 8 x1 4 al4 bx1 0 a 1 0 a 3 cx4 a'n...
a8 x 9 xl5x 1 2 ll
5 a 2 ba 1 3 x 1 3 x 7 a 6 a 7 a 3 ca 1 a
Then we can use the above linear
orders with the following modifications"
we drop
x 1 0 9 x1 3 , x1 4 and x 1 5 ; and in the first linear order we
place
Case 3:
x
immediately above c.
a, c I\ x < b.
Then we use the linear orders
of case 2 with the following modifications:
x
we drop
and x12; and in the second linear order we place
immediately above
Now, assume
a.
A, = {c,d}
is an antichain.
x
54.
a
x
a
b
b
Then
PIA
~
)
2)
*
,
a
x
3)
or
x
b
x
4)
b
Using elementary facts we can assume that
Case 1:
a, b > c, d.
If
x > c, d,
P =
then
da xa a 7x 7al2 x2ca 1 0x 1 3 x 6 x1 5x 11 x 5 xa 3a6a 15aa 5aa
n
a 2 da 9 ca 3 a 1 0 a8a 14x5 a 5 xllx
n
x 1 a 5x 5 ca 3 a2a1 xx 1 a6x 6 agda
If
14a8 aga 2 ba 3
1 5a
1 5x1 2 al 2 ba 6x 6x 1 3 x 7x1
xa1 3 a7 aaK...
4 a1 3 x1 3a1 5 x1 5 a9 a7 x7 x1 2 xal2baa1 0
d \ x > c, then we can use the same linear orders as
above with the following modifications: we drop
x7 , x12, x1 3 and x 1 5 ; and we put x 6 xl1x 5 x
below
If
c
x n,c, d, then we can use the preceding linear
x
; we put
x5 x
immediately below
first linear order and we place
x
immediately
in the first linear order.
orders with the following modifications:
and
in the third linear order.
n
a4
we drop
d
x6
in the
immediately below
a4
55.
that
a, b > c,
da 4 x 4 xx
C>
x, a
By removing
Case 2:
d.
7 x 1 2 ca10
10
Then
P =
1 3 x6
x 1 5 xlx
we can assume
x, b
and
5a 1 4x 8x 9xx 2 bx 3 x 0 xa3
x 2 dx 9 ca3 x3 x 1 0 a 0 x 8x 1 4x 5x 1 x 1 5xl 2 bx 6x 1 3 x 7 x 1 ax 4 xoxa 4 n
'
0..
x 0 x1x5 ca 3 x 3 x 2 x 1 1 x 6 x 8dx 1 4 x 1 2x 1 3x 1 5x 9x 7 bax1 0 Nxxal
0 a4
that
a
By removing
Case 3:
a, b > c, d.
and then
b, we can assume
Then we can use the same linear
orders as in case 1 with the following modifications:
x1 , x 6, x 7 and x 1 3 ; and we place
we drop
below
b
Case 4:
x
immediately
in the second linear order.
we are left with a poset of
c
By removing
dimension 2.
Proof to Lemma 24:
A U
Let
We can assume that A- a10
of the three elements
to none,
a
A* = {a, b, c, d, x, y}.
< i, J < 7},
{a, b, x}
a 0j
is comparable only to
comparable only to b, a 3 3
as follows:
is comparable
a, a 2 j is
is comparable only to x,
56.
a
a4 j is comparable only to
and
b, a 5 j is comparable
only to a and x, a6 j is comparable only to b and x,
and a
is comparable to all three for
three elements
ai
{c, d, y}
0 < J < 7; of the
aiO is comparable to none,
is comparable only to c, a1 2 is comparable only
to d, a13 is comparable only to y, ai4 is comparable
only to c and d, ai 5 is comparable only to c and y,
a16 is comparable only to d and y, and a
0 < i < 7.
Let
M.
{ai J0 < j < 71
for
0 < i < 7.
to all three for
ordering of
N ibe
a linear ordering of
0 < J < 7.
is comparable
7
<
= {a, b, y}
A
Suppose
{a i0
be a linear
< 7}
and
Let
for
A* = {C,
d, x}.
Then by Corollary 20 and Lemma 13 we can assume that both
a, b, y > c, d, x.
are antichains and
L
Let
= dxM 0M 1 Ma xM5M7aM6M2bM3
L2 = dcM 2 xM 3 M 6 M 7 g4
bM 5 M 1 aM0 Z
L3
L
as
xdN 1 N 5 cN4N 7 N 6 N 3 yN 2 N0 ba
= xNoN 3 cN 2 N 6 dN 7 N 5 yN4 Njba.
L1
K2
Let
with the following changes:
immediately above
Let
and
y
and put
be the same as
L2
a4 5
K1
be the same
put
a3 4
just above
with the following
a 5 4.
57.
modifications:
a2 3
put
a5 2
immediately below
at the bottom.
233
Let
K
L4
a4 5
right below
put
d.
and put
a2 3
necessary for
M,
a5 0
a52
Let
with the following changes:
a25
put
Now suppose
A
*
b
just below
a 3 2.
Tt is also
to be the last two elements of
the bottom element of
N5 *
with
3
be the same as
a34
to be the first element of
two elements of
L
right above
K4
immediately above
a4 3 a4 0
and
be the same as
the following modifications:
and put
a4 3
N2
and
Then
= {a, b, c,
M 5,
a2 5 a 4 5
a3 2
to be
to be the last
P = K1
K2
d}
A* = {x, y}.
and
K30 K4.
Again we can assume both are antichains and
a, b, c, d > x, y.
Let
Li = yMMlM4 xM 5 M7 aM 6 M 2 bM3 dc,
L2 = yM 2 xM 3 M 6 M7 M4 bM 5 M1 aM 0 cd,
L3 = xN0N N4 yN 5 N7 cN 6 N 2 dN 3 ba
L
L
= xN 2 N 3 N 6 N 7 N4 dN 5 N1 cN0 .
and
Let
with the following changes:
above
a55
and
be the same as
a33a35
L2 .
Let
between
K3
K
be the same as
put
c
a4 4 a 4 2
and
c.
be the same as
immediately
Let
L3
K2
58.
with the following modifications:
immediately below
Let
K4
put
a3 5
a2 2
be the same as
and put
L4
immediately below
bottom.
It is necessary for
in
Then
M5
*
P = K1
K2
put
a4
a3 3
at the bottom.
with the following changes:
a2 4
a51
K3
and
a4 2
at the
to be below
K .
a55
59.
Ref erences
1.
K. Baker, P. Fishburn and F. Roberts, "Partial Orders
of Dimension 2, Interval Orders and Interval Graphs,"
RAND memorandum, p. 4376 (1970),
2.
G.
Birkhoff,
LAttice Theory,
Amer.
Math.
Soc.
Colloquium Publs., Vol. 25 (1967)
3.
K. Bogart, "Maximum Dimensional Partially Ordered
Sets I", J. Discrete Math.
4.
To appear
K. Bogart and W. Trotter, "Maximal Dimensional
Partially Ordered Sets II",
J. Discrete Math.
To appear
5.
R. P. Dilworth, "A Decomposition Theorem for
Partially Ordered Sets", Annals. Math. 51 (1950),
pp. 161-166
6.
B. Dushnik and E. Miller, "Partially Ordered Sets",
Amer. J. Math. 63 (1941) pp. 600-610
7.
T. Hiraguchi, "On the Dimension of Partially Ordered
Sets",
8.
E.
Sci. Rep.
Szpilr&Jn,
Kanazawa Univ.
1 (1951)
pp.
77-94
"Sur l'Extension de li6rdre Partiel",
Fund, Math, 16 (1930), pp. 386-379
60.
Biographical Sketch
The author was born in Allentown, Pennsylvania
on 9 July 1 9 48.
He graduated from William Allen High
School there in June 1966.
At the end of that month he
entered the United States Naval Academy from which he
graduated four years later with majors in Physics and
Mathematics.
Since that time he has been a Lieutenant
in the U.S. Marine Corps studying at M.T.T.