Maximal Chains and Antichains
in Finite Partially Ordered Sets
Dwight Duffus
Mathematics and Computer Science Department
Emory University
Atlanta GA USA
The 24th Clemson Conference
on Discrete Mathematics and Algorithms
22 October 2009
Outline
1
Classic Problems and Results on [Maximal] Chains and Antichains
Outline
1
Classic Problems and Results on [Maximal] Chains and Antichains
2
Maximal Antichains in the Boolean Lattice 2n
Outline
1
Classic Problems and Results on [Maximal] Chains and Antichains
2
Maximal Antichains in the Boolean Lattice 2n
Pairwise disjoint maximal antichains in 2n
Outline
1
Classic Problems and Results on [Maximal] Chains and Antichains
2
Maximal Antichains in the Boolean Lattice 2n
Pairwise disjoint maximal antichains in 2n
Fibres in 2n
Outline
1
Classic Problems and Results on [Maximal] Chains and Antichains
2
Maximal Antichains in the Boolean Lattice 2n
Pairwise disjoint maximal antichains in 2n
Fibres in 2n
3
Maximal Chains and Antichains in Finite Partially Ordered Sets
Outline
1
Classic Problems and Results on [Maximal] Chains and Antichains
2
Maximal Antichains in the Boolean Lattice 2n
Pairwise disjoint maximal antichains in 2n
Fibres in 2n
3
Maximal Chains and Antichains in Finite Partially Ordered Sets
Conditions on chain size yield pwd maximal antichains
Outline
1
Classic Problems and Results on [Maximal] Chains and Antichains
2
Maximal Antichains in the Boolean Lattice 2n
Pairwise disjoint maximal antichains in 2n
Fibres in 2n
3
Maximal Chains and Antichains in Finite Partially Ordered Sets
Conditions on chain size yield pwd maximal antichains
Conditions on antichain size yield pwd maximal chains
Classic problems and results
Classic problems and results
Dilworth’s Max-Min Theorem [1942]:
A partially ordered set of finite width has a partition into width-many
chains.
Classic problems and results
Dilworth’s Max-Min Theorem [1942]:
A partially ordered set of finite width has a partition into width-many
chains.
Dedekind’s Problem [1897]:
Determine the cardinality of FD(n) the free distributive lattice on n
generators;
Classic problems and results
Dilworth’s Max-Min Theorem [1942]:
A partially ordered set of finite width has a partition into width-many
chains.
Dedekind’s Problem [1897]:
Determine the cardinality of FD(n) the free distributive lattice on n
generators; that is, enumerate the antichains in the Boolean lattice 2n of
all subsets of an n-set.
Classic problems and results
Dilworth’s Max-Min Theorem [1942]:
A partially ordered set of finite width has a partition into width-many
chains.
Dedekind’s Problem [1897]:
Determine the cardinality of FD(n) the free distributive lattice on n
generators; that is, enumerate the antichains in the Boolean lattice 2n of
all subsets of an n-set.
Sperner’s Theorem [1928]:
The maximum size of an antichain in 2n is
n
bn/2c
.
Classic problems and results
Proof of Sperner’s Theorem
Classic problems and results
Proof of Sperner’s Theorem
Call a chain X0 ≺ . . . ≺ Xh symmetric if |X0 | + |Xh | = n.
Classic problems and results
Proof of Sperner’s Theorem
Call a chain X0 ≺ . . . ≺ Xh symmetric if |X0 | + |Xh | = n.
2n has a partition into symmetric chains [SCD]:
Classic problems and results
Proof of Sperner’s Theorem
Call a chain X0 ≺ . . . ≺ Xh symmetric if |X0 | + |Xh | = n.
2n has a partition into symmetric chains [SCD]:
Xh + { n+1}
Xh−1 +{ n+1 }
Xh
X
h−1
X0 + { n+1 }
X0
Classic problems and results
2
n
Classic problems and results
2
n
n
Each symmetric chain contains one set of size bn/2c so there are bn/2c
chains in a SCD C(n). Any antichain A intersects any chain in at most 1
element.
Classic problems and results
2
n
n
Each symmetric chain contains one set of size bn/2c so there are bn/2c
chains in a SCD C(n). Any antichain A intersects any chain in at most 1
element. Thus:
X
n
|A| =
|A ∩ C| ≤
.
bn/2c
C∈C(n)
Classic problems and results
Simple questions about maximal chains and antichains
Classic problems and results
Simple questions about maximal chains and antichains
How many are there?
Classic problems and results
Simple questions about maximal chains and antichains
How many are there?
I
For chains, this is an irritating but elementary combinatorial
enumeration problem.
Classic problems and results
Simple questions about maximal chains and antichains
How many are there?
I
For chains, this is an irritating but elementary combinatorial
enumeration problem.
I
For maximal chains this is not so interesting: n!
Classic problems and results
Simple questions about maximal chains and antichains
How many are there?
I
For chains, this is an irritating but elementary combinatorial
enumeration problem.
I
For maximal chains this is not so interesting: n!
I
For antichains, this is Dedekind’s problem.
Classic problems and results
Simple questions about maximal chains and antichains
How many are there?
I
For chains, this is an irritating but elementary combinatorial
enumeration problem.
I
For maximal chains this is not so interesting: n!
I
For antichains, this is Dedekind’s problem.
I
For maximal antichains, this seems hard.
Classic problems and results
Simple questions about maximal chains and antichains
How many are there?
I
For chains, this is an irritating but elementary combinatorial
enumeration problem.
I
For maximal chains this is not so interesting: n!
I
For antichains, this is Dedekind’s problem.
I
For maximal antichains, this seems hard.
For particular classes of partially ordered sets,
Classic problems and results
Simple questions about maximal chains and antichains
How many are there?
I
For chains, this is an irritating but elementary combinatorial
enumeration problem.
I
For maximal chains this is not so interesting: n!
I
For antichains, this is Dedekind’s problem.
I
For maximal antichains, this seems hard.
For particular classes of partially ordered sets,
I
Are there interesting ways to generate maximal antichains?
Classic problems and results
Simple questions about maximal chains and antichains
How many are there?
I
For chains, this is an irritating but elementary combinatorial
enumeration problem.
I
For maximal chains this is not so interesting: n!
I
For antichains, this is Dedekind’s problem.
I
For maximal antichains, this seems hard.
For particular classes of partially ordered sets,
I
Are there interesting ways to generate maximal antichains?
I
Are there large families of pairwise disjoint maximal antichains?
Antichains in 2n
Pairwise disjoint maximal antichains in 2n
A family of maximal antichains
Antichains in 2n
Pairwise disjoint maximal antichains in 2n
A family of maximal antichains
Let S ⊆ [n] have 2k − 1 elements, k < n/2 and let AS be the set of all
k-subsets of S and their complements:
Antichains in 2n
Pairwise disjoint maximal antichains in 2n
A family of maximal antichains
Let S ⊆ [n] have 2k − 1 elements, k < n/2 and let AS be the set of all
k-subsets of S and their complements:AS is a maximal antichain in 2n .
Antichains in 2n
Pairwise disjoint maximal antichains in 2n
A family of maximal antichains
Let S ⊆ [n] have 2k − 1 elements, k < n/2 and let AS be the set of all
k-subsets of S and their complements:AS is a maximal antichain in 2n .
T
Maximality:
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U
S
S
1
0
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
2k−1
T
U
k
Antichains in 2n
Pairwise disjoint maximal antichains in 2n
A family of maximal antichains
Let S ⊆ [n] have 2k − 1 elements, k < n/2 and let AS be the set of all
k-subsets of S and their complements:AS is a maximal antichain in 2n .
T
Maximality:
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U
Any set A which does not have k elements
in common with S must miss at least k
elements of S and so must be contained in
S
S
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2k−1
T
U
k
some T for a k− subset T of S
Antichains in 2n
Pairwise disjoint maximal antichains in 2n
A family of maximal antichains
Let S ⊆ [n] have 2k − 1 elements, k < n/2 and let AS be the set of all
k-subsets of S and their complements:AS is a maximal antichain in 2n .
T
Maximality:
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U
Any set A which does not have k elements
in common with S must miss at least k
elements of S and so must be contained in
S
S
Antichain:
11
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00
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00
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00
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00
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00
11
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00
11
00
11
00
11
00
11
00
11
00
11
00
11
00
11
2k−1
T
U
some T for a k− subset T of S
k
Antichains in 2n
Pairwise disjoint maximal antichains in 2n
A family of maximal antichains
Let S ⊆ [n] have 2k − 1 elements, k < n/2 and let AS be the set of all
k-subsets of S and their complements:AS is a maximal antichain in 2n .
T
Maximality:
11
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11
U
S
S
Antichain:
0
1
1
0
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
2k−1
T
U
Any set A which does not have k elements
in common with S must miss at least k
elements of S and so must be contained in
some T for a k− subset T of S
Given sets U and T shown, they must
intersect, so that U cannot be a subset of T
k
Antichains in 2n
Pairwise disjoint maximal antichains in 2n
A family of maximal antichains
Let S ⊆ [n] have 2k − 1 elements, k < n/2 and let AS be the set of all
k-subsets of S and their complements:AS is a maximal antichain in 2n .
T
Maximality:
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U
S
S
Antichain:
0
1
1
0
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
2k−1
T
U
Any set A which does not have k elements
in common with S must miss at least k
elements of S and so must be contained in
some T for a k− subset T of S
Given sets U and T shown, they must
intersect, so that U cannot be a subset of T
k
Consequence: There is a family of more than 1.067422n of pairwise
disjoint maximal antichains in 2n .
Antichains in 2n
Fibres in 2n
Sets that intersect all maximal antichains
Antichains in 2n
Fibres in 2n
Sets that intersect all maximal antichains
Call F ⊆ 2n a fibre of 2n if for every maximal antichain M, F ∩ M =
6 ∅.
Given A ∈ 2n , let
C(A) = {B ∈ 2n | A ⊆ B or B ⊆ A}
Antichains in 2n
Fibres in 2n
Sets that intersect all maximal antichains
Call F ⊆ 2n a fibre of 2n if for every maximal antichain M, F ∩ M =
6 ∅.
Given A ∈ 2n , let
C(A) = {B ∈ 2n | A ⊆ B or B ⊆ A}
and note that for all A, C(A) is a fibre.
Antichains in 2n
Fibres in 2n
Sets that intersect all maximal antichains
Call F ⊆ 2n a fibre of 2n if for every maximal antichain M, F ∩ M =
6 ∅.
Given A ∈ 2n , let
C(A) = {B ∈ 2n | A ⊆ B or B ⊆ A}
and note that for all A, C(A) is a fibre.
Let f (n) be the minimum size of a fibre in 2n
Antichains in 2n
Fibres in 2n
Sets that intersect all maximal antichains
Call F ⊆ 2n a fibre of 2n if for every maximal antichain M, F ∩ M =
6 ∅.
Given A ∈ 2n , let
C(A) = {B ∈ 2n | A ⊆ B or B ⊆ A}
and note that for all A, C(A) is a fibre.
Let f (n) be the minimum size of a fibre in 2n
For any A of size n/2, C(A) shows that f (n) ≤ 2n/2 .
Antichains in 2n
Fibres in 2n
Sets that intersect all maximal antichains
Call F ⊆ 2n a fibre of 2n if for every maximal antichain M, F ∩ M =
6 ∅.
Given A ∈ 2n , let
C(A) = {B ∈ 2n | A ⊆ B or B ⊆ A}
and note that for all A, C(A) is a fibre.
Let f (n) be the minimum size of a fibre in 2n
For any A of size n/2, C(A) shows that f (n) ≤ 2n/2 .
(Duffus, Sands, Winkler (1990)) Every fibre in 2n contains at least
n!/2n−1 distinct maximal chains.
Antichains in 2n
Fibres in 2n
Sets that intersect all maximal antichains
Call F ⊆ 2n a fibre of 2n if for every maximal antichain M, F ∩ M =
6 ∅.
Given A ∈ 2n , let
C(A) = {B ∈ 2n | A ⊆ B or B ⊆ A}
and note that for all A, C(A) is a fibre.
Let f (n) be the minimum size of a fibre in 2n
For any A of size n/2, C(A) shows that f (n) ≤ 2n/2 .
(Duffus, Sands, Winkler (1990)) Every fibre in 2n contains at least
n!/2n−1 distinct maximal chains.
(Luczak (1996)) 2n/3 ≤ f (n)
Antichains in 2n
Fibres in 2n
Sets that intersect all maximal antichains
Call F ⊆ 2n a fibre of 2n if for every maximal antichain M, F ∩ M =
6 ∅.
Given A ∈ 2n , let
C(A) = {B ∈ 2n | A ⊆ B or B ⊆ A}
and note that for all A, C(A) is a fibre.
Let f (n) be the minimum size of a fibre in 2n
For any A of size n/2, C(A) shows that f (n) ≤ 2n/2 .
(Duffus, Sands, Winkler (1990)) Every fibre in 2n contains at least
n!/2n−1 distinct maximal chains.
(Luczak (1996)) 2n/3 ≤ f (n)
Improve these bounds on f (n):
2n/3 ≤ f (n) ≤ 2n/2
Chains and antichains in general
Conditions on chain size yield pwd maximal antichains
Chains and antichains in general
Conditions on chain size yield pwd maximal antichains
Chain length and pairwise disjoint maximal antichains
Chains and antichains in general
Conditions on chain size yield pwd maximal antichains
Chain length and pairwise disjoint maximal antichains
length(P) = l(P) = 5
P
5
For any finite partially ordered
set P, if all maximal chains in
P have the same size, say n,
then P has n pairwise disjoint
maximal antichains, namely,
its levels.
P4
P3
P2
P1
P0
Chains and antichains in general
Conditions on chain size yield pwd maximal antichains
Chain length and pairwise disjoint maximal antichains
length(P) = l(P) = 5
P
5
For any finite partially ordered
set P, if all maximal chains in
P have the same size, say n,
then P has n pairwise disjoint
maximal antichains, namely,
its levels.
P4
P3
P2
P1
P0
If we want 2 disjoint maximal antichains, we need only avoid 1-element
chains - AKA isolated elements. But to ensure 3 pwd maximal antichains,
we must restrict the sizes of maximal chains more carefully.
Chains and antichains in general
!1
!2
..
.
!
!n − 1
n ! !
!! !! n
n−1
..
.
!
2
!
1
P (n, 3)
Conditions on chain size yield pwd maximal antichains
Chains and antichains in general
!1
!2
..
.
!
!n − 1
n ! !
!! !! n
n−1
..
.
!
2
!
1
P (n, 3)
Conditions on chain size yield pwd maximal antichains
P(n, 3) has maximal chains of sizes n and
2n − 2, and no 3 pwd maximal antichains.
Chains and antichains in general
!1
!2
..
.
!
!n − 1
n ! !
!! !! n
n−1
..
.
!
2
!
1
P (n, 3)
Conditions on chain size yield pwd maximal antichains
P(n, 3) has maximal chains of sizes n and
2n − 2, and no 3 pwd maximal antichains.
However, this is as bad as things can get.
Chains and antichains in general
!1
!2
..
.
!
!n − 1
n ! !
!! !! n
n−1
..
.
!
2
!
1
P (n, 3)
Conditions on chain size yield pwd maximal antichains
P(n, 3) has maximal chains of sizes n and
2n − 2, and no 3 pwd maximal antichains.
However, this is as bad as things can get.
For all n ≥ 3, if every maximal chain C of
a partially ordered set P satisfies
n ≤ |C | ≤ 2n − 3
then P has 3 pwd maximal antichains.
Chains and antichains in general
!1
!2
..
.
!
!n − 1
n ! !
!! !! n
n−1
..
.
!
2
!
1
Conditions on chain size yield pwd maximal antichains
P(n, 3) has maximal chains of sizes n and
2n − 2, and no 3 pwd maximal antichains.
However, this is as bad as things can get.
For all n ≥ 3, if every maximal chain C of
a partially ordered set P satisfies
n ≤ |C | ≤ 2n − 3
then P has 3 pwd maximal antichains.
P (n, 3)
In fact, there is a condition that yields k pwd maximal antichains, though
it took some time to find it.
Chains and antichains in general
Conditions on chain size yield pwd maximal antichains
Theorem [Duffus and Sands - 2009]
Let n and k be integers with 3 ≤ k ≤ n and let P be a finite partially
ordered set. If every maximal chain C of P satisfies
n ≤ |C | ≤ n +
n−k
k −2
then P contains k pairwise disjoint maximal antichains.
Chains and antichains in general
Conditions on chain size yield pwd maximal antichains
Theorem [Duffus and Sands - 2009]
Let n and k be integers with 3 ≤ k ≤ n and let P be a finite partially
ordered set. If every maximal chain C of P satisfies
n ≤ |C | ≤ n +
n−k
k −2
then P contains k pairwise disjoint maximal antichains.
Alternative formulation:
Let the shortest maximal chains of P have m elements and the longest
have M elements. Then P contains at least
m−2
2+
M −m+1
pairwise disjoint maximal antichains.
Chains and antichains in general
Conditions on chain size yield pwd maximal antichains
The theorem is sharp: for 4 ≤ k ≤ n, P(n, k) has all maximal chains with
size between n and n + n−k
k−2 + 1, and no k pwd maximal antichains.
Chains and antichains in general
Conditions on chain size yield pwd maximal antichains
The theorem is sharp: for 4 ≤ k ≤ n, P(n, k) has all maximal chains with
size between n and n + n−k
k−2 + 1, and no k pwd maximal antichains.
n !
n !
n !
n !
n !
n !
n !
n !
n !
1 !
1 !
1 !
1 !
1 !
1 !
1 !
!
! ck−1
!
!n
! c
!
!
k−2 !
!
!
!
!
!
!
!
!
!!
!
!
!
!
!
!!!(k−3)n" n − 1
!
!
!
!
c
!
!
!
!
i+2
k−2
!c !
!
!
!!
!
i+1
!
!
!
!
!!!(i+1)n"
!
!c3 !
2 !
!
!!! in "
k−2
!c2 !
!
!!
k−2
1 !c1 !
!!! 2n "
!!! n "
k−2
C
1
k−2
1 !
C2
C3
···
Ci+1
P (n, k)
Ci+2
···
Ck−2
1 !
Ck−1
Chains and antichains in general
Idle thought . . .
Conditions on antichain size yield pwd maximal chains
Chains and antichains in general
Conditions on antichain size yield pwd maximal chains
Idle thought . . .
What about the dual question? Can we guarantee the existence of k pwd
maximal chains by bounding the sizes of maximal antichains?
Chains and antichains in general
Conditions on antichain size yield pwd maximal chains
Idle thought . . .
What about the dual question? Can we guarantee the existence of k pwd
maximal chains by bounding the sizes of maximal antichains?
It’s true that
Chains and antichains in general
Conditions on antichain size yield pwd maximal chains
Idle thought . . .
What about the dual question? Can we guarantee the existence of k pwd
maximal chains by bounding the sizes of maximal antichains?
It’s true that
if every maximal antichain contains at least 2 elements then the
partially ordered set has at least 2 pwd maximal chains,
if all maximal antichains have size n then the Dilworth decomposition
yields n pwd maximal chains,
Chains and antichains in general
Conditions on antichain size yield pwd maximal chains
Idle thought . . .
What about the dual question? Can we guarantee the existence of k pwd
maximal chains by bounding the sizes of maximal antichains?
It’s true that
if every maximal antichain contains at least 2 elements then the
partially ordered set has at least 2 pwd maximal chains,
if all maximal antichains have size n then the Dilworth decomposition
yields n pwd maximal chains,
there are several famous examples of valid statements with valid
duals, and
Chains and antichains in general
Conditions on antichain size yield pwd maximal chains
Idle thought . . .
What about the dual question? Can we guarantee the existence of k pwd
maximal chains by bounding the sizes of maximal antichains?
It’s true that
if every maximal antichain contains at least 2 elements then the
partially ordered set has at least 2 pwd maximal chains,
if all maximal antichains have size n then the Dilworth decomposition
yields n pwd maximal chains,
there are several famous examples of valid statements with valid
duals, and
the partially ordered sets P(n, k) are all 2-dimensional.
Chains and antichains in general
Conditions on antichain size yield pwd maximal chains
Idle thought . . .
What about the dual question? Can we guarantee the existence of k pwd
maximal chains by bounding the sizes of maximal antichains?
It’s true that
if every maximal antichain contains at least 2 elements then the
partially ordered set has at least 2 pwd maximal chains,
if all maximal antichains have size n then the Dilworth decomposition
yields n pwd maximal chains,
there are several famous examples of valid statements with valid
duals, and
the partially ordered sets P(n, k) are all 2-dimensional.
Meaning?
Chains and antichains in general
Conditions on antichain size yield pwd maximal chains
Given a partially ordered set P of dimension 2, that is, whose order is the
intersection of 2 linear orders ≤1 and ≤2 , there is a complementary
ordered set Q – its ordering is ≤1 ∩ ≤d2 , where ≤d2 is the dual of ≤2 :
Chains and antichains in general
Conditions on antichain size yield pwd maximal chains
Given a partially ordered set P of dimension 2, that is, whose order is the
intersection of 2 linear orders ≤1 and ≤2 , there is a complementary
ordered set Q – its ordering is ≤1 ∩ ≤d2 , where ≤d2 is the dual of ≤2 :
!1
!2
..
.
!
!n − 1
n ! !
!! !! n
n−1
..
.
!
2
!
1
P (n, 3)
Chains and antichains in general
Conditions on antichain size yield pwd maximal chains
Given a partially ordered set P of dimension 2, that is, whose order is the
intersection of 2 linear orders ≤1 and ≤2 , there is a complementary
ordered set Q – its ordering is ≤1 ∩ ≤d2 , where ≤d2 is the dual of ≤2 :
!1
!2
..
.
!
!n − 1
n ! !
!! !! n
n−1
..
.
!
2
!
1
P (n, 3)
P’s max chains = Q’s max antichains
1 2 ··· n − 1
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1
2
··· n − 1
Q(n, 3)
P’s max antichains = Q’s max chains
Chains and antichains in general
Conditions on antichain size yield pwd maximal chains
Observation
Given 3 ≤ k ≤ n and a 2-dimensional partially ordered set Q all of whose
maximal antichains A satisfy
n−k
,
n ≤ |A| ≤ n +
k −2
Q has k pwd maximal chains. Morever, the ordered sets Q(n, k) show that
the result is sharp.
Chains and antichains in general
Conditions on antichain size yield pwd maximal chains
Observation
Given 3 ≤ k ≤ n and a 2-dimensional partially ordered set Q all of whose
maximal antichains A satisfy
n−k
,
n ≤ |A| ≤ n +
k −2
Q has k pwd maximal chains. Morever, the ordered sets Q(n, k) show that
the result is sharp.
Question
For 3 ≤ k, what are the maximal intervals Ik so that whenever the size of
every maximal antichain of an ordered set P lies in Ik , P contains k pwd
maximal chains?
Chains and antichains in general
Conditions on antichain size yield pwd maximal chains
Observation
Given 3 ≤ k ≤ n and a 2-dimensional partially ordered set Q all of whose
maximal antichains A satisfy
n−k
,
n ≤ |A| ≤ n +
k −2
Q has k pwd maximal chains. Morever, the ordered sets Q(n, k) show that
the result is sharp.
Question
For 3 ≤ k, what are the maximal intervals Ik so that whenever the size of
every maximal antichain of an ordered set P lies in Ik , P contains k pwd
maximal chains?
Surely, we thought, the intervals are smaller in the general case.
Chains and antichains in general
To our surprise . . .
Conditions on antichain size yield pwd maximal chains
Chains and antichains in general
Conditions on antichain size yield pwd maximal chains
To our surprise . . .
Theorem [Howard and Trotter - 2009]
Let n and k be integers with 3 ≤ k ≤ n and let P be a finite partially
ordered set. If every maximal antichain A of P satisfies
n ≤ |A| ≤ n +
n−k
k −2
then P contains k pairwise disjoint maximal chains.
Moreover, the result remains sharp, due to Q(n, k), the complementary
(2-dimensional) partially ordered set to P(n, k).