ANTICHAINS AND CHAINS OF FINITE POSETS.
BENJAMIN IRIARTE
1. Introduction and motivation.
The goal for the project will be, firstly, to understand the concepts
of the theory of antichains and chains of finite posets as first developed
by Greene and Kleitman in [2] and [3]. The first step in our work will
be, in particular, to prove the following result, which is also the subject
of a more recent survey by Fomin and Britz [1]:
Theorem 1.1. Let P be a finite poset with ]P = n. Take i ∈ P. For
all positive integers i, let di be the maximal number of elements of P
contained in a disjoint union of i chains in P , and define d0 = 0. Let
also gi be the maximal number of elements of P contained in a disjoint
union of i antichains in P for i > 0 and g0 = 0. Define δi = di − di−1
and γi = gi − gi−1 for all positive integers i. Then, the sequences
δ = {δi } and γ = {γi } are non-increasing. Furthermore, the partitions
of n defined by δ and γ are conjugate to each other.
This first manuscript will be devoted entirely to present a complete
self-contained proof of Theorem 1.1, which will rely on the ideas developed by Greene and Kleitman. Further sections (not yet written) will
hopefully contain new results.
Now, Theorem 1.1 generalizes the results about shapes of permutations presented in the Appendix of Chapter 7 in Stanley’s EC2 [5]. The
latter can be studied using the machinery of RSK, which is fundamentally well understood. One of the questions one would like to answer
in this classical setting is the following:
Suppose w ∈ Sn and consider the word expression for w. Let δ = {δi }
be such that δ1 + δ2 + · · · + δk is the length of the longest subsequence
of w which can be written as a disjoint union of k disjoint increasing
subsequences, for all k ∈ P. We would like to know under what precise
conditions on w can we find, for each k, k disjoint increasing subsequences of w with lengths δ1 , δ2 , . . . , δk respectively. Also, it might be
interesting to know when exactly is δ strictly decreasing.
This raises the same question for the theory of antichains and chains
of posets. Hence, using the notation of Theorem 1.1, we want to know
1
2
BENJAMIN IRIARTE
sufficient or at least necessary conditions on P so that, for each k, there
exist k disjoint chains (resp. antichains) with cardinalities δ1 , . . . , δk
(resp. γ1 , . . . , γk ) respectively. On the other hand, we would like to
know when δ and γ are decreasing. Answering these questions would
be the final goal for the project. However, although the theory and
key concepts related to our problem are easily accessible and elementary, a complete solution may probably require advanced mathematical
techniques.
It is nonetheless very interesting and important to notice that there
are very few actual computational examples of the theory. In particular, it is not known what the partitions δ and γ are for some well
known and concrete families of posets. Calculating these partitions for
the boolean algebras Bn , the posets of order ideals of boolean algebras
J(Bn ), the partition lattices Π(n), the posets P ar(n) of partitions of n
ordered by dominance, and several other examples many of which can
be found in the exercises of Chapter 3 in Stanley’s EC1 [4], might lead
to very useful and novel insights, and none of these computations have
been carried up except for the case of Bn .
2. The lattice of Sperner k-families.
Let’s begin with our study. The first fundamental goal will be to
prove Theorem 1.1. This will take a substantial amount of previous
work, and we will be able to present a proof only at the end of Section 4.
We begin with a couple of definitions. In all that follows, let P be a
fixed finite poset.
Definition 2.1. Let F1 denote the set of antichains of P . For A, B ∈
F1 , we will write A ≤ B if and only if for all a ∈ A, there exists some
b ∈ B such that a ≤ b.
Clearly, F1 and J(P ) are isomorphic posets, so F1 is a distributive
lattice. Actually, the operations of join and meet in F1 are the ones
derived from J(P ) by looking at maximal elements of order ideals.
Definition 2.2. A subset A ⊆ P is called a k-family if it is a disjoint
union of k antichains A1 ≤ A2 ≤ · · · ≤ Ak of P . Let Fk denote the set
of k-families of P . For A, B ∈ Fk , we will write A ≤ B if and only if
Ai ≤ Bi , for all i ∈ [k].
There is a potential flaw in our definition since, in principle, for a kfamily A given as the disjoint union of some antichains A1 ≤ · · · ≤ Ak ,
it might be the case that the sets Ai are not uniquely determined.
However, notice that if we consider the restriction PA of P to A, then
ANTICHAINS AND CHAINS OF FINITE POSETS.
3
Ak is necessarily the set of maximal elements of PA . Now, by removing the sets of maximal elements successively, one set at a time, this
observation actually shows that all the sets A1 , . . . , Ak are uniquely
determined.
The natural question to ask is what nice properties, if any, does Fk
have as a poset. As we will see, Fk is a lattice.
For A, B ∈ Fk , let A ∨ B be the k-family A1 ∨ B1 ≤ A2 ∨ B2 ≤ · · · ≤
Ak ∨ Bk . Under the stated order, the minimal element of Fk is given by
k empty antichains. Hence, we want to prove that A ∨ B is indeed well
defined. The inequalities in the definition are clear. Thinking from the
viewpoint of order ideals, it is also clear that (Ai ∨ Bi ) ∩ (Aj ∨ Bj ) = ∅
if i 6= j. Finally, the join condition is clear from the fact that F1 is a
lattice and the definition of Fk . We obtain:
Lemma 2.1. The poset of k-families Fk of P is lattice.
Of main importance for our purposes are certain special k-families.
We present them in a separate definition.
Definition 2.3. Let dk = max ]A and δk = dk − dk−1 . A k-family
A∈Fk
A ∈ Fk is called a Sperner k-family if and only if ]A = dk . Let SF k
be the restriction of the poset Fk to Sperner k-families.
It is true that the numbers dk coincide with the numbers given in
Theorem 1.1, as we will see later when we discuss the binary operator
?. Given this definition, we naturally would like to understand the
behavior of Sperner k-families under the join operation defined for Fk .
In order to do this successfully, we first define a new operation on
antichains.
Definition 2.4. For A, B ∈ F1 , let A?B = ((A∪B)−(A∨B))∪(A∩B).
In words, A ? B is the set of non-maximal elements of A ∪ B together
with A ∩ B.
This new operation on antichains has some useful properties. The
first easy property states that (A ? B) ∈ F1 whenever A, B ∈ F1 .
Let’s now study how the operation behaves for k-families, where for
A, B ∈ Fk we naturally define A ? B to be the disjoint union (A1 ?
B1 ) ∪ · · · ∪ (Ak ? Bk ). For completeness, we have to verify the following
fact.
Lemma 2.2. We have (Ai ? Bi ) ∩ (Aj ? Bj ) = ∅ whenever i 6= j.
Proof. Suppose on the contrary that the statement is false. Without
loss of generality, that would imply that there exists some x ∈ Ai ∩ Bj
with i < j. This x has to be a non-maximal element of Ai ∪ Bi in Ai ,
4
BENJAMIN IRIARTE
so x < b for some b ∈ Bi . On the other hand, as Bi ≤ Bj there exists
some b0 ∈ Bj such that b < b0 , so x < b0 . This cannot hold since both
x and b0 are in Bj .
Corollary 2.1. The lattice Fk is closed under the operation ?.
Proof. Let A, B ∈ Fk . By Lemma 2.2, we know that A ? B is the
disjoint union of k antichains. By considering the restriction PA?B of
P to A ? B and removing the sets of maximal elements successively one
at a time, we can see that A ? B is indeed a k-family for some disjoint
antichains C1 ≤ · · · ≤ Ck .
The proof of Corollary 2.1 gives the reason why the numbers dk
coincide with the ones defined in Theorem 1.1. We
We can actually say more about the k-familiy A ? B in Corollary 2.1.
Lemma 2.3. Let A, B ∈ Fk with A 6= B. Then (A ? B) < A, B.
Proof. The main ingredients of the proof are the following facts: (Ai ∧
Bi ) ≤ (Aj ∧ Bj ) for i ≤ j and (Ai ? Bi ) ⊆ (Ai ∧ Bi ) for all i ∈ [k].
Let C1 ≤ C2 ≤ · · · ≤ Ck be a decomposition of A ? B as a disjoint
union of k antichains. Suppose that for some i ∈ [k − 1] we have that
Ci 6≤ (Ai ∧ Bi ). Then, for some ci ∈ Ci it must also be true that
ci ∈ (Aj ∧ Bj ) with j > i. Let ci < ci+1 < · · · < ck be a list of
elements of P such that cm ∈ Cm for all m with i ≤ m ≤ k. For
all i ≤ m1 < m2 ≤ k we must have that cm1 ∈ (Aj1 ∧ Bj1 ) and
cm2 ∈ (Aj2 ∧ Bj2 ) imply j1 < j2 . We obtain a contradiction because
j > i.
Hence, Ci ≤ (Ai ∧ Bi ) ≤ Ai , Bi for all i ∈ [k]. As A 6= B, the
inequality is strict for some choice of i.
Now we are equipped with all the necessary tools to study the poset
SF k .
Lemma 2.4. Let A, B ∈ Fk . Then ]A + ]B = ](A ∨ B) + ](A ? B).
Proof. The inequality is a consequence of Lemma 2.2, the discussion
before Lemma 2.1, and the fact that the stated equality holds if A, B ∈
F1 .
Proposition 2.1. Let A, B ∈ SF k . Then, we have A ∨ B ∈ SF k .
Proof. Using Lemma 2.4 we obtain the equality 2dk = ](A ∨ B) +
](A ? B). Hence, the bounds ](A ∨ B) ≤ dk and ](A ? B) ≤ dk are
attained in both cases by the definition of dk and the fact that A ? B
is a k-family.
ANTICHAINS AND CHAINS OF FINITE POSETS.
5
The proof of Proposition 2.1 shows that SF k is closed under the
operation ?. Finally, we want to establish the fact that SF k is a lattice,
so it remains only to find a minimal element for it. The poset SF k is
finite. Suppose that A, B ∈ SF k are minimal elements. If A 6= B, then
Lemma 2.3 and the closed-ness of SF k under the operation ? yield a
contradiction.
Theorem 2.1. The poset SF k is a lattice.
3. Saturated partitions.
We begin by defining the main concept of this section.
Definition 3.1. Let C = P
{C1 , C2 , . . . , Cm } be a partition of P into
nonempty chains. If dk = m
i=1 min{]Ci , k}, then we say that C is a
k-saturated partition of P .
The first observation is that
P for any partition C of P into nonempty
chains, we have that dk ≤ m
i=1 min{]Ci , k}. This is a consequence
of the fact that any k-family A can intersect a chain of P at most k
times, so the right hand side of the inequality accounts for at least
all the elements of A. Also, if we let Uk be the set of elements of P
contained in a chain in C of length < k and
Pm vk be the number of chains
in C of length ≥ k, then ]Uk + kvk = i=1 min{]Ci , k}. Hence, C is
k-saturated if and only if dk = ]Uk + kvk .
Lemma 3.1. Let C be a saturated k-partition of P with vi chains of
length ≥ i for all positive integers i. Then, δk+1 ≤ vk ≤ δk .
Proof. For all such i, let Ui be the set of elements in chains of length
< i. We have that ]Uk+1 − ]Uk is the number of elements in a chain
in C of length k and vk − vk+1 is the number of chains in C of length
k. Hence, k(vk − vk+1 ) = ]Uk+1 − ]Uk so ]Uk + kvk = ]Uk+1 + kvk+1 .
Adding vk to the left hand side and vk+1 to the right hand side we
obtain ]Uk + (k + 1)vk ≥ ]Uk+1 + (k + 1)vk+1 ≥ dk+1 . Analogously, we
have ]Uk + (k − 1)vk ≥ dk−1 .
As C is k-saturated, then ]Uk + kvk = dk . The result then follows
from the three relations ]Uk + (k − 1)vk ≥ dk−1 , ]Uk + kvk = dk and
]Uk + (k + 1)vk ≥ dk+1 .
We are now ready to prove the two main technical results of this
section. We present them as propositions.
Proposition 3.1. Let C be k-saturated and suppose δk = δk+1 . Then
C is (k + 1)-saturated.
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BENJAMIN IRIARTE
Proof. From the proof of Lemma 3.1 we know that dk = ]Uk+1 + kvk+1
and dk+1 ≤ ]Uk + (k + 1)vk . Using the fact that δk = vk this forces
dk+1 = ]Uk + (k + 1)vk . The observation after Definition 3.1 shows
that dk+1 ≤ ]Uk+1 + (k + 1)vk+1 . Subtracting the relations dk+1 ≤
]Uk+1 + (k + 1)vk+1 and dk = ]Uk+1 + kvk+1 we obtain vk ≤ vk+1 . On
the other hand, it is clear that vk ≥ vk+1 , so vk = vk+1 . Hence, from the
equalities dk = ]Uk + kvk = ]Uk+1 + kvk+1 , we obtain Uk+1 = Uk . The
equality dk+1 = ]Uk +(k+1)vk then becomes dk+1 = ]Uk+1 +(k+1)vk+1 ,
so C is (k + 1)-saturated.
Proposition 3.2. Suppose δk > δk+1 . Then, there exist an element
x ∈ P such that x ∈ A and x ∈ A0 for all A ∈ SF k and all A0 ∈ SF k+1 .
Proof. Let 1̂ and 0̂ be the maximal and minimal elements of the lattice
SF k , respectively. Let B = 1̂k ∩ 0̂k . We divide the proof into several
steps.
Step 1: We have B 6= ∅.
Suppose that B = ∅. Then, 0̂ ∪ 1̂k is a (k + 1)-family with ]0̂ + ]1̂k =
dk + ]1̂k elements. Now, 1̂ − 1̂k is a (k − 1)-family, so ](1̂ − 1̂k ) =
]1̂ − ]1̂k = dk − ]1̂k ≤ dk−1 and δk ≤ ]1̂k . By the statement, we then
have δk+1 < ]1̂k . Hence, dk + ]1̂k > dk + δk+1 = dk+1 , so 0̂ ∪ 1̂ is a
(k + 1)-family with > dk+1 elements, contradiction.
Step 2: If A ∈ SF k+1 , then B ≤ Ak+1 .
Let A0 = A − A1 , so A0 is the k-family given by the disjoint union of
antichains A2 ≤ A3 ≤ · · · ≤ Ak+1 . By Lemma 2.4, we have ](1̂ ∨ A0 ) ≥
]A0 . If ](1̂ ∨ A0 ) > ]A0 , then, as A1 and 1̂ ∨ A0 are disjoint, we have that
A1 ∪ (1̂ ∨ A0 ) is a (k + 1)-family with more than ]A1 + ]A0 = ]A = dk+1
elements, contradiction. Hence, ](1̂ ∨ A0 ) = ]A0 . By Lemma 2.4, we
get ](1̂ ? A0 ) = dk , so (1̂ ? A0 ) ∈ SF k . As 0̂ ≤ (1̂ ? A0 ) ≤ 1̂, we have
0̂k ≤ (1̂ ? A0 )k ≤ 1̂k , and so B ⊆ (1̂ ? A0 )k . By the proof of Lemma 2.3,
we know that (1̂ ? A0 )k ≤ (1̂k ∧ A0k ) = (1̂k ∧ Ak+1 ). Thus, B ≤ Ak+1 .
Step 3: If 1̂ ≤ A ∈ SF k+1 and Ak+1 ∩ B = ∅, then (A − Ak+1 ) ∈ SF k
and so B ⊆ Ak .
By hypothesis, we have 0̂k ≤ 1̂k ≤ Ak+1 and Ak+1 ∩ B = ∅. Clearly,
we then have (0̂k − B) ∩ Ak+1 = ∅, showing that actually 0̂k ∩ Ak+1 = ∅.
Now, A − Ak+1 is a k-family, so ]A − ]Ak+1 = dk+1 − ]Ak+1 ≤ dk and
then ]Ak+1 ≥ δk+1 . The set 0̂ ∪ Ak+1 is then a (k + 1)-family of size
≥ δk+1 + dk = dk+1 , so it must actually be a Sperner (k + 1)-family and
]Ak+1 = δk+1 . Hence, ](A − Ak+1 ) = ]A − ]Ak+1 = dk+1 − δk+1 = dk ,
completing the proof.
Step 4: Let A ∈ SF k+1 . If B ∩ Ak+1 = ∅, then B ⊆ Ak .
ANTICHAINS AND CHAINS OF FINITE POSETS.
7
Let A+ = A∨ 1̂. Then, as A? 1̂ is a k-family and A+ is a (k+1)-family,
so that ]A+ ≤ dk+1 and ](A ? 1̂) ≤ dk , we actually have by Lemma 2.4
that ]A+ = dk+1 , so A+ ∈ SF k+1 . Suppose (B ∩ A+
k+1 ) 6= ∅ and take
+
+
some b ∈ (B ∩ Ak+1 ). We have Ak+1 = (1̂k ∨ Ak+1 ) ⊆ (1̂k ∪ Ak+1 ), so
by hypothesis b ∈ (1̂k − Ak+1 ) and b 6≤ a for any a ∈ Ak+1 . However,
by Step 2 we have B ≤ Ak+1 , so there must exist some a ∈ Ak+1
+
with b ≤ a, contradiction. Hence, (B ∩ A+
k+1 ) = ∅. Also, 1̂ ≤ A
by definition. Therefore, by Step 3 we obtain that B ⊆ A+
k . Now,
+
Ak = (Ak ∨ 1̂k−1 ) ⊆ (Ak ∪ 1̂k+1 ) and we know that B ⊆ 1̂k . Then, as
(1̂k ∩ 1̂k−1 ) = ∅, we get B ⊆ Ak , as we wanted.
Step 5: If for some A ∈ SF k+1 we have B ∩ Ak+1 6= ∅, let A∗ be a
maximal in SF k+1 with this property. Let C = B ∩ A∗k+1 . Then, for
all A ∈ SF k+1 we have that either C ⊆ Ak+1 or C ⊆ Ak .
As in the proof of Step 4, we know (1̂ ∨ A∗ ) ∈ SF k+1 . Since C
occurs in both 1̂k and A∗k+1 , it occurs in (1̂ ∨ A∗ )k+1 . Since A∗ is
maximal with the property that (A∗k+1 ∩ B) 6= ∅ and A∗ ≤ (1̂ ∨ A∗ ), we
must have A∗ = (1̂ ∨ A∗ ) and so 1̂ ≤ A∗ . Now, let A be an arbitrary
Sperner (k + 1)-family. If A ≤ A∗ , then, by Step 2 we have that
B ≤ Ak+1 ≤ A∗k+1 . As the set C occurs in both B and A∗k+1 , we must
have necessarily that C ⊆ Ak+1 . On the other hand, if A 6≤ A∗ , then
we know that (A ∨ A∗ ) > A∗ . Hence, by the maximality property of
A∗ , it must be the case that (A ∨ A∗ ) and B are disjoint sets. As
1̂ ≤ A∗ ≤ (A ∨ A∗ ), then, by Step 3 we obtain immediately that
B ⊆ (A ∨ A∗ )k . But (A ∨ A∗ )k = (Ak ∨ A∗k ) ⊆ (Ak ∪ A∗k ) and C ⊆ A∗k+1
implies (C ∩ A∗k ) = ∅, so C ⊆ Ak and we are done.
To finish the proof of Proposition 3.2, we note that B is a subset of all
Sperner k-families of P , since for an arbitrary A ∈ SF k , 0̂k ≤ Ak ≤ 1̂k
and B = (0̂k ∩ 1̂k ) necessarily imply that B ⊆ Ak . Now, Step 4 and
Step 5 provide the remaining parts to complete the proof.
In order for all the previous results to be of any use, we need to start
proving the existence of k-saturated partitions of P . The first step in
this direction is establishing the case k = 1.
Lemma 3.2. There exists a 1-saturated partition of P .
Proof. Let A be an antichain of P of maximal size. We proceed by
induction on ]P − ]A. In the case ]P − ]A = 0, P is itself an antichain
and A is a 1-saturated partition of P . Suppose that the result holds
for all posets P 0 for which ]P 0 − ]A = i ≥ 0. Assume that ]P − ]A =
(i + 1) > 0. There must exist either a maximal element or a minimal
element x ∈ P such that x 6∈ A. Otherwise, we would have ]P −]A = 0.
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BENJAMIN IRIARTE
Suppose without loss of generality that x is maximal in P . Let Px be
the restriction of P to P − {x}. Then, we have ]Px − ]A = i and
A is an antichain of Px of maximal size, so there exists a 1-saturated
partition Cx = {C1 , . . . , Cm } of Px . As x is maximal in P , x is in P
greater than all the elements in one of the chains in Cx , say C1 . Then,
C = {C1 ∪ {x}, C2 , . . . , Cm } is a 1-saturated partition of P , clearly. We now have all the necessary technical tools to prove the main
theorem of this section.
Theorem 3.1. Let k be a positive integer. Then, there exists a partition of P into chains which is both k-saturated and (k + 1)-saturated.
Proof. We proceed by induction on k and the size of P . Note that if
]P = 1, then there exists trivially a partition of P which is k-saturated
simultaneously for all positive integers k. Suppose that for all posets
P 0 of size i ≥ 1, there exists a partition of P 0 which is both k-saturated
and (k + 1)-saturated, for all positive integers k. Let P be of size i + 1.
We will assume the existence of a k-saturated partition of P and then
construct a partition of P into chains that is both k-saturated and
(k + 1)-saturated. Lemma 3.2 provides the existence of a 1-saturated
partition of P . Note that, per Lemma 3.1, we can assume δk ≥ δk+1 .
If δk = δk+1 , then any k-saturated partition is (k + 1)-saturated by
Proposition 3.1, and we are done immediately.
If δk > δk+1 , per Proposition 3.2, there exists an element x ∈ P
such that x ∈ A and x ∈ A0 for all A ∈ SF k and all A0 ∈ SF k+1 .
If we consider the restriction Px of P to P − {x}, and we let dx,k be
the maximal size of a k-family of Px and dx,k+1 be the maximal size
of a (k + 1)-family of Px , we then obtain dx,k = dk − 1 and dx,k+1 =
dk+1 − 1. By induction, we can find a partition
x = {C1 , . . . , Cm }
PC
m
of Px into nonempty
chains
such
that
d
−
1
=
k
i=1 min{]Ci , k} and
Pm
dk+1 − 1 = i=1 min{]Ci , k + 1}.
The punchline of the proof is then to notice that the partition C =
{C0 , C1 , C2 , . . . , Cm } of P into nonempty chains, with block C0 = {x},
is simultaneously a k-saturated and (k + 1)-saturated partition of P .
Hence, this completes the proof.
A very nice application of Theorem 3.1 and Lemma 3.1 is finally
presented in the following result.
Theorem 3.2. The sequence δ associated to P satisfies that δ1 ≥ δ2 ≥
δ3 ≥ δ4 ≥ . . . , so it defines a partition of ]P .
ANTICHAINS AND CHAINS OF FINITE POSETS.
9
4. Antichains and partitions.
In this section, we will derive Theorem 1.1. To begin, we recall some
classic notation.
Definition 4.1. Let n be a positive integer and let λ be a partition of
n. We will denote by λ̄ the partition of n conjugate to λ.
Also, we need to define the analogous notion of k-families for chains.
We do this briefly.
Definition 4.2. A subset C ⊆ P will be called an h-cofamily of P if
C is the disjoint union of h antichains in P .
Now, it will be useful to introduce a function on partitions of P ,
which behaves well with respect to saturation.
Definition 4.3. Let C = {C1 , C2 , . . . , Cm } be a partition of P into
nonempty chains, and suppose that C1 , C2 , . . . , Ch` are precisely the
chains inSC of length ≥ `, for some positive integer `. If we define
`
Ci , then the partition C` of P with blocks C1 , C2 , . . . , Ch`
S` = P − hi=1
and all the singleton blocks coming from elements of S` , will be called
the `-deformation of C.
For a set S ⊆ P , we will write C = {C1 , C2 , . . . , Ch ; S} to represent
a partition of P into chains where the set of blocks corresponding to
S are the singletons of elements of S. There is an observation worth
mentioning at this point. Notice that, if C is a k-saturated partition of
P and ` ≤ k, then the `-deformation C` of C is still k-saturated. Note
also that we have the equality dk = ]Sk + khk . Now we are ready to
prove interesting results.
Lemma 4.1. Let C = {C1 , C2 , . . . , Ch ; S} be a k-saturated partition of
P , with ]Ci ≥ k for all i ∈ [h]. Then,
(i): the set C = C1 ∪C2 ∪· · ·∪Ch is an h-cofamily of P of maximal
size among all h-cofamilies of P , and
(ii): gh = δ̄1 + δ̄2 + · · · + δ̄h .
Proof. To prove (i), suppose, on the contrary, that there exists an hcofamily C 0 = C10 ∪ C20 ∪ · · · ∪ Ch0 of P with ]C 0 > ]C. Let S 0 = P − C 0 .
Then, we have ]S 0 < ]S,Pand the partitions C0 = {C10 , C20 , . . . , Ch0 ; S 0 }
and C of P satisfy that C 0 ∈C0 min{k, ]C 0 } ≤ kh + ]S 0 < kh + ]S =
P
C∈C min{k, ]C} = dk , which contradicts the observation after Definition 3.1.
To prove (ii), we now know that gh = ]C, but ]C = ]P − ]S =
]P − dk + kh = ]P − (δ1 + δ2 + · · · + δk ) + kh. Let l be such that δl > 0
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BENJAMIN IRIARTE
but δl+1 = 0. Then, ]P − (δ1 + δ2 + · · · + δk ) + kh = δk+1 + · · · + δl + kh,
so gh = ]C = δk+1 + · · · + δl + kh. However, by Lemma 3.1, we have
that δk ≥ h ≥ δk+1 , so thinking of the Young diagram for δ we see that
δk+1 + · · · + δl + kh = δ̄1 + δ̄2 + · · · + δ̄h , hence completing the proof. Corollary 4.1. Let C = {C1 , C2 , . . . , Ch ; S} be a partition of P into
chains. If h = δk for some positive integer k, then, gh = δ̄1 +δ̄2 +· · ·+δ̄h .
Proof. By Theorem 3.1, there exists a partition C = {C1 , . . . , Cm }
of P which is both k-saturated and (k + 1)-saturated. Let C` =
{C1 , . . . , Ch` ; S` } denote the `-deformation of C, and let n` denote the
number of chains in C of length precisely `.
In the case k = 1, consider the partition C1 , which is then 1saturated. We have the equality δ1 = d1 = ]S1 + h1 = h1 and by
Lemma 4.1 (ii), we are done.
For k ≥ 1, by assumption, we have that dk+1 = ]Sk+1 + (k + 1)hk+1
and dk = ]Sk + khk , so δk+1 = (]Sk+1 − ]Sk ) + hk+1 − k(hk+1 − hk ) =
knk + hk+1 − knk = hk+1 . As the (k + 1)-deformation of C is (k + 1)saturated, by part (ii) of Lemma 4.1, the proof is complete.
Now we are almost ready to prove Theorem 1.1, we only need one
more key result.
Proposition 4.1. For all positive integers h, we have that gh ≤ δ̄1 +
δ̄2 + · · · + δ̄h .
Proof. Let C = C1 ∪ C2 ∪ · · · ∪ Ch be an h-cofamily of P of maximal
size. Let S = P − C. Consider the partition of P into chains C =
{C1 , . . . , Ch ; S}. If h = δ1 , then by Corollary 4.1, we are done. If
h > δ1 , then we have immediately that gh = ]P = δ̄1 + δ̄2 + · · · + δ̄δ1 +
· · · + δ̄h . Hence, suppose that h < δ1 . Then, for all positive k, we
know that dk ≤ ]S + kh, so gh = ]C = ]P − ]S ≤ ]P − (dk ) + kh =
P − (δ1 + δ2 + · · · + δk ) + kh. Let l be such that δl = 0 but δl−1 > 0.
We have that P − (δ1 + δ2 + · · · + δk ) + kh = δk+1 + · · · + δl + kh, so
gh ≤ δk+1 + · · · + δl + kh, for all positive integers k < l. Let k ∗ be such
that δk∗ > h ≥ δk∗ +1 , so that k ∗ < l. Then, thinking of the Young
diagram of δ, we can see that δk∗ +1 + · · · + δl + k ∗ h = δ̄1 + δ̄2 + · · · + δ̄h .
This completes the proof.
We have finally arrived to a point were we can prove Theorem 1.1.
This will be the end of our introduction to the subject of chains and
antichains of finite posets. Further sections added to this work will be
devoted to original or novel results.
Proof of Theorem 1.1. Concretely, we want to prove that gh = δ̄1 + δ̄2 +
· · · + δ̄h for all positive integers h. Assume, by induction, that this is
ANTICHAINS AND CHAINS OF FINITE POSETS.
11
true for all posets of size ]P − 1. If h = δk for some k or if h > δ1 , then
we are done by Corollary 4.1 and the proof of Proposition 4.1. Hence,
we can assume that δk > h > δk+1 for some k ≥ 1. By Proposition 3.2,
there exists an element x ∈ P such that x is contained in all k-families
of P of size dk , and also in all (k+1)-families of P of size dk+1 . Let Px be
the restriction of P to P −{x}. For all positive integers i, let dx,i be the
maximal size of an i-family of Px , and let dx,0 = 0. Analogously, define
δx,i as the difference dx,i − dx,i−1 , for all i ≥ 0. We obtain dx,k = dk − 1
and dx,k+1 = dk+1 − 1. Hence, δx,k+1 = δk+1 . Furthermore, we have
that δx,k is equal to either δk or δk − 1.
We claim that δ̄1 + δ̄2 + · · · + δ̄h = δ̄x,1 + δ̄x,2 + · · · + δ̄x,h .
To see this, notice that as h > δk+1 = δx,k+1 and δx,k ≥ δk − 1 ≥ h,
then we have that δi ≥ h if and only if δx,i ≥ h, which holds if and
only if i ≤ k, for all positive integers i. Let l and lx be such that δl > 0
but δ = 0, and δx,lx > 0 but δx,lx +1 = 0. Then, δ̄1 + δ̄2 + · · · + δ̄h =
Plx
Pl l+1
i=1 min{h, δx,i } =
i=1 min{h, δi } = kh+]P −dk = kh+]Px −dx,k =
δ̄x,1 + δ̄x,2 + · · · + δ̄x,h . By the induction hypothesis, we can find an hcofamily of Px of size δ̄x,1 + δ̄x,2 + · · · + δ̄x,h , but this h-cofamily is also
an h-cofamily of P , and so gh ≥ δ̄x,1 + δ̄x,2 +· · ·+ δ̄x,h = δ̄1 + δ̄2 +· · ·+ δ̄h .
Now, using Proposition 4.1, the proof is complete.
We hope to add more sections here.
References
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[4] Richard P. Stanley, Enumerative Combinatorics, Vol. 1.
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[5] Richard P. Stanley, Enumerative Combinatorics, Vol. 2. Cambridge University
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