Partially Ordered Sets (Posets) - University of Iceland › lhk1 › fletta › fletta03.pdf · 2008-05-07 · Chapter 3 Partially Ordered Sets (Posets) 3.1 Posets (3.1) De nition

Chapter 3

Partially Ordered Sets (Posets)

3.1 Posets

(3.1) De�nition (3.1.1.) A partially ordered set (poset) is a pair (P,≤P )where P is a set and ≤P is a binary relation on P satisfying

(1) for all x ∈ P , x ≤P x (reflexivity);

(2) if x ≤P y and y ≤P x, then x = y (anti-symmetry);

(3) if x ≤P y and y ≤P z, then x ≤P z (transitivity).

Where there is no possibility of confusion, we will write ≤ instead of ≤P . Also

x ≥ y means y ≤P xx < y means x ≤p y and x 6= y

Two elements x, y ∈ P are comparable if x ≤P y or y ≤P x. Otherwise, they arecalled incomparable.

(3.2) Example (3.1.2.)

(1) For n ∈ N let P = {i ∈ N : 1 ≤ i ≤ n}. Given x, y ∈ P , de�ne x ≤P y if x ≤ y.

1 < 2 < 3 < · · · < (n− 1) < n

Each pair of elements is comparable. (Totally ordered set.)

(2) Given n ∈ N, let P = {A : A ⊆ {1, . . . , n}}. For x, y ∈ P , de�ne x ≤P y if x ⊆ y.In this case (P,≤P ) is the poset of all subsets of {1, . . . , n} ordered by inclusion.

It is denoted by Bn.

n = 2 : P = {∅, {1}, {2}, {1, 2}} ∅ ≤P {1} ≤P {1, 2}∅ ≤P {2} ≤P {1, 2}

(3) Given n ∈ N let P = {i ∈ N : i divides n}. For x, y ∈ P de�ne x ≤P y if xdivides y. E.g. if n = 18, then P = {1, 2, 3, 6, 9, 18} and

1 ≤P 1, 1 ≤P 2, 1 ≤P 3, . . . , 1 ≤P 18 6 ≤P 6, 6 ≤P 182 ≤P 2, 2 ≤P 6, 2 ≤P 18 9 ≤P 9, 9 ≤P 183 ≤P 6, 3 ≤P 9, 3 ≤P 18 18 ≤P 18

53

CHAPTER 3. PARTIALLY ORDERED SETS (POSETS)

Here (P,≤P ) is denoted by Dn.

(4) Given n ∈ N let P = { partitions of {1, . . . , n}}. For x, y ∈ P de�ne x ≤P y ifevery block of x is contained in a block of y. If n = 4, we have x = {13, 2, 4}and y = {13, 24} ∈ P . Since every block of x is contained in a block of y, wehave x ≤P y. We say that x is a refinement of y. Here (P,≤P ) is the poset of

partitions of {1, . . . , n} ordered by re�nement and is denoted by Πn.

(3.3) De�nition There are two types of subposets of (P,≤P ).

� An induced subposet (Q,≤Q) of (P,≤P ) is such that Q ⊆ P and x ≤Q y ⇔x ≤P y. E.g. choose Q = {1, 2, 3} ⊆ {1, 2, 3, . . .} in example (3) above.

� A weak subposet of (P,≤P ) is (Q,≤Q) where Q ⊆ P and if x ≤Q y thenx ≤P y.

When we say subposet, we mean an induced subposet.

For x, y ∈ P where (P,≤) is a poset, and x ≤ y, the interval

[x, y] = {z ∈ P : x ≤ z ≤ y}

is a subposet. The smallest interval is [x, x] = {x}.

(3.4) De�nition If every interval [x, y] of (P,≤P ) is �nite, then P is called a locallyfinite poset.

If Q is a subposet of P , then Q is called convex if y ∈ Q whenever x < y < z in Pand x, z ∈ Q. Interval posets are always convex.

Open interval:

(x, y) = {z ∈ P : x < z < y}

(3.5) De�nition Given x, y ∈ P , we say that y covers x if x < y and there doesnot exist z ∈ P such that x < z < y.

(3.6) Remark A locally �nite poset is completely determined by its cover relations.

(3.7) De�nition TheHasse diagram of a �nite poset P is the graph with verticesx ∈ P and

� if x < y, then y is drawn above x in the diagram;

� if y covers x, then there is an edge between x and y in the diagram.

(3.8) Example If P = {a, b, c, d, e, f} and

a < b, a < c, a < d, b < e, e < f, c < f, d < f

then the Hasse diagram of P is

•f

0000000~~~~

•e•c •d

��•b @@@@

•a

54


(3.9) Example All di�erent (non-isomorphic) posets on 4-elements

• • • ••• •

•

• •• •

•

��00000•

• •

•00000 •

��•

•

•• ••

•@@@@ •

• ••

@@@@ •~~~~

• •

• @@@ •~~~••

•• @@@

~~~• •

•~~~~ @@@@

•@@@@ •

~~~~

•

•9999

•9999 •

��•

•�� 9999

• •9999

•

•~~~~ @@@@

• • •• • ••

~~~~@@@@

••••

(3.10) Question Why are•

��00000

• @@@~~~• •

and• @@@

~~~• @@@ •

~~~•

not in the above diagrams?

(3.11) Example (From example 3.1.2.)

(1) If n = 4 and P = {1, 2, 3, 4}, the Hasse diagram of the total order on P is

•4•3•2•1

(2) B4 = (P,≤) where P = {A : A ⊆ {1, 2, 3, 4}} and elements are ordered byinclusion.

P = {∅, 1, 2, 3, 4, 12, . . . , 134, 1234}The Hasse diagram:

•

ooooooooooooooo

~~~~~~~~~

@@@@@@@@@

OOOOOOOOOOOOOOO1234

•

~~~~~~~~~

UUUUUUUUUUUUUUUUUUUUUU123 •

ooooooooooooooo


iiiiiiiiiiiiiiiiiiiiii

ooooooooooooooo

''OOOOOOOOOOOOOOO134 •

~~~~~~~~~

@@@@@@@@@234

•

@@@@@@@@@

OOOOOOOOOOOOOOO12 •13

UUUUUUUUUUUUUUUUUUUUUU •

~~~~~~~~~


ooooooooooooooo 23 •

iiiiiiiiiiiiiiiiiiiiii 24 •

ooooooooooooooo

~~~~~~~~~ 34

•

OOOOOOOOOOOOOOO1 •

@@@@@@@@@2 •

~~~~~~~~~ 3 •

ooooooooooooooo 4

•∅

~~~~~~~~~

(3) Consider D12 = (P,≤) where P = {i : i divides 12} = {1, 2, 3, 4, 6, 12}. x ≤ y inP if x divides y.

1 < 2, 3, 4, 6, 12 4 < 122 < 4, 6, 12 6 < 12

3 < 6, 12

Hasse diagram for D12:

•~~~~~

12

@@@@@

•4 •oooooooo 6

•2 @@@@@ •

3~~~~~

•1

~~~~~

55


(4) Π3 = (P,≤) where P is the set of all partitions of {1, 2, 3},

P = {{123}=x1

, {1, 23}=x2

, {2, 13}=x3

, {3, 12}=x4

, {1, 2, 3}=x5

}

Thenx5 < x1, x2, x3, x4 x2, x3, x4 < x1

Hasse diagram:

{123}

ssssssss

KKKKKKKK

{1, 23}

KKKKKKKK{2, 13} {3, 12}

ssssssss

{1, 2, 3}

(3.12) De�nition Given poset (P,≤), if there exists an element a ∈ P such thata ≤ x for all x ∈ P , then a is a smallest element of the poset P and is denoted 0.Similarly, if there is a largest element a such that x ≤ a for all x ∈ P , then a iswritten 1.

(3.13) De�nition A chain is a poset in which every 2 elements are comparable.

� A subset C of a poset P is a chain if C is a chain when regarded as a subposetof P .

� A chain C in a poset is saturated if it is not amissing any elements within it,i.e. 6 ∃z ∈ P − C such that x < z < y for some x, y ∈ C� In a locally �nite poset, a chain x0 < x1 < · · · < xn is saturated ⇔ xi+1 covers xifor all 0 ≤ i < n.

(3.14) De�nition The length of a �nite chain is `(C) = |C| − 1.The length (rank) of a �nite poset P is

`(P ) = max{`(C) : C chain in P}

Length of the interval [x, y] in P is `(x, y).A chain C in a poset P is maximal if no more elements can be added to it.

(3.15) Example

P : •f@@@~~~•e

~~~ •g@@@•d

~~~ •h@@@

•c•b•a

`(P ) = 5

C1 = {a, b, c, d, e, f} is a maximal chain, `(C1) = 5C2 = {h, g, f} is a maximal chain, `(C2) = 2

56


(3.16) De�nition If every maximal chain of P has the same length n, then P is

graded of length n.

Notice that if P is graded, then there is a unique rank function ρ : P → {0, 1, . . . , n}such that

� ρ(x) = 0 if x is a minimal element of P ,

� ρ(y) = ρ(x) + 1 if y covers x in P .

If ρ(x) = i for x ∈ P , then we say element x has rank i.

If x ≤ y, then `(x, y) = ρ(y)− ρ(x).

(3.17) De�nition If P is a graded poset of rank n, then the polynomial

F (P, q) =n∑i=0

piqi,

where pi = # elements of rank i in P , is called the rank generating function

of P .

(3.18) Remark All posets in Example 3.1.2. were graded.

(3.19) Example Let P be the poset with Hasse diagram

rank:

4 •�� <<<<

3 •�� •

<<<<��

2 •<<<<

SSSSSSSSS •�� <<<< •

��

1 •<<<< •

��

0 •

p0 = 1, p1 = 2, p2 = 3, p3 = 2, p4 = 1; F (P, q) = 1 + 2q + 3q2 + 2q3 + q4

(3.20) Example Consider poset P on {1, . . . , n} with 1 < 2 < . . . < n. One maximalchain ⇒ graded. `(C) = n− 1, ρ(x) = x− 1, ρ(P ) = n− 1. ♦

(3.21) Example The poset Bn = (P,≤), where

P = {A : A ⊆ {1, . . . , n}}, x ≤ y ⇔ x ⊆ y

A chain in Bn is a sequence

{A1, . . . , Ak} such that A1 ⊂ A2 ⊂ · · · ⊂ Ak ⊆ {1, . . . , n}

Maximal chain e.g. C:

∅ < {1} < {1, 2} < {1, 2, 3} < . . . < {1, . . . , n}

We see that`(C) = n+ 1− 1 = n

57


which applies to all maximal chains. Thus the poset Bn is graded of length n.

Note that if x ∈ P , then ρ(x) = |x| (no. of elements in x), e.g. ρ({1, . . . , n}) = n. Thus

F (Bn, q) =n∑i=0

piqi =

n∑i=0

(n

i

)qi = (1 + q)n

(3.22) Example Dn = (Pn,≤) where Pn = {1 ≤ i ≤ n : i divides n}. If x, y ∈ Pn,then x ≤ y ⇔ x divides y.

For n = 12 we have 12 = 2231 := pα11 pα2

2 ,

D12 : •~~~~~

12=2231

@@@@@

•4=2230 •oooooooo 6=2131

•2=2130

@@@@@ •3=2031

~~~~~

•1=2030

~~~~~

and |D12| = (2 + 1)(1 + 1) = 6.

Claim Poset Dn is graded.

Proof. If n = pα11 pα2

2 · · · pαmm where p1, . . . , pn are primes, then

Pn = {pβ11 · · · p

βmm : 0 ≤ β1 ≤ α1, . . . , 0 ≤ βm ≤ αm}

and |Pn| = (1 + α1)(1 + α2) · · · (1 + αm). If x ∈ Pn and x = pβ11 · · · p

βmm , then ρ(x) =

β1 + · · ·+ βm. In particular ρ(Pn) = α1 + α2 + · · ·+ αm. �

We have

F (Dn, q) =α1+···+αm∑

j=0

pjqj =

α1∑β1=0

· · ·αm∑βm=0

1qβ1+···+βm =

α1∑β1=0

qβ1

· · · αm∑βm=0

qβm

=(qα1+1 − 1q − 1

)· · ·(qαm+1 − 1q − 1

)

(3.23) De�nition A multichain of a poset P is a chain in which elements areallowed to be repeated.

(3.24) Example Poset P with Hasse diagram:

•e

0000000~~~~

•c•d

��•b @@@@

•a

C = {a, b, b, b, c, c, e} is a multichain. Length of C ′ = {x0 ≤ x1 ≤ · · · ≤ xn} is n. Hence,`(C) = 7− 1 = 6. ♦

58


(3.25) De�nition An antichain of a poset P is a subset A ⊆ P such that everypair of elements of A are incomparable.

(3.26) Example Continuing last example,

A = {c, d} A = {b, d} A = ∅ A = {a}A = {b} A = {c} A = {d} A = {e}

are the antichains of P . ♦

(3.27) De�nition An order ideal (downset) of poset P is a subset I ⊆ P suchthat if x ∈ I and y ≤P x, then y ∈ I.

A dual order ideal (upset) is a subset I ⊆ P such that is x ∈ I and x ≤P y, theny ∈ I.

(3.28) Example (cf. (3.24)) I1 = {a, b, c, d} is a downset of P and not an upset.

I2 = {b, c, d, e} is an upset of P and not a downset. ♦

For a �nite poset P , there is a 1-1 correspondence between antichains of P and downsetsof P :

(3.29) Proposition

� If A is antichain of P (�nite) then

I(A) = {x : x ≤ y and y ∈ A}

is an order ideal.

� If I is an order ideal, then

A(I) = { maximal x ∈ I}

is an antichain.

(3.30) Example (cf. (3.24))

Antichain (A) Order ideal (I)

∅ ∅{a} {a}{b} {a, b}{c} {a, b, c}{d} {a, d}{e} {a, b, c, d, e}{b, d} {a, b, d}{c, d} {a, b, c, d}

59


(3.31) Remark The set of all order ideals (downsets) of a poset P , when ordered byinclusion, forms a poset J(P ).

(3.32) Example Continuing the last example,

{a, b, c, d, e}

{a, b, c, d}

nnnnnnnnPPPPPPPP

{a, b, c} {a, b, d}

ffffffffffffffffff

{a, b}PPPPPPPPP {a, d}

nnnnnnnnn

{a}

∅

i.e. •1

•~~~~~

@@@@@

• •ooooooooo

•@@@@@ •

~~~~~

•

•0

(3.33) Notation If A = {x1, . . . , xk} is an antichain, then we write 〈x1, . . . , xk〉 for theorder ideal corresponding to A.

(3.34) De�nition Two posets (P. ≤P ) and (Q,≤Q) are isomorphic, P ∼= Q, ifthere exists order preserving bijection ϕ : P → Q such that

x ≤P y ⇔ ϕ(x) ≤Q ϕ(y)

(3.35) De�nition The dual of a poset P is P ∗ such that

x ≤P y ⇔ y ≤P ∗ x

(3.36) Remark The Hasse diagram of P ∗ = the Hasse diagram of P upside down.

3.2 Lattices

(3.37) De�nition Given x, y ∈ P (a poset), an element z ∈ P is called

� an upper bound of x and y if x ≤ z and y ≤ z;� a lower bound of x and y if z ≤ x and z ≤ y.

If z is an upper bound of x and y such that z ≤ w for all other upper bounds w of x andy, then z is called a least upper bound.

Similarly, if z is a lower bound of x and y such that w ≤ z for all other lower bounds wof x and y, then z is called a greatest lower bound.

� If the least upper bound of x and y exists, then it is denoted x ∨ y (�x join y�, �xsup y�, �join together�).

� If the greatest lower bound of x and y exists, then it is denoted x∧ y (�x meet y�,�x inf y�, �where they meet�).

60


(3.38) Example

•h

•~~~~~

g

@@@@@

•e •ooooooooo f

•c

@@@@@ •d

~~~~~

•b

•a

f is the least upper bound of c and d.b is the greatest lower bound of c and d.

c ∨ d = f, c ∧ d = b

(3.39) De�nition If P is a poset and for all x, y ∈ P , x∨ y and x∧ y exist, then P iscalled a lattice.

(3.40) Remark In a lattice (L,≤) many properties such as associativity and commu-tativity are true for the ∨ and ∧ operations.

(3.41) Example

••a •

~~~~ @@@@

• •is not a lattice since a ∨ x and a ∧ x does not exist for any x ∈ P \ {a}

•b•a

is a lattice, since a ≤ b and b ≤ b, giving a ∨ b = b and a ∧ b = a

•~~~~

c

@@@@

•a

~~~~ •b

@@@@ is not a lattice, since a ∧ b is unde�ned.

(3.42) Example Out of the 16 di�erent posets on 4 element (see Example 3.9), only2 are lattices:

•~~~~ @@@@

•@@@@ •

~~~~

•and

••••

(3.43) Remark All �nite lattices (|L| <∞) have a 0 and a 1.

If |L| is in�nite, then this need not be true.

(3.44) Example If L = (Z2,≤) where (x, y) ≤ (x′, y′) ⇔ x ≤ x′ and y ≤ y′, wehave:

��

��

��

��

��

��

��

��

��

��@

@@@@@@@

@@@@@@@@

@@@@@@@@

@@@@@@@@

@@@@@@@@

@@@@@@@@

@@@@@@@@

@@@@@@@@

@@@@@@@@

@@@@@@@@

which has no 0 and no 1. ♦

61


(3.45) De�nition If P is a poset and every x, y ∈ P have a meet, x ∧ y, then we saythat P is a meet-semilattice.

Simlarly, if P is a poset and x ∨ y exists for all x, y ∈ P , then we say that P is ajoin-semilattice.

(3.46) Useful Theorem If P is a �nite meet (join) semilattice with a 1 (0), then Pis a lattice.

All di�erent lattices on 5 points:

•••••

••

~~~~ @@@@

•@@@@ •

~~~~

•

•~~~~ @@@@

•@@@@ •

~~~~

••

•~~~~ @@@@

•@@@@ • •

~~~~

•

•~~~

00000•

•

��• @@@•

They are all graded, except the last one.

(3.47) De�nition A lattice is an algebra L = (L,∨,∧) satisfying, for x, y, z ∈ L,

(1) x ∧ x = x and x ∨ x = x;

(2) x ∧ y = y ∧ x and x ∨ y = y ∨ x;(3) x ∧ (y ∧ z) = (x ∧ y) ∧ z and x ∨ (y ∨ z) = (x ∨ y) ∨ z;(4) x ∧ (x ∨ y) = x and x ∨ (x ∧ y) = x.

3.3 Modular Lattices

(3.48) De�nition A �nite lattice L is called modular if it is graded and the rankfunction ρ satis�es

ρ(x) + ρ(y) = ρ(x ∧ y) + ρ(x ∨ y), ∀x, y ∈ L (3.1)

In the case where one is considering (L,∨,∧) as an algebra, this condition is equivalentto (L,∨,∧) also satisfying

(5) for all x, y, z ∈ L with x ≤ z, x ∨ (y ∧ z) = (x ∨ y) ∧ z.

(3.49) Example Consider lattice L with Hasse diagram

•xxxxxx

g

FFFFFF

•d

4444 •

e

4444 •f

•b

4444 •c

•a

.

It is graded with ρ(a) = 0, ρ(b) = ρ(c) = 1, . . .. Notice that ρ(d) = ρ(f) = 2, ρ(d ∨ f) =ρ(g) = 3, ρ(d ∧ f) = ρ(a) = 0. Thus

ρ(d) + ρ(f) = 4 6= 3 = ρ(d ∨ f) + ρ(d ∧ f)

62


So L is not modular.

Alternatively, in terms of condition (5), choose x = b, y = f, z = d. Then x ≤ z and

x ∨ (y ∧ z) = b ∨ (f ∧ d) = b ∨ a = b 6= d = g ∧ d = (b ∨ f) ∧ d = (x ∨ y) ∧ z

(3.50) Proposition The total order P = ({1, . . . , n},≤) is a modular lattice.

Proof. (P,∨,∧) is a lattice with x ∨ y = max(x, y) and x ∧ y = min(x, y). It is easy tocheck that conditions (1) through (4) hold. The lattice is graded with rank ρ(x) = x−1.For x, y ∈ P condition (3.1) holds since

x+ y = min(x, y) + max(x, y), ∀x, y ∈ P

(3.51) Proposition The poset Bn = ({1, . . . , n},≤) is a modular lattice, where x ≤Pif x ⊆ y. (Bn is the poset of all subsets of {1, . . . , n} ordered by inclusion.)

Proof. If we set x ∧ y = x ∩ y and x ∨ y = x ∪ y for x, y ∈ Bn then (Bn,∨,∧) satis�esconditions (1) through (4), and thus is a lattice. The lattice Bn is graded with rankρ(x) = |x|. Since x, y ⊆ {1, . . . , n} we know

|x|+ |y| = |x ∩ y|+ |x ∪ y|

i.e.ρ(x) + ρ(y) = ρ(x ∧ y) + ρ(x ∨ y)

thereby satisfying condition (3.1). Alternatively, if x, y, z ∈ Bn and x ≤ z, then x, y, z ⊆{1, . . . , n} and x ⊆ z. So

x ∨ (y ∧ z) = x ∪ (y ∩ z) = (x ∪ y) ∩ (x ∪ z) = (x ∪ y) ∩ (z)= (x ∨ y) ∧ z

i.e. condition (5) is true. �

Let Vn(q) be the n-dimensional vector space over the �eld Fq (or GF (q)). De�ne Ln(q)to be the poset of subspaces of Vn(q) where x ≤ y if x is a subspace of y.

(3.52) Proposition Ln(q) is a modular lattice.

Proof. It is easy to see that ∅ is the 0 of Ln(q) and that Vn(q) is the 1 of Ln(q). Ifx, y ∈ Ln(q), then one can easily argue that x ∧ y = x ∩ y, which is also a subspace ofVn(q). So Ln(q) is a meet-semilattice. Since it is �nite, by the useful theorem (3.46) it isa lattice.

The lattice Ln(q) is graded with rank ρ(x) = dim(x), the dimension of the subspace x.

If x, y ∈ Ln(q), then x ∨ y = x+ y = {u + v : u ∈ x,v ∈ y}. From linear algebra, oneknows

dimx+ dim y = dim(x ∩ y) + dim(x+ y)

Henceρ(x) + ρ(y) = ρ(x ∧ y) + ρ(x ∨ y)

63


3.4 Distributive Lattices

(3.53) De�nition A �nite lattice (L,∨,∧) is called distributive if for all x, y, z ∈ L,

(6') x ∨ (y ∧ z) = (x ∨ y) ∧ (x ∨ z),

or

(6�) x ∧ (y ∨ z) = (x ∧ y) ∨ (x ∧ z).

imply one another

(3.54) Proposition If L is a �nite distributive lattice, then L is modular.

Proof. If (L,∨,∧) is a distributive lattice, then for all x, y, z ∈ L

x ∨ (y ∧ z) = (x ∨ y) ∧ (x ∨ z)

If x ≤ z, then x ∨ z = z. So

x ∨ (y ∧ z) = (x ∨ y) ∧ z

and condition (5) holds. �

(3.55) Proposition The lattice Bn is distributive.

Proof. From Prop. (3.51) we know Bn is a modular lattice. In order to show it is dis-tributive, we must show either (6') or (6�) holds.

If x, y, z ∈ Bn, then

x ∨ (y ∧ z) = x ∪ (y ∩ z)= (x ∪ y) ∩ (x ∪ z) (since ∩& ∪ distributive)= (x ∨ y) ∧ (x ∨ z)

Therefore (6') is true and Bn is distributive. �

(3.56) Proposition The lattice Ln(q) is not distributive.

Proof. Later... �

(3.57) Theorem (Fundamental theorem of �nite distributive lattices, FTFDL) If L is

a �nite distributive lattice, then there exists a poset P such that L ∼= J(P ).

3.5 Incidence Algebra of a Poset

(3.58) De�nition Let P be a locally �nite poset. The set of intervals of P is

Int(P ) = {[x, y] : x, y ∈ P, x ≤ y}

64


(3.59) De�nition The incidence algebra of P , denoted I(P ), is the algebra ofall functions f : Int(P )→ C, with multiplication

(fg)(x, y) =∑x≤z≤y

f(x, z)g(z, y), where f(x, y) := f([x, y])

(3.60) Remark It is easy to see that the identity of this algebra is

1(x, y) = δ(x, y) =

{1 if x = y

0 if x < y

(3.61) Proposition Let f ∈ I(P ). The following statements are equivalent:

(1') f has left inverse;

(1�) f has right inverse;

(1� ') f has a 2-sided inverse;

(2) f(x, x) 6= 0 for all x ∈ P .

Proof. fg = δ = 1 is equivalent to

� fg(x, x) = 1 for all x ∈ P ;� fg(x, y) = 0 for all x < y ∈ P .

Notice that

(fg)(x, y) =∑x≤z≤y

f(x, z)g(z, y)

= f(x, x)g(x, y) +∑x<z≤y

f(x, z)g(z, y) = 0 ∀x < y

which is equivalent to

g(x, y) = −f(x, x)−1∑x<z≤y

f(x, z)g(z, y)

So this �value� is OK so long as f(x, x) 6= 0. Thus

f has a right inverse ⇔ f(x, x) 6= 0 ∀x ∈ P

Applying the same argument to hf = δ shows

f has a left inverse ⇔ f(x, x) 6= 0 ∀x ∈ P

So

f has a right inverse ⇔ f(x, x) 6= 0 ∀x ∈ P ⇔ f has a left inverse

By looking at the covers of x, one sees that f(x, x)−1 depends only on [x, y] and conse-quently g(x, y) = h(x, y) = f(x, y)−1. �

65


3.6 The Zeta Function

(3.62) De�nition The zeta function is

ζ : I(P )→ C, ζ(x, y) = 1

i.e. ζ of every interval is equal to 1.

(3.63) Remark Notice that

ζ2(x, y) = ζζ(x, y) =∑x≤z≤y

ζ(x, z)ζ(z, y)

=∑x≤z≤y

1 = |[x, y]|

= |{z ∈ P : x ≤ z and z ≤ y}|

Consider k ∈ N:

ζk(x, y) =∑

x=x0≤x1≤···≤xk=yζ(x0, x1)ζ(x1, x2) · · · ζ(xk−1, xk) =

∑x=x0≤x1≤···≤xk=y

1

= |{(x0, . . . , xk) : x = x0, y = xk, x0 ≤ x1 ≤ · · · ≤ xk}|

= the number of multichains of length k from x to y

(3.64) Proposition If k ∈ N, then (ζ − 1)k(x, y) is the number of chains of length kfrom x to y.

Proof. Notice �rst that

(ζ − 1)(x, y) = ζ(x, y)− 1(x, y) =

{0 if x = y

1 if x < y

Using this,

(ζ − 1)k(x, y) =∑

x=x0≤x1≤···≤xk=y(ζ − 1)(x0, x1) · (ζ − 1)(x1, x2) · · · (ζ − 1)(xk−1, xk)

=∑

x=x0<x1<···<xk=y1

= number of chains in P of length k from x to y

(3.65) Proposition The function (2− ζ)−1 ∈ I(P ) counts the number of chains in an

interval.

Proof. First

(2− ζ)(x, y) = 2δ(x, y)− ζ(x, y) =

{1 if x = y

−1 if x < y

66


Since (2− ζ)(x, x) = 1 for all x ∈ P , the inverse (2− ζ)−1 exists.

Fix x, y ∈ P with x ≤ y and let l be the length of the longest chain in [x, y]. From theprevious proposition we know that

(ζ − 1)l+k(u, v) = 0 for all x ≤ u ≤ v ≤ y and k ∈ N

Using this we have

(2− ζ)[1 + (ζ − 1) + · · ·+ (ζ − 1)l

](u, v)

=(1− (ζ − 1)

)[1 + (ζ − 1) + · · ·+ (ζ − 1)l

](u, v)

=(1 + (ζ − 1) + · · ·+ (ζ − 1)l − (ζ − 1)− · · · − (ζ − 1)l − (ζ − 1)l+1

)(u, v)

=(1− (ζ − 1)l+1)(u, v)

= 1(u, v)= δ(u, v)

Thus(2− ζ)−1 = 1 + (ζ − 1) + · · ·+ (ζ − 1)l

on all intervals of [x, y]. This gives

(2− ζ)−1(x, y) = total number of chains in [x, y]

3.7 The Möbius Inversion Formula

The zeta function ζ satis�es ζ(x, x) = 1 for all x ∈ P and so, ζ−1 exists.

(3.66) De�nition The inverse µ = ζ−1 of P is called the Möbius function of Pand is given by

µ(x, x) = 1 for all x ∈ P

µ(x, y) = −∑x≤z<y

µ(x, z) for all x < y ∈ P

(3.67) Example Let P be the poset with Hasse diagram

•

��f

@@@@

•~~~~

e

@@@@

•b @@@@ •c •

dnnnnnnn

•a

Intervals of P :

[a, a], [b, b], . . . , [f, f ][a, b], [a, c], [a, d], [a, e], [a, f ][b, f ], [c, e], [c, f ][d, e], [d, f ][e, f ]

67


So

µ(a, a) = µ(b, b) = · · · = µ(f, f) = 1

µ(a, b) = −∑a≤z<b

µ(a, z) = −µ(a, a) = −1

µ(a, c) = −µ(a, a) = −1 = µ(a, d)µ(a, e) = −µ(a, a)− µ(a, c)− µ(a, d) = −1− (−1)− (−1) = 1µ(a, f) = −µ(a, a)− µ(a, b)− µ(a, c)− µ(a, d)− µ(a, e) = −1− 3(−1)− 1 = 1µ(b, f) = −µ(b, b) = −1µ(c, e) = −µ(c, c) = −1µ(c, f) = −µ(c, c)− µ(c, e) = −1− (−1) = 0µ(d, e) = −1µ(d, f) = 0µ(e, f) = −µ(e, e) = −1

The point of all this:

(3.68) Proposition (Möbius Inversion Formula, M.I.F.) Let P be a poset in which

every order ideal 〈x〉 is �nite. Let f, g : P → C. Then

g(x) =∑y≤x

f(y), ∀x ∈ P ⇔ f(x) =∑y≤x

g(y)µ(y, x), ∀x ∈ P

(3.69) Lemma Let µ be the Möbius function of a poset P . If a < b, then

µ(a, b) = −∑a<z≤b

µ(z, b)

Proof. From the de�nition of µ we have µ(a, a) = 1 and

µ(a, b) = −∑a≤z<b

µ(a, z), if a < b

Notice that if b covers a, a < b, then

µ(a, b) = −µ(a, a) = −1 = −µ(b, b)

By induction of the longest chain in [a, b], the result then holds.

Alternatively: all functions in I(P ) remain the same when we move from P to P ∗ (dualposet). So µP ∗(y, x) in P ∗ = µ(x, y) in P . �

Proof (of M.I.F.). �⇒�: Let f : P → C and de�ne

g(x) =∑y≤x

f(y) ∀x ∈ P

68


Then the sum∑y≤x

µ(y, x)g(y) =∑y≤x

µ(y, x)∑z≤y

f(z) =∑z≤x

f(z)∑z≤y≤x

µ(y, x)

= f(x)µ(x, x) +∑z<x

f(z)∑z≤y≤x

µ(y, x) = f(x) +∑z<x

f(z) · (0)

= f(x) ∀x ∈ P

�⇐�: Handled in exactly the same way. �

The dual form of the M.I.F. is also useful:

(3.70) Proposition (Möbius inversion formula, dual form) Let P be a poset in which

every dual order ideal is �nite. Let f, g : P → C. Then

g(x) =∑x≤y

f(y) ∀x ∈ P ⇔ f(x) =∑x≤y

µ(x, y)g(y) ∀x ∈ P

3.8 Applications of the Möbius Inversion Formula

(3.71) Example

(1) Consider the total order on P = (N,≤). For each i ∈ N, 〈i〉 = {1 ≤ j ≤ i} is�nite. The Möbius function of P :

µ(i, i) = 1 i ≥ 1µ(i, i+ 1) = −µ(i, i) = −1 i ≥ 1µ(i, i+ k) = 0 k > 1, i ≥ 1

If f, g : P → C, then M.I.F. gives

g(x) =∑y≤x

f(y) =x∑y=1

f(y)

i�

f(x) =∑y≤x

µ(y, x)g(y) =x∑y=1

µ(y, x)g(y) = µ(x, x)g(x) + µ(x− 1, x)g(x− 1)

= g(x)− g(x− 1)

(2) Let X1, . . . , Xn ⊆ S be �nite sets and let P be the poset of all intersections ofthese sets, ordered by inclusion. Also let 1 = X1 ∪ · · · ∪Xn ∈ P . We want to givean expression for |X1∪ · · · ∪Xn| in terms og the sizes |Xi| and their intersections.De�ne two functions on P :

g(x) = # elements in set x

f(x) = # elements in set x that do not belong to any set x′ < x

69


g(x) =∑y≤x

f(x)

Notice that

f(1) = 0

so

0 = f(1) =∑y≤1

g(y)µ(y, 1) by M.I.F.

Thus

0 =∑y≤1

|y|µ(y, 1) = |1|µ(1, 1) +∑y<1

|y|µ(y, 1)

i.e.

|X1 ∪ · · · ∪Xn| = −∑y<1

|y|µ(y, 1)

If we let n = 3:

•

~~~~~~X1∪X2∪X3=1

@@@@@@

•β1=X1

@@@@@@ •

~~~~~~β2=X2

@@@@@@ •

~~~~~~ β3=X3

•α1=X1∩X2

@@@@@@ •α2=X1∩X3•α3=X2∩X3

~~~~~~

•X1∩X2∩X3=0

~~~~~~

then

µ(βi, 1) = −1, µ(αi, 1) = 1, µ(0, 1) = −1

Thus

|X1 ∪X2 ∪X3| = −|X1|µ(X1, 1)− |X2|µ(X2, 1)− |X3|µ(X3, 1)

− |X1 ∩X2|µ(X1 ∩X2, 1)− · · · − |X2 ∩X3|µ(X2 ∩X3, 1)

− |X1 ∩X2 ∩X3|µ(X1 ∩X2 ∩X3, 1)= |X1|+ |X2|+ |X3| − |X1 ∩X2| − |X1 ∩X3| − |X2 ∩X3|

+ |X1 ∩X2 ∩X3|

(3) The poset (Bn ≤) and the Principle of Inclusion-Exclusion:

Lemma Let P = (Bn,≤) be the poset of all subsets of {1, . . . , n} ordered by

inclusion. For x ≤ y ∈ Bnµ(x, y) = (−1)|y|−|x|

Proof. Let x, y ∈ Bn with x ≤ y and r = ρ(y) − ρ(x) = |y| − |x|. It is an easyexercise to see that the interval [x, y] of Bn is isomorphic to the poset (Br,≤).(Consider the map ψx : [x, y]→ Br, ψx(z) = z − x.)

For n = 1:•{1}=1

•∅⇒ µ(0, 1) = (−1)1−0.

For n = 2: Same.

70


By induction on n, the above map ψx shows the result to be true for all [x, y] 6=[0, 1]. For [0, 1],

µ(0, 1) = −∑x<1

µ(0, x) = −∑x<1

(−1)|x| = −n−1∑i=0

(n

i

)(−1)i

= − [(1− 1)n − (−1)n] = (−1)n

So, Möbius inversion for Bn tells us that for f, g : Bn → C

g(x) =∑y≤x

f(y) ∀x ∈ Bn ⇔ f(x) =∑y≤x

(−1)|x|−|y|g(y) ∀x ∈ Bn

g(x) =∑x≤y

f(y) ∀x ∈ Bn ⇔ f(x) =∑x≤y

(−1)|y|−|x|g(y) ∀x ∈ Bn

An example of a (f, g) pair on Bn:Let D(n) = # derangements of {1, . . . , n} = #{π ∈ Sn : π1 6= 1, . . . , πn 6= n}(no �xed points). E.g. D(0) = 1, D(1) = 0, D(2) = 1, D(3) = 2. For π ∈ Sn letfix(π) = {i : πi = i}. De�ne

f(T ) = |{π ∈ Sn : fix(π) = T}|

g(T ) = |{π ∈ Sn : fix(π) ⊇ T}|

for all T ⊆ {1, . . . , n}, T ∈ Bn. (Note: g(T ) = (n − |T |)! ). We want to �ndf(∅) = f(0).Since

g(T ) =∑T⊆T ′

f(T ′) ∀T ∈ Bn

the D.M.I.F. gives

f(T ) =∑T⊆T ′

(−1)|T′|−|T |g(T ′) ∀T ∈ Bn

Thus

f(∅) =∑∅⊆T ′

(−1)|T′|(n− |T ′|)! =

n∑i=0

(n

i

)(−1)i(n− i)!

=n∑i=0

n!(n− i)!i!

(−1)i(n− i)! = n!n∑i=0

(−1)i

i!

= n!(

1− 11

+12!− 1

3!+ · · ·+ (−1)n

n!

)= n!

(12!− 1

3!+ · · ·+ (−1)n

n!

)E.g. for n = 3: 3!( 1

2! −13!) = 3− 1 = 2.

As another example, recall problem sheet 5, q. 3: Given X ⊆ {1, . . . , n− 1}, let

α(X) = |{π ∈ Sn : desπ ⊆ X}|

71


It was shown that if X = {x1, . . . , xk}, then

α(X) =(

n

x1, x2 − x1, . . . , n− xk

)What is

β(X) = |{π ∈ Sn : desπ = X}| ?

From the Möbius Inversion Formula: α, β : Bn−1 → C

α(X) =∑Y⊆X

β(Y ) for all X ⊆ {1, . . . , n− 1}

and thus

β(X) =∑Y⊆X

α(Y )µ(Y,X) =∑Y⊆X

α(Y )(−1)|X|−|Y |

=∑

1≤i1<i2<···<ij≤k(−1)k−j

(n

xi1 , xi2 − xi1 , . . . , n− xik

), k = |X|

(4) Classical Möbius function in Number theory: (Dn,≤) is the lattice of divisorsof n.

D12 : •~~~~~

12=2231

@@@@@

•4=2230 •oooooooo 6=2131

•2=2130

@@@@@ •3=2031

~~~~~

•1=2030

~~~~~

µ(1, 1) = 1 = µ(2, 2) = µ(3, 3) = . . .

µ(1, 2) = −1 = µ(1, 3)µ(1, 2× 2) = −µ(1, 1)− µ(1, 2) = 0µ(1, 2× 3) = −µ(1, 1)− µ(1, 2)− µ(1, 3) = 1µ(1, 2× 2× 3) = −µ(1, 1)− µ(1, 2)− µ(1, 3)

−µ(1, 2× 2)− µ(1, 2× 3)= 1− 1 = 0

Lemma Given x, y ∈ Dn, x ≤ y, then

µ(y/x) = µ(x, y) =

(−1)t if y/x is a product of t distinct primes

1 if x = y

0 otherwise.

Set D = in�nite lattice of divisors of the integers; �divisor� lattice. Then

{µ(1), µ(2), µ(3), µ(4), µ(5), µ(6), µ(7), . . .} = {1,−1,−1, 0,−1, 1,−1, . . .}

From this, if

g(m) =∑d|m

f(d) for all integers m,

then

f(m) =∑d|m

g(d)µ(m/d) ∀m ∈ N

72


As an example, let f(x) = x. Then

g(m) =∑d|m

d = sum of divisors of m

and

m = f(m) =∑d|m

g(d)µ(m/d) .

Or if f(x) = 1, then

g(m) =∑d|m

1 = # of divisors of m

and

1 = f(m) =∑d|m

g(d)µ(m/d)

(5) The Möbius function of Ln(q) (lattice of subspaces of Vn(q)) and Πn (lattice ofpartitions of {1, . . . , n}):

� for Ln(q),

µ(0, 1) = (−1)nq(n2)

� for Πn,

µ(0, 1) = (−1)n−1(n− 1)!

3.9 The Möbius Function of Lattices

In particular cases the Möbius function of a lattice may be easy to calculate.

(3.72) Lemma (3.9.1.) Let L be a �nite lattice with 1 covering {x1, . . . , xk} and 0covered by {y1, . . . , ym}.

If x1 ∧ x2 ∧ . . . ∧ xk 6= 0 then µ(0, 1) = 0

If y1 ∨ y2 ∨ . . . ∨ ym 6= 1 then µ(0, 1) = 0

(3.73) Remark The elements {y1, . . . , ym} covering 0 are called atoms. The elements{x1, . . . , xk} covered by 1 are called coatoms.

(3.74) Lemma (3.9.2.) Let L be a �nite distributive lattice. For x, y ∈ L,

µ(x, y) =

(−1)`(x,y) if [x, y] is a Boolean algebra

(i.e. ∼= Bk for some k)0 otherwise

73


Proof. Recall that if L is distributive, then L ∼= J(P ) for some poset P . If x, y ∈ L withx ≤ y, then there are antichains X,Y in P such that y corresponds to the order ideal〈Y 〉 and X corresponds to the order ideal 〈X〉. So, since x ≤ y, 〈X〉 ⊆P 〈Y 〉 and Y \Xis an antichain of P . Thus in J(P ) the sublattice [〈X〉, 〈Y 〉] ∼= [0, 〈Y \ X〉]. Using thisargument it is clear that

〈Y 〉 is the join of covers of 〈X〉 ⇔ [x, y] is a boolean algebra

From the previous lemma (3.72), if 〈Y 〉 is not the join of the covers of 〈X〉, thenµ(x, y) = 0. Hence the result. �

(3.75) Lemma (3.9.3.) Let P be a �nite poset with a 0 and 1. Let ci = # chains

0 = x0 < x1 < · · · < xi = 1 of length i. Then

µ(0, 1) = c0 − c1 + c2 − c3 + c4 − · · ·

(Related to the Euler characteristic).

Proof. The function µ = (1 + (ζ − 1))−1, so

µ(0, 1) = (1 + (ζ − 1))−1(0, 1) = (1− (ζ − 1) + (ζ − 1)2 − (ζ − 1)3 + · · · )(0, 1)

= 1(0, 1)− (ζ − 1)(0, 1) + (ζ − 1)2(0, 1)− · · ·= c0 − c1 + c2 − c3 + · · ·

since ci = (ζ − 1)i(0, 1) by Prop. (3.64). �

3.10 Young's lattice

From chapter 1, we have

Par(n) = set of partitions of the integer n

De�nePar = Par(0) ∪ Par(1) ∪ · · ·

Then

Par(0) = ∅ Par(3) = {3, 21, 111}Par(1) = {1} Par(4) = {4, 31, 22, 211, 1111}Par(2) = {2, 11} Par(5) = {5, 41, 32, 311, 221, 2111, 11111}

The sets Par(n) are easily visualised as Young diagrams. Given λ, µ ∈ Par, de�ne λ ⊆ µif the shape of λ is contained in the shape of µ. For example

32 = ⊆ = 4211

74


but

21 = 6⊆ = 111

Let Y = (Par,⊆) be the poset on Par with this inclusion/containment order

•111

•21

ppppppppppp •3

•11

NNNNNNNNNNN

ppppppppppp •

NNNNNNNNNNN2

ppppppppppp

•

NNNNNNNNNNN1

ppppppppppp

•∅

(3.76) Theorem Y is a lattice.

Proof. Given λ = (λ1, λ2, . . .), µ = (µ1, µ2, . . .) ∈ Y, the smallest shape (diagram) thatcontains both λ and µ is

λ ∨ µ = (max(λ1, µ1),max(λ2, µ2), . . .)

and similarlyλ ∧ µ = (min(λ1, µ1), (λ2, µ2), . . .)

It is straightforward to check that conditions (1)�(4) hold for (Y,∨,∧). �

(3.77) Remark The rank function ρ : Y → N is given by

ρ(λ) = λ1 + λ2 + · · ·

By noticing that

|{λ ∈ Par : ρ(λ) = n}| = p(n) = # partitions of the integer n

the rank generating function of Y is

F (Y, q) =+∞∑n=0

p(n)qn =+∞∏n=0

11− qn

(see problem sheet 4).

(3.78) Proposition Maximal chains [0, λ] in Y, where ρ(λ) = n, are in 1-1 correspon-

dence with all standard Young tableaux of shape λ (SYTλ).

(3.79) Remark Draw the corresponding Young diagram at each element in Young'slattice. Choose your element λ in the lattice. As you proceed from 0 up to λ, in eachstep i, put i in the box that was added to the Young diagram from last step, e.g.

∅ → 1 → 12→ 1 3

2→

1 324

= λ

if we choose λ = 211 and our route is 0→ 1→ 11→ 21→ 211. (Draw this!)

75


Proof. Let 0 = y0 <· y1 <· · · · <· yn = λ be a maximal chain. Since yi+1 has exactlyone more cell than yi and yi ⊂ yi+1, label the cell yi+1 − yi with i + 1. Do this for all0 ≤ i < n. The resulting labeling of the Young diagram is a standard Young tableauxof shape λ. The inverse map is easy to describe. A consequence of this is the number ofmaximal chains in [0, λ] is fλ (calculated by the hook-length formula). Also the numberof maximal chains [0, λ] for all λ where ρ(λ) = n, is equal to In, the number of involutions

{π ∈ Sn : π2 = id}

(3.80) De�nition The dominance order (Par(n),≤) is such that if λ, µ ∈ Par(n)with λ = (λ1, λ2, . . .) and µ = (µ1, µ2, . . .) then we de�ne

λ ≤ µ ⇔ λ1 + · · ·+ λi ≤ µ1 + · · ·+ µi ∀i

3.11 Linear extensions

(3.81) De�nition Let (P,≤P ) be a poset with |P | = n. A linear extension ofP is a bijective order preserving map

σ : P → {1, . . . , n}

such that

if σ−1(i) <P σ−1(j) then i < j.

The number of linear extensions of P is denoted e(P ).

(3.82) Example Consider P: •c@@@@ •d

•a •b

If σ = (σ(a), σ(b), σ(c), σ(d)) = (1, 2, 3, 4) then σ(P ) is •3@@@@ •4

•1 •2and σ is a linear

extension of P .

However if σ is chosen to be (σ(a), σ(b), σ(c), σ(d)) = (3, 2, 1, 4) then σ(P ) is •1@@@@ •4

•3 •2which is not a linear extension.

All linear extensions of P :

•3@@@@ •4

•1 •2•3

@@@@ •4•2 •1

•4@@@@ •3

•1 •2•4

@@@@ •3•2 •1

•4@@@@ •2

•3 •1

σ=(1,2,3,4) σ=(2,1,3,4) σ=(1,2,4,3) σ=(2,1,4,3) σ=(3,1,4,2)

Thus e(P ) = 5. ♦

76


(3.83) Remark

P = • • · · · •︸︷︷︸n

e(P ) = n!

P =

••...

••

n e(P ) = 1

(3.84) Proposition

e(P ) = # maximal chains in J(P )

Proof. Homework. �

(3.85) De�nition If a poset P on elements {1, . . . , n} is such that i <P j ⇒ i < j,then P is called a natural poset on {1, . . . , n}. In this case, each linear extension ofP may be identi�ed with a permutation (σ−1(1), σ−1(2), . . . , σ−1(n)) ∈ Sn.

(3.86) De�nition If P is a natural poset on {1, . . . , n}, then the set of permutationsσ−1 corresponding to linear extensions of P is denoted L(P ) and is called the Jordan-Hölder set of P .

(3.87) Example For P = •3@@@@ •4

•1 •2, L(P ) = {1234, 2134, 1243, 2143, 2413}. ♦

3.12 Rank-selection

Let P be a �nite graded poset of rank n with a 0 and 1. Let ρ be the rank function of P .

(3.88) De�nition Let S ⊆ {0, . . . , n} and PS = {x ∈ P : ρ(x) ∈ S}. PS is called theS-rank-selected subposet of P . Let α(P, S) = # maximal chains in PS . SinceP has a 0 and 1, the quantity α(P, S) can be restricted to S ⊆ {1, . . . , n− 1}.

De�ne for S ⊆ {0, . . . , n},

β(P, S) =∑T⊆S

(−1)|S|−|T |α(P, T ) .

The Möbius inversion formula gives

α(P, S) =∑T⊆S

β(P, T ) .

77


(3.89) Proposition Let P be a �nite poset and L = J(P ) with ρ(L) = n. For S ⊆{1, . . . , n− 1},

β(L, S) = #{π ∈ L(P ) : Des(π) = S}

Proof. Suppose S = {a1, . . . , ak} ⊆ {1, . . . , n − 1}. The structure of J(P ) tells us thatα(L, S) is the number of chains I1, . . . , Ik of order ideals of P such that |Ii| = ai for alli = 1, . . . , k. For such a chain

� arrange the elements of I1 in increasing order;

� arrange the elements of I2 − I1 in increasing order;

�...

� arrange the elements of P − Ik in increasing order.

This maps chains I ∈ LS → π ∈ L(P ) and this map is bijective. Also Des(π) ⊆ S.

If we de�neγ(L, S) = {π ∈ L(P ) : Des(π) = S}

thenα(L, S) =

∑T⊆S

γ(L, S)

By comparison, we have that γ(L, S) = β(L, S). �

(3.90) Proposition

β(Bn, S) = #{π ∈ Sn : Des(π) ∈ Sn}

Proof. Let P be the n element antichain

P : • • · · · •︸︷︷︸n

⇒ J(P ) = Bn

From the previous proposition, and using the fact that L(P ) = Sn, the result follows.�

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Partially Ordered Sets (Posets) - University of Iceland › lhk1 › fletta › fletta03.pdf · 2008-05-07 · Chapter 3 Partially Ordered Sets (Posets) 3.1 Posets (3.1) De nition