Partially Ordered Sets
Theorem 3.12 Let S and T be two finite sets. We have
3.5 Locally Finite Posets and Möbius Functions
Definition 3.19 A poset(S,∞)is locally finite if every closed interval of(S,∞)is finite.
Example 3.20 The poset(N,∞)is locally finite. Indeed, if[p,q]is a closed interval of this poset, then[p,q]is a finite set that consists of q−p+1 natural numbers.
Example 3.21 The poset(N,ι)introduced in Example 1.27 is locally finite. Indeed, if p divides q, then[p,q]is a finite set that contain all multiples of p that divide q. functions. Their convolution product is the function h:S×S−∈Rdefined by
h(x,y)= associative onA(S,∞)and its unit element is the Kronecker functionkdefined by
k(x,y)=
108 3 Combinatorics (because if u >y we have f(u,y)=0)
=
u∪[x,z]
y∪[x,z]
e(x,u)f(u,y)g(y,z).
On the other hand, we can write
(e↔(f ↔g))(x,z)=
u∪[x,z]
e(x,u)(f ↔g)(u,z)
=
u∪[x,z]
e(x,u)
y∪[u,z]
f(u,y)g(y,z)
=
u∪[x,z]
e(x,u)
y∪[x,z]
f(u,y)g(y,z), (because if u >y we have f(u,y)=0)
for x,z∪ S, which shows that↔is associative.
If f ∪A(S,∞)and x ∞y, then we can write (f ↔k)(x,y)=
z∪[x,y]
f(x,z)k(z,y)= f(x,y)
for x,y ∪ S. Thus, f ↔k = f . A similar argument shows thatk↔ f = f . This allows us to conclude thatkis indeed the unit with respect to the↔operation.
LetI(S,∞) = {[x,y] | x,y ∪ S and x ∞ y} ⊆ {∅}be the set of intervals of the poset(S,∞)to which we add the empty set. A useful point of view (see [1]) is to regard the incidence algebra of(S,∞)as consisting of formal sums of the form {f(x,y)· [x,y] | [x,y] ∪I(S,∞)− {∅}}. Define the product of two intervals as
[x,y][u, v] =
[x, v] if y=u,
∅ otherwise.
Further, we assume that the product of formal sums is distributive with respect to addition of these sums. Let f,g∪A(S,∞)be two functions and let f andˆ gˆbe their corresponding formal sums,
fˆ=
{f(x,y)· [x,y] | [x,y] ∪I(S,∞)}, ˆ
g=
{g(u, v)· [u, v] | [u, v] ∪I(S,∞)}.
Then, it is immediate that
fˆgˆ(x,z)=
x∞y∞z
f(x,y)g(y,z)[x,z],
so the usual product of the formal sums fˆgˆcorresponds to the convolution product of f andg.
Theorem 3.24 Let(S,∞)be a locally finite poset. A function f ∪A(S,∞)has an inverse relative to the operation↔if and only if f(x,x)∅=0 for every x ∪S.
Proof Suppose that there exists an inverse f⊃of f (that is, f ↔ f⊃ = f⊃↔ f =k) which yields (f ↔ f⊃)(x,x) = k(x,x) = 1 for every x. Since (f ↔ f⊃)(x,x) =
z∪[x,x] f(x,z)f⊃(z,x)= f(x,x)f⊃(x,x), it follows that f(x,x)∅=0.
To prove the converse implication, we first show the existence of a left inverse of f ; that is, a function f⊃ : S×S ∈ Rsuch that f⊃↔ f = k. For x ∞ y, we must have
z∪[x,y] f⊃(x,z)f(z,y) = k(x,y). This implies f⊃(x,x)f(x,x) =1
and
z∪[x,y] f⊃(x,z)f(z,y)=0 if x ∅=y. Thus, we must have f⊃(x,x)= 1
f(x,x), (3.9)
f⊃(x,y)= − 1 f(y,y)
z∪[x,y)
f⊃(x,z)f(z,y), (3.10)
when x ∞y and
f⊃(x,y)=0,
when x ∅∞ y. Equalities (3.9) and (3.10) give an inductive definition of f⊃ because the poset(S,∞)is locally finite.
To verify that f⊃is a left inverse of f , suppose that x<y. Then, (f⊃↔ f)(x,y)=
z∪[x,y]
f⊃(x,z)f(z,y)
=
z∪[x,y)
f⊃(x,z)f(z,y)+ f⊃(x,y)f(y,y)=0.
If x =y, then(f⊃↔ f)(y,y)=1 and x∅∞y implies(f⊃↔ f)(x,y)=0. Therefore, f⊃↔ f =k.
The function f⊃is also a right inverse of f . Let h= f↔f⊃. We have shown above that every function ofA(S,∞)has a left inverse, so let h⊃ be the left inverse of h.
Thus, we have f ↔ f⊃ = h = k↔h =(h⊃↔h)↔h = h⊃↔(f ↔ f⊃)↔(f ↔ f⊃)= h⊃↔f↔k↔f⊃=h⊃↔f ↔f⊃=h⊃↔h=k, which proves that f⊃is also a right inverse of f . Thus, f⊃is the inverse of f .
110 3 Combinatorics
If the inverse of f ∪A(S,∞)exists, we denote it by the common notation f−1. Corollary 3.25 Let(S,∞)be a locally finite poset and letIA(S,∞)be the set of invertible functions ofA(S,∞). Then(IA(S,∞),{k,↔,−1})is a group.
Proof This is a mere restatement of Theorem 3.24.
Let(S,∞)be a locally finite poset and let θ : S×S −∈ Rbe the Riemann function defined by
θ(x,y)=
1 if x ∞y, 0 otherwise,
for x,y ∪ S. Clearly,θ ∪ A(S,∞), so the functionθ−1exists by Corollary 3.25.
This inverse, known as the Möbius function, is denoted byμand its values can be computed from Equalities (3.9) and (3.10) as
μ(x,x)= 1
θ(x,x) =1, μ(x,y)= −
z∪[x,y)
μ(x,z)θ(z,y)= −
z∪[x,y)
μ(x,z),
for x <y; for x∅∞y, we haveμ(x,y)=0.
Example 3.26 For the poset({1,2,3,4,5,6,7,8},ι)introduced in Example 2.13 the Möbius function is given by
μ(1,1)=1,
μ(1,2)=μ(1,3)=μ(1,5)=μ(1,7)= −1, μ(1,4)= −μ(1,1)−μ(1,2)=0,
μ(1,6)= −μ(1,1)−μ(1,2)−μ(1,3)= −1+1+1=1, μ(1,8)= −μ(1,1)−μ(1,2)−μ(1,4)= −1+1+0=0, and μ(2,2)=1,μ(2,4)= −1,μ(2,6)= −1,
μ(2,8)= −μ(2,2)−μ(2,4)= −1+1=0, μ(3,3)=1,μ(3,6)= −1,
μ(4,4)=μ(5,5)=μ(6,6)=μ(7,7)=μ(8,8)=1. For all other pairs(p,q)with(p,q)∅∪ιwe haveμ(p,q)=0.
The special role played byμis discussed next.
Theorem 3.27 (Möbius Inversion Theorem) Let(S,∞)be a locally finite poset that has the least element 0. If f,g : S −∈Rare two real-valued functions such thatg(x)=
0∞z∞x f(z), then f(x)=
0∞z∞xg(z)μ(z,x)for x∪ S.
Proof Starting from the functions f,g : S −∈ R, define the functions F,G ∪ A(S,∞)by
F(0,x)= f(x),G(0,x)=g(x) F(u,x)=G(u,x)=0, if u>0. The equalityg(x)=
0∞z∞x f(z)can be written as G(0,x)=
0∞z∞x
F(0,z)θ(z,x),
whereθis Riemann’s function. We also have G(u,x)=
u∞z∞x F(u,z)θ(z,x)for u >0 because in this case G(u,x)=0 and F(u,z)=0. Thus, G=F↔θ. Sinceμ is the inverse ofθinIA(S,∞), it follows that F=G↔μ. Consequently,
f(x)=F(0,x)=
0∞z∞x
G(0,z)μ(z,x)
=
0∞z∞x
g(z)μ(z,x),
which is the desired equality.
Now let(S,∞)be a poset that has the greatest element 1. By applying the Möbius inversion theorem to its dual(S,∞−1)=(S,)we obtain the following dual form of the theorem.
Theorem 3.28 (Möbius Dual Inversion Theorem) Let(S,∞)be a locally finite poset that has the greatest element 1. If f,g:S−∈Rare two real-valued functions such that
g(x)=
x∞z∞1
f(z),
then
f(x)=
x∞z∞1
g(z)μ(z,x)
for x ∪S.
Proof This statement follows immediately from Theorem 3.27.
Example 3.29 Let M be a finite set and let(P(M),∧)be the poset of all its subsets.
The Möbius function of this poset is given by
112 3 Combinatorics the equality above. Suppose that the equality holds for sets that differ by fewer than n elements and that|B| − |A| =n. Then, by the definition of the Möbius function,
It is interesting to observe that the principle of inclusion-exclusion can be obtained also from the Möbius dual inversion theorem. Let A0, . . . ,An−1 be n finite sets, where n 2, S = n−1
i=0 Ai, and I be a subset of the set{0, . . . ,n −1}. The complement of I ,{0, . . . ,n−1} −I is denoted byI .¯
Let BIbe the subset of S that consists of those elements that belong to every one of the sets Aiwith i ∪ I and to no other sets. Clearly, we have
Conversely, let x ∪⎭
which proves Equality (3.11). This allows us to write
By the Möbius dual inversion theorem (Theorem 3.28) applied to the poset (P({0, . . . ,n−1}),∧), we have intersection of an empty collection of subsets of a set S equals S. Thus,
Example 3.30 Let n be a natural number such that n 2. Using the inclusion-exclusion principle we can compute the numberν(n)of positive integers that are less than n and are relatively prime with n; that is, the number of integers r such that 1∞r ∞n such that gcd{n,r} =1.
114 3 Combinatorics
Data miners should be aware of what is known today as Ramsey theory because this family of combinatorial results establishes that data sets that are sufficiently large contain spurious patterns whose existence is caused by the sheer size of the data set and do not represent “significant” structures from a data mining point of view.
We begin with a set of basic terms of Ramsey theory.
Definition 3.31 Let C = {c1, . . . ,ck}be a finite set referred to as the set of colors.
A C-coloring of a set S is a mapping f : S −∈ C. The set f−1(c)is the set of elements of S colored by c.
A subset T of S is monochromatic in the color ci if f(t)=ci for every t ∪T . A subset W of S is f -monochromatic if it is monochromatic in some color ci. Clearly, every set of the form f−1(c)for a C-coloring of S is f -monochromatic.
Recall that the set of subsets of size q of a set S is denoted byPq(S).
Theorem 3.32 Let S be a finite set, q a positive natural number, f :S−∈ {c1,c2} a coloring of the set S such that every set inPq(S)is f -monochromatic, and a1and a2be two natural numbers not less than 2.