m¶bius functions of posets i: introduction to partially ordered sets
TRANSCRIPT
![Page 1: M¶bius Functions of Posets I: Introduction to Partially Ordered Sets](https://reader035.vdocuments.mx/reader035/viewer/2022071600/613d28a6736caf36b75a02b3/html5/thumbnails/1.jpg)
Mobius Functions of Posets I: Introduction toPartially Ordered Sets
Bruce SaganDepartment of MathematicsMichigan State University
East Lansing, MI [email protected]
www.math.msu.edu/˜sagan
June 25, 2007
![Page 2: M¶bius Functions of Posets I: Introduction to Partially Ordered Sets](https://reader035.vdocuments.mx/reader035/viewer/2022071600/613d28a6736caf36b75a02b3/html5/thumbnails/2.jpg)
Motivating Examples
Poset Basics
Isomorphism and Products
![Page 3: M¶bius Functions of Posets I: Introduction to Partially Ordered Sets](https://reader035.vdocuments.mx/reader035/viewer/2022071600/613d28a6736caf36b75a02b3/html5/thumbnails/3.jpg)
Outline
Motivating Examples
Poset Basics
Isomorphism and Products
![Page 4: M¶bius Functions of Posets I: Introduction to Partially Ordered Sets](https://reader035.vdocuments.mx/reader035/viewer/2022071600/613d28a6736caf36b75a02b3/html5/thumbnails/4.jpg)
Example A: Combinatorics.
Given a set, S, let
#S = |S| = cardinality of S.
The Principle of Inclusion-Exclusion or PIE is a very useful toolin enumerative combinatorics.
Theorem (PIE)Let U be a finite set and U1, . . . , Un ⊆ U.
|U −n⋃
i=1
Ui | = |U| −∑
1≤i≤n
|Ui |+∑
1≤i<j≤n
|Ui ∩ Uj |
− · · ·+ (−1)n|n⋂
i=1
Ui |.
![Page 5: M¶bius Functions of Posets I: Introduction to Partially Ordered Sets](https://reader035.vdocuments.mx/reader035/viewer/2022071600/613d28a6736caf36b75a02b3/html5/thumbnails/5.jpg)
Example A: Combinatorics.Given a set, S, let
#S = |S| = cardinality of S.
The Principle of Inclusion-Exclusion or PIE is a very useful toolin enumerative combinatorics.
Theorem (PIE)Let U be a finite set and U1, . . . , Un ⊆ U.
|U −n⋃
i=1
Ui | = |U| −∑
1≤i≤n
|Ui |+∑
1≤i<j≤n
|Ui ∩ Uj |
− · · ·+ (−1)n|n⋂
i=1
Ui |.
![Page 6: M¶bius Functions of Posets I: Introduction to Partially Ordered Sets](https://reader035.vdocuments.mx/reader035/viewer/2022071600/613d28a6736caf36b75a02b3/html5/thumbnails/6.jpg)
Example A: Combinatorics.Given a set, S, let
#S = |S| = cardinality of S.
The Principle of Inclusion-Exclusion or PIE is a very useful toolin enumerative combinatorics.
Theorem (PIE)Let U be a finite set and U1, . . . , Un ⊆ U.
|U −n⋃
i=1
Ui | = |U| −∑
1≤i≤n
|Ui |+∑
1≤i<j≤n
|Ui ∩ Uj |
− · · ·+ (−1)n|n⋂
i=1
Ui |.
![Page 7: M¶bius Functions of Posets I: Introduction to Partially Ordered Sets](https://reader035.vdocuments.mx/reader035/viewer/2022071600/613d28a6736caf36b75a02b3/html5/thumbnails/7.jpg)
Example A: Combinatorics.Given a set, S, let
#S = |S| = cardinality of S.
The Principle of Inclusion-Exclusion or PIE is a very useful toolin enumerative combinatorics.
Theorem (PIE)Let U be a finite set and U1, . . . , Un ⊆ U.
|U −n⋃
i=1
Ui | = |U| −∑
1≤i≤n
|Ui |+∑
1≤i<j≤n
|Ui ∩ Uj |
− · · ·+ (−1)n|n⋂
i=1
Ui |.
![Page 8: M¶bius Functions of Posets I: Introduction to Partially Ordered Sets](https://reader035.vdocuments.mx/reader035/viewer/2022071600/613d28a6736caf36b75a02b3/html5/thumbnails/8.jpg)
Example B: Theory of Finite Differences.
LetZ≥0 = the nonnegative integers.
If one takes a function f : Z≥0 → R then there is an analogue ofthe derivative, namely the difference operator
∆f (n) = f (n)− f (n − 1)
(where f (−1) = 0 by definition). There is also an analogue ofthe integral, namely the summation operator
Sf (n) =n∑
i=0
f (i).
The Fundamental Theorem of the Difference Calculus or FTDCis as follows.
Theorem (FTDC)If f : Z≥0 → R then
∆Sf (n) = f (n).
![Page 9: M¶bius Functions of Posets I: Introduction to Partially Ordered Sets](https://reader035.vdocuments.mx/reader035/viewer/2022071600/613d28a6736caf36b75a02b3/html5/thumbnails/9.jpg)
Example B: Theory of Finite Differences.Let
Z≥0 = the nonnegative integers.
If one takes a function f : Z≥0 → R then there is an analogue ofthe derivative, namely the difference operator
∆f (n) = f (n)− f (n − 1)
(where f (−1) = 0 by definition). There is also an analogue ofthe integral, namely the summation operator
Sf (n) =n∑
i=0
f (i).
The Fundamental Theorem of the Difference Calculus or FTDCis as follows.
Theorem (FTDC)If f : Z≥0 → R then
∆Sf (n) = f (n).
![Page 10: M¶bius Functions of Posets I: Introduction to Partially Ordered Sets](https://reader035.vdocuments.mx/reader035/viewer/2022071600/613d28a6736caf36b75a02b3/html5/thumbnails/10.jpg)
Example B: Theory of Finite Differences.Let
Z≥0 = the nonnegative integers.
If one takes a function f : Z≥0 → R then there is an analogue ofthe derivative, namely the difference operator
∆f (n) = f (n)− f (n − 1)
(where f (−1) = 0 by definition).
There is also an analogue ofthe integral, namely the summation operator
Sf (n) =n∑
i=0
f (i).
The Fundamental Theorem of the Difference Calculus or FTDCis as follows.
Theorem (FTDC)If f : Z≥0 → R then
∆Sf (n) = f (n).
![Page 11: M¶bius Functions of Posets I: Introduction to Partially Ordered Sets](https://reader035.vdocuments.mx/reader035/viewer/2022071600/613d28a6736caf36b75a02b3/html5/thumbnails/11.jpg)
Example B: Theory of Finite Differences.Let
Z≥0 = the nonnegative integers.
If one takes a function f : Z≥0 → R then there is an analogue ofthe derivative, namely the difference operator
∆f (n) = f (n)− f (n − 1)
(where f (−1) = 0 by definition). There is also an analogue ofthe integral, namely the summation operator
Sf (n) =n∑
i=0
f (i).
The Fundamental Theorem of the Difference Calculus or FTDCis as follows.
Theorem (FTDC)If f : Z≥0 → R then
∆Sf (n) = f (n).
![Page 12: M¶bius Functions of Posets I: Introduction to Partially Ordered Sets](https://reader035.vdocuments.mx/reader035/viewer/2022071600/613d28a6736caf36b75a02b3/html5/thumbnails/12.jpg)
Example B: Theory of Finite Differences.Let
Z≥0 = the nonnegative integers.
If one takes a function f : Z≥0 → R then there is an analogue ofthe derivative, namely the difference operator
∆f (n) = f (n)− f (n − 1)
(where f (−1) = 0 by definition). There is also an analogue ofthe integral, namely the summation operator
Sf (n) =n∑
i=0
f (i).
The Fundamental Theorem of the Difference Calculus or FTDCis as follows.
Theorem (FTDC)If f : Z≥0 → R then
∆Sf (n) = f (n).
![Page 13: M¶bius Functions of Posets I: Introduction to Partially Ordered Sets](https://reader035.vdocuments.mx/reader035/viewer/2022071600/613d28a6736caf36b75a02b3/html5/thumbnails/13.jpg)
Example C: Number Theory
If d , n ∈ Z then write d |n if d divides evenly into n. Thenumber-theoretic Mobius function is µ : Z>0 → Z defined as
µ(n) =
{0 if n is not square free,(−1)k if n = product of k distinct primes.
The importance of µ lies in the number-theoretic MobiusInversion Theorem or MIT.
Theorem (Number Theory MIT)Let f , g : Z>0 → R satisfy
f (n) =∑d |n
g(d)
for all n ∈ Z>0. Then
g(n) =∑d |n
µ(n/d)f (d).
![Page 14: M¶bius Functions of Posets I: Introduction to Partially Ordered Sets](https://reader035.vdocuments.mx/reader035/viewer/2022071600/613d28a6736caf36b75a02b3/html5/thumbnails/14.jpg)
Example C: Number TheoryIf d , n ∈ Z then write d |n if d divides evenly into n.
Thenumber-theoretic Mobius function is µ : Z>0 → Z defined as
µ(n) =
{0 if n is not square free,(−1)k if n = product of k distinct primes.
The importance of µ lies in the number-theoretic MobiusInversion Theorem or MIT.
Theorem (Number Theory MIT)Let f , g : Z>0 → R satisfy
f (n) =∑d |n
g(d)
for all n ∈ Z>0. Then
g(n) =∑d |n
µ(n/d)f (d).
![Page 15: M¶bius Functions of Posets I: Introduction to Partially Ordered Sets](https://reader035.vdocuments.mx/reader035/viewer/2022071600/613d28a6736caf36b75a02b3/html5/thumbnails/15.jpg)
Example C: Number TheoryIf d , n ∈ Z then write d |n if d divides evenly into n. Thenumber-theoretic Mobius function is µ : Z>0 → Z defined as
µ(n) =
{0 if n is not square free,(−1)k if n = product of k distinct primes.
The importance of µ lies in the number-theoretic MobiusInversion Theorem or MIT.
Theorem (Number Theory MIT)Let f , g : Z>0 → R satisfy
f (n) =∑d |n
g(d)
for all n ∈ Z>0. Then
g(n) =∑d |n
µ(n/d)f (d).
![Page 16: M¶bius Functions of Posets I: Introduction to Partially Ordered Sets](https://reader035.vdocuments.mx/reader035/viewer/2022071600/613d28a6736caf36b75a02b3/html5/thumbnails/16.jpg)
Example C: Number TheoryIf d , n ∈ Z then write d |n if d divides evenly into n. Thenumber-theoretic Mobius function is µ : Z>0 → Z defined as
µ(n) =
{0 if n is not square free,(−1)k if n = product of k distinct primes.
The importance of µ lies in the number-theoretic MobiusInversion Theorem or MIT.
Theorem (Number Theory MIT)Let f , g : Z>0 → R satisfy
f (n) =∑d |n
g(d)
for all n ∈ Z>0. Then
g(n) =∑d |n
µ(n/d)f (d).
![Page 17: M¶bius Functions of Posets I: Introduction to Partially Ordered Sets](https://reader035.vdocuments.mx/reader035/viewer/2022071600/613d28a6736caf36b75a02b3/html5/thumbnails/17.jpg)
Example C: Number TheoryIf d , n ∈ Z then write d |n if d divides evenly into n. Thenumber-theoretic Mobius function is µ : Z>0 → Z defined as
µ(n) =
{0 if n is not square free,(−1)k if n = product of k distinct primes.
The importance of µ lies in the number-theoretic MobiusInversion Theorem or MIT.
Theorem (Number Theory MIT)Let f , g : Z>0 → R satisfy
f (n) =∑d |n
g(d)
for all n ∈ Z>0. Then
g(n) =∑d |n
µ(n/d)f (d).
![Page 18: M¶bius Functions of Posets I: Introduction to Partially Ordered Sets](https://reader035.vdocuments.mx/reader035/viewer/2022071600/613d28a6736caf36b75a02b3/html5/thumbnails/18.jpg)
Mobius inversion over partially ordered sets is important for thefollowing reasons.
1. It unifies and generalizes the three previous examples.
2. It makes the number-theoretic definition transparent.
3. It encodes topological information about partially orderedsets.
4. It can be used to solve combinatorial problems.
![Page 19: M¶bius Functions of Posets I: Introduction to Partially Ordered Sets](https://reader035.vdocuments.mx/reader035/viewer/2022071600/613d28a6736caf36b75a02b3/html5/thumbnails/19.jpg)
Mobius inversion over partially ordered sets is important for thefollowing reasons.
1. It unifies and generalizes the three previous examples.
2. It makes the number-theoretic definition transparent.
3. It encodes topological information about partially orderedsets.
4. It can be used to solve combinatorial problems.
![Page 20: M¶bius Functions of Posets I: Introduction to Partially Ordered Sets](https://reader035.vdocuments.mx/reader035/viewer/2022071600/613d28a6736caf36b75a02b3/html5/thumbnails/20.jpg)
Mobius inversion over partially ordered sets is important for thefollowing reasons.
1. It unifies and generalizes the three previous examples.
2. It makes the number-theoretic definition transparent.
3. It encodes topological information about partially orderedsets.
4. It can be used to solve combinatorial problems.
![Page 21: M¶bius Functions of Posets I: Introduction to Partially Ordered Sets](https://reader035.vdocuments.mx/reader035/viewer/2022071600/613d28a6736caf36b75a02b3/html5/thumbnails/21.jpg)
Mobius inversion over partially ordered sets is important for thefollowing reasons.
1. It unifies and generalizes the three previous examples.
2. It makes the number-theoretic definition transparent.
3. It encodes topological information about partially orderedsets.
4. It can be used to solve combinatorial problems.
![Page 22: M¶bius Functions of Posets I: Introduction to Partially Ordered Sets](https://reader035.vdocuments.mx/reader035/viewer/2022071600/613d28a6736caf36b75a02b3/html5/thumbnails/22.jpg)
Mobius inversion over partially ordered sets is important for thefollowing reasons.
1. It unifies and generalizes the three previous examples.
2. It makes the number-theoretic definition transparent.
3. It encodes topological information about partially orderedsets.
4. It can be used to solve combinatorial problems.
![Page 23: M¶bius Functions of Posets I: Introduction to Partially Ordered Sets](https://reader035.vdocuments.mx/reader035/viewer/2022071600/613d28a6736caf36b75a02b3/html5/thumbnails/23.jpg)
Outline
Motivating Examples
Poset Basics
Isomorphism and Products
![Page 24: M¶bius Functions of Posets I: Introduction to Partially Ordered Sets](https://reader035.vdocuments.mx/reader035/viewer/2022071600/613d28a6736caf36b75a02b3/html5/thumbnails/24.jpg)
A partially ordered set or poset is a set P together with a binaryrelation ≤ such that for all x , y , z ∈ P:
1. (reflexivity) x ≤ x ,
2. (antisymmetry) x ≤ y and y ≤ x implies x = y ,
3. (transitivity) x ≤ y and y ≤ z implies x ≤ z.
Given any poset notation, if we wish to be specific about theposet P involved, we attach P as a subscript. For example,using ≤P for ≤. We also adopt the usual conventions forinequalities. For example, x < y means x ≤ y and x 6= y .If x , y ∈ P then x is covered by y or y covers x , written x � y , ifx < y and there is no z with x < z < y . The Hasse diagram ofP is the (directed) graph with vertices P and an edge from x upto y if x � y .
![Page 25: M¶bius Functions of Posets I: Introduction to Partially Ordered Sets](https://reader035.vdocuments.mx/reader035/viewer/2022071600/613d28a6736caf36b75a02b3/html5/thumbnails/25.jpg)
A partially ordered set or poset is a set P together with a binaryrelation ≤ such that for all x , y , z ∈ P:
1. (reflexivity) x ≤ x ,
2. (antisymmetry) x ≤ y and y ≤ x implies x = y ,
3. (transitivity) x ≤ y and y ≤ z implies x ≤ z.
Given any poset notation, if we wish to be specific about theposet P involved, we attach P as a subscript. For example,using ≤P for ≤. We also adopt the usual conventions forinequalities. For example, x < y means x ≤ y and x 6= y .If x , y ∈ P then x is covered by y or y covers x , written x � y , ifx < y and there is no z with x < z < y . The Hasse diagram ofP is the (directed) graph with vertices P and an edge from x upto y if x � y .
![Page 26: M¶bius Functions of Posets I: Introduction to Partially Ordered Sets](https://reader035.vdocuments.mx/reader035/viewer/2022071600/613d28a6736caf36b75a02b3/html5/thumbnails/26.jpg)
A partially ordered set or poset is a set P together with a binaryrelation ≤ such that for all x , y , z ∈ P:
1. (reflexivity) x ≤ x ,
2. (antisymmetry) x ≤ y and y ≤ x implies x = y ,
3. (transitivity) x ≤ y and y ≤ z implies x ≤ z.
Given any poset notation, if we wish to be specific about theposet P involved, we attach P as a subscript. For example,using ≤P for ≤. We also adopt the usual conventions forinequalities. For example, x < y means x ≤ y and x 6= y .If x , y ∈ P then x is covered by y or y covers x , written x � y , ifx < y and there is no z with x < z < y . The Hasse diagram ofP is the (directed) graph with vertices P and an edge from x upto y if x � y .
![Page 27: M¶bius Functions of Posets I: Introduction to Partially Ordered Sets](https://reader035.vdocuments.mx/reader035/viewer/2022071600/613d28a6736caf36b75a02b3/html5/thumbnails/27.jpg)
A partially ordered set or poset is a set P together with a binaryrelation ≤ such that for all x , y , z ∈ P:
1. (reflexivity) x ≤ x ,
2. (antisymmetry) x ≤ y and y ≤ x implies x = y ,
3. (transitivity) x ≤ y and y ≤ z implies x ≤ z.
Given any poset notation, if we wish to be specific about theposet P involved, we attach P as a subscript. For example,using ≤P for ≤. We also adopt the usual conventions forinequalities. For example, x < y means x ≤ y and x 6= y .If x , y ∈ P then x is covered by y or y covers x , written x � y , ifx < y and there is no z with x < z < y . The Hasse diagram ofP is the (directed) graph with vertices P and an edge from x upto y if x � y .
![Page 28: M¶bius Functions of Posets I: Introduction to Partially Ordered Sets](https://reader035.vdocuments.mx/reader035/viewer/2022071600/613d28a6736caf36b75a02b3/html5/thumbnails/28.jpg)
A partially ordered set or poset is a set P together with a binaryrelation ≤ such that for all x , y , z ∈ P:
1. (reflexivity) x ≤ x ,
2. (antisymmetry) x ≤ y and y ≤ x implies x = y ,
3. (transitivity) x ≤ y and y ≤ z implies x ≤ z.
Given any poset notation, if we wish to be specific about theposet P involved, we attach P as a subscript. For example,using ≤P for ≤.
We also adopt the usual conventions forinequalities. For example, x < y means x ≤ y and x 6= y .If x , y ∈ P then x is covered by y or y covers x , written x � y , ifx < y and there is no z with x < z < y . The Hasse diagram ofP is the (directed) graph with vertices P and an edge from x upto y if x � y .
![Page 29: M¶bius Functions of Posets I: Introduction to Partially Ordered Sets](https://reader035.vdocuments.mx/reader035/viewer/2022071600/613d28a6736caf36b75a02b3/html5/thumbnails/29.jpg)
A partially ordered set or poset is a set P together with a binaryrelation ≤ such that for all x , y , z ∈ P:
1. (reflexivity) x ≤ x ,
2. (antisymmetry) x ≤ y and y ≤ x implies x = y ,
3. (transitivity) x ≤ y and y ≤ z implies x ≤ z.
Given any poset notation, if we wish to be specific about theposet P involved, we attach P as a subscript. For example,using ≤P for ≤. We also adopt the usual conventions forinequalities. For example, x < y means x ≤ y and x 6= y .
If x , y ∈ P then x is covered by y or y covers x , written x � y , ifx < y and there is no z with x < z < y . The Hasse diagram ofP is the (directed) graph with vertices P and an edge from x upto y if x � y .
![Page 30: M¶bius Functions of Posets I: Introduction to Partially Ordered Sets](https://reader035.vdocuments.mx/reader035/viewer/2022071600/613d28a6736caf36b75a02b3/html5/thumbnails/30.jpg)
A partially ordered set or poset is a set P together with a binaryrelation ≤ such that for all x , y , z ∈ P:
1. (reflexivity) x ≤ x ,
2. (antisymmetry) x ≤ y and y ≤ x implies x = y ,
3. (transitivity) x ≤ y and y ≤ z implies x ≤ z.
Given any poset notation, if we wish to be specific about theposet P involved, we attach P as a subscript. For example,using ≤P for ≤. We also adopt the usual conventions forinequalities. For example, x < y means x ≤ y and x 6= y .If x , y ∈ P then x is covered by y or y covers x , written x � y , ifx < y and there is no z with x < z < y .
The Hasse diagram ofP is the (directed) graph with vertices P and an edge from x upto y if x � y .
![Page 31: M¶bius Functions of Posets I: Introduction to Partially Ordered Sets](https://reader035.vdocuments.mx/reader035/viewer/2022071600/613d28a6736caf36b75a02b3/html5/thumbnails/31.jpg)
A partially ordered set or poset is a set P together with a binaryrelation ≤ such that for all x , y , z ∈ P:
1. (reflexivity) x ≤ x ,
2. (antisymmetry) x ≤ y and y ≤ x implies x = y ,
3. (transitivity) x ≤ y and y ≤ z implies x ≤ z.
Given any poset notation, if we wish to be specific about theposet P involved, we attach P as a subscript. For example,using ≤P for ≤. We also adopt the usual conventions forinequalities. For example, x < y means x ≤ y and x 6= y .If x , y ∈ P then x is covered by y or y covers x , written x � y , ifx < y and there is no z with x < z < y . The Hasse diagram ofP is the (directed) graph with vertices P and an edge from x upto y if x � y .
![Page 32: M¶bius Functions of Posets I: Introduction to Partially Ordered Sets](https://reader035.vdocuments.mx/reader035/viewer/2022071600/613d28a6736caf36b75a02b3/html5/thumbnails/32.jpg)
Example: The Chain.
The chain of length n is Cn = {0, 1, . . . , n} with the usual ≤ onthe integers.
C3=
s0
s1
s2
s3
![Page 33: M¶bius Functions of Posets I: Introduction to Partially Ordered Sets](https://reader035.vdocuments.mx/reader035/viewer/2022071600/613d28a6736caf36b75a02b3/html5/thumbnails/33.jpg)
Example: The Chain.The chain of length n is Cn = {0, 1, . . . , n} with the usual ≤ onthe integers.
C3=
s0
s1
s2
s3
![Page 34: M¶bius Functions of Posets I: Introduction to Partially Ordered Sets](https://reader035.vdocuments.mx/reader035/viewer/2022071600/613d28a6736caf36b75a02b3/html5/thumbnails/34.jpg)
Example: The Chain.The chain of length n is Cn = {0, 1, . . . , n} with the usual ≤ onthe integers.
C3=
s0
s1
s2
s3
![Page 35: M¶bius Functions of Posets I: Introduction to Partially Ordered Sets](https://reader035.vdocuments.mx/reader035/viewer/2022071600/613d28a6736caf36b75a02b3/html5/thumbnails/35.jpg)
Example: The Chain.The chain of length n is Cn = {0, 1, . . . , n} with the usual ≤ onthe integers.
C3=
s0
s1
s2
s3
![Page 36: M¶bius Functions of Posets I: Introduction to Partially Ordered Sets](https://reader035.vdocuments.mx/reader035/viewer/2022071600/613d28a6736caf36b75a02b3/html5/thumbnails/36.jpg)
Example: The Chain.The chain of length n is Cn = {0, 1, . . . , n} with the usual ≤ onthe integers.
C3=
s0
s1
s2
s3
![Page 37: M¶bius Functions of Posets I: Introduction to Partially Ordered Sets](https://reader035.vdocuments.mx/reader035/viewer/2022071600/613d28a6736caf36b75a02b3/html5/thumbnails/37.jpg)
Example: The Chain.The chain of length n is Cn = {0, 1, . . . , n} with the usual ≤ onthe integers.
C3=
s0
s1
s2
s3
![Page 38: M¶bius Functions of Posets I: Introduction to Partially Ordered Sets](https://reader035.vdocuments.mx/reader035/viewer/2022071600/613d28a6736caf36b75a02b3/html5/thumbnails/38.jpg)
Example: The Chain.The chain of length n is Cn = {0, 1, . . . , n} with the usual ≤ onthe integers.
C3=
s0
s1
s2
s3
![Page 39: M¶bius Functions of Posets I: Introduction to Partially Ordered Sets](https://reader035.vdocuments.mx/reader035/viewer/2022071600/613d28a6736caf36b75a02b3/html5/thumbnails/39.jpg)
Example: The Boolean Algebra.
The Boolean algebra is
Bn = {S : S ⊆ {1, 2, . . . , n}}
partially ordered by S ≤ T if and only if S ⊆ T .
B3 =
t∅QQ
��
��
��t t t{1} {2} {3}Q
QQQt{1, 2}
��
��
��
QQt{1, 3}
��
��
��t{2, 3}�
��
���
QQt{1, 2, 3}
Note that B3 looks like a cube.
![Page 40: M¶bius Functions of Posets I: Introduction to Partially Ordered Sets](https://reader035.vdocuments.mx/reader035/viewer/2022071600/613d28a6736caf36b75a02b3/html5/thumbnails/40.jpg)
Example: The Boolean Algebra.The Boolean algebra is
Bn = {S : S ⊆ {1, 2, . . . , n}}
partially ordered by S ≤ T if and only if S ⊆ T .
B3 =
t∅QQ
��
��
��t t t{1} {2} {3}Q
QQQt{1, 2}
��
��
��
QQt{1, 3}
��
��
��t{2, 3}�
��
���
QQt{1, 2, 3}
Note that B3 looks like a cube.
![Page 41: M¶bius Functions of Posets I: Introduction to Partially Ordered Sets](https://reader035.vdocuments.mx/reader035/viewer/2022071600/613d28a6736caf36b75a02b3/html5/thumbnails/41.jpg)
Example: The Boolean Algebra.The Boolean algebra is
Bn = {S : S ⊆ {1, 2, . . . , n}}
partially ordered by S ≤ T if and only if S ⊆ T .
B3 =
t∅QQ
��
��
��t t t{1} {2} {3}Q
QQQt{1, 2}
��
��
��
QQt{1, 3}
��
��
��t{2, 3}�
��
���
QQt{1, 2, 3}
Note that B3 looks like a cube.
![Page 42: M¶bius Functions of Posets I: Introduction to Partially Ordered Sets](https://reader035.vdocuments.mx/reader035/viewer/2022071600/613d28a6736caf36b75a02b3/html5/thumbnails/42.jpg)
Example: The Boolean Algebra.The Boolean algebra is
Bn = {S : S ⊆ {1, 2, . . . , n}}
partially ordered by S ≤ T if and only if S ⊆ T .
B3 =
t∅
��
��
��t t t{1} {2} {3}Q
QQQt{1, 2}
��
��
��
QQt{1, 3}
��
��
��t{2, 3}�
��
���
QQt{1, 2, 3}
Note that B3 looks like a cube.
![Page 43: M¶bius Functions of Posets I: Introduction to Partially Ordered Sets](https://reader035.vdocuments.mx/reader035/viewer/2022071600/613d28a6736caf36b75a02b3/html5/thumbnails/43.jpg)
Example: The Boolean Algebra.The Boolean algebra is
Bn = {S : S ⊆ {1, 2, . . . , n}}
partially ordered by S ≤ T if and only if S ⊆ T .
B3 =
t∅QQ
��
��
��t t t{1} {2} {3}
QQt{1, 2}
��
��
��
QQt{1, 3}
��
��
��t{2, 3}�
��
���
QQt{1, 2, 3}
Note that B3 looks like a cube.
![Page 44: M¶bius Functions of Posets I: Introduction to Partially Ordered Sets](https://reader035.vdocuments.mx/reader035/viewer/2022071600/613d28a6736caf36b75a02b3/html5/thumbnails/44.jpg)
Example: The Boolean Algebra.The Boolean algebra is
Bn = {S : S ⊆ {1, 2, . . . , n}}
partially ordered by S ≤ T if and only if S ⊆ T .
B3 =
t∅QQ
��
��
��t t t{1} {2} {3}Q
QQQt{1, 2}
��
��
��
QQt{1, 3}
��
��
��t{2, 3}�
��
���
QQt{1, 2, 3}
Note that B3 looks like a cube.
![Page 45: M¶bius Functions of Posets I: Introduction to Partially Ordered Sets](https://reader035.vdocuments.mx/reader035/viewer/2022071600/613d28a6736caf36b75a02b3/html5/thumbnails/45.jpg)
Example: The Boolean Algebra.The Boolean algebra is
Bn = {S : S ⊆ {1, 2, . . . , n}}
partially ordered by S ≤ T if and only if S ⊆ T .
B3 =
t∅QQ
��
��
��t t t{1} {2} {3}Q
QQQt{1, 2}
��
��
��
QQt{1, 3}
��
��
��t{2, 3}
��
��
��
QQt{1, 2, 3}
Note that B3 looks like a cube.
![Page 46: M¶bius Functions of Posets I: Introduction to Partially Ordered Sets](https://reader035.vdocuments.mx/reader035/viewer/2022071600/613d28a6736caf36b75a02b3/html5/thumbnails/46.jpg)
Example: The Boolean Algebra.The Boolean algebra is
Bn = {S : S ⊆ {1, 2, . . . , n}}
partially ordered by S ≤ T if and only if S ⊆ T .
B3 =
t∅QQ
��
��
��t t t{1} {2} {3}Q
QQQt{1, 2}
��
��
��
QQt{1, 3}
��
��
��t{2, 3}�
��
���
QQt{1, 2, 3}
Note that B3 looks like a cube.
![Page 47: M¶bius Functions of Posets I: Introduction to Partially Ordered Sets](https://reader035.vdocuments.mx/reader035/viewer/2022071600/613d28a6736caf36b75a02b3/html5/thumbnails/47.jpg)
Example: The Boolean Algebra.The Boolean algebra is
Bn = {S : S ⊆ {1, 2, . . . , n}}
partially ordered by S ≤ T if and only if S ⊆ T .
B3 =
t∅QQ
��
��
��t t t{1} {2} {3}Q
QQQt{1, 2}
��
��
��
QQt{1, 3}
��
��
��t{2, 3}�
��
���
QQt{1, 2, 3}
Note that B3 looks like a cube.
![Page 48: M¶bius Functions of Posets I: Introduction to Partially Ordered Sets](https://reader035.vdocuments.mx/reader035/viewer/2022071600/613d28a6736caf36b75a02b3/html5/thumbnails/48.jpg)
Example: The Divisor Lattice.
Given n ∈ Z>0 the corresponding divisor lattice is
Dn = {d ∈ Z>0 : d |n}
partially ordered by c ≤Dn d if and only if c|d .
D18 =
t1@@
@@
��
��t t2 3�
���
@@
@@
��
��t t6 9�
���
@@
@@t18
Note that D18 looks like a rectangle.
![Page 49: M¶bius Functions of Posets I: Introduction to Partially Ordered Sets](https://reader035.vdocuments.mx/reader035/viewer/2022071600/613d28a6736caf36b75a02b3/html5/thumbnails/49.jpg)
Example: The Divisor Lattice.Given n ∈ Z>0 the corresponding divisor lattice is
Dn = {d ∈ Z>0 : d |n}
partially ordered by c ≤Dn d if and only if c|d .
D18 =
t1@@
@@
��
��t t2 3�
���
@@
@@
��
��t t6 9�
���
@@
@@t18
Note that D18 looks like a rectangle.
![Page 50: M¶bius Functions of Posets I: Introduction to Partially Ordered Sets](https://reader035.vdocuments.mx/reader035/viewer/2022071600/613d28a6736caf36b75a02b3/html5/thumbnails/50.jpg)
Example: The Divisor Lattice.Given n ∈ Z>0 the corresponding divisor lattice is
Dn = {d ∈ Z>0 : d |n}
partially ordered by c ≤Dn d if and only if c|d .
D18 =
t1@@
@@
��
��t t2 3�
���
@@
@@
��
��t t6 9�
���
@@
@@t18
Note that D18 looks like a rectangle.
![Page 51: M¶bius Functions of Posets I: Introduction to Partially Ordered Sets](https://reader035.vdocuments.mx/reader035/viewer/2022071600/613d28a6736caf36b75a02b3/html5/thumbnails/51.jpg)
Example: The Divisor Lattice.Given n ∈ Z>0 the corresponding divisor lattice is
Dn = {d ∈ Z>0 : d |n}
partially ordered by c ≤Dn d if and only if c|d .
D18 =
t1
@@
@@
��
��t t2 3�
���
@@
@@
��
��t t6 9�
���
@@
@@t18
Note that D18 looks like a rectangle.
![Page 52: M¶bius Functions of Posets I: Introduction to Partially Ordered Sets](https://reader035.vdocuments.mx/reader035/viewer/2022071600/613d28a6736caf36b75a02b3/html5/thumbnails/52.jpg)
Example: The Divisor Lattice.Given n ∈ Z>0 the corresponding divisor lattice is
Dn = {d ∈ Z>0 : d |n}
partially ordered by c ≤Dn d if and only if c|d .
D18 =
t1@@
@@
��
��t t2 3
��
��
@@
@@
��
��t t6 9�
���
@@
@@t18
Note that D18 looks like a rectangle.
![Page 53: M¶bius Functions of Posets I: Introduction to Partially Ordered Sets](https://reader035.vdocuments.mx/reader035/viewer/2022071600/613d28a6736caf36b75a02b3/html5/thumbnails/53.jpg)
Example: The Divisor Lattice.Given n ∈ Z>0 the corresponding divisor lattice is
Dn = {d ∈ Z>0 : d |n}
partially ordered by c ≤Dn d if and only if c|d .
D18 =
t1@@
@@
��
��t t2 3�
���
@@
@@
��
��t t6 9
��
��
@@
@@t18
Note that D18 looks like a rectangle.
![Page 54: M¶bius Functions of Posets I: Introduction to Partially Ordered Sets](https://reader035.vdocuments.mx/reader035/viewer/2022071600/613d28a6736caf36b75a02b3/html5/thumbnails/54.jpg)
Example: The Divisor Lattice.Given n ∈ Z>0 the corresponding divisor lattice is
Dn = {d ∈ Z>0 : d |n}
partially ordered by c ≤Dn d if and only if c|d .
D18 =
t1@@
@@
��
��t t2 3�
���
@@
@@
��
��t t6 9�
���
@@
@@t18
Note that D18 looks like a rectangle.
![Page 55: M¶bius Functions of Posets I: Introduction to Partially Ordered Sets](https://reader035.vdocuments.mx/reader035/viewer/2022071600/613d28a6736caf36b75a02b3/html5/thumbnails/55.jpg)
Example: The Divisor Lattice.Given n ∈ Z>0 the corresponding divisor lattice is
Dn = {d ∈ Z>0 : d |n}
partially ordered by c ≤Dn d if and only if c|d .
D18 =
t1@@
@@
��
��t t2 3�
���
@@
@@
��
��t t6 9�
���
@@
@@t18
Note that D18 looks like a rectangle.
![Page 56: M¶bius Functions of Posets I: Introduction to Partially Ordered Sets](https://reader035.vdocuments.mx/reader035/viewer/2022071600/613d28a6736caf36b75a02b3/html5/thumbnails/56.jpg)
In a poset P, a minimal element is x ∈ P such that there is noy ∈ P with y < x .
A maximal element is x ∈ P such that thereis no y ∈ P with y > x .
t ttt t
@@
@@
u v
w
x yExample. The poset on the left hasminimal elements u and v ,and maximal elements x and y .
A poset has a zero if it has a unique minimal element, 0. Aposet has a one if it has a unique maximal element, 1. A posetif bounded if it has both a 0 and a 1.
Example. Our three fundamental examples are bounded:
0Cn = 0, 1Cn = n, 0Bn = ∅, 1Bn = {1, . . . , n}, 0Dn = 1, 1Dn = n.
If x ≤ y in P then the corresponding closed interval is
[x , y ] = {z : x ≤ z ≤ y}.Open and half-open intervals are defined analogously. Notethat [x , y ] is a poset in its own right and it has a zero and a one:
0[x ,y ] = x , 1[x ,y ] = y .
![Page 57: M¶bius Functions of Posets I: Introduction to Partially Ordered Sets](https://reader035.vdocuments.mx/reader035/viewer/2022071600/613d28a6736caf36b75a02b3/html5/thumbnails/57.jpg)
In a poset P, a minimal element is x ∈ P such that there is noy ∈ P with y < x . A maximal element is x ∈ P such that thereis no y ∈ P with y > x .
t ttt t
@@
@@
u v
w
x yExample. The poset on the left hasminimal elements u and v ,and maximal elements x and y .
A poset has a zero if it has a unique minimal element, 0. Aposet has a one if it has a unique maximal element, 1. A posetif bounded if it has both a 0 and a 1.
Example. Our three fundamental examples are bounded:
0Cn = 0, 1Cn = n, 0Bn = ∅, 1Bn = {1, . . . , n}, 0Dn = 1, 1Dn = n.
If x ≤ y in P then the corresponding closed interval is
[x , y ] = {z : x ≤ z ≤ y}.Open and half-open intervals are defined analogously. Notethat [x , y ] is a poset in its own right and it has a zero and a one:
0[x ,y ] = x , 1[x ,y ] = y .
![Page 58: M¶bius Functions of Posets I: Introduction to Partially Ordered Sets](https://reader035.vdocuments.mx/reader035/viewer/2022071600/613d28a6736caf36b75a02b3/html5/thumbnails/58.jpg)
In a poset P, a minimal element is x ∈ P such that there is noy ∈ P with y < x . A maximal element is x ∈ P such that thereis no y ∈ P with y > x .
t ttt t
@@
@@
u v
w
x yExample. The poset on the left hasminimal elements u and v ,
and maximal elements x and y .
A poset has a zero if it has a unique minimal element, 0. Aposet has a one if it has a unique maximal element, 1. A posetif bounded if it has both a 0 and a 1.
Example. Our three fundamental examples are bounded:
0Cn = 0, 1Cn = n, 0Bn = ∅, 1Bn = {1, . . . , n}, 0Dn = 1, 1Dn = n.
If x ≤ y in P then the corresponding closed interval is
[x , y ] = {z : x ≤ z ≤ y}.Open and half-open intervals are defined analogously. Notethat [x , y ] is a poset in its own right and it has a zero and a one:
0[x ,y ] = x , 1[x ,y ] = y .
![Page 59: M¶bius Functions of Posets I: Introduction to Partially Ordered Sets](https://reader035.vdocuments.mx/reader035/viewer/2022071600/613d28a6736caf36b75a02b3/html5/thumbnails/59.jpg)
In a poset P, a minimal element is x ∈ P such that there is noy ∈ P with y < x . A maximal element is x ∈ P such that thereis no y ∈ P with y > x .
t ttt t
@@
@@
u v
w
x yExample. The poset on the left hasminimal elements u and v ,and maximal elements x and y .
A poset has a zero if it has a unique minimal element, 0. Aposet has a one if it has a unique maximal element, 1. A posetif bounded if it has both a 0 and a 1.
Example. Our three fundamental examples are bounded:
0Cn = 0, 1Cn = n, 0Bn = ∅, 1Bn = {1, . . . , n}, 0Dn = 1, 1Dn = n.
If x ≤ y in P then the corresponding closed interval is
[x , y ] = {z : x ≤ z ≤ y}.Open and half-open intervals are defined analogously. Notethat [x , y ] is a poset in its own right and it has a zero and a one:
0[x ,y ] = x , 1[x ,y ] = y .
![Page 60: M¶bius Functions of Posets I: Introduction to Partially Ordered Sets](https://reader035.vdocuments.mx/reader035/viewer/2022071600/613d28a6736caf36b75a02b3/html5/thumbnails/60.jpg)
In a poset P, a minimal element is x ∈ P such that there is noy ∈ P with y < x . A maximal element is x ∈ P such that thereis no y ∈ P with y > x .
t ttt t
@@
@@
u v
w
x yExample. The poset on the left hasminimal elements u and v ,and maximal elements x and y .
A poset has a zero if it has a unique minimal element, 0.
Aposet has a one if it has a unique maximal element, 1. A posetif bounded if it has both a 0 and a 1.
Example. Our three fundamental examples are bounded:
0Cn = 0, 1Cn = n, 0Bn = ∅, 1Bn = {1, . . . , n}, 0Dn = 1, 1Dn = n.
If x ≤ y in P then the corresponding closed interval is
[x , y ] = {z : x ≤ z ≤ y}.Open and half-open intervals are defined analogously. Notethat [x , y ] is a poset in its own right and it has a zero and a one:
0[x ,y ] = x , 1[x ,y ] = y .
![Page 61: M¶bius Functions of Posets I: Introduction to Partially Ordered Sets](https://reader035.vdocuments.mx/reader035/viewer/2022071600/613d28a6736caf36b75a02b3/html5/thumbnails/61.jpg)
In a poset P, a minimal element is x ∈ P such that there is noy ∈ P with y < x . A maximal element is x ∈ P such that thereis no y ∈ P with y > x .
t ttt t
@@
@@
u v
w
x yExample. The poset on the left hasminimal elements u and v ,and maximal elements x and y .
A poset has a zero if it has a unique minimal element, 0. Aposet has a one if it has a unique maximal element, 1.
A posetif bounded if it has both a 0 and a 1.
Example. Our three fundamental examples are bounded:
0Cn = 0, 1Cn = n, 0Bn = ∅, 1Bn = {1, . . . , n}, 0Dn = 1, 1Dn = n.
If x ≤ y in P then the corresponding closed interval is
[x , y ] = {z : x ≤ z ≤ y}.Open and half-open intervals are defined analogously. Notethat [x , y ] is a poset in its own right and it has a zero and a one:
0[x ,y ] = x , 1[x ,y ] = y .
![Page 62: M¶bius Functions of Posets I: Introduction to Partially Ordered Sets](https://reader035.vdocuments.mx/reader035/viewer/2022071600/613d28a6736caf36b75a02b3/html5/thumbnails/62.jpg)
In a poset P, a minimal element is x ∈ P such that there is noy ∈ P with y < x . A maximal element is x ∈ P such that thereis no y ∈ P with y > x .
t ttt t
@@
@@
u v
w
x yExample. The poset on the left hasminimal elements u and v ,and maximal elements x and y .
A poset has a zero if it has a unique minimal element, 0. Aposet has a one if it has a unique maximal element, 1. A posetif bounded if it has both a 0 and a 1.
Example. Our three fundamental examples are bounded:
0Cn = 0, 1Cn = n, 0Bn = ∅, 1Bn = {1, . . . , n}, 0Dn = 1, 1Dn = n.
If x ≤ y in P then the corresponding closed interval is
[x , y ] = {z : x ≤ z ≤ y}.Open and half-open intervals are defined analogously. Notethat [x , y ] is a poset in its own right and it has a zero and a one:
0[x ,y ] = x , 1[x ,y ] = y .
![Page 63: M¶bius Functions of Posets I: Introduction to Partially Ordered Sets](https://reader035.vdocuments.mx/reader035/viewer/2022071600/613d28a6736caf36b75a02b3/html5/thumbnails/63.jpg)
In a poset P, a minimal element is x ∈ P such that there is noy ∈ P with y < x . A maximal element is x ∈ P such that thereis no y ∈ P with y > x .
t ttt t
@@
@@
u v
w
x yExample. The poset on the left hasminimal elements u and v ,and maximal elements x and y .
A poset has a zero if it has a unique minimal element, 0. Aposet has a one if it has a unique maximal element, 1. A posetif bounded if it has both a 0 and a 1.
Example. Our three fundamental examples are bounded:
0Cn = 0, 1Cn = n, 0Bn = ∅, 1Bn = {1, . . . , n}, 0Dn = 1, 1Dn = n.
If x ≤ y in P then the corresponding closed interval is
[x , y ] = {z : x ≤ z ≤ y}.Open and half-open intervals are defined analogously. Notethat [x , y ] is a poset in its own right and it has a zero and a one:
0[x ,y ] = x , 1[x ,y ] = y .
![Page 64: M¶bius Functions of Posets I: Introduction to Partially Ordered Sets](https://reader035.vdocuments.mx/reader035/viewer/2022071600/613d28a6736caf36b75a02b3/html5/thumbnails/64.jpg)
In a poset P, a minimal element is x ∈ P such that there is noy ∈ P with y < x . A maximal element is x ∈ P such that thereis no y ∈ P with y > x .
t ttt t
@@
@@
u v
w
x yExample. The poset on the left hasminimal elements u and v ,and maximal elements x and y .
A poset has a zero if it has a unique minimal element, 0. Aposet has a one if it has a unique maximal element, 1. A posetif bounded if it has both a 0 and a 1.
Example. Our three fundamental examples are bounded:
0Cn = 0,
1Cn = n, 0Bn = ∅, 1Bn = {1, . . . , n}, 0Dn = 1, 1Dn = n.
If x ≤ y in P then the corresponding closed interval is
[x , y ] = {z : x ≤ z ≤ y}.Open and half-open intervals are defined analogously. Notethat [x , y ] is a poset in its own right and it has a zero and a one:
0[x ,y ] = x , 1[x ,y ] = y .
![Page 65: M¶bius Functions of Posets I: Introduction to Partially Ordered Sets](https://reader035.vdocuments.mx/reader035/viewer/2022071600/613d28a6736caf36b75a02b3/html5/thumbnails/65.jpg)
In a poset P, a minimal element is x ∈ P such that there is noy ∈ P with y < x . A maximal element is x ∈ P such that thereis no y ∈ P with y > x .
t ttt t
@@
@@
u v
w
x yExample. The poset on the left hasminimal elements u and v ,and maximal elements x and y .
A poset has a zero if it has a unique minimal element, 0. Aposet has a one if it has a unique maximal element, 1. A posetif bounded if it has both a 0 and a 1.
Example. Our three fundamental examples are bounded:
0Cn = 0, 1Cn = n,
0Bn = ∅, 1Bn = {1, . . . , n}, 0Dn = 1, 1Dn = n.
If x ≤ y in P then the corresponding closed interval is
[x , y ] = {z : x ≤ z ≤ y}.Open and half-open intervals are defined analogously. Notethat [x , y ] is a poset in its own right and it has a zero and a one:
0[x ,y ] = x , 1[x ,y ] = y .
![Page 66: M¶bius Functions of Posets I: Introduction to Partially Ordered Sets](https://reader035.vdocuments.mx/reader035/viewer/2022071600/613d28a6736caf36b75a02b3/html5/thumbnails/66.jpg)
In a poset P, a minimal element is x ∈ P such that there is noy ∈ P with y < x . A maximal element is x ∈ P such that thereis no y ∈ P with y > x .
t ttt t
@@
@@
u v
w
x yExample. The poset on the left hasminimal elements u and v ,and maximal elements x and y .
A poset has a zero if it has a unique minimal element, 0. Aposet has a one if it has a unique maximal element, 1. A posetif bounded if it has both a 0 and a 1.
Example. Our three fundamental examples are bounded:
0Cn = 0, 1Cn = n, 0Bn = ∅,
1Bn = {1, . . . , n}, 0Dn = 1, 1Dn = n.
If x ≤ y in P then the corresponding closed interval is
[x , y ] = {z : x ≤ z ≤ y}.Open and half-open intervals are defined analogously. Notethat [x , y ] is a poset in its own right and it has a zero and a one:
0[x ,y ] = x , 1[x ,y ] = y .
![Page 67: M¶bius Functions of Posets I: Introduction to Partially Ordered Sets](https://reader035.vdocuments.mx/reader035/viewer/2022071600/613d28a6736caf36b75a02b3/html5/thumbnails/67.jpg)
In a poset P, a minimal element is x ∈ P such that there is noy ∈ P with y < x . A maximal element is x ∈ P such that thereis no y ∈ P with y > x .
t ttt t
@@
@@
u v
w
x yExample. The poset on the left hasminimal elements u and v ,and maximal elements x and y .
A poset has a zero if it has a unique minimal element, 0. Aposet has a one if it has a unique maximal element, 1. A posetif bounded if it has both a 0 and a 1.
Example. Our three fundamental examples are bounded:
0Cn = 0, 1Cn = n, 0Bn = ∅, 1Bn = {1, . . . , n},
0Dn = 1, 1Dn = n.
If x ≤ y in P then the corresponding closed interval is
[x , y ] = {z : x ≤ z ≤ y}.Open and half-open intervals are defined analogously. Notethat [x , y ] is a poset in its own right and it has a zero and a one:
0[x ,y ] = x , 1[x ,y ] = y .
![Page 68: M¶bius Functions of Posets I: Introduction to Partially Ordered Sets](https://reader035.vdocuments.mx/reader035/viewer/2022071600/613d28a6736caf36b75a02b3/html5/thumbnails/68.jpg)
In a poset P, a minimal element is x ∈ P such that there is noy ∈ P with y < x . A maximal element is x ∈ P such that thereis no y ∈ P with y > x .
t ttt t
@@
@@
u v
w
x yExample. The poset on the left hasminimal elements u and v ,and maximal elements x and y .
A poset has a zero if it has a unique minimal element, 0. Aposet has a one if it has a unique maximal element, 1. A posetif bounded if it has both a 0 and a 1.
Example. Our three fundamental examples are bounded:
0Cn = 0, 1Cn = n, 0Bn = ∅, 1Bn = {1, . . . , n}, 0Dn = 1,
1Dn = n.
If x ≤ y in P then the corresponding closed interval is
[x , y ] = {z : x ≤ z ≤ y}.Open and half-open intervals are defined analogously. Notethat [x , y ] is a poset in its own right and it has a zero and a one:
0[x ,y ] = x , 1[x ,y ] = y .
![Page 69: M¶bius Functions of Posets I: Introduction to Partially Ordered Sets](https://reader035.vdocuments.mx/reader035/viewer/2022071600/613d28a6736caf36b75a02b3/html5/thumbnails/69.jpg)
In a poset P, a minimal element is x ∈ P such that there is noy ∈ P with y < x . A maximal element is x ∈ P such that thereis no y ∈ P with y > x .
t ttt t
@@
@@
u v
w
x yExample. The poset on the left hasminimal elements u and v ,and maximal elements x and y .
A poset has a zero if it has a unique minimal element, 0. Aposet has a one if it has a unique maximal element, 1. A posetif bounded if it has both a 0 and a 1.
Example. Our three fundamental examples are bounded:
0Cn = 0, 1Cn = n, 0Bn = ∅, 1Bn = {1, . . . , n}, 0Dn = 1, 1Dn = n.
If x ≤ y in P then the corresponding closed interval is
[x , y ] = {z : x ≤ z ≤ y}.Open and half-open intervals are defined analogously. Notethat [x , y ] is a poset in its own right and it has a zero and a one:
0[x ,y ] = x , 1[x ,y ] = y .
![Page 70: M¶bius Functions of Posets I: Introduction to Partially Ordered Sets](https://reader035.vdocuments.mx/reader035/viewer/2022071600/613d28a6736caf36b75a02b3/html5/thumbnails/70.jpg)
In a poset P, a minimal element is x ∈ P such that there is noy ∈ P with y < x . A maximal element is x ∈ P such that thereis no y ∈ P with y > x .
t ttt t
@@
@@
u v
w
x yExample. The poset on the left hasminimal elements u and v ,and maximal elements x and y .
A poset has a zero if it has a unique minimal element, 0. Aposet has a one if it has a unique maximal element, 1. A posetif bounded if it has both a 0 and a 1.
Example. Our three fundamental examples are bounded:
0Cn = 0, 1Cn = n, 0Bn = ∅, 1Bn = {1, . . . , n}, 0Dn = 1, 1Dn = n.
If x ≤ y in P then the corresponding closed interval is
[x , y ] = {z : x ≤ z ≤ y}.
Open and half-open intervals are defined analogously. Notethat [x , y ] is a poset in its own right and it has a zero and a one:
0[x ,y ] = x , 1[x ,y ] = y .
![Page 71: M¶bius Functions of Posets I: Introduction to Partially Ordered Sets](https://reader035.vdocuments.mx/reader035/viewer/2022071600/613d28a6736caf36b75a02b3/html5/thumbnails/71.jpg)
In a poset P, a minimal element is x ∈ P such that there is noy ∈ P with y < x . A maximal element is x ∈ P such that thereis no y ∈ P with y > x .
t ttt t
@@
@@
u v
w
x yExample. The poset on the left hasminimal elements u and v ,and maximal elements x and y .
A poset has a zero if it has a unique minimal element, 0. Aposet has a one if it has a unique maximal element, 1. A posetif bounded if it has both a 0 and a 1.
Example. Our three fundamental examples are bounded:
0Cn = 0, 1Cn = n, 0Bn = ∅, 1Bn = {1, . . . , n}, 0Dn = 1, 1Dn = n.
If x ≤ y in P then the corresponding closed interval is
[x , y ] = {z : x ≤ z ≤ y}.Open and half-open intervals are defined analogously.
Notethat [x , y ] is a poset in its own right and it has a zero and a one:
0[x ,y ] = x , 1[x ,y ] = y .
![Page 72: M¶bius Functions of Posets I: Introduction to Partially Ordered Sets](https://reader035.vdocuments.mx/reader035/viewer/2022071600/613d28a6736caf36b75a02b3/html5/thumbnails/72.jpg)
In a poset P, a minimal element is x ∈ P such that there is noy ∈ P with y < x . A maximal element is x ∈ P such that thereis no y ∈ P with y > x .
t ttt t
@@
@@
u v
w
x yExample. The poset on the left hasminimal elements u and v ,and maximal elements x and y .
A poset has a zero if it has a unique minimal element, 0. Aposet has a one if it has a unique maximal element, 1. A posetif bounded if it has both a 0 and a 1.
Example. Our three fundamental examples are bounded:
0Cn = 0, 1Cn = n, 0Bn = ∅, 1Bn = {1, . . . , n}, 0Dn = 1, 1Dn = n.
If x ≤ y in P then the corresponding closed interval is
[x , y ] = {z : x ≤ z ≤ y}.Open and half-open intervals are defined analogously. Notethat [x , y ] is a poset in its own right and it has a zero and a one:
0[x ,y ] = x , 1[x ,y ] = y .
![Page 73: M¶bius Functions of Posets I: Introduction to Partially Ordered Sets](https://reader035.vdocuments.mx/reader035/viewer/2022071600/613d28a6736caf36b75a02b3/html5/thumbnails/73.jpg)
Example: The Chain.In C9 we have the interval [4, 7]:
s4
s5
s6
s7
This interval looks like C3.
![Page 74: M¶bius Functions of Posets I: Introduction to Partially Ordered Sets](https://reader035.vdocuments.mx/reader035/viewer/2022071600/613d28a6736caf36b75a02b3/html5/thumbnails/74.jpg)
Example: The Chain.In C9 we have the interval [4, 7]:
s4
s5
s6
s7
This interval looks like C3.
![Page 75: M¶bius Functions of Posets I: Introduction to Partially Ordered Sets](https://reader035.vdocuments.mx/reader035/viewer/2022071600/613d28a6736caf36b75a02b3/html5/thumbnails/75.jpg)
Example: The Chain.In C9 we have the interval [4, 7]:
s4
s5
s6
s7
This interval looks like C3.
![Page 76: M¶bius Functions of Posets I: Introduction to Partially Ordered Sets](https://reader035.vdocuments.mx/reader035/viewer/2022071600/613d28a6736caf36b75a02b3/html5/thumbnails/76.jpg)
Example: The Boolean Algebra.In B7 we have the interval [{3}, {2, 3, 5, 6}]:
t{3}Q
QQQ
��
��
��t t t{2, 3} {3, 5} {3, 6}Q
QQQt{2, 3, 5}
��
��
��
QQt{2, 3, 6}
��
��
��t{3, 5, 6}�
��
���
QQt{2, 3, 5, 6}
Note that this interval looks like B3.
![Page 77: M¶bius Functions of Posets I: Introduction to Partially Ordered Sets](https://reader035.vdocuments.mx/reader035/viewer/2022071600/613d28a6736caf36b75a02b3/html5/thumbnails/77.jpg)
Example: The Boolean Algebra.In B7 we have the interval [{3}, {2, 3, 5, 6}]:
t{3}
��
��
��t t t{2, 3} {3, 5} {3, 6}Q
QQQt{2, 3, 5}
��
��
��
QQt{2, 3, 6}
��
��
��t{3, 5, 6}�
��
���
QQt{2, 3, 5, 6}
Note that this interval looks like B3.
![Page 78: M¶bius Functions of Posets I: Introduction to Partially Ordered Sets](https://reader035.vdocuments.mx/reader035/viewer/2022071600/613d28a6736caf36b75a02b3/html5/thumbnails/78.jpg)
Example: The Boolean Algebra.In B7 we have the interval [{3}, {2, 3, 5, 6}]:
t{3}Q
QQQ
��
��
��t t t{2, 3} {3, 5} {3, 6}
QQt{2, 3, 5}
��
��
��
QQt{2, 3, 6}
��
��
��t{3, 5, 6}�
��
���
QQt{2, 3, 5, 6}
Note that this interval looks like B3.
![Page 79: M¶bius Functions of Posets I: Introduction to Partially Ordered Sets](https://reader035.vdocuments.mx/reader035/viewer/2022071600/613d28a6736caf36b75a02b3/html5/thumbnails/79.jpg)
Example: The Boolean Algebra.In B7 we have the interval [{3}, {2, 3, 5, 6}]:
t{3}Q
QQQ
��
��
��t t t{2, 3} {3, 5} {3, 6}Q
QQQt{2, 3, 5}
��
��
��
QQt{2, 3, 6}
��
��
��t{3, 5, 6}
��
��
��
QQt{2, 3, 5, 6}
Note that this interval looks like B3.
![Page 80: M¶bius Functions of Posets I: Introduction to Partially Ordered Sets](https://reader035.vdocuments.mx/reader035/viewer/2022071600/613d28a6736caf36b75a02b3/html5/thumbnails/80.jpg)
Example: The Boolean Algebra.In B7 we have the interval [{3}, {2, 3, 5, 6}]:
t{3}Q
QQQ
��
��
��t t t{2, 3} {3, 5} {3, 6}Q
QQQt{2, 3, 5}
��
��
��
QQt{2, 3, 6}
��
��
��t{3, 5, 6}�
��
���
QQt{2, 3, 5, 6}
Note that this interval looks like B3.
![Page 81: M¶bius Functions of Posets I: Introduction to Partially Ordered Sets](https://reader035.vdocuments.mx/reader035/viewer/2022071600/613d28a6736caf36b75a02b3/html5/thumbnails/81.jpg)
Example: The Boolean Algebra.In B7 we have the interval [{3}, {2, 3, 5, 6}]:
t{3}Q
QQQ
��
��
��t t t{2, 3} {3, 5} {3, 6}Q
QQQt{2, 3, 5}
��
��
��
QQt{2, 3, 6}
��
��
��t{3, 5, 6}�
��
���
QQt{2, 3, 5, 6}
Note that this interval looks like B3.
![Page 82: M¶bius Functions of Posets I: Introduction to Partially Ordered Sets](https://reader035.vdocuments.mx/reader035/viewer/2022071600/613d28a6736caf36b75a02b3/html5/thumbnails/82.jpg)
Example: The Divisor Lattice.In D80 we have the interval [2, 40]:
t2@@
@@
��
��t t10 4�
���
@@
@@
��
��t t20 8�
���
@@
@@t40
Note that this interval looks like D18.
![Page 83: M¶bius Functions of Posets I: Introduction to Partially Ordered Sets](https://reader035.vdocuments.mx/reader035/viewer/2022071600/613d28a6736caf36b75a02b3/html5/thumbnails/83.jpg)
Example: The Divisor Lattice.In D80 we have the interval [2, 40]:
t2
@@
@@
��
��t t10 4�
���
@@
@@
��
��t t20 8�
���
@@
@@t40
Note that this interval looks like D18.
![Page 84: M¶bius Functions of Posets I: Introduction to Partially Ordered Sets](https://reader035.vdocuments.mx/reader035/viewer/2022071600/613d28a6736caf36b75a02b3/html5/thumbnails/84.jpg)
Example: The Divisor Lattice.In D80 we have the interval [2, 40]:
t2@@
@@
��
��t t10 4
��
��
@@
@@
��
��t t20 8�
���
@@
@@t40
Note that this interval looks like D18.
![Page 85: M¶bius Functions of Posets I: Introduction to Partially Ordered Sets](https://reader035.vdocuments.mx/reader035/viewer/2022071600/613d28a6736caf36b75a02b3/html5/thumbnails/85.jpg)
Example: The Divisor Lattice.In D80 we have the interval [2, 40]:
t2@@
@@
��
��t t10 4�
���
@@
@@
��
��t t20 8
��
��
@@
@@t40
Note that this interval looks like D18.
![Page 86: M¶bius Functions of Posets I: Introduction to Partially Ordered Sets](https://reader035.vdocuments.mx/reader035/viewer/2022071600/613d28a6736caf36b75a02b3/html5/thumbnails/86.jpg)
Example: The Divisor Lattice.In D80 we have the interval [2, 40]:
t2@@
@@
��
��t t10 4�
���
@@
@@
��
��t t20 8�
���
@@
@@t40
Note that this interval looks like D18.
![Page 87: M¶bius Functions of Posets I: Introduction to Partially Ordered Sets](https://reader035.vdocuments.mx/reader035/viewer/2022071600/613d28a6736caf36b75a02b3/html5/thumbnails/87.jpg)
Example: The Divisor Lattice.In D80 we have the interval [2, 40]:
t2@@
@@
��
��t t10 4�
���
@@
@@
��
��t t20 8�
���
@@
@@t40
Note that this interval looks like D18.
![Page 88: M¶bius Functions of Posets I: Introduction to Partially Ordered Sets](https://reader035.vdocuments.mx/reader035/viewer/2022071600/613d28a6736caf36b75a02b3/html5/thumbnails/88.jpg)
If P is a poset then x , y ∈ P have a greatest lower bound ormeet if there is an element x ∧ y in P such that
1. x ∧ y ≤ x and x ∧ y ≤ y ,
2. if z ≤ x and z ≤ y then z ≤ x ∧ y .
Also x , y ∈ P have a least upper bound or join if there is anelement x ∨ y in P such that
1. x ∨ y ≥ x and x ∨ y ≥ y ,
2. if z ≥ x and z ≥ y then z ≥ x ∧ y .
We say P is a lattice if every x , y ∈ P have both a meet and ajoin.Example.
1. Cn is a lattice with i ∧ j = min{i , j} and i ∨ j = max{i , j}.2. Bn is a lattice with S ∧ T = S ∩ T and S ∨ T = S ∪ T .
3. Dn is a lattice with c ∧d = gcd{c, d} and c ∨d = lcm{c, d}.
![Page 89: M¶bius Functions of Posets I: Introduction to Partially Ordered Sets](https://reader035.vdocuments.mx/reader035/viewer/2022071600/613d28a6736caf36b75a02b3/html5/thumbnails/89.jpg)
If P is a poset then x , y ∈ P have a greatest lower bound ormeet if there is an element x ∧ y in P such that
1. x ∧ y ≤ x and x ∧ y ≤ y ,
2. if z ≤ x and z ≤ y then z ≤ x ∧ y .
Also x , y ∈ P have a least upper bound or join if there is anelement x ∨ y in P such that
1. x ∨ y ≥ x and x ∨ y ≥ y ,
2. if z ≥ x and z ≥ y then z ≥ x ∧ y .
We say P is a lattice if every x , y ∈ P have both a meet and ajoin.Example.
1. Cn is a lattice with i ∧ j = min{i , j} and i ∨ j = max{i , j}.2. Bn is a lattice with S ∧ T = S ∩ T and S ∨ T = S ∪ T .
3. Dn is a lattice with c ∧d = gcd{c, d} and c ∨d = lcm{c, d}.
![Page 90: M¶bius Functions of Posets I: Introduction to Partially Ordered Sets](https://reader035.vdocuments.mx/reader035/viewer/2022071600/613d28a6736caf36b75a02b3/html5/thumbnails/90.jpg)
If P is a poset then x , y ∈ P have a greatest lower bound ormeet if there is an element x ∧ y in P such that
1. x ∧ y ≤ x and x ∧ y ≤ y ,
2. if z ≤ x and z ≤ y then z ≤ x ∧ y .
Also x , y ∈ P have a least upper bound or join if there is anelement x ∨ y in P such that
1. x ∨ y ≥ x and x ∨ y ≥ y ,
2. if z ≥ x and z ≥ y then z ≥ x ∧ y .
We say P is a lattice if every x , y ∈ P have both a meet and ajoin.Example.
1. Cn is a lattice with i ∧ j = min{i , j} and i ∨ j = max{i , j}.2. Bn is a lattice with S ∧ T = S ∩ T and S ∨ T = S ∪ T .
3. Dn is a lattice with c ∧d = gcd{c, d} and c ∨d = lcm{c, d}.
![Page 91: M¶bius Functions of Posets I: Introduction to Partially Ordered Sets](https://reader035.vdocuments.mx/reader035/viewer/2022071600/613d28a6736caf36b75a02b3/html5/thumbnails/91.jpg)
If P is a poset then x , y ∈ P have a greatest lower bound ormeet if there is an element x ∧ y in P such that
1. x ∧ y ≤ x and x ∧ y ≤ y ,
2. if z ≤ x and z ≤ y then z ≤ x ∧ y .
Also x , y ∈ P have a least upper bound or join if there is anelement x ∨ y in P such that
1. x ∨ y ≥ x and x ∨ y ≥ y ,
2. if z ≥ x and z ≥ y then z ≥ x ∧ y .
We say P is a lattice if every x , y ∈ P have both a meet and ajoin.Example.
1. Cn is a lattice with i ∧ j = min{i , j} and i ∨ j = max{i , j}.2. Bn is a lattice with S ∧ T = S ∩ T and S ∨ T = S ∪ T .
3. Dn is a lattice with c ∧d = gcd{c, d} and c ∨d = lcm{c, d}.
![Page 92: M¶bius Functions of Posets I: Introduction to Partially Ordered Sets](https://reader035.vdocuments.mx/reader035/viewer/2022071600/613d28a6736caf36b75a02b3/html5/thumbnails/92.jpg)
If P is a poset then x , y ∈ P have a greatest lower bound ormeet if there is an element x ∧ y in P such that
1. x ∧ y ≤ x and x ∧ y ≤ y ,
2. if z ≤ x and z ≤ y then z ≤ x ∧ y .
Also x , y ∈ P have a least upper bound or join if there is anelement x ∨ y in P such that
1. x ∨ y ≥ x and x ∨ y ≥ y ,
2. if z ≥ x and z ≥ y then z ≥ x ∧ y .
We say P is a lattice if every x , y ∈ P have both a meet and ajoin.Example.
1. Cn is a lattice with i ∧ j = min{i , j} and i ∨ j = max{i , j}.2. Bn is a lattice with S ∧ T = S ∩ T and S ∨ T = S ∪ T .
3. Dn is a lattice with c ∧d = gcd{c, d} and c ∨d = lcm{c, d}.
![Page 93: M¶bius Functions of Posets I: Introduction to Partially Ordered Sets](https://reader035.vdocuments.mx/reader035/viewer/2022071600/613d28a6736caf36b75a02b3/html5/thumbnails/93.jpg)
If P is a poset then x , y ∈ P have a greatest lower bound ormeet if there is an element x ∧ y in P such that
1. x ∧ y ≤ x and x ∧ y ≤ y ,
2. if z ≤ x and z ≤ y then z ≤ x ∧ y .
Also x , y ∈ P have a least upper bound or join if there is anelement x ∨ y in P such that
1. x ∨ y ≥ x and x ∨ y ≥ y ,
2. if z ≥ x and z ≥ y then z ≥ x ∧ y .
We say P is a lattice if every x , y ∈ P have both a meet and ajoin.Example.
1. Cn is a lattice with i ∧ j = min{i , j} and i ∨ j = max{i , j}.2. Bn is a lattice with S ∧ T = S ∩ T and S ∨ T = S ∪ T .
3. Dn is a lattice with c ∧d = gcd{c, d} and c ∨d = lcm{c, d}.
![Page 94: M¶bius Functions of Posets I: Introduction to Partially Ordered Sets](https://reader035.vdocuments.mx/reader035/viewer/2022071600/613d28a6736caf36b75a02b3/html5/thumbnails/94.jpg)
If P is a poset then x , y ∈ P have a greatest lower bound ormeet if there is an element x ∧ y in P such that
1. x ∧ y ≤ x and x ∧ y ≤ y ,
2. if z ≤ x and z ≤ y then z ≤ x ∧ y .
Also x , y ∈ P have a least upper bound or join if there is anelement x ∨ y in P such that
1. x ∨ y ≥ x and x ∨ y ≥ y ,
2. if z ≥ x and z ≥ y then z ≥ x ∧ y .
We say P is a lattice if every x , y ∈ P have both a meet and ajoin.
Example.
1. Cn is a lattice with i ∧ j = min{i , j} and i ∨ j = max{i , j}.2. Bn is a lattice with S ∧ T = S ∩ T and S ∨ T = S ∪ T .
3. Dn is a lattice with c ∧d = gcd{c, d} and c ∨d = lcm{c, d}.
![Page 95: M¶bius Functions of Posets I: Introduction to Partially Ordered Sets](https://reader035.vdocuments.mx/reader035/viewer/2022071600/613d28a6736caf36b75a02b3/html5/thumbnails/95.jpg)
If P is a poset then x , y ∈ P have a greatest lower bound ormeet if there is an element x ∧ y in P such that
1. x ∧ y ≤ x and x ∧ y ≤ y ,
2. if z ≤ x and z ≤ y then z ≤ x ∧ y .
Also x , y ∈ P have a least upper bound or join if there is anelement x ∨ y in P such that
1. x ∨ y ≥ x and x ∨ y ≥ y ,
2. if z ≥ x and z ≥ y then z ≥ x ∧ y .
We say P is a lattice if every x , y ∈ P have both a meet and ajoin.Example.
1. Cn is a lattice with i ∧ j = min{i , j} and i ∨ j = max{i , j}.
2. Bn is a lattice with S ∧ T = S ∩ T and S ∨ T = S ∪ T .
3. Dn is a lattice with c ∧d = gcd{c, d} and c ∨d = lcm{c, d}.
![Page 96: M¶bius Functions of Posets I: Introduction to Partially Ordered Sets](https://reader035.vdocuments.mx/reader035/viewer/2022071600/613d28a6736caf36b75a02b3/html5/thumbnails/96.jpg)
If P is a poset then x , y ∈ P have a greatest lower bound ormeet if there is an element x ∧ y in P such that
1. x ∧ y ≤ x and x ∧ y ≤ y ,
2. if z ≤ x and z ≤ y then z ≤ x ∧ y .
Also x , y ∈ P have a least upper bound or join if there is anelement x ∨ y in P such that
1. x ∨ y ≥ x and x ∨ y ≥ y ,
2. if z ≥ x and z ≥ y then z ≥ x ∧ y .
We say P is a lattice if every x , y ∈ P have both a meet and ajoin.Example.
1. Cn is a lattice with i ∧ j = min{i , j} and i ∨ j = max{i , j}.2. Bn is a lattice with S ∧ T = S ∩ T and S ∨ T = S ∪ T .
3. Dn is a lattice with c ∧d = gcd{c, d} and c ∨d = lcm{c, d}.
![Page 97: M¶bius Functions of Posets I: Introduction to Partially Ordered Sets](https://reader035.vdocuments.mx/reader035/viewer/2022071600/613d28a6736caf36b75a02b3/html5/thumbnails/97.jpg)
If P is a poset then x , y ∈ P have a greatest lower bound ormeet if there is an element x ∧ y in P such that
1. x ∧ y ≤ x and x ∧ y ≤ y ,
2. if z ≤ x and z ≤ y then z ≤ x ∧ y .
Also x , y ∈ P have a least upper bound or join if there is anelement x ∨ y in P such that
1. x ∨ y ≥ x and x ∨ y ≥ y ,
2. if z ≥ x and z ≥ y then z ≥ x ∧ y .
We say P is a lattice if every x , y ∈ P have both a meet and ajoin.Example.
1. Cn is a lattice with i ∧ j = min{i , j} and i ∨ j = max{i , j}.2. Bn is a lattice with S ∧ T = S ∩ T and S ∨ T = S ∪ T .
3. Dn is a lattice with c ∧d = gcd{c, d} and c ∨d = lcm{c, d}.
![Page 98: M¶bius Functions of Posets I: Introduction to Partially Ordered Sets](https://reader035.vdocuments.mx/reader035/viewer/2022071600/613d28a6736caf36b75a02b3/html5/thumbnails/98.jpg)
Outline
Motivating Examples
Poset Basics
Isomorphism and Products
![Page 99: M¶bius Functions of Posets I: Introduction to Partially Ordered Sets](https://reader035.vdocuments.mx/reader035/viewer/2022071600/613d28a6736caf36b75a02b3/html5/thumbnails/99.jpg)
For posets P and Q, an order preserving map is f : P → Q with
x ≤P y =⇒ f (x) ≤Q f (y).
An isomorphism is a bijection f : P → Q such that both f andf−1 are order preserving. In this case P and Q are isomorphic,written P ∼= Q.
PropositionIf i ≤ j in Cn then [i , j] ∼= Cj−i .If S ⊆ T in Bn then [S, T ] ∼= B|T−S|.If c|d in Dn then [c, d ] ∼= Dd/c .
Proof for Cn. Define f : [i , j] → Cj−i by f (k) = k − i . Then f isorder preserving since
k ≤ l =⇒ k − i ≤ l − i =⇒ f (k) ≤ f (l).
Also f is bijective with inverse f−1(k) = k + i . It is easy to checkthat f−1 is order preserving.Exercise. Prove the other two parts of the Proposition.
![Page 100: M¶bius Functions of Posets I: Introduction to Partially Ordered Sets](https://reader035.vdocuments.mx/reader035/viewer/2022071600/613d28a6736caf36b75a02b3/html5/thumbnails/100.jpg)
For posets P and Q, an order preserving map is f : P → Q with
x ≤P y =⇒ f (x) ≤Q f (y).
An isomorphism is a bijection f : P → Q such that both f andf−1 are order preserving. In this case P and Q are isomorphic,written P ∼= Q.
PropositionIf i ≤ j in Cn then [i , j] ∼= Cj−i .If S ⊆ T in Bn then [S, T ] ∼= B|T−S|.If c|d in Dn then [c, d ] ∼= Dd/c .
Proof for Cn. Define f : [i , j] → Cj−i by f (k) = k − i . Then f isorder preserving since
k ≤ l =⇒ k − i ≤ l − i =⇒ f (k) ≤ f (l).
Also f is bijective with inverse f−1(k) = k + i . It is easy to checkthat f−1 is order preserving.Exercise. Prove the other two parts of the Proposition.
![Page 101: M¶bius Functions of Posets I: Introduction to Partially Ordered Sets](https://reader035.vdocuments.mx/reader035/viewer/2022071600/613d28a6736caf36b75a02b3/html5/thumbnails/101.jpg)
For posets P and Q, an order preserving map is f : P → Q with
x ≤P y =⇒ f (x) ≤Q f (y).
An isomorphism is a bijection f : P → Q such that both f andf−1 are order preserving. In this case P and Q are isomorphic,written P ∼= Q.
PropositionIf i ≤ j in Cn then [i , j] ∼= Cj−i .
If S ⊆ T in Bn then [S, T ] ∼= B|T−S|.If c|d in Dn then [c, d ] ∼= Dd/c .
Proof for Cn. Define f : [i , j] → Cj−i by f (k) = k − i . Then f isorder preserving since
k ≤ l =⇒ k − i ≤ l − i =⇒ f (k) ≤ f (l).
Also f is bijective with inverse f−1(k) = k + i . It is easy to checkthat f−1 is order preserving.Exercise. Prove the other two parts of the Proposition.
![Page 102: M¶bius Functions of Posets I: Introduction to Partially Ordered Sets](https://reader035.vdocuments.mx/reader035/viewer/2022071600/613d28a6736caf36b75a02b3/html5/thumbnails/102.jpg)
For posets P and Q, an order preserving map is f : P → Q with
x ≤P y =⇒ f (x) ≤Q f (y).
An isomorphism is a bijection f : P → Q such that both f andf−1 are order preserving. In this case P and Q are isomorphic,written P ∼= Q.
PropositionIf i ≤ j in Cn then [i , j] ∼= Cj−i .If S ⊆ T in Bn then [S, T ] ∼= B|T−S|.
If c|d in Dn then [c, d ] ∼= Dd/c .
Proof for Cn. Define f : [i , j] → Cj−i by f (k) = k − i . Then f isorder preserving since
k ≤ l =⇒ k − i ≤ l − i =⇒ f (k) ≤ f (l).
Also f is bijective with inverse f−1(k) = k + i . It is easy to checkthat f−1 is order preserving.Exercise. Prove the other two parts of the Proposition.
![Page 103: M¶bius Functions of Posets I: Introduction to Partially Ordered Sets](https://reader035.vdocuments.mx/reader035/viewer/2022071600/613d28a6736caf36b75a02b3/html5/thumbnails/103.jpg)
For posets P and Q, an order preserving map is f : P → Q with
x ≤P y =⇒ f (x) ≤Q f (y).
An isomorphism is a bijection f : P → Q such that both f andf−1 are order preserving. In this case P and Q are isomorphic,written P ∼= Q.
PropositionIf i ≤ j in Cn then [i , j] ∼= Cj−i .If S ⊆ T in Bn then [S, T ] ∼= B|T−S|.If c|d in Dn then [c, d ] ∼= Dd/c .
Proof for Cn. Define f : [i , j] → Cj−i by f (k) = k − i . Then f isorder preserving since
k ≤ l =⇒ k − i ≤ l − i =⇒ f (k) ≤ f (l).
Also f is bijective with inverse f−1(k) = k + i . It is easy to checkthat f−1 is order preserving.Exercise. Prove the other two parts of the Proposition.
![Page 104: M¶bius Functions of Posets I: Introduction to Partially Ordered Sets](https://reader035.vdocuments.mx/reader035/viewer/2022071600/613d28a6736caf36b75a02b3/html5/thumbnails/104.jpg)
For posets P and Q, an order preserving map is f : P → Q with
x ≤P y =⇒ f (x) ≤Q f (y).
An isomorphism is a bijection f : P → Q such that both f andf−1 are order preserving. In this case P and Q are isomorphic,written P ∼= Q.
PropositionIf i ≤ j in Cn then [i , j] ∼= Cj−i .If S ⊆ T in Bn then [S, T ] ∼= B|T−S|.If c|d in Dn then [c, d ] ∼= Dd/c .
Proof for Cn. Define f : [i , j] → Cj−i by f (k) = k − i .
Then f isorder preserving since
k ≤ l =⇒ k − i ≤ l − i =⇒ f (k) ≤ f (l).
Also f is bijective with inverse f−1(k) = k + i . It is easy to checkthat f−1 is order preserving.Exercise. Prove the other two parts of the Proposition.
![Page 105: M¶bius Functions of Posets I: Introduction to Partially Ordered Sets](https://reader035.vdocuments.mx/reader035/viewer/2022071600/613d28a6736caf36b75a02b3/html5/thumbnails/105.jpg)
For posets P and Q, an order preserving map is f : P → Q with
x ≤P y =⇒ f (x) ≤Q f (y).
An isomorphism is a bijection f : P → Q such that both f andf−1 are order preserving. In this case P and Q are isomorphic,written P ∼= Q.
PropositionIf i ≤ j in Cn then [i , j] ∼= Cj−i .If S ⊆ T in Bn then [S, T ] ∼= B|T−S|.If c|d in Dn then [c, d ] ∼= Dd/c .
Proof for Cn. Define f : [i , j] → Cj−i by f (k) = k − i . Then f isorder preserving since
k ≤ l =⇒ k − i ≤ l − i =⇒ f (k) ≤ f (l).
Also f is bijective with inverse f−1(k) = k + i . It is easy to checkthat f−1 is order preserving.Exercise. Prove the other two parts of the Proposition.
![Page 106: M¶bius Functions of Posets I: Introduction to Partially Ordered Sets](https://reader035.vdocuments.mx/reader035/viewer/2022071600/613d28a6736caf36b75a02b3/html5/thumbnails/106.jpg)
For posets P and Q, an order preserving map is f : P → Q with
x ≤P y =⇒ f (x) ≤Q f (y).
An isomorphism is a bijection f : P → Q such that both f andf−1 are order preserving. In this case P and Q are isomorphic,written P ∼= Q.
PropositionIf i ≤ j in Cn then [i , j] ∼= Cj−i .If S ⊆ T in Bn then [S, T ] ∼= B|T−S|.If c|d in Dn then [c, d ] ∼= Dd/c .
Proof for Cn. Define f : [i , j] → Cj−i by f (k) = k − i . Then f isorder preserving since
k ≤ l =⇒ k − i ≤ l − i =⇒ f (k) ≤ f (l).
Also f is bijective with inverse f−1(k) = k + i .
It is easy to checkthat f−1 is order preserving.Exercise. Prove the other two parts of the Proposition.
![Page 107: M¶bius Functions of Posets I: Introduction to Partially Ordered Sets](https://reader035.vdocuments.mx/reader035/viewer/2022071600/613d28a6736caf36b75a02b3/html5/thumbnails/107.jpg)
For posets P and Q, an order preserving map is f : P → Q with
x ≤P y =⇒ f (x) ≤Q f (y).
An isomorphism is a bijection f : P → Q such that both f andf−1 are order preserving. In this case P and Q are isomorphic,written P ∼= Q.
PropositionIf i ≤ j in Cn then [i , j] ∼= Cj−i .If S ⊆ T in Bn then [S, T ] ∼= B|T−S|.If c|d in Dn then [c, d ] ∼= Dd/c .
Proof for Cn. Define f : [i , j] → Cj−i by f (k) = k − i . Then f isorder preserving since
k ≤ l =⇒ k − i ≤ l − i =⇒ f (k) ≤ f (l).
Also f is bijective with inverse f−1(k) = k + i . It is easy to checkthat f−1 is order preserving.
Exercise. Prove the other two parts of the Proposition.
![Page 108: M¶bius Functions of Posets I: Introduction to Partially Ordered Sets](https://reader035.vdocuments.mx/reader035/viewer/2022071600/613d28a6736caf36b75a02b3/html5/thumbnails/108.jpg)
For posets P and Q, an order preserving map is f : P → Q with
x ≤P y =⇒ f (x) ≤Q f (y).
An isomorphism is a bijection f : P → Q such that both f andf−1 are order preserving. In this case P and Q are isomorphic,written P ∼= Q.
PropositionIf i ≤ j in Cn then [i , j] ∼= Cj−i .If S ⊆ T in Bn then [S, T ] ∼= B|T−S|.If c|d in Dn then [c, d ] ∼= Dd/c .
Proof for Cn. Define f : [i , j] → Cj−i by f (k) = k − i . Then f isorder preserving since
k ≤ l =⇒ k − i ≤ l − i =⇒ f (k) ≤ f (l).
Also f is bijective with inverse f−1(k) = k + i . It is easy to checkthat f−1 is order preserving.Exercise. Prove the other two parts of the Proposition.
![Page 109: M¶bius Functions of Posets I: Introduction to Partially Ordered Sets](https://reader035.vdocuments.mx/reader035/viewer/2022071600/613d28a6736caf36b75a02b3/html5/thumbnails/109.jpg)
If P and Q are posets, then their product is
P ×Q = {(a, x) : a ∈ P, x ∈ Q}
partially ordered by
(a, x) ≤P×Q (b, y) ⇐⇒ a ≤P b and x ≤Q y .
Example.
tt
a
b
×
ttt
x
y
z
=
t(a, x)@@
@@
��
��t t(b, x) (a, y)�
���
@@
@@
��
��t t(b, y) (a, z)�
���
@@
@@t(b, z)
∼= D18
![Page 110: M¶bius Functions of Posets I: Introduction to Partially Ordered Sets](https://reader035.vdocuments.mx/reader035/viewer/2022071600/613d28a6736caf36b75a02b3/html5/thumbnails/110.jpg)
If P and Q are posets, then their product is
P ×Q = {(a, x) : a ∈ P, x ∈ Q}
partially ordered by
(a, x) ≤P×Q (b, y) ⇐⇒ a ≤P b and x ≤Q y .
Example.
tt
a
b
×
ttt
x
y
z
=
t(a, x)@@
@@
��
��t t(b, x) (a, y)�
���
@@
@@
��
��t t(b, y) (a, z)�
���
@@
@@t(b, z)
∼= D18
![Page 111: M¶bius Functions of Posets I: Introduction to Partially Ordered Sets](https://reader035.vdocuments.mx/reader035/viewer/2022071600/613d28a6736caf36b75a02b3/html5/thumbnails/111.jpg)
If P and Q are posets, then their product is
P ×Q = {(a, x) : a ∈ P, x ∈ Q}
partially ordered by
(a, x) ≤P×Q (b, y) ⇐⇒ a ≤P b and x ≤Q y .
Example.
tt
a
b
×
ttt
x
y
z
=
t(a, x)@@
@@
��
��t t(b, x) (a, y)�
���
@@
@@
��
��t t(b, y) (a, z)�
���
@@
@@t(b, z)
∼= D18
![Page 112: M¶bius Functions of Posets I: Introduction to Partially Ordered Sets](https://reader035.vdocuments.mx/reader035/viewer/2022071600/613d28a6736caf36b75a02b3/html5/thumbnails/112.jpg)
If P and Q are posets, then their product is
P ×Q = {(a, x) : a ∈ P, x ∈ Q}
partially ordered by
(a, x) ≤P×Q (b, y) ⇐⇒ a ≤P b and x ≤Q y .
Example.
tt
a
b
×
ttt
x
y
z
=
t(a, x)@@
@@
��
��t t(b, x) (a, y)�
���
@@
@@
��
��t t(b, y) (a, z)�
���
@@
@@t(b, z)
∼= D18
![Page 113: M¶bius Functions of Posets I: Introduction to Partially Ordered Sets](https://reader035.vdocuments.mx/reader035/viewer/2022071600/613d28a6736caf36b75a02b3/html5/thumbnails/113.jpg)
If P is a poset then let Pn =
n︷ ︸︸ ︷P × · · · × P.
PropositionFor the Boolean algebra: Bn
∼= (C1)n.
If the prime factorization of n is n = pm11 · · ·pmk
k , then for thedivisor lattice: Dn
∼= Cm1 × · · · × Cmk .
Proof for Bn. Since C1 = {0, 1}, we define a mapf : Bn → (C1)
n by
f (S) = (b1, b2, . . . , bn) where bi =
{1 if i ∈ S,0 if i 6∈ S.
for 1 ≤ i ≤ n. To show f is order preserving supposef (S) = (b1, . . . , bn) and f (T ) = (c1, . . . , cn). Now S ≤ T in Bn
means S ⊆ T . Equivalently, i ∈ S implies i ∈ T for every1 ≤ i ≤ n. So for each 1 ≤ i ≤ n we have bi ≤ ci in C1. Butthen (b1, . . . , bn) ≤ (c1, . . . , cn) in (C1)
n, i.e. f (S) ≤ f (T ).Constructing f−1 is done in the obvious way. The proof that f−1
is order preserving is just the proof for f read backwards.Exercise. Prove the statement for Dn.
![Page 114: M¶bius Functions of Posets I: Introduction to Partially Ordered Sets](https://reader035.vdocuments.mx/reader035/viewer/2022071600/613d28a6736caf36b75a02b3/html5/thumbnails/114.jpg)
If P is a poset then let Pn =
n︷ ︸︸ ︷P × · · · × P.
PropositionFor the Boolean algebra: Bn
∼= (C1)n.
If the prime factorization of n is n = pm11 · · ·pmk
k , then for thedivisor lattice: Dn
∼= Cm1 × · · · × Cmk .
Proof for Bn. Since C1 = {0, 1}, we define a mapf : Bn → (C1)
n by
f (S) = (b1, b2, . . . , bn) where bi =
{1 if i ∈ S,0 if i 6∈ S.
for 1 ≤ i ≤ n. To show f is order preserving supposef (S) = (b1, . . . , bn) and f (T ) = (c1, . . . , cn). Now S ≤ T in Bn
means S ⊆ T . Equivalently, i ∈ S implies i ∈ T for every1 ≤ i ≤ n. So for each 1 ≤ i ≤ n we have bi ≤ ci in C1. Butthen (b1, . . . , bn) ≤ (c1, . . . , cn) in (C1)
n, i.e. f (S) ≤ f (T ).Constructing f−1 is done in the obvious way. The proof that f−1
is order preserving is just the proof for f read backwards.Exercise. Prove the statement for Dn.
![Page 115: M¶bius Functions of Posets I: Introduction to Partially Ordered Sets](https://reader035.vdocuments.mx/reader035/viewer/2022071600/613d28a6736caf36b75a02b3/html5/thumbnails/115.jpg)
If P is a poset then let Pn =
n︷ ︸︸ ︷P × · · · × P.
PropositionFor the Boolean algebra: Bn
∼= (C1)n.
If the prime factorization of n is n = pm11 · · ·pmk
k , then for thedivisor lattice: Dn
∼= Cm1 × · · · × Cmk .
Proof for Bn. Since C1 = {0, 1}, we define a mapf : Bn → (C1)
n by
f (S) = (b1, b2, . . . , bn) where bi =
{1 if i ∈ S,0 if i 6∈ S.
for 1 ≤ i ≤ n. To show f is order preserving supposef (S) = (b1, . . . , bn) and f (T ) = (c1, . . . , cn). Now S ≤ T in Bn
means S ⊆ T . Equivalently, i ∈ S implies i ∈ T for every1 ≤ i ≤ n. So for each 1 ≤ i ≤ n we have bi ≤ ci in C1. Butthen (b1, . . . , bn) ≤ (c1, . . . , cn) in (C1)
n, i.e. f (S) ≤ f (T ).Constructing f−1 is done in the obvious way. The proof that f−1
is order preserving is just the proof for f read backwards.Exercise. Prove the statement for Dn.
![Page 116: M¶bius Functions of Posets I: Introduction to Partially Ordered Sets](https://reader035.vdocuments.mx/reader035/viewer/2022071600/613d28a6736caf36b75a02b3/html5/thumbnails/116.jpg)
If P is a poset then let Pn =
n︷ ︸︸ ︷P × · · · × P.
PropositionFor the Boolean algebra: Bn
∼= (C1)n.
If the prime factorization of n is n = pm11 · · ·pmk
k , then for thedivisor lattice: Dn
∼= Cm1 × · · · × Cmk .
Proof for Bn.
Since C1 = {0, 1}, we define a mapf : Bn → (C1)
n by
f (S) = (b1, b2, . . . , bn) where bi =
{1 if i ∈ S,0 if i 6∈ S.
for 1 ≤ i ≤ n. To show f is order preserving supposef (S) = (b1, . . . , bn) and f (T ) = (c1, . . . , cn). Now S ≤ T in Bn
means S ⊆ T . Equivalently, i ∈ S implies i ∈ T for every1 ≤ i ≤ n. So for each 1 ≤ i ≤ n we have bi ≤ ci in C1. Butthen (b1, . . . , bn) ≤ (c1, . . . , cn) in (C1)
n, i.e. f (S) ≤ f (T ).Constructing f−1 is done in the obvious way. The proof that f−1
is order preserving is just the proof for f read backwards.Exercise. Prove the statement for Dn.
![Page 117: M¶bius Functions of Posets I: Introduction to Partially Ordered Sets](https://reader035.vdocuments.mx/reader035/viewer/2022071600/613d28a6736caf36b75a02b3/html5/thumbnails/117.jpg)
If P is a poset then let Pn =
n︷ ︸︸ ︷P × · · · × P.
PropositionFor the Boolean algebra: Bn
∼= (C1)n.
If the prime factorization of n is n = pm11 · · ·pmk
k , then for thedivisor lattice: Dn
∼= Cm1 × · · · × Cmk .
Proof for Bn. Since C1 = {0, 1}, we define a mapf : Bn → (C1)
n by
f (S) = (b1, b2, . . . , bn) where bi =
{1 if i ∈ S,0 if i 6∈ S.
for 1 ≤ i ≤ n.
To show f is order preserving supposef (S) = (b1, . . . , bn) and f (T ) = (c1, . . . , cn). Now S ≤ T in Bn
means S ⊆ T . Equivalently, i ∈ S implies i ∈ T for every1 ≤ i ≤ n. So for each 1 ≤ i ≤ n we have bi ≤ ci in C1. Butthen (b1, . . . , bn) ≤ (c1, . . . , cn) in (C1)
n, i.e. f (S) ≤ f (T ).Constructing f−1 is done in the obvious way. The proof that f−1
is order preserving is just the proof for f read backwards.Exercise. Prove the statement for Dn.
![Page 118: M¶bius Functions of Posets I: Introduction to Partially Ordered Sets](https://reader035.vdocuments.mx/reader035/viewer/2022071600/613d28a6736caf36b75a02b3/html5/thumbnails/118.jpg)
If P is a poset then let Pn =
n︷ ︸︸ ︷P × · · · × P.
PropositionFor the Boolean algebra: Bn
∼= (C1)n.
If the prime factorization of n is n = pm11 · · ·pmk
k , then for thedivisor lattice: Dn
∼= Cm1 × · · · × Cmk .
Proof for Bn. Since C1 = {0, 1}, we define a mapf : Bn → (C1)
n by
f (S) = (b1, b2, . . . , bn) where bi =
{1 if i ∈ S,0 if i 6∈ S.
for 1 ≤ i ≤ n. To show f is order preserving supposef (S) = (b1, . . . , bn) and f (T ) = (c1, . . . , cn).
Now S ≤ T in Bn
means S ⊆ T . Equivalently, i ∈ S implies i ∈ T for every1 ≤ i ≤ n. So for each 1 ≤ i ≤ n we have bi ≤ ci in C1. Butthen (b1, . . . , bn) ≤ (c1, . . . , cn) in (C1)
n, i.e. f (S) ≤ f (T ).Constructing f−1 is done in the obvious way. The proof that f−1
is order preserving is just the proof for f read backwards.Exercise. Prove the statement for Dn.
![Page 119: M¶bius Functions of Posets I: Introduction to Partially Ordered Sets](https://reader035.vdocuments.mx/reader035/viewer/2022071600/613d28a6736caf36b75a02b3/html5/thumbnails/119.jpg)
If P is a poset then let Pn =
n︷ ︸︸ ︷P × · · · × P.
PropositionFor the Boolean algebra: Bn
∼= (C1)n.
If the prime factorization of n is n = pm11 · · ·pmk
k , then for thedivisor lattice: Dn
∼= Cm1 × · · · × Cmk .
Proof for Bn. Since C1 = {0, 1}, we define a mapf : Bn → (C1)
n by
f (S) = (b1, b2, . . . , bn) where bi =
{1 if i ∈ S,0 if i 6∈ S.
for 1 ≤ i ≤ n. To show f is order preserving supposef (S) = (b1, . . . , bn) and f (T ) = (c1, . . . , cn). Now S ≤ T in Bn
means S ⊆ T .
Equivalently, i ∈ S implies i ∈ T for every1 ≤ i ≤ n. So for each 1 ≤ i ≤ n we have bi ≤ ci in C1. Butthen (b1, . . . , bn) ≤ (c1, . . . , cn) in (C1)
n, i.e. f (S) ≤ f (T ).Constructing f−1 is done in the obvious way. The proof that f−1
is order preserving is just the proof for f read backwards.Exercise. Prove the statement for Dn.
![Page 120: M¶bius Functions of Posets I: Introduction to Partially Ordered Sets](https://reader035.vdocuments.mx/reader035/viewer/2022071600/613d28a6736caf36b75a02b3/html5/thumbnails/120.jpg)
If P is a poset then let Pn =
n︷ ︸︸ ︷P × · · · × P.
PropositionFor the Boolean algebra: Bn
∼= (C1)n.
If the prime factorization of n is n = pm11 · · ·pmk
k , then for thedivisor lattice: Dn
∼= Cm1 × · · · × Cmk .
Proof for Bn. Since C1 = {0, 1}, we define a mapf : Bn → (C1)
n by
f (S) = (b1, b2, . . . , bn) where bi =
{1 if i ∈ S,0 if i 6∈ S.
for 1 ≤ i ≤ n. To show f is order preserving supposef (S) = (b1, . . . , bn) and f (T ) = (c1, . . . , cn). Now S ≤ T in Bn
means S ⊆ T . Equivalently, i ∈ S implies i ∈ T for every1 ≤ i ≤ n.
So for each 1 ≤ i ≤ n we have bi ≤ ci in C1. Butthen (b1, . . . , bn) ≤ (c1, . . . , cn) in (C1)
n, i.e. f (S) ≤ f (T ).Constructing f−1 is done in the obvious way. The proof that f−1
is order preserving is just the proof for f read backwards.Exercise. Prove the statement for Dn.
![Page 121: M¶bius Functions of Posets I: Introduction to Partially Ordered Sets](https://reader035.vdocuments.mx/reader035/viewer/2022071600/613d28a6736caf36b75a02b3/html5/thumbnails/121.jpg)
If P is a poset then let Pn =
n︷ ︸︸ ︷P × · · · × P.
PropositionFor the Boolean algebra: Bn
∼= (C1)n.
If the prime factorization of n is n = pm11 · · ·pmk
k , then for thedivisor lattice: Dn
∼= Cm1 × · · · × Cmk .
Proof for Bn. Since C1 = {0, 1}, we define a mapf : Bn → (C1)
n by
f (S) = (b1, b2, . . . , bn) where bi =
{1 if i ∈ S,0 if i 6∈ S.
for 1 ≤ i ≤ n. To show f is order preserving supposef (S) = (b1, . . . , bn) and f (T ) = (c1, . . . , cn). Now S ≤ T in Bn
means S ⊆ T . Equivalently, i ∈ S implies i ∈ T for every1 ≤ i ≤ n. So for each 1 ≤ i ≤ n we have bi ≤ ci in C1.
Butthen (b1, . . . , bn) ≤ (c1, . . . , cn) in (C1)
n, i.e. f (S) ≤ f (T ).Constructing f−1 is done in the obvious way. The proof that f−1
is order preserving is just the proof for f read backwards.Exercise. Prove the statement for Dn.
![Page 122: M¶bius Functions of Posets I: Introduction to Partially Ordered Sets](https://reader035.vdocuments.mx/reader035/viewer/2022071600/613d28a6736caf36b75a02b3/html5/thumbnails/122.jpg)
If P is a poset then let Pn =
n︷ ︸︸ ︷P × · · · × P.
PropositionFor the Boolean algebra: Bn
∼= (C1)n.
If the prime factorization of n is n = pm11 · · ·pmk
k , then for thedivisor lattice: Dn
∼= Cm1 × · · · × Cmk .
Proof for Bn. Since C1 = {0, 1}, we define a mapf : Bn → (C1)
n by
f (S) = (b1, b2, . . . , bn) where bi =
{1 if i ∈ S,0 if i 6∈ S.
for 1 ≤ i ≤ n. To show f is order preserving supposef (S) = (b1, . . . , bn) and f (T ) = (c1, . . . , cn). Now S ≤ T in Bn
means S ⊆ T . Equivalently, i ∈ S implies i ∈ T for every1 ≤ i ≤ n. So for each 1 ≤ i ≤ n we have bi ≤ ci in C1. Butthen (b1, . . . , bn) ≤ (c1, . . . , cn) in (C1)
n, i.e. f (S) ≤ f (T ).
Constructing f−1 is done in the obvious way. The proof that f−1
is order preserving is just the proof for f read backwards.Exercise. Prove the statement for Dn.
![Page 123: M¶bius Functions of Posets I: Introduction to Partially Ordered Sets](https://reader035.vdocuments.mx/reader035/viewer/2022071600/613d28a6736caf36b75a02b3/html5/thumbnails/123.jpg)
If P is a poset then let Pn =
n︷ ︸︸ ︷P × · · · × P.
PropositionFor the Boolean algebra: Bn
∼= (C1)n.
If the prime factorization of n is n = pm11 · · ·pmk
k , then for thedivisor lattice: Dn
∼= Cm1 × · · · × Cmk .
Proof for Bn. Since C1 = {0, 1}, we define a mapf : Bn → (C1)
n by
f (S) = (b1, b2, . . . , bn) where bi =
{1 if i ∈ S,0 if i 6∈ S.
for 1 ≤ i ≤ n. To show f is order preserving supposef (S) = (b1, . . . , bn) and f (T ) = (c1, . . . , cn). Now S ≤ T in Bn
means S ⊆ T . Equivalently, i ∈ S implies i ∈ T for every1 ≤ i ≤ n. So for each 1 ≤ i ≤ n we have bi ≤ ci in C1. Butthen (b1, . . . , bn) ≤ (c1, . . . , cn) in (C1)
n, i.e. f (S) ≤ f (T ).Constructing f−1 is done in the obvious way. The proof that f−1
is order preserving is just the proof for f read backwards.
Exercise. Prove the statement for Dn.
![Page 124: M¶bius Functions of Posets I: Introduction to Partially Ordered Sets](https://reader035.vdocuments.mx/reader035/viewer/2022071600/613d28a6736caf36b75a02b3/html5/thumbnails/124.jpg)
If P is a poset then let Pn =
n︷ ︸︸ ︷P × · · · × P.
PropositionFor the Boolean algebra: Bn
∼= (C1)n.
If the prime factorization of n is n = pm11 · · ·pmk
k , then for thedivisor lattice: Dn
∼= Cm1 × · · · × Cmk .
Proof for Bn. Since C1 = {0, 1}, we define a mapf : Bn → (C1)
n by
f (S) = (b1, b2, . . . , bn) where bi =
{1 if i ∈ S,0 if i 6∈ S.
for 1 ≤ i ≤ n. To show f is order preserving supposef (S) = (b1, . . . , bn) and f (T ) = (c1, . . . , cn). Now S ≤ T in Bn
means S ⊆ T . Equivalently, i ∈ S implies i ∈ T for every1 ≤ i ≤ n. So for each 1 ≤ i ≤ n we have bi ≤ ci in C1. Butthen (b1, . . . , bn) ≤ (c1, . . . , cn) in (C1)
n, i.e. f (S) ≤ f (T ).Constructing f−1 is done in the obvious way. The proof that f−1
is order preserving is just the proof for f read backwards.Exercise. Prove the statement for Dn.