Part 10
Partial Orders
Printed version of the lecture Discrete Mathematics on 30. September 2009
Tommy R. Jensen, Department of Mathematics, KNU
10.1
Contents
1
Relations
1
2
Partially Ordered Sets
2
3
Chains and Antichains
4
4
Dilworth’s Theorem
5
5
Conclusion
6
10.2
1
Relations
Relations
Definition
If X is any set, then the set of ordered pairs from X,
X × X = {(a, b) : a, b ∈ X}
is the set of all pairs (a, b) of elements of X.
Ordered pairs have the property a 6= b ⇒ (a, b) 6= (b, a).
Definition
A relation on X is any subset of X × X.
If the relation on X is R ⊆ X × X, then we write aRb to mean the same as (a, b) ∈ R.
Example
The relation ⊆ on X is defined by
a ⊆ b if and only if a is a subset of b.
It is not usual to write (a, b) ∈⊆ if a ⊆ b (but not wrong).
10.3
1
Properties of relations
Definition
Let R be a relation on a set X.
• R is reflexive if xRx for all x ∈ X.
• R is irreflexive if x6Rx ((x, x) is not in R) for all x ∈ X.
• R is symmetric if xRy implies yRx for all x, y ∈ X.
• R is antisymmetric if xRy implies y6Rx for all x, y ∈ X with x 6= y.
• R is transitive if xRy and yRz imply xRz for all x, y, z ∈ X.
Example. ⊆ is reflexive, antisymmetric and transitive.
The proper subset relation ⊂ is irreflexive, antisymmetric and transitive.
The equality relation = is reflexive, symmetric, antisymmetric and transitive.
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10.4
Partially Ordered Sets
Partial orders
Definition
A partial order on X is a reflexive, antisymmetric and transitive relation on X.
A strict partial order on X is an irreflexive, antisymmetric and transitive relation.
Instead of writing R for a relation which is a partial order, we usually write ≤ if it is not the partial order
⊆.
If x ≤ y and x 6= y, then we write x < y.
We mostly write < for a strict partial order, unless it is the relation ⊂ .
Definition
A pair (X, ≤) where X is a set and ≤ is a partial order on X is called a partially ordered set, or sometimes
poset.
10.5
Total orders
Definition
If R is any relation on a set X then two elements x and y of X are comparable in R if xRy or yRx.
Definition
A partial order ≤ on X is a total order if any two elements of X are comparable.
Theorem 4.5.1
If X is a finite set of size n, then every total order of X is of the form x1 < x2 < · · · < xn , where x1 , x2 , . . . , xn
is some ordering (permutation) of the elements of X.
Definition
A partial order which is a total order is also called a linear order.
10.6
Diagrams of partially ordered sets
Definition
If (X, ≤) is a partially ordered set, and x, y are two elements of X, then x is covered by y, and we write
x <c y, if
• x < y, and
• if x ≤ z ≤ y, then z = x or z = y.
A Hasse diagram of a poset (X, ≤) is a drawing in the plane, with one point px for each element x of
X, such that
if x is covered by y, then py is drawn above px
and the two points px , py are connected by a line.
10.7
2
Examples of Hasse diagrams
The cube as Hasse diagram of the subset relation
10.8
Examples of Hasse diagram
Hasse diagram of a semantic relation
10.9
Extensions
Definition
If R1 and R2 are relations, and if R1 ⊆ R2 , then R2 is called an extension of R1 .
In particular, if ≤1 and ≤2 are partial orders on X, then ≤2 is an extension of ≤1 if
x ≤1 y implies x ≤2 y for all x and y in X.
10.10
Theorem on linear extensions
Definition
A linear extension of a partial order is an extension which is a linear order.
Theorem 4.5.2
Let (X, ≤) be a finite partially ordered set. Then there exists a linear extension of (X, ≤).
Example. For the cube
we have a linear extension 0/ ≤ {x} ≤ {y} ≤ {z} ≤ {x, y} ≤ {x, z} ≤ {y, z} ≤ {x, y, z}.
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10.11
3
Chains and Antichains
Chains and antichains
Definition
Let (X, ≤) be a partially ordered set.
A chain in (X, ≤) is a subset C of X such that
for all x, y in C : x ≤ y or y ≤ x.
So all elements of a chain C are comparable.
An antichain in (X, ≤) is a subset A of X such that
for all x, y in A with x 6= y : x 6≤ y and y 6≤ x.
So no two different elements of an antichain A are comparable.
If C is a chain and A an antichain, then
|A ∩C| ≤ 1.
10.12
Maximal and minimal elements
Definition
A maximal element of a partially ordered set (X, ≤) is an element a such that
a 6< x for all x ∈ X.
A minimal element is an element b such that
x 6< b for all x ∈ X.
The set of maximal elements of a partially ordered set is an antichain.
The set of minimal elements of a partially ordered set is an antichain.
10.13
Partitioning a poset into antichains
Theorem 5.6.1
Let (X, ≤) be a partially ordered set with exactly r elements in its largest chain.
Then X can be partitioned into r antichains, but no fewer.
Example with r = 4.
10.14
Construction of a partition into r antichains
Start with a partially ordered set (X, ≤) in which the size of a longest chain is equal to r.
1. Find the set of all minimal elements, and let it become one of the antichains A of the partition.
2. Remove all the elements of A from the partially ordered set.
3. If there are any elements left, then go to (1). Otherwise stop.
The last element of every longest chain is a minimal element.
Therefore the length of a longest chain always decreases by one each time step (2) gets performed.
After r steps the length of a longest chain becomes zero, and we have all elements of X in the r antichains.
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10.15
4
Dilworth’s Theorem
Dilworth’s theorem
Theorem 5.6.2
Let (X, ≤) be a finite partially ordered set with exactly m elements in its largest antichain.
Then X can be partitioned into m chains, but no fewer.
Example with m = 3.
10.16
Proof of Dilworth’s theorem
Proof of Theorem 5.6.2 by induction on n = |X|
The base case n = 0
This case is trivially true.
The induction step
Assume n > 0, and the statement is true for all partially ordered sets with n − 1 elements.
Let a be a maximal element of (X, ≤).
Let (X 0 , ≤0 ) be the partially ordered set that we obtain by removing a from (X, ≤).
Then X 0 has size n − 1, so the statement of Dilworth’s theorem is true for (X 0 , ≤0 ).
Let k be the size of a largest antichain in (X 0 , ≤0 ).
Then we can find chains C1 ,C2 , . . . ,Ck in (X 0 , ≤0 ) such that every element of X 0 = X \ {a} belongs to
exactly one of them.
10.17
Proof of Dilworth’s theorem
The induction step, continued
For each i = 1, 2, . . . , k let xi be the largest element of Ci that belongs to an antichain Ai of size k in (X 0 , ≤0 ).
Let A = {x1 , x2 , . . . , xk }.
We can show that A is an antichain in (X 0 , ≤0 ).
Take any xi , x j in A with i 6= j, and let y be an element of Ai ∩C j .
Then y is an element of C j that also belongs to the antichain Ai of size k.
We have chosen x j as the largest element of C j that belongs to an antichain of size k.
So it follows that y ≤ x j .
Both y and xi belong to the antichain Ai . This implies y 6≤ xi .
Now y ≤ x j and y 6≤ xi together imply x j 6≤ xi , by transitivity.
Since no two elements of A are comparable, this shows that A is an antichain.
10.18
5
Proof of Dilworth’s theorem
The induction step, continued
We have found an antichain A = {x1 , x2 , . . . , xk } in (X 0 , ≤0 ) consisting of the largest elements of chains
C1 ,C2 , . . . ,Ck that also belong to antichains of size k in (X 0 , ≤0 ).
Case 1. Assume a ≥ xi for some i, where 1 ≤ i ≤ k.
Let C be the chain {a} ∪ {x ∈ Ci : x ≤ xi }.
Let (X”, ≤ ”) be the partially ordered set obtained by removing all of C from (X, ≤).
Then a largest antichain in (X”, ≤ ”) has size k − 1, since every largest antichain in (X 0 , ≤0 ) contains an
element of {x ∈ Ci : x ≤ xi } by the choice of xi .
It follows by induction that (X”, ≤ ”) has a partitioning into k − 1 chains.
When we add the chain C to this partitioning, we get a partitioning of X into k chains.
So the proof is finished in the case a ≥ xi for some i.
10.19
Proof of Dilworth’s theorem
The induction step, final case
Case 2. Assume a 6≥ xi for all i = 1, 2, . . . , k.
We chose a maximal in (X, ≤), so xi 6≥ a is also true for every i = 1, 2, . . . , k.
This implies that A ∪ {a} is an antichain in (X, ≤) of size k + 1.
We partition X into the k + 1 chains C1 ,C2 , . . . ,Ck , {a}.
This finishes the proof also in Case 2.
10.20
5
Conclusion
Conclusion
This ends the lecture!
10.21
Next time:
The Inclusion-Exclusion Principle
6
10.22
7