Introduction to ProofsIntroduction to Proofs Combinatorics Preview May 2014 Lambert Introduction to...

Introduction to Proofs

Combinatorics Preview

May 2014

Lambert Introduction to Proofs

Relations on Sets

Definition

A relation R from a set A to a set B is a subset of A× B. In thespecial case where R is a subset of A× A, we that R is a relationon A.

Example

Let R be the relation “is greater than” on the set A = {1, 2, 3, 4}.Find R.

Example

Let A be the set of all integers, and let R be the relation on Adefined by

R = {(a, b) ∈ A× A|a + b is a multiple of 3}.

Give examples of some elements in R.


Properties of Relations

Definition

A relation R on a set A is reflexive if for all x ∈ A, we find(x , x) ∈ R.

Definition

A relation R on a set A is called irreflexive if for all x ∈ A, we find(x , x) 6∈ R.

Definition

A relation R on a set A is symmetric if (a, b) ∈ R implies(b, a) ∈ R for a, b ∈ A.


More properties of relations

Definition

A relation R on a set A is asymmetric if (a, b) ∈ R implies(b, a) 6∈ R for a, b ∈ A.

Definition

A relation R on a set A is antisymmetric if (a, b) ∈ R and(b, a) ∈ R implies a = b for a, b ∈ A.

Definition

A relation R on a set A is transitive if (a, b) ∈ R and (b, c) ∈ Rimplies (a, c) ∈ R for a, b, c ∈ A.

Definition

A relation R on a set A is negatively transitive if (a, b) 6∈ R and(b, c) 6∈ R implies (a, c) 6∈ R for a, b, c ∈ A.


Examples

Example

Does a relation on a set that is neither reflexive nor irreflexiveexist?

Example

Give an example of a relation on R that is asymmetric andirreflexive. Can we say anything else about the example justprovided.

Example

Give an example of a relation that is transitive and symmetric.


Theorem 5

Theorem

A relation is irreflexive, transitive, and antisymmetric if and only ifit is transitive and asymmetric

Proof.

Suppose that R is a relation on the set A with the relation beingirreflexive, transitive, and antisymmetric. Suppose(a, b), (b, a) ∈ R. Then a = b, but (a, a) 6∈ R since R is irreflexive.Thus, (a, b), (b, a) ∈ R can not both hold, which yields R isasymmetric.Conversely, suppose R is transitive and asymmetric. As a result, Ris vacuously antisymmetric. We also have asymmetry implyingirreflexivity, since (a, a) ∈ R implies that (a, a) ∈ R and (a, a) ∈ R,which can not happen in the case of asymmetry.


Necessity

Question

Is it necessary the R is irreflexive in the previous theorem. In otherwords, does there exist a relation on a set that is antisymmetricand transitive that is not transitive and asymmetric?

Answer

Let R be the relation on Z given by (a, b) ∈ R if a|b. This providesus with an example that irreflexive is necessary in the previoustheorem.


Partially ordered sets

Definition

A partially ordered set (also called poset) is a set S combined witha relation R such that R is reflexive, transitive, and antisymmetric.

Example

Consider the relation R on the set Z given by (a, b) ∈ R if andonly if a|b where a, b ∈ Z.

Example

Let A be a nonempty set. Consider the relation R on the set P(A)given by (X ,Y ) ∈ R if and only if X ⊆ Y for X ,Y ∈ P(A).


Applications

Let A ={Savannah, Atlanta, New York, Chicago, Denver, San Francisco}.Three people in class must rank these cities from 1 to 6 (with 1being the favorite). We shall form a relation R on the set A byplacing (a, b) ∈ R if for all three rankings we find the ranking of ais less than or equal to the ranking of b where a and b are cities inA. Determine if this ranking system forms a poset.


Total Order

Definition

Let R be a relation on a set A. If for any a, b ∈ A, either(a, b) ∈ R or (b, a) ∈ R, then the partial order is called a totalorder, or a linear order.

Example

Consider the relation “is greater than” on the set of realnumbers. This is an example of a total ordering.

Consider the lexicographical ordering on the set of orderedpairs in R2. This is another example of a total ordering.


Chains and Antichains

Definition

If a subset S of a partially ordered set A is totally ordered, then wecall S a chain.

Definition

A subset S of a partially ordered set A is called an antichain if theelements of S are pairwise incomparable.

Example

Consider the relation R on the set A = {1, 2, · · · , 16} formed byhaving (a, b) ∈ R if a divides b. Find examples of chains andantichains.

Example

Use the relation “≤” on the set A = {1, 2, · · · , 10} by comparingonly even numbers together and odd numbers together.


Partitions

Definition

A partition of a set A is a collection S of nonempty subsets of Asuch that every element of A belongs to exactly one subset of S .

Example

Consider the set A = {a, b, c , d , e, f }. Find two disjoint partitionsof A, called S1 and S2, such that |S1| = |S2| = 3 and for everyX ∈ S1 and Y ∈ S2 we find |X | 6= |Y |.


Partitions and Chains

Example

Consider the relation R on the set A = {1, 2, · · · , 16} formed byhaving (a, b) ∈ R if a divides b. Find the minimum number ofchains necessary to form a partition of A.

Example

Consider the relation R on the set A = {1, 2, · · · , 16} formed byhaving (a, b) ∈ R if a divides b. Find the maximum number ofelements in an antichain of A.


Dilworth’s Theorem

Dilworth’s Theorem

Let P be a partially ordered finite set. The minimum number m ofchains that form a partition of P is equal to the maximum numberM of elements in an antichain of P.

The Dual of Dilworth’s Theorem [Mirsky, 1971]

Let P be a partially ordered set. If P possesses no chain of m + 1elements, then P is the union of m antichains.


Proof of the Dual of Dilworth’s Theorem

Proof.

We shall use induction for this proof. Notice for m = 1, we findthere is no chain with more than 1 element. This coincides with anantichain for the whole set P, so our base case holds. Assume form ≥ 2 and assume that the the theorem holds for m− 1. Let P bea partially ordered set that has no chain of m + 1 elements. Let Sbe the set of maximal elements of each chain in P. We claim Sforms an antichain. Now we consider the set P − S . Supposex1 < x2 < · · · < xm were a chain in P − S . Then this would alsobe a maximal chain in P and we would have xm ∈ S , which is acontradiction. Hence P − S has no chain of m elements. Byapplying the induction hypothesis, P − S is the union of m − 1antichains. As a result, (P − S)∪ S is the union of m antichains aswe desired.


Applications of Dilworth’s Theorem

Sequence Size

Let a1, a2, · · · , an2+1 be a permutation of the integers1, 2, · · · , n2 + 1. Then this sequence has a subsequence of lengthn + 1 that is monotone.

Proof.

Assume there does not exist a subsequence of length n + 1 that ismonotone. We shall create a relation R by placing (ai , aj) ∈ R ifai ≤ aj and i ≤ j . Since each chain can have at most n elements(otherwise we would have an increasing sequence of size n + 1 ormore), then the minimum size of a partition must contain at leastn + 1 elements (if there were only n in the partition we would haveonly n2 elements accounted for). Dilworth’s theorem implies thereexists an antichain with n + 1 elements. For an antichain to becreated, notice that ai > aj for i ≤ j . This means we have adecreasing sequence of size n + 1 as we desired.


Application

Counting Subsets

Let the sets Ai , 1 ≤ i ≤ k, be distinct subsets of {1, 2, · · · , n}.Suppose Ai ∩ Aj 6= ∅ for all i and j . Show that k ≤ 2n−1 and givean example where equality holds.


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