Equivalence relations (CSCI 2824, Spring 2015)

Revision

Equivalence Relation

Equivalence Classes

Example-1

Modular-Congruences

In this lecture, we will revise some of the concepts on relations that we covered previously.

Reflexive, Symmetric and Transitive Relations.
Partial and Total Orders.

Then we will look into equivalence relations and equivalence classes. We will see how an equivalence on a set partitions the set into equivalence classes.

Let us make sure we understand key concepts before we move on. To do so, take five minutes to solve the following problems on your own. You are welcome to discuss your solutions with me after class.

All these problems concern a set .
Relation $R_1 = { (1,1), (2,2), (3,3), (4,4) }$ .
Relation $R_2 = { (1,1), (2,2), (2,3), (3,2) }$ .
Relation $R_3 = { (3,2), (2,3) }$ .
Relation $R_4 = { (1,1), (1,2), (2,2), (1,3), (3,3), (1,4), (4,4) }$ .
Relation $R_5 = { (1,2), (2,3), (1,3), (2,2) }$ .
Relation $R_6 = { }$ (empty relation).

An equivalence relation over a set is a reflexive, symmetric and transitive relation.

Question

Which of the following are examples of equivalence relations over A = {1,2,3,4} .

$R_1 = { }$ .
$R_2 = {(1,1), (1,2), (2,1), (2,2), (3,3), (4,4) }$ .
$R_3 = { (1,1), (1,2), (2,1), (2,2), (3,1), (3,2), (1,3), (2,3), (3,3), (4,4) }$ .

Just check that the relations above are reflexive, symmetric and transitive.

Answer

fails to be reflexive.
is an equivalence relation.
is also an equivalence relation.

The relation R = { (n,n) | n in mathbb{N} } is an equivalence relation.

The Cartesian product of any set with itself is a relation R = A times A . All possible tuples exist in . This relation is also an equivalence.

For any number , we have an equivalence relation $R_p = { (m,n) | m - n mod p = 0 }$ .

Often we denote (m,n) in R_p by the notation m equiv n mod p (read as and are congruent modulo ).

Verify that R_p is an equivalence for any .

Reflexivity:

For all n in N , we have that n - n mod p = 0 . Therefore (n,n) in R_p .

Symmetry:

Clearly if m - n mod p = 0 then n-m mod p = 0 .

Transitivity:

If m- n mod p = 0 and n - k mod p = 0 then displaystyle{ (m - k) mod p = ((m-n) + (n-k)) mod p = (m-n) mod p + (n-k) mod p = 0 } .

Reflexive?

Symmetric?

Transitive?

Irreflexive?

Anti-Symmetric?

We now look at how equivalence relation on partitions the original set .

Let us take the set A = {1,2,3,4} . And the equivalence $R_1 = { (1,1), (1,2), (2,1), (2,2), (3,3), (4,4) }$ .

Let us collect everything that is equivalent to. This gives us the set $S_1 = {1,2}$ . Collecting everything equivalent to again gives us S_1 . Collecting everything that is equivalent to gives us {3} and similarly for , we get {4} .

Therefore, R_1 is said to induce the following partition of the set :

{ {1,2}, {3}, {4} }

Question

What is the partition induced by the equivalence: $R_2 = { (1,1), (1,2), (2,1), (2,2), (3,1), (3,2), (1,3), (2,3), (3,3), (4,4) }$ ?

To answer this, consider:

$[1]_{R_2} = { 1, 2, 3 }$
.

The partition induced by R_2 is therefore: { { 1,2,3}, {4} } .

Notation

The equivalence class of an element a in A under an equivalence relation is denoted as [a]_R .

$[a]_R = { b | (a,b) in R$

Claim-1

If (a,b) in R then [a]_R = [b]_R .

Proof

Let (a,b) in R . We show that [a]_R subseteq [b]_R and vice versa, [b]_R subseteq [a]_R .

To show that [a]_R subseteq [b]_R , let x in [a]_R . We have (x,a) in R . Since (a,b) in R , we conclude by transitivity of that (x,b) in R . Therefore x in [b]_R .

The proof of [b]_R subseteq [a]_R is very similar. Therefore [a]_R = [b]_R represent the same equivalence classes.

We now show that two equivalence classes are either the same or disjoint.

Claim-2

Whenever [a]_R cap [b]_R not= emptyset then [a]_R = [b]_R .

{Proof} Let c in [a]_R cap [b]_R . We have (c,a) in R and (c,b) in R . Therefore by symmetry and transitivity of , we conclude that (a,b) in R . Therefore [a]_R = [b]_R .

This shows that given any set , and equivalence relation , the equivalence classes

${ [a]_R | a in A }$

is a partition of the set :

If , we conclude that .
$bigcup_{ain A} [a]_R = A$ .