Equivalence Classes

[Harish Kumar]

I am not able to understand clearly the concept of equivalence classes. Is it mandatory that for a particular equivalence relation in Set A, all the equivalence classes are either mutually disjoint or equal? Or are they mutually disjoint for all possible E-relations for a set A(or only on specific E-relation?)?

 

Secondly,

 

In the equivalence relation:

 

Let S be the set of all integers and let n > 1 be a fixed integer. Define for a,b belonging to S, a~b, if a-b is a multiple of n.

 

How is it, in the case above, the equivalence class of a consists of all integers of the form a + kn, where k=0,1,-1,2,-2,3,-3,...;

there are n distinct equivalence classes, namely cl(0), cl(1),...cl(n-1), when clearly cl(2), cl(4) have elements in common?

Edited November 18, 2010 by Harish Kumar

[ajb]

Is it mandatory that for a particular equivalence relation in Set A, all the equivalence classes are either mutually disjoint or equal?

 

This follows from the definition of an equivalence relation. Every element of our set A belongs to one and only one equivalence class specified by an equivalence relation.

[Harish Kumar]

This follows from the definition of an equivalence relation. Every element of our set A belongs to one and only one equivalence class specified by an equivalence relation.

 

 

But then if we create a relation, a~b, a,b belong to Integers, where every member is divisible by 2, aren't those members also present in the relation "divisible by four" ? So how come we say they are mutually disjoint?

[Harish Kumar]

This is an example from I N Herstein's Topics in Algebra:

 

Let S be a set of all Integers. Given a,b belongs to S, define a~b if a-b is an even integer.

 

He goes on to say that in this case, the equivalence class of A consists of all the integers of the form a+2m, where m=0,+-1,+-2,+3...; and in this example there are only two distinct equivalence classes,namely cl(0) and cl(1).

 

Clearly cl(0) and cl(2) will have many elements in common.

 

So is not the "mutually disjoint" condition broken in this case, since the intersection of Sets cl(0) and cl(2) is not a null set?

 

Can you please clarify?

[Harish Kumar]

Nobody in the mathematics department?

[uncool]

This is an example from I N Herstein's Topics in Algebra:

 

Let S be a set of all Integers. Given a,b belongs to S, define a~b if a-b is an even integer.

 

He goes on to say that in this case, the equivalence class of A consists of all the integers of the form a+2m, where m=0,+-1,+-2,+3...; and in this example there are only two distinct equivalence classes,namely cl(0) and cl(1).

 

Clearly cl(0) and cl(2) will have many elements in common.

 

So is not the "mutually disjoint" condition broken in this case, since the intersection of Sets cl(0) and cl(2) is not a null set?

 

Can you please clarify?

They are not mutually disjoint because the two are exactly the same.

 

Let a be in cl(0). We will prove that it is in cl(2):

 

a is in cl(0) because there is an m such that 0 + 2m = a, or a = 2m. Then a - 2 = 2m - 2 = 2(m - 1).

Therefore, a is in cl(2).

 

Similarly, we can prove the converse, so the two are the same.

=Uncool-

[Harish Kumar]

They are not mutually disjoint because the two are exactly the same.

 

Let a be in cl(0). We will prove that it is in cl(2):

 

a is in cl(0) because there is an m such that 0 + 2m = a, or a = 2m. Then a - 2 = 2m - 2 = 2(m - 1).

Therefore, a is in cl(2).

 

Similarly, we can prove the converse, so the two are the same.

=Uncool-

 

Thank you very much uncool. And sorry for the delayed response. I was revisiting my college maths and I was tied up with too much work at the workplace. I appreciate your clarification very much.