How Does Equivalence Class Equality Work in Set Theory?

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SUMMARY

The discussion centers on proving that if \( b \) is in the equivalence class of \( a \) under an equivalence relation \( R \) on set \( A \), then the equivalence classes of \( a \) and \( b \) are equal, denoted as \( [[a]] = [[b]] \). Participants clarify that to establish this equality, one must demonstrate that any member of one equivalence class is also a member of the other. Key points include the definitions of reflexivity, symmetry, and transitivity as they pertain to equivalence relations, and the importance of correctly identifying elements versus sets in the proof process.

PREREQUISITES
  • Understanding of equivalence relations, including reflexivity, symmetry, and transitivity.
  • Familiarity with set notation and the concept of equivalence classes.
  • Basic proof techniques in mathematics, particularly set theory.
  • Ability to differentiate between elements of a set and subsets.
NEXT STEPS
  • Study the properties of equivalence relations in detail, focusing on reflexivity, symmetry, and transitivity.
  • Learn how to construct proofs involving set equality and membership.
  • Explore examples of equivalence classes in various mathematical contexts.
  • Review common pitfalls in mathematical proofs, particularly in set theory and relations.
USEFUL FOR

Students of mathematics, particularly those studying set theory and equivalence relations, as well as educators looking for insights into common misunderstandings in proofs involving equivalence classes.

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Homework Statement



Prove the following statement:
Let R be an equivalence relation on set A. If b is in the equivalence class of a, denoted [[a]] then [[a]]=[].

Homework Equations



[[a]], [[a]]=[]; definition of equivalence: a relation R on a set A that is reflexive, symmetric and transitive is an equivalence relation.

The Attempt at a Solution



Consider an element b in set {x in S| x R a} denoted by the equivalence relation [[a]]. If b is in this set, it is an element of the equivalence class. It follows that if b is an element of [[a]], then it must be an equivalence relation and is reflexive such that [[a]] R b. It is also symmetric, such that [[a]] R b = b R [[a]], and transitive such that for any c in [[a]], if [[a]] R b and b R c then [[a]] R c. Thus, b is also an equivalence relation and the set {x in S|x R b} may be denoted by []. Hence, [[a]] = [].

I'm not sure if I was on track here or not. I feel as if I went in a circle without actually proving anything. I'm not sure exactely what else to do, but I thought maybe to show they are equal, I have to express the sets as equal somehow. Or would I go about it by saying that [[a]] = [] if a = b? Any critique of this proof is highly welcome -- I'm pretty new at this whole thing. :redface:
 
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You can't write [[a]]Rb. The left hand side is an equivalence class, and the RHS is an element. ?R? only makes sense if both sides are in S. You also at one point say that b is an element of S and b is an equivalence relation.

To show that []=[[a]] or indeed any set eqaulity all you need to do is demonstrate that if x is [] it is in [[a]], and if y is in [[a]] it is in []. These are immediate from the definitions, and the hypothesis that b is in [[a]]. For we want to take x in [], i.e. xRb, and show that this implies x is in [[a]], i.e. that xRa. So, if xRb, and bRa, then does xRa?
 
quasar_4 said:

Homework Statement



Prove the following statement:
Let R be an equivalence relation on set A. If b is in the equivalence class of a, denoted [[a]] then [[a]]=[].

Homework Equations



[[a]], [[a]]=[]; definition of equivalence: a relation R on a set A that is reflexive, symmetric and transitive is an equivalence relation.

The Attempt at a Solution



Consider an element b in set {x in S| x R a} denoted by the equivalence relation [[a]]. If b is in this set, it is an element of the equivalence class. It follows that if b is an element of [[a]], then it must be an equivalence relation and is reflexive such that [[a]] R b.

You mean a R b.

It is also symmetric, such that [[a]] R b = b R [[a]], and transitive such that for any c in [[a]], if [[a]] R b and b R c then [[a]] R c.
Again, no. [[a]] is a subset of A, not a member. You mean: if a R b and b R c, then a R c.

Thus, b is also an equivalence relation and the set {x in S|x R b} may be denoted by []. Hence, [[a]] = [].

No, b is definitely NOT an equivalence relation! b is a member of set A and an equivalence relation is a subset of Rx R.

[/quote]I'm not sure if I was on track here or not. I feel as if I went in a circle without actually proving anything. I'm not sure exactely what else to do, but I thought maybe to show they are equal, I have to express the sets as equal somehow. Or would I go about it by saying that [[a]] = [] if a = b? Any critique of this proof is highly welcome -- I'm pretty new at this whole thing. :redface:[/QUOTE]

You want to show that [[a]]= [] which means you want to show two sets are equal. You do that by showing any member of one is a member of the other. If c is in [[a]], then c R a. Since you are told that b is in [[a]], b R a also. What does that tell you about c and b? Going the other way, while you were told that b is in [[a]], you are NOT told directly that a is in []. You will have to prove that.
 

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