How Can Equivalence Relations Determine Equal Classes?

[Tokenfreak]

I am trying to prove this as I am practicing for a test but I am pretty much clueless on this problem:

Prove that if ~ is an equivalence relation on a set s and [a] denotes the equivalence class of a in s under ~, then a ~ b if and only if [a] = .


If anyone can give me some points on how to approach or start this problem it would be great. Thanks.

 

[SteveL27]

I am trying to prove this as I am practicing for a test but I am pretty much clueless on this problem:

Prove that if ~ is an equivalence relation on a set s and [a] denotes the equivalence class of a in s under ~, then a ~ b if and only if [a] = .


If anyone can give me some points on how to approach or start this problem it would be great. Thanks.

It falls directly out of the definition of equivalence relation, so it's tricky to think of a hint. But what happens if [a] \neq ? Then there must either be an element in ____ that's not in ______ or vice versa. Then what?

 

[Tokenfreak]

So would it be safe to assume that an equivalence relation ~ on a set s is a relation satisfying a,b in s. If [a] != , then there must be an element in a that's not in b or vice versa. Therefore, this contradicts that [a] = .Is this close? I am pretty much clueless on this proof.

 

[SteveL27]

So would it be safe to assume that an equivalence relation ~ on a set s is a relation satisfying a,b in s.

No. You need to go back to your book and read what an equivalence relation is.

Aren't a and b assumed to be elements of s? So a,b in s is true of all a and b in s. Has nothing to do with equivalence relations.

If [a] != , then there must be an element in a that's not in b or vice versa. Therefore, this contradicts that [a] = .

Well yes, if you assume [a] != then that contradicts [a] = . Isn't that always the case no matter what [a] and are?

I think you need to read your text and/or class notes to understand what an equivalence relation is.