Equivalnce relation proof

[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.

 

[blue_raver22]

First, let's review the definition of an equivalence relation. An equivalence relation is a relation on a set that is reflexive, symmetric, and transitive. This means that for any elements a, b, and c in the set, the following properties hold:

1. Reflexivity: a ~ a (every element is related to itself)
2. Symmetry: if a ~ b, then b ~ a (the relation is bidirectional)
3. Transitivity: if a ~ b and b ~ c, then a ~ c (the relation is transitive)

Now, to prove the statement given, we need to show that if a ~ b, then [a] = , and vice versa.

Assuming a ~ b, we know that this relation satisfies the properties of an equivalence relation. This means that [a] and are both equivalence classes under the relation ~. By definition, an equivalence class is a set of elements that are related to each other by the given relation. In this case, [a] and contain all the elements that are related to a and b, respectively.

If a ~ b, then we know that a and b are related, and therefore, they must be in the same equivalence class. This means that [a] = , since both sets contain the same elements.

Conversely, if [a] = , then we know that both sets contain the same elements. This implies that a and b are related under the relation ~, since they are both in the same equivalence class. Therefore, a ~ b.

In conclusion, we have shown that if a ~ b, then [a] = , and vice versa. This proves the statement given in the problem.