Equality of Equivalence Classes

[wolfmanzak]

Homework Statement

Suppose is an equivalence relation on a set S. If $$a \sim b$$ for some $$a,b \in S$$,then $$E_{a}=E_{b}$$

Homework Equations

The Attempt at a Solution

Assume $$a \sim b$$ for some $$a,b \in S$$. Pick $$x \in (a,b)$$. For $$a \in S$$ the equivalence class of a can be written as $$\{x \in S | a \sim x\}$$. For $$b \in S$$ the equivalence class of b is the set $$\{x \in S | b \sim x\}$$.

Here is where I am a little stuck, I'm not sure if picking x in (a,b) is even possible or the right way to start this problem. I just don't know how to start the problem, if I know how to start it, I am pretty sure I can use the properties of Equivalence classes/relations and their definitions to show that the equivalence classes are equal but I need a good starting point.

Any and all help is much appreciated. Thanks in advance.

 

[lanedance]

by the transivity of the equivalence relation, you should be able to show any element of one set is in the other

 

[wolfmanzak]

I know it's much simpler than I'm making it. I know that $$a \sim b$$ from the problem statement, what method or steps do I take to go from there to show that $$E_{a}=E_{b}$$?

 

[lanedance]

consider anyelement from Ea, and try and show it is in Eb an vice versa, then you're done

 

[wolfmanzak]

So if I pick $$z \in E_{a}$$ this would mean that because a~z and that b~a(by symmetry of equivalence classes)we would have b~a and a~z and thus b~z(transitivity) and thus z is in the equivalence class of b. Would this be enough to show that the two equivalence classes of a and b respectively were equal? Or would I need to "go the other way" as well?

 

[HallsofIvy]

I would recommend a proof by contraction.

Suppose a~ b but $E_a\ne E_b$. Then either:
1) There exist x in $E_a$ that is not in $E_b$ or
2) There exist x in $E_b$ that is not in [$E_a$.

In other words
1) There exist x that is equivalent to a but not to b or
2) There exist x that is equivalent to b but not to a.

The transitive law shows both of those are impossible.