# Equivalence Principle: A hint on how to start

dpa
Equivalence Principle: A hint on how to start!

Hi, I have no idea where to start.

1. Statement Problem
Let X be a non empty set with a equivalence relation ~ on it. Prove that for all x,y$\in$X,
[x]=[y] if and only if x~y.

## Homework Equations

For the Equivalence Relation to exist, it must be transitive, reflexive and symmetric.

## The Attempt at a Solution

I have no idea where to start. May be,
~ exists means that, x=y. But is self evident.
How do I prove the "only If" part as well?

Thank You.

voko

What do the square brackets mean in [x] = [y]?

dpa

Hi voko,

I did not understand that either. That is the exact statement in the Homework question. I assumed it simply meant x=y.
Can you help me if it is x=y?

Thank You.

voko

This cannot be proven for simple equality, because I can easily give you a counter-example.

Can the square brackets mean the "equivalence class" of x?

dpa

Yes, that is it. Still, I have no idea how to prove equivalent classes as equal.

dpa

I will try it now.
Thank You.
dpa

voko

Start by proving the "if" part. Use the definition of the equivalence class and the properties of equivalence.

dpa

Here is my solution,

From the definition of equivalence class, we suppose any z such that,
[x]={z$\in$Xlx~z} holds. - - - - - - - - - - - - - - - - -(i)

for every such z,
since,
x~y and x~z=z~x [property of symmetry]
we write from law of transitivity,
z~y exists.
Thus,
we can now define,
[y]={z$\in$Xly~z}, which is true.
can also be written as
[y]={z$\in$Xlx~z} - - - - - - - - - - - - - - - - - - - - - - - - -(ii)
Thus from i and ii, we can write that
[x]=[y]

Now we show that it holds true only when x~y by method of contradiction (how?).
Suppose there exists w such that x~w, but x~y is false,
but what next?

Do I say since x~y does not exist,
we cannot write,
[y]={z$\in$Xly~z} exists for every
[x]={z$\in$Xlx~z}
??
Any more hints?

dpa

I got it. I have to prove if a=>b and then if b=>a.
Is first part correct then?

voko

The first part seems OK. The second part, prove by contradiction. Suppose there are x and y such that [x] = [y], but x not ~y.

dpa

is it better now?

To Prove [x]=[y] iff x~y
Here, First we take x~y and prove [x]=[y]. Then we take x is not ~y but [x]=[y] and arrive at x~y.

Firstly,
For any element zEX and zE[x] we know there exists a relation z~x
Sincem z~x and x~y, we can write from law of transitivity,
z~y which implies zE[y].

Thus we can write from zE[x] and zE[y] that [x]=[y].

Next, we assume x is not ~y but [x]=[y] exists.
Here, [x]=[y] implies there exists an element z where zE[x] and z~x.
There also exists zE[y] for which z~y exists.
Thus from law of transitivity, and from z~y and z~x, we can write,
x~y.

Thus Prooved.

voko

In the second part of the proof, instead of saying "there also exists zE[y]" you should say "because [x] = [y], zE[y]".

Otherwise the proof is good.

dpa

Someone said zE[x] and zE[y] cannot imply [x]=[y]. It is indeed true if equivalent relation does not exist.
What happens when the equivalence relation x~y does exist as in above example. So, is the first part of proof correct in the light of this comment?
I am totally new and just had one/two class on this topic.
Thank You.

voko

The first part of the proof is correct. You take ANY element from [x] and show that it also exists in [y]. You could equally take ANY element for [y] and likewise show it is in [x]. So each contains all the elements of the other.