# Equivalence Relation, prove dom(R) = range(R) = X

1. Oct 8, 2011

### Ceci020

1. The problem statement, all variables and given/known data
Given:
R is an equivalence relation over a nonempty set X

Prove:
dom(R) = X
and range(R) = X

2. Relevant equations

3. The attempt at a solution
I have the following thoughts:
Since R is an equivalence relation over X by hypothesis, R satisfies:
Reflexivity: <x,x> belongs to R
Symmetry: <x,y> belongs to R, and <y,x> belongs to R
Transitivity: <x,y> belongs to R, <y,z> belongs to R, and <x,z> belongs to R

with x, y, z E X

Base on definition of domain and range of a relation R over a set X, I have:
dom(R) = {x E X : there exists y belongs to Y such that <x,y> E R}
range(R) = {y E Y : there exists x belongs to X such that <x,y> E R}

What I'm confused is that I don't know how to connect my ideas together. The properties that R satisfies is with x, y, and z E X. And R is a subset of X x X. There is no Y whatsoever. So what should I do (or say) next to come to the conclusion?

2. Oct 8, 2011

### Dick

Reflexive is ALL you need. <x,x> is in R for ALL x in X, yes?

3. Oct 9, 2011

### Ceci020

I'm sorry but can you explain a little bit more?

I think that since for all x in X, <x,x> is always in R holds, then according to the conditions of the domain, there indeed exists x E X such that there also exists a Y (namely, y = x) that makes <x,y> E R. But how about the range?

thank you

4. Oct 9, 2011

### Dick

If <a,b> is in R, then 'a' is in the domain and 'b' is in the range, right?