A proving question?

Question

We assume that Y is a non-empty set and we say that ~ be an equivalence relation on Y. We let W be the set of distinct equivalence classes given by ~. prove that elements of W are disjoint and their union is all of Y.

please help. thank you.

Answer 1

Lolita almost got it right...but she didn't prove what she said she would.

Since an equivalence relation is reflexive, all points of Y are in some equivalence class.

Let's take two equivalence classes, A and B, and assume for contradiction that they are not disjoint. Then A intersect B is nonempty. x belongs to A, and x belongs to B for some x. Then x~a for all a in A, and x~b for all b in B. Since an equivalence relation is symmetric and transitive (we have to flip one to use transitivity), a~b for all a in A and all b in B. But since A is the set of all things equivalent to a, then B is contained in A. Similarly, A is contained in B. Thus, A=B. So there is no element of W which intersects any other element.

This means that the elements of W are disjoint. Since everything in Y is an element of some element of W, the union of all elements in W is Y.

Answer 2

k

take A,B two elements in W,
this means that A, B are two subsets of Y.
we will prove that A intersection B is the empty set. For that we assume the opposite, i.e. that there exists
x in A intersection B.
Then x ~ a, for every a in A
and x~ b, for every a in A,
but then b~x, so we have the following b~x, and x~a , therefore
b~a, but this happens for every a in A and every b in B
In other words, A=B.
So if A int B is not empty then A=B,
therefore if A is not equal to B then A int B = empty set.

the union of all the elements in W is equal to Y, because each element y in Y, has the property that y~y, so, it belongs to some element of W.

be aware, the zex guy is not right