1. ## Equivalence classes

If you have a fixed equivalence relation ~ on a set S, then for any a contained in S the set:
[a] = { b in S : b~a } is called the equivalence class of a.

1) every element in S belongs to atleast one equivalence class because the relation is necessarily reflexive.

2)No element can belong to two different equivalence classes.

Thirdly, i get the impression that if u take into account both 1) and 2) this amounts to saying that every element belongs to exactly one equivalence class. If this is the case then what is the use of there being such a thing as equivalence classes if their only members are the elements themselves...i.e. for all a in S, [a] = {a} ?????

2. Also, on the matter of equivalence relations, my textbook states that any partition on a set determines an equivalence relation on it. What exactly does this mean? (I understand what partitions and equivalence relations are but could someone show what they're saying a little more formally?)

3. Originally Posted by Obstacle1
If you have a fixed equivalence relation ~ on a set S, then for any a contained in S the set:
[a] = { b in S : b~a } is called the equivalence class of a.

1) every element in S belongs to atleast one equivalence class because the relation is necessarily reflexive.

2)No element can belong to two different equivalence classes.
This is the fundamental theorem of equivalence classes. It is used all the time in advanced mathematics.

It is a theorem that partitions a non-empty set. Hence it takes all the elements of the set and puts them into disjoint sets.

For example consider:
Z={0,1,-1,2,-2,3,-3,...}

And define x~y if and only if x=y(mod 2).
So pick any number, say 0:
Now find all y so that: 0 = y (mod 2).
Those are: 0 (itself), 2, -2, 4, -4, ....

Now pick any number which is not much those just listed above. Because if you do then the ones that are equivalent to it is going to be the same class of numbers.

Say you pick 1 (for it is not in that set)/
Now find all y so that: 1 = y(mod 2).
Those are: 1(itself), -1,3,-3,.....

As you can see we have completely depleted all the elements of the set. And divided the set into two disjoint sets:
{0,2,-2,...}
{1,-1,3,-3,...}

4. Originally Posted by ThePerfectHacker
This is the fundamental theorem of equivalence classes. It is used all the time in advanced mathematics.

It is a theorem that partitions a non-empty set. Hence it takes all the elements of the set and puts them into disjoint sets.

For example consider:
Z={0,1,-1,2,-2,3,-3,...}

And define x~y if and only if x=y(mod 2).
So pick any number, say 0:
Now find all y so that: 0 = y (mod 2).
Those are: 0 (itself), 2, -2, 4, -4, ....

Now pick any number which is not much those just listed above. Because if you do then the ones that are equivalent to it is going to be the same class of numbers.

Say you pick 1 (for it is not in that set)/
Now find all y so that: 1 = y(mod 2).
Those are: 1(itself), -1,3,-3,.....

As you can see we have completely depleted all the elements of the set. And divided the set into two disjoint sets:
{0,2,-2,...}
{1,-1,3,-3,...}

Thanks. That has made things a lot clearer. Unfortunately i think i was confused just by the explanation i had read in my textbook. This often seems to be the case with me, is there anything besides repeated practice a wise mathematician such as yourself could recommend to me to avoid becoming confused by textbook explanations in the future?

5. Originally Posted by Obstacle1
Also, on the matter of equivalence relations, my textbook states that any partition on a set determines an equivalence relation on it. Could someone show what they're saying a little more formally?
There is a two-way relationship between a partition of a set and an equivalence relation on the same set.
Given a relation ~ on a set S then the collection of equivalence classes determined by ~ forms a partition of S. You did that above.

Now start with a partition $\Pi$ of S. $\Pi$ is a collection of non-empty subsets of S, their union is S. Define a relation on S as followers: $\left( {x,y} \right) \in \rho$ if and only if $x\& y$ belong to the same cell in $\Pi$. It is easy to see that because each $x \in S$ is in some cell of $\Pi$ then $\rho$ is reflexive. And we know that ‘same as’ is symmetric and transitive. There $\rho$ is an equivalence relation.