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I am currently reading about the subject given in the title of this thread. The definition they give for equivalence classes in my textbook is a rather ostentatious in its wording, so I just want to make certain that I am understanding it properly. They say to let R be an equivalence relation on a set A, meaning that this this particular relation is reflexive, symmetric, and transitive, right? Essentially the rest of it seems to say that you can partition off the elements that make the relation reflexive, thereby creating a subset of the relation R. Does that seem right?

I could really use some help, thank you!

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Reflexive, symmetric and transitive - that is the standard (not to say: only) definition of equivalence relation. An equivalence also defines a partition (and vice versa), but I am not sure what you mean by partitioning off the elements that make the relation refelxive ... –  Hagen von Eitzen Nov 7 '12 at 19:10
    
Think of students in a lecture room and divide them into rows! –  Ronnie Brown Nov 8 '12 at 10:48
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They say to let R be an equivalence relation on a set A, meaning that this this particular relation is reflexive, symmetric, and transitive, right?

Yes, any relation that satisfies these properties is by definition an equivalence relation. It is called an equivalence relation because it satisfies the fundamental properties of what it means for elements in a set to be "equal".

Essentially the rest of it seems to say that you can partition off the elements that make the relation reflexive, thereby creating a subset of the relation R.

The important thing to understand is that it partitions up the set into disjoint (non-overlapping) subsets.

Let $X$ be a set of people standing in a crowded room, and define an equivalence relation $R$ on $X$ by saying that for any two people $x, y \in X$, $xRy$ if and only if $x$ and $y$ have first names beginning with the same letter. Then you divide up all the people in the room to non-overlapping subsets: $X_{a} \subset X$ people with first names beginning with $a$, $X_{b} \subset X$ beginning with $b$ , and so on.

Then you can write $X$ as the union of partitions defined by the relation $R$:

$$X = X_{a} \cup X_{b} \cup \dots \cup X_{y} \cup X_{z}.$$

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And, importantly, the $X_i$ are "pairwise" disjoint: every element of $X$ is in one, and only one $X_i$. –  amWhy Nov 7 '12 at 19:20
    
Oh, so the only thing that I got wrong was by saying that the equivalence class was a subset of the relation, but it is actually a subset of the set that R is a relation on? –  Mack Nov 7 '12 at 19:21
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@Emack: Yes!! note my comment, as well. Every element is in one and only one equivalence class. –  amWhy Nov 7 '12 at 19:22
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Yes. For example, consider the integers under the relation of equality modulo 3. This basically divides the integer set into 3 equivalence classes:

$\{3k, 3k+1, 3k+2\}_{k \in \mathbb{Z}}$

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The point is that equivalence relations are essentially the same as partitions. Indeed, the equivalence classes of an equivalence relation form a partition. Conversely, a partition forms an equivalence relation by relating elements if they lie in the same component of the partition.

For example, "the same parity" is an equivalence relation on the integers, i.e $\rm\:j \equiv k\iff 2\:|\:j-k.\:$ The equivalence classes of evens $\, {\cal E}\ = 2\,\Bbb Z\,$ and odds $\,{\cal O} = 1+2\,\Bbb Z\,$ form a partition ot the integers $\, {\cal E} \cup\, {\cal O} = \Bbb Z\,$ and $\:{\cal E\cap O} = \emptyset.$ Conversely, given the partition into even and odds, we obtain the parity equivalence relation by defining integers to be equivalent if they lie in the same component, i.e. they are both even, or both odd.

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I really liked the example about the even and odd numbers, it really made the conception even more palpable. Thank you! –  Mack Nov 7 '12 at 20:16
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