# Find the class of equivalence of a element of a given equivalence relation.

Yesterday on my Abstract Algebra course, we were having a problem with equivalence relations. We had a given set: $$A = \{a, b, c\}$$

We found all the partitions of $A$, and one of them was: $$P = \{ \{a\} , \{b, c\} \}$$

Then we built an equivalence relation $S$ from this partition, where two elements are in equivalence relation if $a$ and $b$ belong to the same cell.

So the relation of equivalence is: $$S = \{ (a,a) , (b,b) , (c,c) , (b,c) , (c,b) \}$$

After this the professor, without explaining anything wrote:

The class of equivalence of $(b,c)$: $[(b,c)] = \{ (b,c) , (c,b) \}$

So can anyone explain this last line? Because I don't understand it.

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Wouldn't it make more sense to say that the class of equivalence of $b$, $[b] = \{b,c\}$? The equivalence relation was defined on elements of $A$, not on pairs of elements from $A$. – Andrew Parker Feb 27 '12 at 20:38
well that's all the problem. I'm trying to figure it out and yet no clarity. I thought it like this too, but i don't know. – cprogcr Feb 27 '12 at 20:39

When you have an equivalence relation $R$ on a set $X$, and an element $x\in X$, you can talk about the equivalence class of $x$ (relative to $R$), which is the set $$[x] = \{y\in X\mid (x,y)\in R\} = \{y\in X\mid (y,x)\in R\} = \{y\in X\mid (x,y),(y,x)\in R\}.$$
But I note that your professor did not say "equivalence class", he said "Class of Equivalence". That suggests he may be refering to some other (defined) concept. I would suggest that you talk to your professor directly and ask him what he means by "Class of Equivalence", and whether it is the same thing as "equivalence class"; explain what your understanding of "equivalence class" is, and why you would be confused if he said "The equivalence class of $(b,c)$ is $[(b,c)]={(b,c),(c,b)}$" (at least, I would be confused because in order to talk about "the equivalence class of $(b,c)$", I would need some equivalence relation defined on some set that contains $(b,c)$, and we don't seem to have that on hand).