# Equal equivalence classes proof

Let there be two sets $A$ and $B$ and let their Cartesian product be $A{\times}B$. Let there be an equivalence relation $R:R\,{\subset}\,A{\times}B$. Let's define an equivalence class now: ${\lbrack}{\lbrack}a{\rbrack}{\rbrack}_R=\{b\,{\in}\,B:(a,b)\,{\in}\,R\}$

Let $[[x]]_R$ and $[[y]]_R$ be two such equivalence classes.

We know that $[[x]]_R=[[y]]_R$. How to use this to prove that $x\,R\,y$?

• If $R$ is just a "relation" then we don't know it. But if it is an equivalence relation, then we do. – GEdgar Nov 12 '15 at 15:03
• How? Yes, it's supposed to be an equivalence relation. – user285146 Nov 12 '15 at 15:04
• Use the definition of $[[a]]_R$ given, and the definition of "equivalence relation". – GEdgar Nov 12 '15 at 15:10

1. Equivalence relation is a subset of $A\times B$, not an element of it, so $R\notin A\times B$ but $R\subset A\times B$.
2. Equivalence relations are only defined in case $B = A$. An easy reminder is: any point must be equivalent to itself, so $(a,a)\in R$ for all $a\in A$. Of course that implies $R\subset A\times A$ (or at least $A\subset B$).
3. Since $(a,a)\in R$ for any $a$, $a \in [a]$. So if $[a] = [b]$ then $b\in [b]$ implies that $b\in [a]$, hence by definition $(a,b)\in R$.
• What's your source on the equivalence relations being defined only if $A=B$? The source I'm using says nothing about that, it just says that they must be transitive, reflexive and symmetric. – user285146 Nov 12 '15 at 20:29
• Well, they can't be reflexive, aRa, unless a is in both A and B and $R \subset of (A \cap B) x (A \cap B)$. – fleablood Nov 12 '15 at 20:44
• What do you mean by $of(A∩B)x(A∩B)$? – user285146 Nov 13 '15 at 21:41
• Ilya has a point. If $R \subset of A \times B$ then aRa implies $a \in A$ and $a \in B$ so $a \in A \cap B$ so R is a subset of the cross product of the intersections. $R \subset A \cap B \times A \cap B$. – fleablood Nov 13 '15 at 22:18
Let $b \in [[x]]_R = [[y]]_R$. Then $b R x$ and $b R y$. By symmetry and transitivity $x R y$.