Equal equivalence classes proof

Question

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 $x\,R\,y$?

Answer 1

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 $(a,b)\in R$.

Answer 2

Let $b \in [[x]]_R = [[y]]_R$. Then $b R x$ and $b R y$. By symmetry and transitivity $x R y$.