Let $R$ be an equivalence relation on a set $A$, $a,b \in A$. Prove $[a] = [b]$ iff $aRb$.

Question

Hello I need help with the proof strategy for this problem.

Let $R$ be an equivalence relation on a set $A$ and let $a,b \in A$. Prove that $[a] = [b]$ if and only if $aRb$.

Answer 1

So we want to prove that $[a] = [b] \Rightarrow aRb$, and $aRb \Rightarrow [a] = [b]$.

For the first part, $[a] = \{x \mathop \lvert aRx\}$, and by reflexivity, $a \in [a]$. But since $[a] = [b]$, $a \in[b]$. And so by definition, $bRa$, and by symmetry $aRb$.

For the second part, you need to show that $aRb$ implies that $[a] = [b]$, and I think the best way to do that is to take an arbitrary element $x \in [a]$, and use the definition of $[a]$ and transitivity to show that $x \in [b]$, and vice versa, thus showing that the two sets are equal.

Answer 2

Suppose $[a] = [b]$. By reflexivity, $b \in [b]$ so $b \in [a]$, which implies $a Rb$. That's one implication. To show the converse, assume $aRb$ and then show that $[a] \subset [b]$ and $[b] \subset [a]$. So let $x \in [a]$. Then.....