Show that $s_1(\text{equivalence class}) = s_2(\text{equivalence class})$ iff $s_1\mathrel{R}s_2$.

Question

Let $R$ be an equivalence relation on $S$. Show that for all $s_1, s_2$ elements of $S$ we have $s_1(\text{equivalence class}) = s_2(\text{equivalence class})$ iff $s_1\mathrel{R}s_2$.

I understand that this is a iff proof, so we have to show both ways
$s_1(\text{equivalence class}) = s_2(\text{equivalence class})\implies s_1\mathrel{R}s_2$ and
$s_1\mathrel{R}s_2$ $\implies$ $s_1(\text{equivalence class}) = s_2(\text{equivalence class})$.

But I'm having trouble finding the connections in each case.

Answer 1

Questions like this can be tricky if you're not used to "definition chasing." I'll demonstrate how to organize this type of proof. What you write as $s_1(\text{equivalence class})$ is generally written $[s_1]$. I'll write it that way in this answer.

Def. If $S$ is a set, $\text{R}$ is a relation on $S$, and $x\in S$, the equivalence class of $s$ under $\text{R}$, denoted $[s]$ (or $[s]_\text{R}$, when unclear), is the set of elements related to $s$ under $\text{R}$. In other words, $$[s]_\text{R}:=\{x\in S:s\text{ R }x\}.$$

Assume that $[s_1]=[s_2]$. You know from the inclusion $[s_1]\subseteq [s_2]$ that $s_2\text{ R }s_1$, and you know from $[s_1]\supseteq [s_2]$ that $s_1\text{ R }s_2$.

Conversely, suppose that $s_1\text{ R }s_2$. By definition, for every $x\in [s_1]$, $s_1\text{ R }x$. Since $\text{R}$ is an equivalence relation, it is symmetric, so $x\text{ R }s_1$. We also have that $\text{R}$ is transitive, so $x\text{ R }s_1$ and $s_1\text{ R }s_2$ implies that $x\text{ R }s_2$. Therefore, $s_2\text{ R }x$, so $x\in [s_2]$. From this, we have that $[s_1]\subset [s_2]$. An identical argument shows that $[s_2]\subset [s_1]$, so $[s_1]=[s_2]$.