Equivalence Class holds Equivalent Elements

Theorem
Let $\mathcal R$ be an equivalence relation on a set $S$. Then:


 * $\left({x, y}\right) \in \mathcal R \iff \left[\!\left[{x}\right]\!\right]_\mathcal R = \left[\!\left[{y}\right]\!\right]_\mathcal R$

Proof

 * First we prove that $\left({x, y}\right) \in \mathcal R \implies \left[\!\left[{x}\right]\!\right]_\mathcal R = \left[\!\left[{y}\right]\!\right]_\mathcal R$.

Suppose $\left({x, y}\right) \in \mathcal R: x, y \in S$.

Then:

So $\left[\!\left[{x}\right]\!\right]_\mathcal R \subseteq \left[\!\left[{y}\right]\!\right]_\mathcal R$.

Now:

... so we have shown that $\left({x, y}\right) \in \mathcal R \implies \left[\!\left[{x}\right]\!\right]_\mathcal R = \left[\!\left[{y}\right]\!\right]_\mathcal R$.


 * Next we prove that $\left[\!\left[{x}\right]\!\right]_\mathcal R = \left[\!\left[{y}\right]\!\right]_\mathcal R \implies \left({x, y}\right) \in \mathcal R$:

By definition of set equality, $\left[\!\left[{x}\right]\!\right]_{\mathcal R} = \left[\!\left[{y}\right]\!\right]_\mathcal R$ means $\left({x \in \left[\!\left[{x}\right]\!\right]_\mathcal R \iff x \in \left[\!\left[{y}\right]\!\right]_\mathcal R}\right)$.

So by definition of equivalence class$\left({y, x}\right) \in \mathcal R$.

Hence by definition of equivalence relation: $\mathcal R$ is symmetric, $\left({x, y}\right) \in \mathcal R$.

So we have shown that $\left[\!\left[{x}\right]\!\right]_\mathcal R = \left[\!\left[{y}\right]\!\right]_\mathcal R \implies \left({x, y}\right) \in \mathcal R$.


 * Thus, we have:

So by Material Equivalence, $\left({x, y}\right) \in \mathcal R \iff \left[\!\left[{x}\right]\!\right]_\mathcal R = \left[\!\left[{y}\right]\!\right]_\mathcal R$.