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