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

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


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

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


 * Thus, we have:

$$ $$

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