Union of Equivalences/Proof 1

Theorem
The union of two equivalence relations is not necessarily an equivalence relation itself.

Proof
This can be shown by giving an example.

Let $S = \left\{{a, b, c}\right\}$, and let $\mathcal R_1$ and $\mathcal R_2$ be equivalences on $S$ such that:


 * $\left[\!\left[{a}\right]\!\right]_{\mathcal R_1} = \left[\!\left[{b}\right]\!\right]_{\mathcal R_1} = \left\{{a, b}\right\}$
 * $\left[\!\left[{c}\right]\!\right]_{\mathcal R_1} = \left\{{c}\right\}$
 * $\left[\!\left[{a}\right]\!\right]_{\mathcal R_2} = \left\{{a}\right\}$
 * $\left[\!\left[{b}\right]\!\right]_{\mathcal R_2} = \left[\!\left[{c}\right]\!\right]_{\mathcal R_2} = \left\{{b, c}\right\}$

Let $\mathcal R_3 = \mathcal R_1 \cup \mathcal R_2$.

Then:

However:

So $\mathcal R_3$ is not transitive, and therefore $\mathcal R_3 = \mathcal R_1 \cup \mathcal R_2$ is not an equivalence relation.