Union of Equivalences

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.