Prove that union of two equivalences relations is also equivalence relation if and only if that union equals to thier compositions

Question

Let R$_1$, $R_2$ be two equivalence relations on X. Prove that R$_1$ $\bigcup$ R$_2$ is an equivalence relation if and only if R$_1$ $\bigcup$ R$_2$ = R$_1$ $\circ$ R$_2$.

Answer 1

Hints:

1. We always have $R_1\cup R_2\subseteq R_1\circ R_2$.

   $R_1=R_1\circ\mathrm{Id} \subseteq R_1\circ R_2$, similarly $R_2\subseteq R_1\circ R_2$.

2. Any equivalence relation that contains $R_1\cup R_2\,$ must contain $R_1\circ R_2$ as well.

   Because of transitivity: suppose $R_1\cup R_2\subseteq E$, and let $a\, R_1\circ R_2\, c$, then there's a $b$ in between, and $a\, E\, b\, E\, c$ hence $a\, E\, c$ holds.

3. If $R_1\cup R_2=R_1\circ R_2$ then also $R_2\circ R_1=(R_1\circ R_2)^{op}=R_1\cup R_2$.
   You can apply it to show that $R_1\cup R_2$ is transitive.

   $R_1\circ R_2\circ R_1\circ R_2\ =\ R_1\circ R_1\circ R_2\circ R_2\ \subseteq\ R_1\circ R_2$.