Prove that the union of two equivalence relations on the same set an equivalence relation iff? Let $R$ and $E$ be equivalence relations on set $A$.
Prove that $R\cup E$ is an equivalence relation on set A  iff for all $a\in A$,  $[a]_{R} \subseteq [a]_{E}$ OR $[a]_{E} \subseteq [a]_{R}$.
Someone can help me here ?
I don't have any idea.
Thanks. 
 A: Suppose $R\cup E$ is an equivalence relation, and suppose $[a]_R\not\subset[a]_E,$ i.e. $\exists b\in A$ such that $(a,b)\in R$ but $(a,b)\not\in E.$ Then $\forall c\in A$ such that $(a, c)\in E,$ as $(c, a), (a, b)\in R\cup E,$ we must have $(c, b)\in R\cup E,$ thus $(c, b)\in R$ (if $(c,b)\in E,$ then $(a,b)\in E$). This shows that $(a,c)\in R.$ Thus $[a]_E\subset[a]_R.$
Conversely suppose that $[a]_E\subseteq[a]_R$ (the other case is similar). Then, if $(a,b), (b, c)\in R\cup E,$ we can suppose without loss of generality that $(a,b)\in R, (b,c)\in E.$ Then by our hypothesis, $(b,c)\in R.$ Thus $(a,c)\in R\subseteq R\cup E.$ This shows that $R\cup E$ is an equivalence relation (the other two axioms always hold).
Hope this helps.
A: Reflexivity and symmetry are clearly inherited from $R$ and $E$.
If $R\cup E$ is transitive, then since $a\sim_{R\cup E}b$ for all $b\in[a]_R$ and $a\sim_{R\cup E}c$ for all $c\in[a]_E$, also $b\sim_{R\cup E}c$ for all such $b,c$, and the inclusion condition follows.
If $R\cup E$ is not transitive, we have $b\sim_{R\cup E}a$ and $a\sim_{R\cup E}c$ but not $b\sim_{R\cup E}c$, which can only occur if one of $b,c$ is only in $[a]_R$ and the other is only in $[a]_E$, violating the inclusion condition.
