Proved that if $R\cup E$ equivalence relation so $a / E \subseteq a / R$ OR $a / R \subseteq a / E$

Question

Let R, S be equivalence relations on A. Proved that if $R\cup E$ Is an equivalence relation on A So $\Rightarrow$ For all $a\in A$ , $$a / E \subseteq a / R$$ OR $$a / R \subseteq a / E$$

I thought about it this way:

$R\cup E$ Is an equivalence relation and R, S equivalence relations. On $R\cup E$, The only problem comes when talking about transitive. Because Symmetrical and reflexivity is always saved.

So I looked in the transitive relation: If exist $$<a,b> \in R\cup E$$ AND $$<b,c> \in R\cup E$$ So must be $$<a,c> \in R\cup E$$

How I show From here that must be: For all $a\in A$ , $$a / E \subseteq a / R$$ OR $$a / R \subseteq a / E$$

Thank you!

Answer 1

If $S:=R\cup E$ is an equivalence relation on $A$ then $a/S=a/R\cup a/E$ for every $a\in A$.

Fix one $a\in A$ and assume that $p\in a/R-a/E$ and $q\in a/E-a/R$.

Note that $p/S=a/S=q/S$ and $a/R=p/R$ and $a/E=q/E$.

Then $p\in a/S=q/S=q/R\cup q/E$.

Also $p\notin a/E=q/E$ so we conclude that $p\in q/R$.

However, then $q/R=p/R=a/R$ contradicting $q\notin a/R$.

So apparantly it cannot happen that the sets $a/R-a/E$ and $a/E-a/R$ are both non-empty.

Final conclusion: $a/R\subseteq a/E$ or $a/E\subseteq a/R$.