If $R \cup S$ is an equivalence relation on $A$, then $R$ and $S$ are equivalence relations on $A$

Question

Let $A$ be a non-empty set, and let $R$ and $S$ be relations on A.

If $R \cup S$ is an equivalence relation on $A$, then $R$ and $S$ are equivalence relations on $A$.

How can i prove it? I think it is not necessarily true but i dont know how to start.

Answer 1

This is not true. Here's an explicit example. Let $A = \{0, 1\}$, $R = \{(0, 0)\}$, $S = \{(1, 1)\}$. Then $R \cup S = \{(0, 0), (1, 1)\}$, which is an equivalence relation (just equality). Neither $R$ nor $S$ are equivalence relations as neither are reflexive.

Answer 2

For a counter example just remove an essential item from R but keep it in S. And maybe (but not necessarily) vice versa

For example if $T= R\cup S = \mathbb Z \times \mathbb Z$ be the equivalence relationship on $\mathbb Z$ that everything is equivalent to everything (can't get more equivalent than that! $a T b$ for all $a, b \in \mathbb Z$. So $T$ is reflexive because $r T r$ for all $r\in \mathbb Z$). And $T$ is symmetric because $r T s$ and $s T r$ for all $r,s \in \mathbb Z$. And $T$ is transitive because $r Ts; sTt; rTt$ for all $r,s,t\in \mathbb Z$.)

Now let $R = \{(5, 7), (7,2), (2,2)\}$ and $S = (\mathbb Z \times \mathbb Z)\setminus R$.

$R$ is not reflexive because if $r \ne 2$ then $(r,r) \not \in \mathbb R$ (although $(r,r) \in S$). And $S$ is not reflexive because $(2,2) \not \in S$.

And $R$ is not symmetric because $(5,7) \in R$ but $(7,5)\not \in R$. $S$ is not symmetric because $(7,5) \in S$ but $(5, 7) \not \in S$.

$R$ is not transitive because $(5, 7), (7,2) \in R$ but $(5,2) \not \in R$. ANd $S$ is not transitive because $(5,2), (2,7) \in S$ but $(5,7) \not \in R$.