$R_1$ and $R_2$ are equivalence relations on $A$. Prove or disprove:
$R_1 \circ R_2$ is also an equivalence relation.
I check on internet what mean the notation $\circ$. Is defined like that:
$$R \circ S = \left\{(x,z) \in A \times A \mid \exists y \in A: xRy \wedge ySz\right\}$$
For equivalence relation you need show that is reflexive, transitive, symmetric.
I think I find good example that this no work for transitive and symmetric.
Say $A = \left\{0,1,2\right\}$, we say $R_1=\left\{(0,0),(1,1),(2,2),(0,1),(1,0)\right\}$ and $R_2=\left\{(0,0),(1,1),(2,2),(1,2),(2,1)\right\}$.
Then $R_1 \circ R_2= \left\{(0,0),(1,1),(1,2),(2,2),(2,1),(0,1),(0,2),(1,0)\right\}$
This is no symmetric because $(0,2)$ exist but $(2,0)$ no exist.
This is no transitive because $(2,1)$ exist and $(1,0)$ exist, but $(2,0)$ don't exist.
Now only missing is reflexive. I don't know how show it because it must be reflexive. But how show it's reflexive? I show the other correct? Because I ask friend and he say when you find counter example you have proof good.
Pls help need for exam how solve these example.
Best Answer
Your example is correct, and you don't need to do any more work! To show a relation is not an equivalence relation, you only need to show that one of the three conditions (reflexive, symmetric, transitive) fails. To be an equivalence relation, all three need to be true, as as soon as one of them fails, it's not an equivalence relation. So you have shown that $R_1\circ R_2$ is not an equivalence relation in your example.
(In fact, if $R_1$ and $R_2$ are reflexive, then $R_1\circ R_2$ always is reflexive as well. The proof is easy: for any $x\in A$, $(x,x)\in R_1$ and $(x,x)\in R_2$, so $(x,x)\in R_1\circ R_2$ (taking the $y$ in the definition of $R_1\circ R_2$ to be $x$).)